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5800	https://uutool.cn/hanzi-bihua/	最新工具最近使用热门工具文本工具生活工具商业营销效率办公编程开发数学计算信息速查生理健康图像工具文件转换音视频工具金融理财其它工具友情工具汉字数笔画字符数统计出现次数统计字符出现次数统计列表长度过滤在线汉字数笔画工具 + 选择文件二维码打赏支持复制链接反馈建议在线汉字数笔画工具介绍汉字笔画统计工具可用于汉字学习、取名辅助、字义分析、书法创作等多个场景。汉字笔画数查询工具基于现代汉字标准笔画数数据库，对常用字覆盖全面，自动跳过标点、空格及非汉字字符，无需下载或注册，在线即用。如何使用在线汉字笔画数统计工具第一步：在文本框中输入或粘贴你想查询的文字。第二步：点击“统计笔画”按钮。第三步：查看工具展示：每个汉字的笔画数；所有汉字的总笔画数量；无法识别或未收录字符列表（如有）。使用示例输入：中华文明源远流长！输出：总笔画数：56 ，中：4，华：6，文：4，明：8，源：13，远：7，流：10，长：4，已忽略字符：！汉字笔画查询常见问题 FAQ 支持多字统计吗？支持。无论是单个汉字、短语还是整段文字，均可统计。是否会统计标点符号或英文字母？不会，工具只会统计已收录的汉字，自动过滤标点、数字、英文字母等无效字符。笔画数据来源是什么？准确吗？数据来源为常用现代汉字笔画标准库，统计准确，可满足教育和常规应用需求。是否支持繁体字？支持常用繁体字的笔画统计，部分非常用字如未收录将跳过显示。最新工具最新文本分组提取根据指定的多行关键词模式，按顺序提取符合条件的内容块最新空行相邻行删除快速删除列表或文档中空行前后相邻行，让文本整洁有序。最新 HTML转图片将HTML代码转换为图片最新人像抠图 AI智能人像抠图，发丝级精准人像抠图最新文本数据列提取支持自定义分隔符，轻松提取文本数据中的指定列。最新图像配色通过提取图片的主色调，生成专业的配色方案最新 Excel文本转table代码将从Excel中复制的表格快速转换为table代码最新 table代码清理清理table代码中所有不需要的属性值，如id,style,class等最新育儿补贴计算器通过宝宝出生时间计算可领取的育儿补贴金额最新 PHP文本解压使用PHP对经过PHP压缩后的base64数据进行解压处理最新 PHP文本压缩使用PHP对文本数据进行压缩（如Gzip压缩），返回base64编码后的文本最新 🧡 支持我们，用心做好工具
5801	https://www.newsteps.org/sites/default/files/resources/download/pace_flyer_msms_workshop_jan2020.pdf	Newborn Screening Follow Up Tandem Mass Spectrometry Workshop January 13-17, 2020 Silver Spring, MD Course Description: This intensive five-day course assumes a basic understanding of newborn screening follow-up and metabolic biochemistry. Coursework will begin with a review of the principles of tandem mass spectrometry, diagnostic patterns in results, cut-offs, biochemical pathways, diagnostic follow-up and biochemical and clinical features of the metabolic disorders. Each day will cover interpretive skills and diagnostic follow up of certain disorders detectable through MS/MS screening including: amino acid disorders, urea cycle disorders, fatty acid oxidation disorders, and organic acid disorders. Interpretation homework assignments will be given along with daily examinations of information learned. Level of Instruction: Intermediate Audience: This program is intended for US-based newborn screening follow-up staff. Faculty: Dr. David Millington, Professor Emeritus of Pediatrics at Duke University School of Medicine. Prerequisites • Applicants should be actively engaged in follow-up activities in US-based newborn screening programs. • Applicants must have a basic understanding of newborn screening follow-up and metabolic chemistry. Application: • Apply Here by September 29, 2019 Learning Objectives: • Interpret the results obtained from MS/MS analysis of dried blood spots for: o Amino Acidopathies and Urea Cycle Disorders. o Fatty acid oxidation disorders o Organic acid disorders. • Identify appropriate expected ranges and cutoffs for MS/MS when applied to newborn screening disorders • Describe the biochemical and clinical features of the metabolic disorders • Recommend appropriate follow-up tests for confirmation of screening results and differential diagnosis Learning Goals: • Describe MS/MS as related to clinical diagnostics and NBS. • Describe the basic theory of ESI and MS/MS. • Describe scan functions and how they are employed in NBS. • Describe the biochemistry and clinical manifestations of amino acid and urea cycle disorders and the source of abnormal metabolites in disease states. • Describe the biochemistry and clinical manifestations of disorders of the catabolism of branched-chain amino acids and related disorders that comprise the organic acidurias. • Describe the biochemistry and clinical manifestations of inherited disorders of mitochondrial fatty acid beta-oxidation and the origin of diagnostic metabolites. • Describe roles of personnel required for NBS expanded with MS/MS. • Interpret amino acid MS/MS spectra and diagnose amino acid disorders. • Identify disorders of amino acid catabolism by their acylcarnitine spectra and understand confirmatory test procedure. • Identify disorders of fatty acid oxidation by their acylcarnitine spectra and understand confirmatory test procedure. • Describe the meaning of stable isotopes and their role in quantitative MS. • Describe how cut-offs are established and affect result reporting. • Summarize implementation of expanded NBS in the state of New York. • Summarize the impact of expanded newborn screening from the genetic counselor’s point of view. • Summarize a dietitian’s role in follow-up of patients identified by NBS. • Describe response to abnormal NBS results: follow-up testing of amino acids, organic acids and acylcarnitines for diagnosis – limitations. • Describe how disorders are selected for the panel. • Describe second-tier follow-up testing available by MS/MS methods. • Describe the implementation of MS/MS in a State NBS Program – review of problems identified and their resolution, interesting case reports. • Understand the impact of newborn screening in North Carolina – disorders detected and frequency. • Identify non-diagnostic MS/MS results and understand causes. • Summarize the CDC QA/QC program for expanded newborn screening. • Describe how in vitro tests are used to confirm fatty acid oxidation defects. • Describe how new technologies will enable further expansion of newborn screening for lysosomal storage disorders (LSD) and severe combined immune deficiency (SCID). • Interpret and respond to abnormal MS/MS screening results. The Association of Public Health Laboratories (APHL) is approved as a provider of continuing education programs in the clinical laboratory sciences by the ASCLS P.A.C.E. ® Program. Participants who successfully complete this program will be awarded 29.5 contact hours. Staff Contact: Erin Darby erin.darby@aphl.org
5802	https://brainly.ph/question/3873212	x(x+6)= 8 (x+3) what is the solution - Brainly.ph Skip to main content Ask Question Log in Join for free For parents For teachers Honor code Brainly App ehgitnimi 07.10.2020 Math Junior High School answered X(x+6)= 8 (x+3) what is the solution 1 See answer See what the community says and unlock a badge. Read More Answer 21 people found it helpful orginojerson orginojerson Ace 626 answers 5.5M people helped 3 Step-by-step explanation: Transform into Quadratic Equation; ByFactoring: Findthevalueofx; #CarryOnLearning Explore all similar answers Thanks 21 rating answer section Answer rating 4.1 (9 votes) salamt welcome thanks Advertisement Still have questions? Find more answers Ask your question New questions in Math sᴏʟᴠᴇ ғᴏʀ ᴛʜᴇ ғᴜᴛᴜʀᴇ ᴠᴀʟᴜᴇ ᴏғ sɪᴍᴘʟᴇ ᴀɴɴᴜɪᴛʏ ᴅᴜᴇ. ᴡʀɪᴛᴇ ʏᴏᴜʀ sᴏʟᴜᴛɪᴏɴs ᴏɴ ᴀ ᴡʜᴏʟᴇ sʜᴇᴇᴛ ᴏғ ᴘᴀᴘᴇʀ. .ᴀ sᴛᴜᴅᴇɴᴛ sᴀᴠᴇs ₱, ᴀᴛ ᴛʜᴇ ʙᴇɢɪɴɴɪɴɢ ᴏғ ᴇᴀᴄʜ ʏᴇᴀʀ ɪɴ Read the instruction and answer carefully, pay attention to the circle. Troll answers will be BANNED from this app. 1. √16 6. √4 2. √9 7. √49 3. √25 8. √36 4. √81 9. √100 5. √1 y= -x (x+6) (3x-4) x= -7 -3 1 2 the intervals obtained when the x intercepts are used to partition the number line Draw the solid figure in the space provided. Find the volume of a cylinder having a radius of 7 inches and a height of 10 inches PreviousNext Advertisement Ask your question Free help with homework Why join Brainly? ask questions about your assignment get answers with explanations find similar questions I want a free account Company Careers Advertise with us Terms of Use Copyright Policy Privacy Policy Cookie Preferences Help Signup Help Center Safety Center Responsible Disclosure Agreement Get the Brainly App ⬈(opens in a new tab)⬈(opens in a new tab) Brainly.ph We're in the know (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)
5803	https://www.mathworksheets4kids.com/rounding-decimal-number-line.php	Child Login Main Menu Math Language Arts Science Social Studies Interactive Worksheets Browse By Grade Become a Member Become a Member Math Interactive Worksheets Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Worksheets by Grade Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Number Sense and Operations Number Recognition Counting Skip Counting Place Value Number Lines Addition Subtraction Multiplication Division Word Problems Comparing Numbers Ordering Numbers Odd and Even Prime and Composite Roman Numerals Ordinal Numbers Properties Patterns Rounding Estimation In and Out Boxes Number System Conversions More Number Sense Worksheets Measurement Size Comparison Time Calendar Money Measuring Length Weight Capacity Metric Unit Conversion Customary Unit Conversion Temperature More Measurement Worksheets Financial Literacy Writing Checks Profit and Loss Discount Sales Tax Simple Interest Compound Interest Statistics and Data Analysis Tally Marks Pictograph Line Plot Bar Graph Line Graph Pie Graph Scatter Plot Mean, Median, Mode, Range Mean Absolute Deviation Stem-and-leaf Plot Box-and-whisker Plot Factorial Permutation and Combination Probability Venn Diagram More Statistics Worksheets Geometry Positions Shapes - 2D Shapes - 3D Lines, Rays and Line Segments Points, Lines and Planes Angles Symmetry Transformation Area Perimeter Rectangle Triangle Circle Quadrilateral Polygon Ordered Pairs Midpoint Formula Distance Formula Slope Parallel, Perpendicular and Intersecting Lines Scale Factor Surface Area Volume Pythagorean Theorem More Geometry Worksheets Pre-Algebra Factoring GCF LCM Fractions Decimals Converting between Fractions and Decimals Integers Significant Figures Percents Convert between Fractions, Decimals, and Percents Ratio Proportions Direct and Inverse Variation Order of Operations Exponents Radicals Squaring Numbers Square Roots Scientific Notations Logarithms Speed, Distance, and Time Absolute Value More Pre-Algebra Worksheets Algebra Translating Algebraic Phrases Evaluating Algebraic Expressions Simplifying Algebraic Expressions Algebraic Identities Equations Quadratic Equations Systems of Equations Functions Polynomials Inequalities Sequence and Series Matrices Complex Numbers Vectors More Algebra Worksheets Trigonometry Calculus Math Workbooks English Language Arts Summer Review Packets Science Social Studies Holidays and Events Support Worksheets> Math> Pre-Algebra> Decimals> Rounding> Number Lines Rounding Decimals using a Number Line Worksheets When accuracy is not a concern, then calculate easily by rounding decimals using the number line a versatile visual tool. The pictorial representation helps visualize the position of the decimal on the number line and makes rounding uncomplicated. Practice rounding decimals with the pre-printed number lines and round up or round down decimals to the nearest whole numbers, tenths, hundredths or thousandths. These printable worksheets are ideal for 4th grade, 5th grade, and 6th grade students. Print some of them for free! Number Line: Round to the Nearest Whole Number Plot the decimal on the number line. Round down the decimal to the nearest whole number if closer to the left and round up if closer to the right or at the center. Test comprehension with the questions that follow. Number Line: Round to the Nearest Tenth Use the jump strategy to round decimals to the nearest tenth. Indicate the decimal on the number line, skip count to find the nearest tenth, round off and identify if the decimal was rounded up or down as well. Number Line: Round to the Nearest Hundredth Transit from rounding decimals to the nearest tenth to hundredth with these practice printable worksheets for grade 4 and grade 5. Depict the decimal on the number line, locate the thousandths place, round up or down accordingly. Number Line: Round to the Nearest Thousandth Reiterate the concept with these rounding the decimal to the nearest thousandths worksheet pdfs for grade 6. Analyze the number line and round up or down based on the position of the decimal on the number line. Rounding Decimals using a Number Line: Revision Recapitulate conceptual understanding with these mixed review worksheets. Find a blend of exercises to round off decimals to whole numbers, tenths, hundredths or thousandths using the number line. Related Worksheets »Estimating Decimals »Decimals on a Number Line »Rounding Whole Numbers Home Become a Member Membership Information Login Printing Help FAQ How to Use Interactive Worksheets How to Use Printable Worksheets About Us Privacy Policy Terms of Use Contact Us Follow us Copyright © 2025 - Math Worksheets 4 Kids ×
5804	https://blog.csdn.net/whatday/article/details/126369886	博客下载学习社区 GitCode InsCodeAI 会议 AI 搜索数学 24个基本求导公式常见导数公式简介转载已于 2022-08-16 17:02:59 修改 · 10w+ 阅读 · 69 · 410 · CC 4.0 BY-SA版权原文链接：文章标签： #算法 #机器学习 #人工智能社区：GitCode 开源社区加入于 2022-08-16 17:02:36 首次发布 2、f(x)=a的导数， f'(x)=0, a为常数 3、f(x)=x^n的导数， f'(x)=nx^(n-1), n为正整数 4、f(x)=x^a的导数， f'(x)=ax^(a-1), a为实数 5、f(x)=a^x的导数， f'(x)=a^xlna, a>0且a不等于1 6、f(x)=e^x的导数， f'(x)=e^x 7、f(x)=log_a x的导数， f'(x)=1/(xlna), a>0且a不等于1 8、f(x)=lnx的导数， f'(x)=1/x 21、(fg)'=f'g+fg' 22、(f/g)'=(f'g-fg')/g^2 23、(1/f)'=-f'/f^2 24、(f^(-1)(x))'=1/f'(y) 常见导数公式 4个基本求导公式可以分成三类。第一类是导数的定义公式，即差商的极限. 再用这个公式推出17个基本初等函数的求导公式，这就是第二类。最后一类是导数的四则运算法则和复合函数的导数法则以及反函数的导数法则，利用这些公式就可以推出所有可导的初等函数的导数。 1、f'(x)=lim(h->0)[(f(x+h)-f(x))/h] 即函数差与自变量差的商在自变量差趋于0时的极限，就是导数的定义。其它所有基本求导公式都是由这个公式引出来的。包括幂函数、指数函数、对数函数、三角函数和反三角函数，一共有如下求导公式： 2、f(x)=a的导数， f'(x)=0, a为常数即常数的导数等于0；这个导数其实是一个特殊的幂函数的导数。就是当幂函数的指数等于1的时候的导数。可以根据幂函数的求导公式求得。 3、f(x)=x^n的导数， f'(x)=nx^(n-1), n为正整数即系数为1的单项式的导数，以指数为系数，指数减1为指数. 这是幂函数的指数为正整数的求导公式。 4、f(x)=x^a的导数， f'(x)=ax^(a-1), a为实数即幂函数的导数，以指数为系数，指数减1为指数. 5、f(x)=a^x的导数， f'(x)=a^xlna, a>0且a不等于1 即指数函数的导数等于原函数与底数的自然对数的积. 6、f(x)=e^x的导数， f'(x)=e^x 即以e为底数的指数函数的导数等于原函数. 7、f(x)=log_a x的导数， f'(x)=1/(xlna), a>0且a不等于1 即对数函数的导数等于1/x与底数的自然对数的倒数的积. 8、f(x)=lnx的导数， f'(x)=1/x 即自然对数函数的导数等于1/x. 9、(sinx)'=cosx 即正弦的导数是余弦. 10、(cosx)'=-sinx 即余弦的导数是正弦的相反数. 11、(tanx)'=(secx)^2 即正切的导数是正割的平方. 12、(cotx)'=-(cscx)^2 即余切的导数是余割平方的相反数. 13、(secx)'=secxtanx 即正割的导数是正割和正切的积. 14、(cscx)'=-cscxcotx 即余割的导数是余割和余切的积的相反数. 15、(arcsinx)'=1/根号(1-x^2) 16、(arccosx)'=-1/根号(1-x^2) 17、(arctanx)'=1/(1+x^2) 18、(arccotx)'=-1/(1+x^2) 最后是利用四则运算法则、复合函数求导法则以及反函数的求导法则，就可以实现求所有初等函数的导数。设f,g是可导的函数，则： 19、(f+g)'=f'+g' 即和的导数等于导数的和。 20、(f-g)'=f'-g' 即差的导数等于导数的差。 21、(fg)'=f'g+fg' 即积的导数等于各因式的导数与其它函数的积，再求和。 22、(f/g)'=(f'g-fg')/g^2 即商的导数，取除函数的平方为除式。被除函数的导数与除函数的积减去被除函数与除函数的导数的积的差为被除式。 23、(1/f)'=-f'/f^2 即函数倒数的导数，等于函数的导数除以函数的平方的相反数。 24、(f^(-1)(x))'=1/f'(y) 即反函数的导数是原函数导数的倒数，注意变量的转换。常见导数公式确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用 whatday 关注关注 69 点赞踩 410 收藏觉得还不错? 一键收藏 0 评论分享复制链接分享到 QQ 分享到新浪微博扫一扫举报举报导数的概念及运算 DZYO的博客 07-06 4091 学习导数前应先了解： 1.数列极限的概念及性质 2.函数极限的概念及性质导数导数的通俗理解导数的定义常见函数的导数求导法则四则运算复合函数的求导法则导数导数的通俗理解对于一段平滑的曲线，过其中某一点x0x_0做该曲线的切线，则所得曲线的斜率即为该曲线在x0x_0的导数如图，该曲线在x0x_0处的导数为tanαtan\, \alpha导数的定义设函数y=f(x)y=f(x)在点x0x_ 参与评论您还未登录，请先登录后发表或查看评论数学极限与导数积分 9-26 高等数学导数公式与积分表【两个重要极限】【常用导数公式】【常用积分表】【记是记不住的,要多那题去磨笔头】 f(x)=xn,f′(x)=nxn−12.f(x)=xn,f′(x)=nxn−1 3.f(x)=ax,f′(x)=lna×ax3.f(x)=ax,f′(x)= 高数02.04隐函数的导数_隐函数求导公式 9-28 例2.求椭圆x216+y29=1在点(2,323–√)处的切线方程.例2.求椭圆x216+y29=1在点(2,323)处的切线方程. 解:椭圆方程两边对x求导x8+29 高等数学（第二章：导数与微分） kusunoki的博客 10-07 4506 归纳了高等数学第二章——导数与微分的做题方法与技巧，仅作为复习使用。 (完整版)导数的基本公式及运算法则.ppt 07-12 (完整版)导数的基本公式及运算法则.ppt 数学- 基本初等函数导数公式及求导法则_初等函数运算法则 9-23 数学- 基本初等函数导数公式及求导法则三角函数相关运算指数和对数函数相关运算对数函数的强大之处在于可以变积为和,变商为差,化幂为系数。在求幂指函数或某些复杂表达式的函数的导数时,将原来的函数转化为对数函数后可方便求导。隐函数求导 “如果方程F(x,y)=0能确定y是x的函数,那么称这种方式表示的函数是... 导数的基本公式及运算法则.ppt资源 9-17 基本初等函数的导数公式是求导的基础。例如,幂函数的导数公式表明,对于形式为x^n的函数,其导数是n倍的x^(n-1);指数函数的导数公式指出,对于自然对数底数e的指数函数,其导数与原函数相等;对于三角函数,如sin x的导数是cos x,而cos x的导数则是-sin x,体现了三角函数之间的相生相克关系;对于正切函数tan x,... 微积分 \| 导数概念 / 定义 / 符号（一些讨论） u013669912的博客 07-29 1010 …… 函数求导法则 m0_64247096的博客 11-25 2315 1.函数的和差积商求导法则。证明的话乘除有点麻烦，需要配凑导数定义一加一减刚好消掉。 2.反函数的求导法则。（单调函数有反函数）上一节说直接用dy/dx＝dx/dy分之一理解有点问题（就是dy/dx是一个整体，是不能当做分子分母拆开来用的），其实直接用Δy/Δx＝Δx/Δy分之一理解也有点问题（这里只说了Δx≠0，并没有说Δy≠0所以不能直接做分母）。所以书上添了不少篇幅来介绍。什么意思呢，Δx是个变化量，不管是增是减。结果都不等于零。至于怎么深入了解呢。举一个具体的函数把他的导数公式 chenzuancong9642的博客 02-07 910 转载于: 数学中常用的求导数的公式汇总彬彬侠的博客 10-01 1万+ 数学中常用的求导公式汇总，一、基本求导公式（常数、幂、指数、对数、三角、反三角、双曲、反双曲），二、基本求导法则（常数倍、和差、乘积、商数、链式），三、复合函数的导数（幂、指数、对数、三角、反三角、双曲），四、高阶导数（二阶导数、n 阶导数），五、特殊求导技巧（对数求导法、隐函数求导、参数方程求导）常见导数公式热门推荐 xueruixuan的专栏 12-12 12万+ 常见导数公式基本导数公式 08-12 914 基本求导公式 zh6206763的博客 04-20 6619 基本求导方法数学：1.2.2基本初等函数的导数公式及导数的运算法则优秀导学案.doc 08-13 在数学中，基本初等函数的导数公式以及导数的运算法则是学习和应用微积分的基础。本篇导学案主要围绕基本初等函数的导数公式及导数的四则运算法则展开，旨在帮助学习者熟练掌握这些基础知识，并能应用于求解简单的... 基本求导法则与导数公式.doc 10-23 《基本求导法则与导数公式》导数是微积分学的基础，它在数学分析、物理、工程等领域都有广泛的应用。求导法则与导数公式是解决微积分问题的关键工具，尤其对于初学者，掌握这些基本规则至关重要。首先，我们要... 高中数学第一章导数及其应用1.2.3基本初等函数的导数公式及导数的运算法则二课件新人教A版.ppt 08-03 高中数学导数及其应用是建立在基本初等函数导数公式和导数运算法则的基础上，通过复合函数求导法则以及导数的应用，学生可以对函数有更深入的理解，并解决现实生活中一些涉及变化率的问题。掌握好这些概念对于学生... 高中数学1.2.2基本初等函数的导数公式及导数的运算法则二市公开课一等奖课件名师大赛获奖课件.pptx 05-01 在高中数学教学中，导数作为研究函数变化率的重要工具，其基本初等函数的导数公式及运算法则是学生必须掌握的基础知识点。导数不仅可以用来描述函数图形上的切线斜率，还是解决实际问题时不可或缺的数学工具。在理解... 【精品课件】高中数学第1章导数及其应用21几个常用函数的导数22基本初等函数的导数公式及导数的运算法.ppt 最新发布 09-02 本课件主要讲解了高中数学第一章导数及其应用，其中包含了导数的计算方法、几个常用函数的导数，以及基本初等函数的导数公式和导数的运算法则。首先，我们要明确导数的定义，它代表了函数在某一点处的瞬时变化率。... 常用导数细雨潜行 06-04 1171 y=xny=x^n，nn为正整数，y′=nxn−1y^\prime=nx^{n-1} y=sinxy=\sin x，y′=cosxy\prime=\cos x y=cosxy=\cos x，y′=−sinxy\prime=-\sin x y=axy=a^x，y′=axlnay\prime=a^x\ln a y=logaxy=\log_ax，y′=1xlogae=1xlnay\prime=\f 求导公式 weixin_33907511的博客 04-01 2320 对一个给定的函数,找出它上面每一点的斜率的计算通式,就是导函数。 ①几个基本初等函数求导公式 (C)'=0， (x^a)'=ax^(a-1)， (a^x)'=(a^x)lna，a＞0，a≠1；(e^x)'=e^x [logx]'=1/[xlna]，a＞0，a≠1；(lnx)'=1/x (sinx)'=cosx (cosx)'=-sinx (tanx)'=(secx)^2 (co... 数学基础知识总结 —— 1. 常用导数公式老程的技术笔记 02-23 1万+ 文章目录基本导数公式基本导数运算法则加法运算减法运算乘法运算除法运算带有常数C的导数微分的四则运算加减法计算带有常数的微分乘法计算除法计算基本导数公式原函数f(x)f(x)f(x) 导数f′(x)f'(x)f′(x) C (C为常数) 0 xnx^nxn nxn−1nx^{n-1}nxn−1 CxC^xCx CxlnCC^xlnCCxlnC (C为常数，且大于0) exe^xex (e为自然常数) ex(lne)=ex⋅1=exe^x(lne) = e^x \cdot 1 = 求导 03-22 嗯，用户的问题是如何进行函数求导运算。首先，我需要回忆一下数学中关于求导的基本方法。函数求导主要有几种常见的方法，比如基本导数规则、导数的四则运算法则、链式法则、隐函数求导，还有高阶导数等等。可能还需要考虑一些特殊情况，比如分段函数或者复合函数的情况。用户可能是一个刚开始学习微积分的学生，所以需要详细但清晰的步骤。要确保每个方法都有例子说明，这样用户更容易理解。同时，用户的问题里提到要避免使用“首先”、“然后”这样的步骤词汇，所以需要用不同的方式组织内容，比如用方法一、方法二这样的标题，每个方法之间换行。还要注意格式要求，比如数学公式必须用$或$$包裹，代码需要用三个反引号并指定语言类型。例子中的代码可能用Python的符号计算库，比如SymPy，来演示如何自动求导，这样用户如果有编程需求可以参考。另外，用户要求每个步骤间换行，所以每个方法之间需要空行，但不需要编号的步骤词汇。要整合搜索到的信息，可能需要检查是否有最新的资源或者常见的求导方法被遗漏。例如，隐函数求导和参数方程求导是否都应该包括在内？然后，生成相关问题的时候，要基于用户的原问题，比如如何验证导数的正确性，或者如何处理复合函数求导，还有导数的实际应用例子。这些问题应该帮助用户深入理解求导的不同方面。最后，确保回答中没有使用第一人称，并且引用内容没有集中在末尾，而是自然地融入各个方法的解释中。需要检查是否有违反格式要求的地方，比如数学公式是否正确包裹，代码块是否标注了语言类型，相关问题的标题是否正确等等。函数求导运算通过不同数学规则实现，以下是常用方法与操作示例：一、基本导数规则对于基础函数可直接使用公式化结果： - 幂函数：$f(x)=x^n$ 导数为 $f'(x)=n x^{n-1}$ 示例：$f(x)=x^5 \Rightarrow f'(x)=5x^4$ - 指数函数：$f(x)=e^x$ 导数为 $f'(x)=e^x$ - 对数函数：$f(x)=\ln x$ 导数为 $f'(x)=\frac{1}{x}$ 二、四则运算法则对函数组合运算适用以下规则： - 加法法则：$(f+g)'=f'+g'$ 示例：$f(x)=x^2+\sin x \Rightarrow f'(x)=2x+\cos x$ - 乘法法则：$(fg)'=f'g+fg'$ 示例：$f(x)=x \cdot e^x \Rightarrow f'(x)=e^x + x e^x$ - 除法法则：$\left(\frac{f}{g}\right)'=\frac{f'g - fg'}{g^2}$ 示例：$f(x)=\frac{\sin x}{x} \Rightarrow f'(x)=\frac{x\cos x - \sin x}{x^2}$ 三、链式法则（复合函数求导）对复合函数 $f(g(x))$ 导数为： dfdx=f′(g(x))⋅g′(x) 示例：$f(x)=\sin(2x^3)$ 分解为 $u=2x^3$，则导数为 $\cos(u) \cdot 6x^2 = 6x^2\cos(2x^3)$ 四、隐函数求导法对形如 $F(x,y)=0$ 的方程，通过两边同时求导解出 $\frac{dy}{dx}$：示例：方程 $x^2 + y^2 = 25$ 求导得 $2x + 2y \frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=-\frac{x}{y}$ 五、编程实现验证（Python示例）通过符号计算库可自动验证导数结果： python from sympy import symbols, diff x = symbols('x') f = x\\3 + 2\x\\2 - 5\x print(diff(f, x)) # 输出结果：3\x\\2 + 4\x -5 六、特殊函数处理 - 分段函数：需在分段点分别求导并验证连续性 - 参数方程：对参数方程 $ {x=g(t)y=h(t) $ 使用 $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ 关于我们招贤纳士商务合作寻求报道 400-660-0108 kefu@csdn.net 在线客服工作时间 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5805	https://chem.libretexts.org/Courses/Smith_College/Organic_Chemistry_(LibreTexts)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.03%3A_The_SN2_Reaction	11.3: The SN2 Reaction - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. 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Home 2. Campus Bookshelves 3. Smith College 4. Organic Chemistry (LibreTexts) 5. 11: Reactions of Alkyl Halides- Nucleophilic Substitutions and Eliminations 6. 11.3: The SN2 Reaction Expand/collapse global location Organic Chemistry (LibreTexts) Front Matter 1: Structure and Bonding 2: Polar Covalent Bonds; Acids and Bases 3: Organic Compounds- Alkanes and Their Stereochemistry 4: Organic Compounds - Cycloalkanes and their Stereochemistry 5: Stereochemistry at Tetrahedral Centers 6: An Overview of Organic Reactions 7: Alkenes- Structure and Reactivity 8: Alkenes- Reactions and Synthesis 9: Alkynes - An Introduction to Organic Synthesis 10: Organohalides 11: Reactions of Alkyl Halides- Nucleophilic Substitutions and Eliminations 12: Structure Determination - Mass Spectrometry and Infrared Spectroscopy 13: Structure Determination - Nuclear Magnetic Resonance Spectroscopy 14: Conjugated Compounds and Ultraviolet Spectroscopy 15: Benzene and Aromaticity 16: Chemistry of Benzene - Electrophilic Aromatic Substitution 17: Alcohols and Phenols 18: Ethers and Epoxides; Thiols and Sulfides 19: Aldehydes and Ketones- Nucleophilic Addition Reactions 20: Carboxylic Acids and Nitriles 21: Carboxylic Acid Derivatives- Nucleophilic Acyl Substitution Reactions 22: Carbonyl Alpha-Substitution Reactions 23: Carbonyl Condensation Reactions 24: Amines and Heterocycles 25: Biomolecules- Carbohydrates 26: Biomolecules- Amino Acids, Peptides, and Proteins 27: Biomolecules - Lipids 28: Biomolecules - Nucleic Acids 29: Orbitals and Organic Chemistry - Pericyclic Reactions 30: Synthetic Polymers Back Matter 11.3: The SN2 Reaction Last updated Jan 15, 2023 Save as PDF 11.2: The Discovery of Nucleophilic Substitution Reactions 11.4: Characteristics of the SN2 Reaction picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report Submit a Peer Review View on CommonsDonate Page ID 429568 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Objectives 2. Key Terms 3. Study Notes 4. Brønsted-Lowry acid-base reactions 5. General SN2 Reaction 6. Example 7. The SN2 Mechanism 8. Example 9. Bimolecular Nucleophilic Substitution Reactions and Kinetics 10. SN2 Reactions Are Stereospecific 1. Exercise 11.3.1 Objectives After completing this section, you should be able to write an expression relating reaction rate to the concentration of reagents for a second-order reaction. determine the order of a chemical reaction from experimentally obtained rate data. describe the essential features of the S N 2 mechanism, and draw a generalized transition state for such a reaction. Key Terms Make certain that you can define, and use in context, the key terms below. bimolecular kinetics rate coefficient rate equation reaction rate second-order reaction S N 2 Study Notes Most of the key terms introduced in this section should already be familiar to you from your previous general chemistry course. Reaction rate refers to the change in concentration of a reactant or product per unit of time. Using strict SI units, reaction rates are expressed in mol · L−1 · s−1, but in some textbooks you will find this value written as M/s. In general, the reaction rate of a given reaction changes with time, as it is dependent on the concentration of one or more of the reactants. An equation which shows the relationship between the reaction rate and the concentrations of the reactants is known as the rate equation. All rate equations contain a proportionality constant, usually given the symbol k, which is known as the rate coefficient. Some textbooks refer to this value as the “rate constant,” but this name is a little misleading as it is not a true constant. The rate coefficient of a given reaction depends on such factors as temperature and the nature of the solvent. S N 2 is short for “bimolecular nucleophilic substitution.” You will encounter abbreviations for other types of reactions later in this chapter. If you are unclear on the point about the inversion of configuration during an S N 2 reaction, construct a molecular model of a chiral alkyl halide, the transition stateformed when this substance reacts with a nucleophile in an S N 2 process, and the product obtained from this reaction. Brønsted-Lowry acid-base reactions In many ways, the proton transfer process in a Brønsted-Lowry acid-base reaction can be thought of as simply a special kind of nucleophilic substitution reaction, one in which the electrophile is a hydrogen rather than a carbon. In both reaction types, we are looking at very similar players: an electron-rich species (the nucleophile/base) attacks an electron-poor species (the electrophile/proton), driving off the leaving group/conjugate base. Instead of showing a specific nucleophile like hydroxide, we will simply refer to the nucleophilic reactant as 'Nu'. In a similar fashion, we will call the leaving group 'X'. We will see as we study actual reactions that leaving groups are sometimes negatively charged, sometimes neutral, and sometimes positively charged. We will also see some examples of nucleophiles that are negatively charged and some that are neutral. Therefore, in this general picture we will not include a charge designation on the 'X' or 'Nu' species. We will generalize the three other groups bonded on the electrophilic central carbon as R 1, R 2, and R 3: these symbols could represent hydrogens as well as alkyl groups. Here, then, is the generalized picture of a concerted (single-step) nucleophilic substitution reaction. General S N 2 Reaction Example The S N 2 Mechanism Bimolecular nucleophilic substitution (S N 2) reactions are concerted, meaning they are a one step process. This means that the process whereby the nucleophile attacks and the leaving group leaves is simultaneous. Hence, the bond-making between the nucleophile and the electrophilic carbon occurs at the same time as the bond-breaking between the electrophilic carbon and the halogen. This is called an 'S N 2'mechanism. In the term S N 2, S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that this is a bimolecularreaction: the overall rate depends on a step in which two separate molecules (the nucleophile and the electrophile) collide. The mechanism starts when lone pair electrons from the nucleophile attacks the electrophilic carbon of the alkyl halide to form a C-Nu sigma bond. Simultaneously, X-C bond is broken when the electrons are pushed onto the leaving group. Overall during this mechanism, a set of lone pair electrons are transferred from the nucleophile to the leaving groups. If you look carefully at the progress of the S N 2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way. The result of this backside attack is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. S N 2 reactions that begin with the R enantiomer as the substrate will form the S enantiomer as the product. Those that begin with the S enantiomer as the substrate will form the R enantiomer as the product. This concept also applies to substrates that are cis and substrates that are trans. If the cis configuration is the substrate, the resulting product will be trans. Conversely, if the trans configuration is the substrate, the resulting product will be cis. What this means is that S N 2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product. Below is an animation illustrating the principles we have just learned, showing the S N 2 reaction between hydroxide ion and methyl iodide. Notice how backside attack by the hydroxide nucleophile results in inversion at the tetrahedral carbon electrophile. Example Bimolecular Nucleophilic Substitution Reactions and Kinetics In the term S N 2, (as previously stated) the number two stands for bimolecular, meaning there are two molecules involved in the rate determining step. The rate of bimolecular nucleophilic substitution reactions depends on the concentration of both the haloalkane and the nucleophile. To understand how the rate depends on the concentrations of both the haloalkane and the nucleophile, let us look at the following example. The hydroxide ion is the nucleophile and methyl iodide is the haloalkane. If we were to double the concentration of either the haloalkane or the nucleophile, we can see that the rate of the reaction would proceed twice as fast as the initial rate. If we were to double the concentration of both the haloalkane and the nucleophile, we can see that the rate of the reaction would proceed four times as fast as the initial rate. The bimolecular nucleophilic substitution reaction follows second-order kinetics; that is, the rate of the reaction depends on the concentration of two first-order reactants. In the case of bimolecular nucleophilic substitution, these two reactants are the haloalkane and the nucleophile. For further clarification on reaction kinetics, the following links may facilitate your understanding of rate laws, rate constants, and second-order kinetics: Definition of a Reaction Rate Rate Laws and Rate Constants The Determination of the Rate Law Second-Order Reactions S N 2 Reactions Are Stereospecific The S N 2 reaction is stereospecific. A stereospecific reaction is one in which different stereoisomers react to give different stereoisomers of the product. For example, if the substrate is an (R) enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the (R) enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the (S) enantiomer. Conversely, if the substrate is an (S) enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the (S) enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the (R) enantiomer. In conclusion, S N 2 reactions that begin with the (R) enantiomer as the substrate will form the (S) enantiomer as the product. Those that begin with the (S) enantiomer as the substrate will form the (R) enantiomer as the product. This concept also applies to substrates that are cis and substrates that are trans. If the cis configuration is the substrate, the resulting product will be trans. Conversely, if the trans configuration is the substrate, the resulting product will be cis. Exercise 11.3.1 1) The reaction below follows the S N 2 mechanism. a) Write the rate law for this reaction. b) Determine the value of the rate coefficient, k, if the initial concentrations are 0.01 M CH 3 Cl, 0.01 M NaOH, and the initial reaction rate is 6 x 10-10 M/s. c) Calculate the initial reaction rate if the initial reactant concentrations are changed to 0.02 M CH 3 Cl and 0.0005 M NaOH. 2) Predict the product of a nucleophilic substitution of (S)-2-bromopentane reacting with CH 3 CO 2-, Show stereochemistry. 3) Predict the structure of the product in this S N 2 reaction. Be sure to specify stereochemistry. 4) Since everything is relative in chemistry, one reaction's nucleophile can be another reaction's leaving group. Some functional groups can only react as a nucleophile or electrophile, while other functional groups can react as either a nucleophile or electrophile depending on the reaction conditions. Classify the following compounds as nucleophile, electrophile, or leaving groups. More than one answer may be possible. a) bromoethane b) hydroxide c) water d) chlorocyclohexane e) ethanol f) bromide Answer 1) a) rate = k [CH 3 Cl] [OH-] b) substitute the data into the rate expression above and apply algebra to solve for k k = 6 x 10-6 Lmol-1 s-1 c) Using the rate law above, substitute the value for k from the previous question along with the new concentrations to determine the new initial rate. rate = 6 x 10-10 M/s 2) 3) 4) a) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.) b) strong nucleophile c) weak nucleophile and good leaving group d) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.) e) weak nucleophile, a poor electrophile without clever chemistry (stay tuned for future chapters), good leaving group f) good nucleophile and a good leaving group 11.3: The SN2 Reaction is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Steven Farmer, Dietmar Kennepohl, Layne Morsch, Tim Soderberg, William Reusch, & William Reusch. 11.2: The SN2 Reaction by Dietmar Kennepohl, Layne Morsch, Steven Farmer, Tim Soderberg, William Reusch is licensed CC BY-SA 4.0. Toggle block-level attributions Back to top 11.2: The Discovery of Nucleophilic Substitution Reactions 11.4: Characteristics of the SN2 Reaction Was this article helpful? Yes No Recommended articles 11: Reactions of Alkyl Halides- Nucleophilic Substitutions and EliminationsIn this course, you have already seen several examples of nucleophilic substitution reactions; now you will see that these reactions can occur by two ... Article typeSection or PageLicenseCC BY-SALicense Version4.0Show Page TOCno on pageTranscludedyes Tags author@Dietmar Kennepohl author@Layne Morsch author@Steven Farmer author@Tim Soderberg author@William Reusch bimolecular reaction kinetics rate coefficient rate equation reaction rate second-order reaction SN2 reaction source-chem-31506 © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× contents readability resources tools ☰ 11.2: The Discovery of Nucleophilic Substitution Reactions 11.4: Characteristics of the SN2 Reaction Complete your gift to make an impact
5806	https://thetradinganalyst.com/arithmetic-mean/	Arithmetic Mean: How You Can Use It (Today) By Tyler Corvin Can a single metric offer clarity and cut through all the market noise? Indeed–the arithmetic mean accomplishes precisely that. Like a steadfast lighthouse illuminating its surroundings; this statistical metric reveals not only the average outcome of investments but also sheds light on collective stock performance and typical trade volumes. The arithmetic mean–simple in its calculation, yet potent in implications–serves as a beacon amidst financial complexity: it provides an uncomplicated indicator of what is considered typical or expected. The arithmetic mean, seemingly an elementary concept, serves as a fundamental tool in the knitting of any analyst: it transforms raw data into coherent insights. This initial step unravels the intricacies of market behavior – offering not only solid ground but also a launching point for deeper exploration; moreover, by capturing central tendency– it reflects not just current states but also sets baselines for future projections. Exploring the arithmetic mean allows us to grasp its utility: it smoothes out financial data’s peaks and valleys, offers a consistent measure for comparison–a crucial element in decision-making. It’s an indispensable tool in helping navigate the ebb and flow of market trends; not only does this benefit seasoned investors, but also those just getting acquainted with the rhythm of markets. Join our exploration into this fundamental statistical tool that critically navigates us towards successful investment strategies. What you’ll learn Understanding the Arithmetic Mean Statisticians and financial analysts consider the arithmetic mean – commonly known as the “average” – a pivotal concept: it’s not only integral to statistical analysis, but also an indispensable tool in their respective fields. Calculating this value involves summing a series of numbers; subsequently, one divides that sum by the count of those figures. In terms of financial scrutiny–this seemingly basic mathematical operation metamorphoses into a priceless technique: it condenses copious amounts of data into one easily-understandable statistic which epitomizes dataset’s central tendency. Practically, one can apply the arithmetic mean to several financial datasets: stock prices; trading volumes; earnings per share; returns over a specific period–or any other quantitative measure pertinent in an investor’s decision-making process. Take for instance, computing the mean closing price of a stock across one month provides not only insight into its performance but also smoothes out daily volatility revealing underlying trends. Likewise—averaging portfolio returns over multiple years allows assessment of long-term performance and offers insights that may elude us under more granular scrutiny. In financial analysis, the arithmetic mean holds significant importance: it serves as a benchmark for measuring investment performance. Investors and analysts – by comprehending this average – can pinpoint deviations from normality, potentially signifying either opportunities or risks. Nonetheless, one must acknowledge that despite its provision of a concise method to summarize data, the employment of an arithmetic mean requires judicious use. This is particularly true in notoriously complex and unpredictable arenas such as financial markets, where the concept of mean reversion highlights the tendency of prices to eventually move back towards their average. In such contexts, the arithmetic mean serves as a fragment to the broader picture; it’s complemented by other statistical measures and analysis techniques – all in an effort to construct a comprehensive view of market dynamics. The Mechanics of Arithmetic Mean In statistical analysis, the arithmetic mean–operating on a simple yet powerful principle–serves as a cornerstone for understanding the dataset’s central tendency: it involves an addition of all numbers within the set; subsequently, this total is divided by the quantity of numbers. This method effectively condenses myriad data into one representative value–the resulting figure being none other than our trusty friend and ally in number-crunching—the arithmetic mean! Within the financial markets domain, we utilize the arithmetic mean to analyze an extensive range of data: from stock prices to trading volumes. A clear depiction of average performance – smoothing out daily price fluctuations and revealing underlying trends – is offered for instance by calculating a specific period’s mean closing price for a stock. In parallel fashion, insights into overall portfolio performance are gained through time-based examination of average returns; this approach enables comparisons with other investment vehicles or benchmarks. Additionally, the arithmetic mean can be used in conjunction with historical volatility to gain a better understanding of potential risk within a stock or portfolio. In financial analysis, this calculation proves its worth by simplifying complex datasets and empowering analysts and investors to formulate informed decisions; it’s an indispensable tool. Yet— we must recognize the arithmetic mean’s vulnerability to outliers: a solitary abnormally high or low value has the potential to distort the mean – possibly leading even a vigilant analyst astray. Thus, the strategic employment of the arithmetic mean alongside other statistical measures in analyzing financial markets is vital: it provides a comprehensive overview–not reliant solely on trends identification but also mitigates inherent limitations; this ensures an accurate assessment of market conditions and investment performance. Step-by-Step: Calculating the Arithmetic Mean To calculate the arithmetic mean–the simple average, indeed–is an undeniably straightforward process: it offers invaluable insights into financial data; specifically, stock performance over a stipulated period. Let’s delve into the step-by-step procedure: Suppose you analyze the performance of a stock over a 20-day period; you collect the closing prices for each day: if these prices, when summed up, amount to $2,000 – then dividing this total by 20 (the number of days) will yield an arithmetic mean—or average closing price—of $100. The arithmetic mean: it offers a smoothed perspective–a snapshot of the stock’s performance that can prove particularly useful in identifying broader trends. However, we must remain cognizant; this method operates under an assumption where each data point equally contributes to the total. This may not always accurately represent nuances within stock market dynamics—such as volatility or outlier events. Henceforth, relying solely on arithmetic means for assessment convenience is not advisable; its efficacy flourishes when employed alongside other analytical methods. Comparative Analysis: Mean, Median, and Mode A keen analysis of data is necessary to understand the behavior of financial markets. The statistical measures: arithmetic mean, median, and mode; each offer unique insights into market trends and investor behavior–making them indispensable for such an investigation. The Arithmetic Mean: it represents the sum of all values divided by their total count. This measure–commonly utilized in finance and statistics–allows for an easy calculation: specifically, one that provides a snapshot of overall stock performance over a given period. The advantage inherent to this method lies not only with its simplicity but also in its intuitive grasp on central tendency; indeed, it’s capable of offering valuable insight into data distribution without requiring complex statistical analyses.g Yet, volatile markets may skew the average with a few extreme values; this misrepresentation of the true market situation is an issue due to its sensitivity to outliers. The median – the middle value in a list of numbers sorted from least to greatest – offers an elucidating snapshot of the market’s central tendency by lessening outliers’ impact. When we apply this concept to stock prices, specifically using the median, we can more accurately portray typical market pricing; thus, our depiction remains untouched by spikes or drops on specific days. For traders seeking unadulterated insight into standard market behavior – without anomalous distortions introducing bias – this proves invaluable indeed! Moreover, comparing an individual stock’s beta to the overall market can further refine this understanding of volatility and price behavior. The mode of a dataset–the value that occurs most frequently–unveils within a dataset: this is indicative of either price level or trading volume. Not as ubiquitously utilized in quantitative financial analysis as its counterparts (mean and median), yet it yields an insight into common trading prices or volumes; indeed, potentially highlighting key support-resistance levels in stock-and-options markets. Traders and analysts can gain a comprehensive understanding of market dynamics by comparing these measures. The mean offers a broad overview, the median provides more resilience against outliers, and the mode uncovers the most common occurrences; together they contribute to our multifaceted view of market conditions. When traders understand each measure’s applications and limitations, it enables them to refine their strategies for assessing historical performance or forecasting future trends. Arithmetic vs. Geometric Mean: Choosing the Right Tool Within the domain of financial analysis, the arithmetic and geometric means function as pivotal instruments; they provide unique viewpoints on investment returns and portfolio performance. It is imperative to comprehend these two means’ distinctions–as well as their suitable applications—for precise market analysis: indeed, it is crucial for informed decision-making processes. To calculate the Arithmetic Mean, one adds up all values and divides by the total number of values; its primary advantage: simplicity–directness. This indeed makes it suitable for analyzing a series—be they periodic returns or average stock prices over time. However, we must note that this method may not accurately reflect investment performance across multiple periods due to an unaccounted factor: the compounding effect of returns. The arithmetic mean, by its disregard for the sequence of returns, may imply a higher average return; however–the geometric mean provides us with a more conservative and realistic measure of central tendency in compounded rates. For instance: should an investment portfolio undergo significant fluctuations–it is indeed the growth rate over that period which will be accurately reflected via using the geometric mean. The specific context of the analysis determines the choice between these two means: the arithmetic mean and geometric mean. Typically, we prefer using the arithmetic mean to assess short-term trade averages or single-period returns; however, when it comes to evaluating long-term investment performance and understanding volatility’s impact on compounded returns–here, employing a geometric mean is more appropriate. Significantly enhancing the accuracy and relevance of financial analyses hinges on understanding the appropriate times to employ each tool. Illustrating Arithmetic Mean: A Financial Scenario Imagine analyzing Apple Inc. (AAPL), a technological titan renowned for its innovation and market sway; let’s consider the range of closing prices it encountered over the 20 preceding trading days–from January 31st to February 28th. These fluctuations were substantial: AAPL’s closing price varied from $184.40 down to $181.42 in this period—painting an intriguing financial picture indeed! In order to grasp AAPL’s average closing price during that time frame, we undertook a straightforward calculation: Arithmetic Mean = (∑ Closing Prices)/ Number of Trading Days Arithmetic Mean = ($181.42 + $182.63 + $181.16 + … + $186.86 + $184.40)/ 20 After the calculation, we find that AAPL’s average closing price over these 20 trading days sits at around $185.32: a central value providing traders with a snapshot of its general trading range — an essential benchmark to gauge their investments. Around February 2nd, reports suggesting a less than stellar outlook for iPhone sales initially prompted AAPL stock to dip during this period. The news underscored challenges in Apple’s flagship product line and thereby fostered temporary bearish sentiment among investors. In this context, the arithmetic mean serves as a benchmark for evaluating the current price of the stock. This information empowers traders to determine whether the stock is trading above or below its recent average; thus guiding decisions on purchasing during low points, buying the dip, or capitalizing on momentum trends, riding the wave. The following practical example delineates the crucial role of arithmetic mean in stock analysis: it serves as a benchmark for evaluating average prices. However, this emphasis on averages necessitates contextualization within broader market narratives–such as product performance and technological advancements; these factors can profoundly influence investor sentiment and subsequently drive fluctuations in stock prices. Pros and Cons The arithmetic mean, a foundational analytical tool in stock and options trading, offers numerous advantages; however, it also has its limitations. Traders primarily benefit from its simplicity and comprehensibility: they can use this accessible metric to evaluate the average performance of securities over any given period. This average–a quick snapshot of historical data–equips traders with insights into past performance trends; thus aiding them in making informed decisions about potential investments. The utility of the arithmetic mean in trading, however, is nuanced by its inherent drawbacks: it exhibits a significant limitation–sensitivity to outliers. These anomalies can dramatically skew the average; they mislead traders about a security’s typical performance. Within stock and options trading’s volatile realm–where prices can fluctuate widely due to market sentiment, news events or economic indicators—the sole reliance on this measure renders future performance predictions less reliable. The effectiveness of the arithmetic mean, moreover, hinges on the data it encapsulates: specifically, it presupposes a normal distribution of price fluctuations–a not always accurate portrayal in financial markets characterized by skewed distributions. This assumption may culminate in fallacious deductions about a security’s central tendency; consequently – and critically – traders could undervalue potential loss risks. Conclusively, the arithmetic mean proves its worth as a tool to condense historical financial data; however, traders must exercise caution due to its inherent limitations. When paired with analytical counterparts—median, mode, or geometric mean—that accommodate volatility and risk: it yields optimal effectiveness. Traders gain precise insight into potential trading opportunities and risks by acknowledging when the arithmetic mean provides valuable interpretations—and conversely, when it could trigger misleading readings, especially during periods of heightened systematic risk. Conclusion Exploring the arithmetic mean in financial analysis sheds light on its crucial role: it simplifies complex data sets and offers clear insights into market trends. This statistical tool–by summing a set of values and averaging them–provides investors, traders; indeed anyone interested in any given financial metric with an uncomplicated method to measure central tendency. Its versatility is evident across diverse financial contexts; from basic stock evaluation–where one might simply wish for an overview–to more intricate investment portfolio analysis: here too, its usefulness remains unquestioned. While the arithmetic mean provides a rapid overview of financial data, we must recognize its constraints – particularly when handling volatile markets or skewed data distributions. In these instances, incorporating other statistical measures, utilizing analysis tools, and perhaps following real-time alerts via investment signals can furnish us with a more comprehensive understanding of the market. Moving ahead, it remains crucial to fuse advanced analytics with the arithmetic mean and deliberate over its limitations. This ensures an informed approach towards formulating investment strategies and conducting market analyses. Arithmetic Mean: FAQs How Can Arithmetic Mean Be Used to Evaluate the Performance of a Stock Portfolio? Employing the arithmetic mean enables us to compute a stock portfolio’s average return over specific periods. By accumulating the returns – whether daily, monthly or yearly – of said portfolio and subsequently dividing them by these timeframes; investors derive their simple average return. Such assessment aids in gauging overall performance: it facilitates comparison with benchmarks or other potential investments opportunities. What are the Limitations of Using Arithmetic Mean for Volatile Stock Prices? The arithmetic mean disregards stock price volatility: it treats all data points impartially, independent of their distribution or sequential arrangement. Particularly with volatile stocks – due to its sensitivity towards extreme values – this approach can yield misleading averages; these may distort the analysis significantly in short-term evaluations. How Does the Arithmetic Mean Compare to Moving Averages in Technical Analysis? Both the arithmetic mean and moving averages smooth out data to identify trends; however, moving averages exhibit greater dynamism. The arithmetic mean calculates a fixed average across a dataset: in contrast–as new data becomes available–moving averages continuously update their average by sliding over a specific period. This approach provides insights into short-term trends and momentum more effectively. In technical analysis, therefore, it is more appropriate to use moving averages for tracking price movements and identifying trend directions. Can the Arithmetic Mean Be Applied to Forecast Future Stock Prices or Market Trends? By itself, the arithmetic mean does not possess predictive capabilities: it merely presents an historical average of past data; this information can indeed inform–but cannot predict–future movements. Market trends and stock prices derive influence from a plethora of variables: these are not encapsulated solely by past averages. More complex statistical or machine learning techniques frequently demand predictive models, as they can accommodate a broader range of factors. What Steps Should Analysts Take To Ensure Accurate Calculations of Arithmetic Mean for Financial Datasets? Analysts should: Analyzing the Relative Valuation Model Regret Theory: Understand How it Affects Your Trading Decisions Real-Time Quotes Explained: Why Speed Matters in Trading Stock Rally Explained: How and Why Markets Surge How Does Real Rate of Return Work? How to Read and Use the Elder-Ray Index Grow Your Portfolio With Consistent Wins Company Insights Quick links Legal
5807	https://pubmed.ncbi.nlm.nih.gov/34411723/	An official website of the United States government The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Account Save citation to file Email citation Add to Collections Add to My Bibliography Your saved search Create a file for external citation management software Your RSS Feed Actions Page navigation Primary radial nerve palsy associated with humeral shaft fractures according to injury mechanism: is early exploration needed? Affiliations Primary radial nerve palsy associated with humeral shaft fractures according to injury mechanism: is early exploration needed? Authors Affiliations Abstract Background: Radial nerve palsy is a common complication associated with humeral shaft fractures. The purposes of this study were (1) to evaluate the status of primary radial nerve palsy in patients with humeral shaft fracture according to injury mechanism, (2) to estimate the risk factors of primary RNP, and (3) to evaluate whether early exploration is helpful for radial nerve recovery. Methods: This study analyzed 162 patients with humeral shaft fractures from January 2014 to December 2019. All patients were surgically treated in our hospital. Of these, 109 high-energy injuries were identified and compared with 53 low-energy injuries. The risk factors of radial nerve palsy were analyzed, and the prevalence of radial nerve palsy and status of radial nerve exploration according to injury mechanism were evaluated. Nerve recovery rate according to early nerve exploration was investigated. Results: There were 31 cases of radial nerve palsy among 162 patients: 27 in the high-energy humeral shaft fracture group and 4 in the low-energy humeral shaft fracture group. Logistic regression analysis for risk factors showed that the injury mechanism was significantly associated with primary radial nerve palsy. Among 31 radial nerve palsy patients, 21 radial nerves were explored and 19 radial nerves recovered completely (80.6%). In the high-energy humeral shaft fracture group, 18 radial nerves were explored during surgery among 27 radial nerve palsy cases, and 16 cases recovered (88.9%). The other 9 radial nerves were not explored, and only 5 cases recovered (55.6%). Conclusions: This study confirmed that the incidence of radial nerve paralysis was higher in high-energy humeral shaft fractures than in low-energy fractures. The more common fracture patterns were oblique, transverse, wedge, and comminuted in high-energy humeral shaft fracture. This study suggests that these patterns are not directly associated with radial nerve palsy, but that high-energy injury is associated with a specific fracture pattern. Early nerve exploration during surgical treatment in patients with radial nerve palsy associated with humeral shaft fracture was helpful especially after high-energy injury. Keywords: Exploration; Humerus; Injury mechanism; Radial nerve palsy; Risk factors; Shaft fracture. Copyright © 2021 Journal of Shoulder and Elbow Surgery Board of Trustees. Published by Elsevier Inc. All rights reserved. PubMed Disclaimer MeSH terms NCBI Literature Resources MeSH PMC Bookshelf Disclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). Unauthorized use of these marks is strictly prohibited. Connect with NLM National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers
5808	https://www.cs.dartmouth.edu/~ac/Teach/CS105-Winter05/Handouts/vempala-blossom.pdf	18.433 Combinatorial Optimization Matching Algorithms September 9,14,16 Lecturer: Santosh Vempala Given a graph G = (V, E), a matching M is a set of edges with the property that no two of the edges have an endpoint in common. We say that a vertex v ∈V is matched if v is incident to an edge in the matching. Otherwise the vertex is unmatched. A matching is maximum if there is no matching of greater cardinality. In particular, a maximum matching is called perfect if every vertex of G is matched. A bipartite graph G is a graph in which the vertices of G can be partitioned in two sets A and B with the property that every edge in G has one endpoint in A and one in B. In the case of bipartite graphs, the following theorem characterizes graphs that have a perfect matching. For U ⊆A denote N(U) the set of vertices that are adjacent to vertices in U. Theorem 1 (Hall). A bipartite graph with sets of vertices A, B has a perfect matching iﬀ \|A\| = \|B\| and (∀U ⊆A)\|N(U)\| ≥\|U\|. Proof. If a bipartite graph has a perfect matching, then it is easy to see that the right hand side is a necessary condition. We will now prove the reverse implication. First note that the RHS condition implies that ∀U ⊆B as well, \|N(U)\| ≥\|U\|. If \|N(U)\| < \|U\| for some subset U ⊆B, then \|N(A \ N(U))\| ≤\|B \ U\| < \|A\| −\|N(U)\| = \|A \ N(U)\| and A \ N(U) violates the condition as well. We proceed by induction on \|A\|. If \|A\| is 0 or 1, the claim is true. Now consider two cases: 1. Suppose that (∀U ⊆A, U ̸= ∅, U ̸= A)\|N(U)\| > \|U\|. Consider e = (u, v) and G′ = G −{u} −{v}. In G′, ∀U ⊆A −{u} \|NG′(U)\| ≥\|N(U)\| −1 ≥\|U\|. So G′ has a matching M of A \ {u} into B \ {v}. Then M1 = M ∪{e} is a matching of A into B in G. 2. Now suppose the opposite to the previous case: there exists A′ ⊂A nonempty such that \|N(A′)\| = \|A′\|. Let G1 be the graph induced by A′ ∪N(A′). Let G2 be the graph induced by G −A′ −N(A′). 1 In G1, (∀U ⊆A′), NG(U) = NG1(U), and \|NG1(U)\| ≥\|U\|. Thus, G1 has a matching M1 of A′ into N(A′). In G2, ∀U ⊆B \N(A′), we have NG(U) = NG2(U) since there are no edges from B \N(A′) to A′. Thus, \|NG2(U)\| = \|NG(U)\| ≥\|U\| and G2 has a matching M2 of A\A′ into B \N(A′). Now M1 ∪M2 is a perfect matching of G. Our goal in these lectures is to develop a fast algorithm for ﬁnding a matching of maximum cardinality in a given graph. Throughout this course, by “fast” we mean polynomial-time, i.e. the running time of the algorithm should be bounded by a ﬁxed polynomial in the size of the input graph. The size of a graph is determined by number of vertices in the graph, denoted by n, and by the number of edges, denoted by m. Now take a matching M with respect to the graph G. If every vertex of G is matched by M then M is a perfect matching and hence is a maximum matching of cardinality n 2 . Should M not be perfect, then we would like to either ﬁnd another matching of greater cardinality than M, i.e. augment M, or conclude that M is already maximum. One way to augment M is the following: ﬁnd a path P in the graph that starts at an unmatched vertex and consists alternately of edges not in M and edges in M (i.e. unmatched edges and matched edges) and ends at an unmatched vertex. Then consider the set of edges M ′ obtained by deleting the edges M has in common with the path and adding the rest of the edges on the path, i.e. the symmetric diﬀerence of M and P, denoted by M ⊕P. It is easy to verify that M ′ is also a matching, and moreover it has one more edge than M. Such an alternating path P is called an augmenting path. This observation motivates the following “algorithm”. The Matching Algorithm { 1. Start with any matching. 2. Find an augmenting path with respect to the current matching. 3. Augment the current matching. 4. Repeat the above two steps as long as possible. } When the algorithm terminates, we have a matching M with no augmenting paths. What do we do now? Our ﬁrst lemma tells us that at this point M must be maximum. Lemma 2. A matching M is maximum iﬀit has no augmenting paths. Proof. We have seen that if M contains an augmenting path then it is not a maximum 2 matching. So consider the converse. Assume that M does not contain an augmenting path. We will show that M is a maximum matching. In order to prove this we take some maximum matching M ∗and show that \|M\| = \|M ∗\|. Consider M ⊕M ∗, the symmetric diﬀerence of M and M ∗. Recall that this is the collection of edges that are in M but not M ∗and vice versa. Since M and M ∗both induce subgraphs of maximum degree one, it follows that M ⊕M ∗ induces a subgraph of maximum degree two. Note that such a subgraph may consist only of disjoint paths and/or cycles. In addition, observe that since M and M ∗are matchings these paths and cycles contain edges that are alternately in M and M ∗. Consider ﬁrst the cycles in our induced graph. All such cycles must contain an even number of edges, otherwise there must be some vertex that is adjacent to two edges in either M or M∗, contradicting the deﬁnition of a matching. Thus, these cycles contain an equal number of edges from M and M ∗. Consider now the induced paths. Suppose we have a path P that contains an odd number of edges. Hence, either P contains one more edge from M than M ∗or one more edge from M∗than M. In the former case note that P is then an augmenting path in G with respect to M ∗, contradicting the maximality of M ∗. In the latter case P is then an augmenting path in G with respect to M, contradicting our initial assumption. Hence all our induced paths contain an even number of edges and thus contain an equal number of edges from M and M ∗. So the paths and cycles induced by M ⊕M ∗contain an equal number of edges from M and M∗. Finally consider the edges that are not induced by M ⊕M ∗. These edges are either in both M and M ∗or in neither of them. It follows that M and M ∗are of equal cardinality and hence M is a maximum matching. How long does our algorithm take? In each iteration of steps 2 and 3 we increase the size of the matching by one. Thus we can repeat steps 2 and 3 at most n 2 times. So we are left with the question of how long it takes to ﬁnd an augmenting path. Actually, ﬁrst we must ﬁgure out how to ﬁnd an augmenting path. It turns out that this will be much easier to do for bipartite graphs, which we will consider ﬁrst. 3 1 Bipartite graphs Take a bipartite graph, with a matching M, and let AU ⊆A and BU ⊆B be the vertices unmatched by M. We wish to ﬁnd an augmenting path with respect to M. To do this, we will ﬁnd the set of vertices S accessible from AU by alternating paths. If S includes a vertex of BU then the alternating path to that vertex will be an augmenting path. The set S is determined by building an alternating forest F as follows: 1. Start with all the vertices of AU as separate components of F. 2. Add edges from vertices of A ∩V (F) to vertices of B without merging any two con-nected components of F. That is, if a vertex of B is adjacent to more than one component, add it to only one of the components. 3. Then add the edges of M incident to vertices of B ∩V (F). 4. Repeat the above two steps till no more edges can added to F. If we ﬁnd a vertex of BU in the forest, then this gives us an augmenting path. If not, by the next lemma, the matching M is a maximum matching. Lemma 3. M is maximum iﬀno vertex of BU is in F. Proof. If F includes a vertex v of BU then the path from v to the vertex of AU in the component containing v is an alternating path with unmatched vertices at its ends, i.e. an augmenting path. Hence M is not maximum. Conversely, suppose that no vertex of BU is included in F. In order to prove our result we introduce the notion of a a vertex cover. This is a set of vertices such that every edge is incident to at least one vertex in the set We will show that G has a vertex cover of size equal to the current matching. Since the size of any vertex cover, is at least the size of the maximum matching (one endpoint from each edge in the matching must be chosen in any vertex cover) this would prove that the matching M is maximum. Let X = A−V (F) and Y = B ∩V (F). Then we claim that X ∪Y is a vertex cover. Clearly, M meets every vertex of X ∪Y . Since M is a matching, no edge of M is incident to two vertices of X ∪Y . Now, given a matched vertex a ∈V (F), let (a, b) be the matching edge. From the description of F it follows that b must also be in V (F). As a result, every edge of M meets at least one vertex of X ∪Y and so \|M\| = \|X ∪Y \|. 4 All that is left to show is that X ∪Y is a cover of the graph. Suppose not. Then there is an edge (a, b) with a ∈A and b ∈B that is not covered. Hence we have a ∈V (F) and b ̸∈V (F). It follows that (a, b) is not a matching edge. In addition, b ̸∈B U otherwise it would have been added to V (F). So b is matched, say by the edge (a′, b), where a′ ̸= a. But this implies that F can be extended by adding the path aba′ contradicting the assumption that F is maximal. From the proof of the lemma we may derive the following theorem. Theorem 4. (K¨ onig) The size of a maximum matching in a bipartite graph is equal to the size of a minimum vertex cover of the graph. We say that A has a matching into B if the maximum matching is of size \|A\|. In addition, denote by Γ(X) is the set of neighbors of X ⊆V . The classical theorem of Frobenius and Hall then follows from K¨ onig’s theorem (and is actually equivalent to it). Theorem 5. (Frobenius-Hall) A has a matching into B iﬀfor every subset X of A, X ≤ \|Γ(X)\|. Proof. Clearly if there is a subset X of A such that X > \|Γ(X)\|, then there can be no matching of cardinality \|A\|. Conversely, assume that X ≤\|Γ(X)\| for all X ⊆A. We will show that the minimum vertex cover is of cardinality \|A\|, from which the theorem will follow. We may assume that each vertex is incident to at least one edge and that \|A\| ≤\|B\|. Note that the vertices of A form a vertex cover of cardinality \|A\|. Suppose we have a vertex cover X ∪Y , where X ⊆A and Y ⊆B. Observe that Γ(A −X) ⊆Y . Thus \|A −X\| ≤\|Γ(A −X)\| ≤\|Y \| and hence \|X ∪Y \| ≥\|A\| as desired. Theorem 6. A maximum matching can be found in a bipartite graph in O(m√n) time. Proof. It is easy to see that the time spent in ﬁnding an augmenting path is O(m) and the total number of augmentations is at most n 2 . So the total time is O(mn). To improve upon this analysis observe that the algorithm for ﬁnding augmenting paths might ﬁnd more than one path. In this case let us augment on a maximal set of disjoint augmenting paths. With this modiﬁcation we can show that the number of phases (where a phase is the construction of the alternating forest) is O(√n). The key observation, which is left as an exercise, is the following: Observation 7. The length of the shortest augmenting path increases in each phase. 5 Given this observation, after √n phases, the augmenting paths all have length at least 2√n + 1. Now consider any optimal matching M ∗and the symmetric diﬀerence of M and M∗. If M is not maximum then there must be some augmenting paths with respect to M in the symmetric diﬀerence. Since each of these has length at least 2√n + 1 there can only be O(√n) such paths in all (the total number of vertices is n). Thus \|M ∗\| −\|M\| < √n and hence the algorithm will terminate in at most √n more phases. 2 General graphs It is not hard to see that the algorithm from the previous section does not apply to general graphs. The main problem is caused by odd cycles with a maximal number of matching edges, i.e. cycles of length 2k + 1 which contain k matching edges. Such cycles are called blossoms, an example of which is shown in Figure 1, where the matching edges are shown in bold. Figure 1: A Blossom. The next lemma that shows us a way to deal with blossoms is the central idea in Edmonds’ algorithm for ﬁnding a maximum matching in general graphs. Lemma 8. (Cycle Shrinking) Let M be a matching of G and B be a blossom. Further, assume that B is vertex-disjoint from (i.e. has no vertices in common with) the rest of M. Consider the graph G′ obtained by contracting B to a single vertex. Then the matching M ′ of G′ induced by M is maximum in G′ iﬀM is maximum in G. Proof. First suppose that M ′ is not maximum in G′. From Lemma 2 it follows that G′ contains an augmenting path P ′ with respect to M ′. Suppose that P ′ does not intersect the blossom B in G, then P ′ is also an augmenting path in G and hence M is not maximum. So P ′ intersects B in G. In particular, the contracted blossom B must be an end vertex of the path P ′ in G′ since B is vertex-disjoint from M ′. Let P ′ meet B at the vertex v, and let 6 u be the unmatched vertex in the blossom. Let P ′′ be the path from v to u in the blossom that begins with the matching edge incident to v. It is easy to see that P = P ′ ∪P ′′ is then an augmenting path in G and so, again, M is not a maximum matching. Now assume that M is not a maximum matching in G. We will show that M ′ is not a maximum matching in G′. So take an augmenting path P in G. We may assume that P intersects the blossom B, otherwise P is an augmenting path in G′. Note that since B contains only one unmatched vertex, it follows that at least one of the endpoints of P, say w, lies outside B. Let P ′ be the path created by starting at w and following P until it intersects the blossom. Observe that P ′ is an augmenting path in G′ and the result follows. To ﬁnd an augmenting path in a general graph, we will modify the procedure for bipartite graphs, so that it also detects blossoms. If it does, we shrink the blossom and restart on the new graph. Any augmenting path found on the new graph can be easily translated to an augmenting path in the original graph. Further, by the previous lemma, if the matching is maximum in the new graph, then it is also maximum in the original graph. Here is a formal description of the algorithm. Let M be a matching of G and let U be the subset of unmatched vertices (if every vertex is matched then the matching is maximum). We construct a forest F so that it has one connected component for each vertex of U. As before extend F by alternately adding unmatched and matched edges. Then the edges of M that are added to F will be at an odd distance from U. Also, vertices that are at an odd distance from U will have degree two (with one unmatched edge and one matched edge). Let us call such vertices inner vertices and the rest outer vertices. The vertices of U are all outer vertices. Now consider the neighborhood of outer vertices. One of the following four possibilities must arise. 1. If we ﬁnd an outer vertex x incident to a vertex y not in F, then we can add the edges (x, y) and (y, z) to F where (y, z) is an edge of M. 2. If two outer vertices belonging to diﬀerent components are adjacent, then the roots of these components have an augmenting path between them. 3. If two outer vertices x, y in the same component are adjacent, then let C be the cycle formed by the edge (x, y) along with the path from x to y in F. Let P be the path connecting C to the root of the component. First, we can switch the edges of P to obtain a matching M1 of the same size as P. Then C satisﬁes the condition of the 7 cycle shrinking lemma. So we shrink C to a single vertex and get a new graph G′. Now the goal is to ﬁnd an augmenting path in G′. 4. If every outer vertex only has inner vertices as neighbors, then M is already maximum. Too see this suppose F has p inner vertices and q outer vertices. Then q −p = \|U\| since each matched outer vertex is matched with an inner vertex and vice versa. Now if we delete all the inner vertices of F from G, then the outer vertices will all be isolated components. But this means that any matching of G has to miss at least q −p of them, and hence q −p vertices of G. Since M misses exactly q −p vertices, it must be maximum. Thus, from the description of the algorithm, we obtain the lemma below. Lemma 9. At each step of the algorithm, we either increase the size of F, or decrease the size of G or ﬁnd an augmenting path or stop with a maximum matching. Theorem 10. A maximum matching can be found in O(n4) time. Proof. Clearly our algorithm makes less than n augmentations. In addition, we can shrink at most n blossoms before ﬁnding an augmenting path. Finding an augmenting path or a blossom takes O(m) time since in growing a forest we examine each edge at most once. Hence our overall running time is O(mn2) = O(n4). The following theorem can be derived from Edmonds’ algorithm. Theorem 11 (Tutte). A graph G has a perfect matching iﬀfor any subset of vertices X, the number of odd-sized components of the graph G \ X obtained by deleting X from G is at most \|X\|. Proof. The necessity of the right hand condition is clear: if there exists a set of vertices X such that G −X has more that \|X\| odd-sized components, then there aren’t enough vertices in X to match all the odd-sized components, because odd-sized components need an external vertex to be matched and can only be matched with vertices in X. For the suﬃciency, consider the forest in Edmonds’ algorithm at the last step. Denote by X the set of inner vertices, p = \|X\|. Note that the vertices of X haven’t been shrunk, because shrunk vertices have to be unmatched. If we consider G−X, then we get the outer vertices as isolated components (that is how the algorithm terminates: outer vertices only have inner vertices as neighbors). Some of these may correspond to shrunk odd components in the original graph. As in the description of the algorithm, call q the number of outer 8 vertices, so that q −p is the number of unmatched vertices. Because of our hypothesis (applied to the set X), we have at most \|X\| odd components in the original graph, that is, we have at most p outer vertices. In other words, q = p and all vertices are matched. In his paper (called “Paths, Trees and Flowers”) describing this algorithm, Edmonds also deﬁned the notion of polynomial-time algorithms. In the decades since, this notion has come to play a fundamental role in complexity theory. 9
5809	https://www.nature.com/articles/s41598-025-06018-5	Spin conductivity in two-dimensional Heisenberg model on Lieb lattice under magnetic field and spin-orbit interactions \| Scientific Reports Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Thank you for visiting nature.com. You are using a browser version with limited support for CSS. 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Advertisement View all journals Search Search Search articles by subject, keyword or author Show results from Search Advanced search Quick links Explore articles by subject Find a job Guide to authors Editorial policies Log in Explore content Explore content Research articles News & Comment Collections Subjects Follow us on Facebook Follow us on Twitter Sign up for alerts RSS feed About the journal About the journal About Scientific Reports Contact Journal policies Guide to referees Calls for Papers Editor's Choice Journal highlights Open Access Fees and Funding Publish with us Publish with us For authors Language editing services Open access funding Submit manuscript Sign up for alerts RSS feed nature scientific reports articles article Spin conductivity in two-dimensional Heisenberg model on Lieb lattice under magnetic field and spin-orbit interactions Download PDF Download PDF Article Open access Published: 02 July 2025 Spin conductivity in two-dimensional Heisenberg model on Lieb lattice under magnetic field and spin-orbit interactions Farshad Azizi1& Hamed Rezania2 Scientific Reportsvolume 15, Article number:22815 (2025) Cite this article 993 Accesses 1 Citations Metrics details Abstract We investigate here the impact of an out-of-plane magnetic field and spin-orbit interaction on the spin conductivity of the two-dimensional Heisenberg model on a Lieb lattice. In this study, the Hamiltonian of the spin model has been transformed into a strongly interacting bosonic gas using a hard boson transformation. In this transformation, the occupation of a boson at each site is restricted by adding a hard core repulsion. To determine the excitation spectrum of the mapped model, the Green’s function method has been employed. Based on the spectrum of the bosonic gas, the two-particle Green’s function related to the spin conductivity of the two-dimensional Heisenberg model has been calculated. Computational results indicate that with an increase in the strength of the Dzyaloshinskii-Moriya interaction, the peak position in the dynamic spin conductivity shifts to higher frequencies under a constant magnetic field. However, the magnetic field does not affect the peak position of the dynamic spin conductivity. On the other hand, the intensity of dynamic spin conductivity increases with the strength of the Dzyaloshinskii-Moriya interaction. Our findings suggest that for a range of values of Dzyaloshinskii-Moriya interaction strength, the static transverse structure factor continuously decreases with the magnetic field. Additionally, for each value of the magnetic field, the temperature dependence of the static spin conductivity of localized electrons on the lattice exhibits a limited temperature peak. Similar content being viewed by others Spin structure factors of doped monolayer Germanene in the presence of spin-orbit coupling Article Open access 07 April 2021 Spin-orbit coupling effects on transport properties of electronic Lieb lattice in the presence of magnetic field Article Open access 20 May 2022 Biquadratic exchange interactions in two-dimensional magnets Article Open access 09 October 2020 Introduction Studying the Heisenberg model on lattices like the Lieb lattice enhances our understanding of quantum magnetism. This study clarifies the nature of quantum spin liquids, the connection between zero-point motion and spin interactions creates opportunities for investigating novel states of matter that may find use in quantum technologies1,2,3. The possibility of realizing a quantum spin liquid, a magnetically disordered state that preserves the system’s symmetries even at absolute zero temperature, has recently attracted significant attention to quantum magnetism on geometrically two-dimensional frustrated spin systems with (S=1/2)2,3,4,5,6 $$_2$$ .” Proc. Natl. Acad. Sci. 119(51), e2211193119 (2022)."),7, 184426 (2023)."). Due to significant quantum effects on magnetic characteristics that differ significantly from those of three-dimensional antiferromagnets, the low-dimensional spin model with antiferromagnetic interactions received a lot of attention8, 224414 (2009)."),9. Phys. Rev. Mater. 4(10), 104420 (2020)."),10, 184427 (2004)."). The Heisenberg model with first exchange interaction in Lieb lattice is among the most crucial for characterizing these systems11, 063601 (2011)."),12."),13."). Quantum zero-point motion in low-dimensional spin systems can suppress long-range magnetic order, leading to a quantum spin liquid state—a correlated state with topological properties and no symmetry breaking, capable of supporting fractionalized excitations.15, 4138 (2018)."),16, 153–160 (1973)."),17, 423–440 (1974)."),18, 365 (1988)."),19, 12377 (1992)."),20, 207203 (2005)."). Due to the unique features caused by its topology, the two-dimensional Lieb lattice21, a line-centered square lattice, has garnered significant scientific attention. Three atoms make up a square unit cell in this lattice. Additionally, a three-band structure with particle-hole symmetry and a flat band touching two linearly dispersing bands at a Dirac point define this structure’s energy spectrum22,23,24,25,26. Dirac cones are known to produce anomalous behavior, including essentially massless fermions. Flat bands have the potential to raise the superconducting transition temperature23,27 and enable the realization of magnetic order28 due to the fractional quantum Hall effect29,30. The quantum antiferromagnetic (S=1/2) Heisenberg model on the Lieb lattice, a line-centered square lattice, with nearest-neighbor antiferromagnetic exchange coupling, is particularly interesting. In fact, the CuO(_{2}) planes of cuprate superconductors are a common example of a Lieb lattice.31, 2794 (1987)."),32, 770 (1991)."). The Lieb lattice may also be implemented on many physical platforms, including cold atoms11, 063601 (2011)."),33, 995–1001 (2017)."),34, 365–393 (2018)."),35, 260401 (2005).") in optical lattices, according to certain theories11, 063601 (2011)."),36, 041402 (2010)."). Conversely, experimental findings demonstrate that a dispersionless localized flat-band state may be realized in Lieb photonic lattices23, 245504 (2015)."). Numerous spin half compounds that might be well explained by two-dimensional Heisenberg antiferromagnetic models have been studied experimentally37, 170401 (2017)."),38."),39."). There are still unanswered questions in the active field of Hamiltonian research40,41. Experimental techniques on materials with low dimensional electronic systems have validated the findings of theoretical research42 into the transport characteristics of low dimensional magnets43. Prominent spin chain materials with transport and magnetic properties include CaCu({2})O({3})44, 027201 (2007)."), SrCuO$_{2}$ and Sr$_{2}$CuO$_{3}$45, 054412 (2001)."),46, 020405 (2010)."). Numerous investigations have focused on the impact of magnetic fields on transport properties47, 137202 (2008)."). An experimental study has reported the findings on magnetothermal transport in the spin-1/2 chain compound copper pyrazine dinitride. Thermal conductivity experiments at magnetic fields of several Teslas have been reported for this molecule, which has a tiny exchange constant of about 10 K. A theoretical approach has been used to explain these experimental findings48, 064406 (2009)."). Dzyaloshinskii-Moriya spin anisotropy in the exchange coupling between nearest neighbor spins defines antisymmetric anisotropic features that may be used to study the impact of spin-orbit couplings49, 042319 (2009)."),50, 054433 (2020)."),51, 336 (2020)."),52, 2050138 (2020)."),53, 134418 (2020)."),54, 214414 (2008)."). A DM vector perpendicular to the layer introduces the coupling constant strength of the Dzyaloshiskii-Moriya (DM) interaction, resulting in an easy-plane spin anisotropic Hamiltonian for the Lieb lattice. The model Hamiltonian’s _SU_(2) symmetry is broken by such DM interaction, reducing it to _U_(1) symmetry around the _D_ vector55: Direct evidence of the Dzyaloshinskii-Moriya interaction.” Phys. Rev. Lett. 104(13), 137203 (2010)."),56, L100403 (2023)."),57, 014416 (2019)."). We have previously investigated the various properties of different magnetic structures using the Heisenberg model in the presence of the Dzyaloshinskii-Moriya interaction, which represents the spin-orbit coupling58."),59, 1597–1612 (2022)."),60."),61, 081012 (2021)."),62."),63."),64."),65."). In the model Hamiltonian, we have also included a Zeeman component that describes how the localized spins interact with the intensity of the external magnetic field, _B_. A field-induced ferromagnetic state with gapped magnon excitations is the ground state of the model Hamiltonian for applied large magnetic fields (_B_) perpendicular to the Lieb lattice plane66, 10019 (1996)."). The energy difference between the model’s ground state and first excited state disappears at the critical field ($B_{c}$) when the magnetic field is reduced, and a spiral transverse magnetic ordering appears67, 110303 (2014)."),68, 174427 (2004)."). Zhuo et al.69, 054412 (2024).") investigates the thermal Hall effect and topological characteristics of magnons in lattice-trimmed ferromagnets. The authors demonstrate that the system may transition between a trivial insulator phase and a topological magnon insulator using spin-wave theory, with notable changes in thermal Hall conductivity at critical transition points. These results might be investigated in high-temperature superconductors or organic magnetic materials69, 054412 (2024)."). The excitation spectrum is generated by transforming a spin Hamiltonian to a bosonic one in the current work using hard core bosonic representation70,71. As long as the hard core repulsion between particles is included in the bosonic Hamiltonian, such a bosonic transformation for localized electrons on the Lieb lattice is valid71. A hopping term for bosonic particles is added to the main portion of the bosonic model Hamiltonian by an anisotropic exchange interaction brought on by the DM interaction72,73,74. The DM interaction not only modifies critical points and thermodynamic features, but it also produces distinct universal behavior. Heisenberg chain and anisotropic spin ladders are two examples of the transport characteristics of the anisotropic antiferromagnetic Heisenberg model Hamiltonian that we have examined in earlier research.75,76 The goal of this study is to determine how the magnetic field and antisymmetric exchange interaction (DM interaction) affect the static and dynamic spin conductivities of localized spins on a two-dimensional Lieb lattice at a limited temperature in the field-induced gapped spin-polarized phase. In particular, for magnetic fields beyond threshold field (B_{c}), we investigate the impact of magnetic field and Dzyaloshinskii-Moriya interaction strength on the static and dynamic spin conductivities of the Heisenberg model on Lieb lattice. The frequency dependence of the dynamical spin conductivity in the anisotropic Heisenberg model on the Lieb lattice is analyzed for magnetic fields above the critical threshold. The excitation spectrum in terms of many body computations for bosonic gas is provided by the hard core boson transformation for spin operators, which we have implemented77. We have determined the bosonic self-energy to get the spinon dispersion relation using the Brueckner technique77. In order to compute the spin conductivity of the model Hamiltonian including a fictional magnetic field gradient is employed to obtain a spin current. Spin conductivity has been determined using bosonic Green’s function, which is based on the formulation of the Kubo formula for the spin transport characteristics of the Heisenberg magnet78. The final part will go over and examine our findings to demonstrate how the frequency dependence of the dynamical spin conductivity of localized electrons on Lieb lattice is influenced by the magnetic field and the intensity of the Dzyaloshinskii-Moriya interaction. Additionally, the temperature behavior of static spin conductivity has been examined for a range of magnetic field and Dzyaloshinskii-Moriya interaction strength values. Fig. 1 Lieb lattice showing three-site basis in unit cell, with square unit cell so that its constructing vectors are (\textbf{a}{1},\textbf{a}{2}). Full size image Theoretical formalism Our goal in this work was to highlight the spin transport characteristics of the localized electrons on the line-centered square Lieb lattice. To do this, we plotted the Lieb lattice structure in the left panel of Fig.1, which shows that each unit cell has three distinct sublattices (A, B, and H). The lattice site in sublattice H has four nearest neighbor sites in the other sublattices. The constructing vectors unit cell with amplitude a in Lieb lattice have been given by (\textbf{a}{1}=a\textbf{i}) and (\textbf{a}{2}=a\textbf{j}) so that (\textbf{i}) and (\textbf{j}) denote the unit vectors along the x and y axis respectively. The localized electron dynamics have been described using the antiferromagnetic Heisenberg model Hamiltonian, taking into account the antisymmetric anisotropic Dzyaloshinskii-Moria interaction on the Lieb lattice, as follows: $$\begin{aligned} H=J\sum {\langle ij\rangle } \textbf{S}{i}\cdot \textbf{S}{j}+\textbf{D}\cdot \sum {\langle ij\rangle } \textbf{S}{i}\times \textbf{S}{j}-g\mu {B}B\sum {i}S_{i}^{z}, \end{aligned}$$ (1) where the nearest neighbor lattice sites in a Lieb lattice to be denoted by the symbol (\langle ij\rangle). J introduces the intensity of the antiferromagnetic interaction between spins on nearest neighbor lattice sites. The antisymmetric element in the model Hamiltonian with vector (\textbf{D}=(0,0,D)) provides the spin orbit coupling effects, where D denotes the intensity of the Dzyaloshinskii-Moriya (DM) interaction between the spins of localized electrons on nearest neighbor sites. Additionally, the parameters (g\simeq 2.2) and (\mu B) introduce gyromagnetic and Bohr magneton constants, respectively. In order to determine the magnetic field strength, _B, a fixed magnetic field perpendicular to the Lieb lattice plane has been applied. A Zeeman term of this kind is appended to the initial model Hamiltonian. The SU(2) symmetry of the anisotropic Heisenberg model Hamiltonian in Eq. (1) is reduced by the DM interaction and Zeeman terms. We now discuss the translational invariance of Eq. (1), which is essential for the validity of our methodology. The Lieb lattice is a line-centered square lattice with a three-site unit cell (sublattices A, B, and H), and its translational invariance is determined by the periodicity of the lattice and the interactions. To prove translational invariance, we consider the lattice structure and the form of the interactions. As mentioned before, the Lieb lattice is periodic with lattice vectors (\textbf{a}1 = a \textbf{i}) and (\textbf{a}_2 = a \textbf{j}). The Heisenberg term ((J \sum {\langle ij \rangle } \textbf{S}_i \cdot \textbf{S}_j)) is invariant under lattice translations because it involves scalar products of spins at nearest-neighbor sites, which are defined consistently across the lattice. Since the model Hamiltonian has been applied for a crystal, thus Hamiltonian is invariant under translational vector. It means that Hamiltonian commutes with tranlational vector operator. The Zeeman term ((-g \mu _B B \sum _i S_i^z)) is also translationally invariant, as it applies uniformly to all sites. The DM term ((\textbf{D} \cdot \sum _{\langle ij \rangle } \textbf{S}_i \times \textbf{S}_j)) requires careful examination. For a pair of nearest-neighbor sites (i) and (j), the DM interaction is: $$\begin{aligned} \textbf{D} \cdot (\textbf{S}_i \times \textbf{S}_j) = D (S_i^x S_j^y - S_i^y S_j^x), \end{aligned}$$ (2) since (\textbf{D} = (0, 0, D)). This term is antisymmetric under the exchange of (i) and (j), but the sum over all nearest-neighbor pairs (\langle ij \rangle) must be checked for translational invariance. In the Lieb lattice, each site in sublattice H has four nearest neighbors (two in sublattice A and two in sublattice B), and the lattice vectors (\Delta = {0, -\textbf{a}_1}) and (\Delta ' = {0, -\textbf{a}_2}) define the connections. The DM vector (\textbf{D}) is uniform across the lattice, and the lattice’s periodic structure ensures that the interaction is identical for equivalent pairs in different unit cells. Thus, translating the Hamiltonian by a lattice vector (\textbf{a}_1) or (\textbf{a}_2) leaves the Hamiltonian unchanged, confirming translational invariance. Regarding the reviewer’s point about Skyrmions, we acknowledge that DM interactions can stabilize Skyrmion configurations in certain magnetic systems, particularly in non-centrosymmetric lattices or with specific boundary conditions. However, in our study, we focus on the field-induced spin-polarized phase above the critical magnetic field ((B > B_c)), where the ground state is a gapped, fully polarized ferromagnetic state, as noted in Sect.167,68. In this regime, the strong magnetic field aligns spins along the (z)-direction, suppressing Skyrmion formation, which typically occurs in systems with competing interactions or lower fields. Our Fourier transformation is valid in this phase, as the system remains periodic and translationally invariant. In summary, we can mention that the Lieb lattice’s periodicity, defined by lattice vectors (\textbf{a}_1 = a \textbf{i}) and (\textbf{a}_2 = a \textbf{j}), ensures translational invariance of the Hamiltonian (Eq. (1)). The Heisenberg and Zeeman terms are inherently translationally invariant due to their uniform application across lattice sites. The DM interaction, with a uniform (\textbf{D} = (0, 0, D)), preserves this invariance because the nearest-neighbor pairs (\langle ij \rangle) are defined consistently across the lattice, maintaining periodicity. In the field-induced spin-polarized phase ((B > B_c)), the strong magnetic field suppresses non-uniform configurations like Skyrmions, ensuring the validity of the Fourier transformation used in our analysis. Three distinct bosonic operators must be used in a hard core bosonic transformation in order to translate the spin model Hamiltonian to a bosonic one. In unit cell index l with sublattice A, hard core bosonic transformations for spin operators of localized electrons are provided by $$\begin{aligned} S^{+}{l,A}=a{l},\;\;S^{-}{l,A}=a^{\dag }{l} ,\;\;S^{z}{l,A}=1/2-a^{\dag }{l}a_{l}, \end{aligned}$$ (3) where (S^{+}) and (S^{-}) to represent the spin ladder operators. The hard core transformation for localized electrons in unit cell index l on sublattice B can be introduced similarly as $$\begin{aligned} S^{+}{l,B}=b{l},\;\;S^{-}{l,B}=b^{\dag }{l} ,\;\;S^{z}{l,B}=1/2-b^{\dag }{l}b_{l} \end{aligned}$$ (4) Similarly, we have transformations for sublattice H. $$\begin{aligned} S^{+}{l,H}=h{l},\;\;S^{-}{l,H}=h^{\dag }{l} ,\;\;S^{z}{l,H}=1/2-h^{\dag }{l}h_{l} \end{aligned}$$ (5) Additionally, the creation operators for bosonic particles on lattice sites in unit cell index l on sublattice A,B,H are described by (a^{\dag }{l}, b^{\dag }{l}), and (h^{\dag }_{l}). The model Hamiltonian in Eq.(1) is obtained using hard core bosonic transformations as $$\begin{aligned} \mathcal {H}= & \frac{J}{2}\sum {i,\Delta }\Big (S^{+}{i,H}S^{-}{i+\Delta ,A}+S^{-}{i,H}S^{+}{i+\Delta ,A}\Big ) +J\sum {i,\Delta }S^{z}{i,H}S^{z}{i+\Delta ,A} \nonumber \+ & \frac{J}{2}\sum {i,\Delta '}\Big (S^{+}{i,H}S^{-}{i+\Delta ',B}+S^{-}{i,H}S^{+}{i+\Delta ',B}\Big ) +J\sum {i,\Delta '}S^{z}{i,H}S^{z}{i+\Delta ',B} \nonumber \+ & \sum {i,\Delta } \frac{D}{2i}\Big (S^{-}{i,H}S^{-}{i+\Delta ,A}-S^{+}{i,H}S^{-}{i+\Delta ,A}\Big )+ \sum {i,\Delta '} \frac{D}{2i}\Big (S^{-}{i,H}S^{-}{i+\Delta ',B}-S^{+}{i,H}S^{-}{i+\Delta ',B}\Big ) \nonumber \- & g\mu {B}B\sum {i}\Big (S^{z}{i,A}+S^{z}{i,B}+S^{z}_{i,H}\Big ). \end{aligned}$$ (6) (\Delta) and (\Delta ') in the equation above are the translational vectors joining the nearest neighbours unit cells and are provided by $$\begin{aligned} \Delta =\textbf{0},\textbf{a}{1}, \Delta '=\textbf{0},\textbf{a}{2}, \end{aligned}$$ (7) where the vectors connect sublattice H to A and B sites, respectively, consistent with the lattice geometry Fig.1. The Fourier transformations of bosonic operators can be described as $$\begin{aligned} a_\textbf{k}=\frac{1}{\sqrt{N}}\sum \textbf{k}e^{-i\textbf{k}.\textbf{R}{l}}a_{l},\;\; b_\textbf{k}=\frac{1}{\sqrt{N}}\sum \textbf{k}e^{-i\textbf{k}.\textbf{R}{l}}b_{l},\;\; h_\textbf{k}=\frac{1}{\sqrt{N}}\sum \textbf{k}e^{-i\textbf{k}.\textbf{R}{l}}h_{l}. \end{aligned}$$ (8) Following the application of hard core bosonic transformation and the aforementioned definitions, the bilinear component of the bosonic model hamiltonian takes on the following form. $$\begin{aligned} \mathcal {H}{bil}= & \sum \textbf{k}\Big [\Big (g\mu {B}B-2J\Big )h^{\dag }\textbf{k}h_\textbf{k}+ \Big (g\mu {B}B-J\Big )( a^{\dag }\textbf{k}a_\textbf{k}+b^{\dag }\textbf{k}b\textbf{k}) +\phi {1}(\textbf{k})h^{\dag }{\textbf{k}}a_{\textbf{k}}+\phi ^{}{1}(\textbf{k})a^{\dag }{\textbf{k}}h_{\textbf{k}}\nonumber \+ & \phi {2}(\textbf{k})h^{\dag }{\textbf{k}}b_{\textbf{k}}+\phi ^{}{2} (\textbf{k})b^{\dag }{\textbf{k}}h_{\textbf{k}}\Big ],\nonumber \ \phi {1}(\textbf{k})= & \frac{1}{2}(J-iD)(1+e^{-ik{x}a}),\nonumber \ \phi {2}(\textbf{k})= & \frac{1}{2}(J-iD)(1+e^{-ik{y}a}), \end{aligned}$$ (9) where the wave vectors (\textbf{k}) are part of the Lieb lattice’s first Brillouin zone. The model Hamiltonian’s bilinear portion, (\textbf{H}_{bil}(\textbf{k})), may be expressed in matrix form as follows: $$\begin{aligned} \textbf{H}{bil}(\textbf{k})= & \left( \begin{array}{ccc} g\mu {B}B-J & 0 & \phi {1}^{}(\textbf{k}) \ 0 & g\mu {B}B-J & \phi {2}^{}(\textbf{k})\ \phi {1}(\textbf{k})& \phi {2}(\textbf{k}) & g\mu {B}B-2J\ \end{array} \right) . \end{aligned}$$ (10) The non-interacting portion of the model Hamiltonian may be expressed in diagonalized form since the equation values of the model Hamiltonian in Eq.(10) are easily obtained. $$\begin{aligned} \mathcal {H}{bil}= & \sum \textbf{k}\sum ^{3}{j=1} \omega {j}(\textbf{k})\gamma ^{\dag }{j,\textbf{k}}\gamma {j,\textbf{k}}, \nonumber \ \omega {1}(\textbf{k})= & g\mu {B}B-J,\;\; \omega {2}(\textbf{k})=\frac{1}{2}\Big (2g\mu {B}B-3J-\sqrt{4(\|\phi {1}(\textbf{k})\|^{2} +\|\phi {2}(\textbf{k})\|^{2})}+J^{2}\Big ),\nonumber \ \omega {3}(\textbf{k})= & \frac{1}{2}\Big (2g\mu {B}B-3J+\sqrt{4(\|\phi {1}(\textbf{k})\|^{2} +\|\phi {2}(\textbf{k})\|^{2})}+J^{2}\Big ). \end{aligned}$$ (11) An excitation spectrum of non-interacting bosonic particles is introduced in the equation (\omega {j}(\textbf{k})) above. The expansion of the old operators (a^{\dag }\textbf{k}), (b^{\dag }\textbf{k}), and (h^{\dag }\textbf{k}) using the eigenvectors of the matrix form of the bilinear Hamiltonian in Eq.(9) for each eigenvalue can be found in terms of new operators (\gamma ^{\dag }_{j,\textbf{k}}) in Eq.(11). $$\begin{aligned} a^{\dag }\textbf{k}= & u{12}^{}(\textbf{k})\gamma ^{\dag }{1,\textbf{k}} +u{22}^{}(\textbf{k})\gamma ^{\dag }{2,\textbf{k}}+u{32}^{}(\textbf{k})\gamma ^{\dag }{3,\textbf{k}}\nonumber \ b^{\dag }\textbf{k}= & u_{13}^{}(\textbf{k})\gamma ^{\dag }{1,\textbf{k}} +u{23}^{}(\textbf{k})\gamma ^{\dag }{2,\textbf{k}}+u{33}^{}(\textbf{k}) \gamma ^{\dag }{3,\textbf{k}}\nonumber \ h^{\dag }\textbf{k}= & u_{21}^{}(\textbf{k}) \gamma ^{\dag }{2,\textbf{k}}+u{31}^{}(\textbf{k})\gamma ^{\dag }_{3,\textbf{k}}, \end{aligned}$$ (12) In the equation above, the coefficients of expansions, or (u_{ij}(\textbf{k})), are provided by $$\begin{aligned} u_{12}(\textbf{k})= & -\frac{1}{\sqrt{1+\frac{\|\phi {2}(\textbf{k})\|^{2}}{\|\phi {1}(\textbf{k})\|^{2}}}} \frac{\phi {2}(\textbf{k})}{\phi {1}(\textbf{k})},\;\; u_{13}(\textbf{k})=-\frac{1}{\sqrt{1+\frac{\|\phi {2}(\textbf{k})\|^{2}}{\|\phi {1}(\textbf{k})\|^{2}}}},\nonumber \ u_{21}(\textbf{k})= & -\frac{1}{\sqrt{1+\frac{\|\phi {1}(\textbf{k})\|^{2}}{\|\phi {2}(\textbf{k})\|^{2}}+ \frac{(J+\mathcal {C}(\textbf{k}))^{2}}{4\|\phi {2}(\textbf{k})\|^{2}}}} \frac{J+\mathcal {C}(\textbf{k})}{2\phi ^{}{2}(\textbf{k})},\nonumber \ u{22}(\textbf{k})= & \frac{1}{\sqrt{1+\frac{\|\phi {1}(\textbf{k})\|^{2}}{\|\phi {2}(\textbf{k})\|^{2}}+ \frac{(J+\mathcal {C}(\textbf{k}))^{2}}{4\|\phi _{2}(\textbf{k})\|^{2}}}} \frac{\phi ^{}{1}(\textbf{k})}{\phi ^{}{2}(\textbf{k})},\;\;u{23}(\textbf{k})=\frac{1}{\sqrt{1+\frac{\|\phi {1}(\textbf{k})\|^{2}}{\|\phi {2}(\textbf{k})\|^{2}}+ \frac{(J+\mathcal {C}(\textbf{k}))^{2}}{4\|\phi {2}(\textbf{k})\|^{2}}}},\nonumber \ u{31}(\textbf{k})= & -\frac{1}{\sqrt{1+\frac{\|\phi {1}(\textbf{k})\|^{2}}{\|\phi {2}(\textbf{k})\|^{2}}+ \frac{(J-\mathcal {C}(\textbf{k}))^{2}}{4\|\phi _{2}(\textbf{k})\|^{2}}}} \frac{J-\mathcal {C}(\textbf{k})}{2\phi ^{}{2}(\textbf{k})},\nonumber \ u{32}(\textbf{k})= & \frac{1}{\sqrt{1+\frac{\|\phi {1}(\textbf{k})\|^{2}}{\|\phi {2}(\textbf{k})\|^{2}}+ \frac{(J-\mathcal {C}(\textbf{k}))^{2}}{4\|\phi {2}(\textbf{k})\|^{2}}}} \frac{\phi ^{}_{1}(\textbf{k})}{\phi ^{}{2}(\textbf{k})},\;\;u_{33}(\textbf{k})=\frac{1}{\sqrt{1+\frac{\|\phi {1}(\textbf{k})\|^{2}}{\|\phi {2}(\textbf{k})\|^{2}}+ \frac{(J-\mathcal {C}(\textbf{k}))^{2}}{4\|\phi {2}(\textbf{k})\|^{2}}}},\nonumber \ \mathcal {C}(\textbf{k})= & \sqrt{4(\|\phi {1}(\textbf{k})\|^{2} +\|\phi _{2}(\textbf{k})\|^{2})+J^{2}}. \end{aligned}$$ (13) Using the aforementioned expansions, we may determine the matrix elements of Matsubara forms of noninteracting bosonic Green’s function in the manner shown below. $$\begin{aligned} G^{(0)}{aa}(\textbf{k},i\omega {n})= & -\int {0}^{1/(k{B}T)}d\tau e^{i\omega {n}\tau }\langle T(a{\textbf{k}}(\tau )a^{\dag }{\textbf{k}}(0))\rangle \nonumber \= & \frac{\|u{12}(\textbf{k})\|^{2}}{i\omega {n}-\omega {1}(\textbf{k})}+\frac{\|u_{22}(\textbf{k})\|^{2}}{i\omega {n}-\omega {2}(\textbf{k})}+ \frac{\|u_{32}(\textbf{k})\|^{2}}{i\omega {n}-\omega {3}(\textbf{k})},\nonumber \ G^{(0)}{bb}(\textbf{k},i\omega {n})= & -\int {0}^{1/(k{B}T)}d\tau e^{i\omega {n}\tau }\langle T(b{\textbf{k}} (\tau )b^{\dag }{\textbf{k}}(0))\rangle \nonumber \= & \frac{\|u{13}(\textbf{k})\|^{2}}{i\omega {n}-\omega {1}(\textbf{k})}+\frac{\|u_{23}(\textbf{k})\|^{2}}{i\omega {n}-\omega {2}(\textbf{k})}+ \frac{\|u_{33}(\textbf{k})\|^{2}}{i\omega {n}-\omega {3}(\textbf{k})},\nonumber \ G^{(0)}{hh}(\textbf{k},i\omega {n})= & -\int {0}^{1/(k{B}T)}d\tau e^{i\omega {n}\tau }\langle T(h{\textbf{k}}(\tau )h^{\dag }{\textbf{k}}(0))\rangle = \frac{\|u{21}(\textbf{k})\|^{2}}{i\omega {n}-\omega {2}(\textbf{k})}+ \frac{\|u_{31}(\textbf{k})\|^{2}}{i\omega {n}-\omega {3}(\textbf{k})},\nonumber \ G^{(0)}{ha}(\textbf{k},i\omega {n})= & -\int {0}^{1/(k{B}T)}d\tau e^{i\omega {n}\tau }\langle T(h{\textbf{k}}(\tau )a^{\dag }{\textbf{k}}(0))\rangle = \frac{u{21}(\textbf{k})u^{}{22}(\textbf{k})}{i\omega {n}-\omega {2}(\textbf{k})}+ \frac{u{31}(\textbf{k})u^{}{32}(\textbf{k})}{i\omega {n}-\omega {3}(\textbf{k})},\nonumber \ G^{(0)}{ah}(\textbf{k},i\omega {n})= & -\int {0}^{1/(k_{B}T)}d\tau e^{i\omega {n}\tau }\langle T(a{\textbf{k}}(\tau )h^{\dag }{\textbf{k}}(0))\rangle = \frac{u^{}{21}(\textbf{k})u{22}(\textbf{k})}{i\omega {n}-\omega {2}(\textbf{k})}+ \frac{u^{}{31}(\textbf{k})u_{32}(\textbf{k})}{i\omega {n}-\omega {3}(\textbf{k})},\nonumber \ G^{(0)}{hb}(\textbf{k},i\omega {n})= & -\int {0}^{1/(k{B}T)}d\tau e^{i\omega {n}\tau }\langle T(h{\textbf{k}}(\tau )b^{\dag }{\textbf{k}}(0))\rangle = \frac{u{21}(\textbf{k})u^{}{23}(\textbf{k})}{i\omega {n}-\omega {2}(\textbf{k})}+ \frac{u{31}(\textbf{k})u^{}{33}(\textbf{k})}{i\omega {n}-\omega {3}(\textbf{k})},\nonumber \ G^{(0)}{ah}(\textbf{k},i\omega {n})= & -\int {0}^{1/(k_{B}T)}d\tau e^{i\omega {n}\tau }\langle T(a{\textbf{k}}(\tau )h^{\dag }{\textbf{k}}(0))\rangle = \frac{u^{}{21}(\textbf{k})u{23}(\textbf{k})}{i\omega {n}-\omega {2}(\textbf{k})}+ \frac{u^{}{31}(\textbf{k})u_{33}(\textbf{k})}{i\omega {n}-\omega {3}(\textbf{k})},\nonumber \ G^{(0)}{ab}(\textbf{k},i\omega {n})= & -\int {0}^{1/(k{B}T)}d\tau e^{i\omega {n}\tau }\langle T(a{\textbf{k}}(\tau )b^{\dag }{\textbf{k}}(0))\rangle \nonumber \= & \frac{u{12}(\textbf{k})u^{}{13}(\textbf{k})}{i\omega {n}-\omega {1}(\textbf{k})}+\frac{u{22}(\textbf{k})u^{}{23}(\textbf{k})}{i\omega {n}-\omega {2}(\textbf{k})}+ \frac{u{32}(\textbf{k})u^{}{33}(\textbf{k})}{i\omega {n}-\omega {3}(\textbf{k})},\nonumber \ G^{(0)}{ba}(\textbf{k},i\omega {n})= & -\int {0}^{1/(k_{B}T)}d\tau e^{i\omega {n}\tau }\langle T(b{\textbf{k}}(\tau )a^{\dag }_{\textbf{k}}(0))\rangle \nonumber \ = & \frac{u^{}{12}(\textbf{k})u{13}(\textbf{k})}{i\omega {n}-\omega {1}(\textbf{k})}+\frac{u^{}{22}(\textbf{k})u{23} (\textbf{k})}{i\omega {n}-\omega {2}(\textbf{k})}+ \frac{u^{}{32}(\textbf{k})u{33}(\textbf{k})}{i\omega {n}-\omega {3}(\textbf{k})}, \end{aligned}$$ (14) In this case, (\omega {n}=2n\pi k{B}T) represents the bosonic Mastabara’s frequency. In the field-induced spin-polarized phase ((B > B_c)), the magnetization (\langle S_\alpha ^z \rangle) for sublattices (\alpha = A, B, H) is computed using the Green’s function formalism (Eqs. (2–4, 12)). At low temperatures ((k_B T/J = 0.05)) and high magnetic fields, we find (\langle S_\alpha ^z \rangle \approx 1/2), with negligible transverse components due to the dominant Zeeman term. The DM interaction introduces anisotropic hopping in the bosonic Hamiltonian but does not significantly disrupt the polarized state in this regime. An intrasite strong repulsion between bosonic particles must be added in order to maintain the SU(2) spin algebra. One partial excitation spectrum of a hard core bosonic gas with intrasite interaction strength ({\mathcal {U}} \rightarrow \infty) may be used to get the excitation spectrum of the original spin model Hamiltonian. This portion of the model Hamiltonian is produced in terms of the Fourier transformation of bosonic operators as $$\begin{aligned} \mathcal {H}{int}=\mathcal {U}\sum {\textbf{k},\mathbf{k'},\textbf{q}}\Big [a^{\dag }{ \textbf{k}+\textbf{q}}a^{\dag }{\mathbf{k'}-\textbf{q}}a_{\mathbf{k'}}a_{\textbf{k}}+b^{\dag }{ \textbf{k}+\textbf{q}}b^{\dag }{\mathbf{k'}-\textbf{q}}b_{\mathbf{k'}}b_{\textbf{k}}+ h^{\dag }{\textbf{k}+\textbf{q}}h^{\dag }{\mathbf{k'}-\textbf{q}}h_\mathbf{k'}h_\textbf{k}\Big ], \end{aligned}$$ (15) increasing the intensity of the local interaction (\mathcal {U}\rightarrow \infty). A note is in order here. This part of Hamiltonian, i.e. (\mathcal {H}{int}) is manually added to the original model Hamiltonian. In other words, this part is not derived by inserting hard core bosonic transformation into original Heisenberg spin model Hamiltonian in Eq. (1). The purpose of this additional hard core repulsion is preservation of SU(2) spin algebra when the hard core bosonic transformation is applied. Eqs. (3, 4, 5) transform the the components of spin operators to the bosonic particles. The necessary condition for the validation of these transformations is the preservation of SU(2) spin algebra. It means commutaion relations ([S{l,\eta }^{\alpha },S_{l',\eta '}^{\beta }]=i\varepsilon {\alpha \beta \gamma }\delta {ll'}\delta {\eta \eta '}) (where (\alpha ,\beta ,\gamma =x,y,z) are spatial coordinates, (l,l') lattice unit cell indexes, (\eta ,\eta ') leabels sublattice indexes.) should be resulted from bosonic commutaion relations ([a{l},a^{\dag }{l'}]=[b{l},b^{\dag }{l'}]=[h{l},h^{\dag }{l'}]=\delta {ll'}). In order to result SU(2) spin algebra from bosonic algebra, it is necessary one boson sits on each lattice site. In other words, we should apply a hard core repulsion between bosonic particles as added Hamiltonian term (\mathcal {H}{int}=\mathcal {U}\sum {l}\Big [a^{\dag }{ l}a^{\dag }{l}a_{l}a_{l}+b^{\dag }{ l}b^{\dag }{l}b_{l}b_{l}+h^{\dag }{ l}h^{\dag }{l}h_{l}h_{l}\Big ]). After applying Fourier transformations of bosonic operators (based on Eq. (8)) to the hard core bosonic repulsion, we arrive to the Eq. (15). The other quartic term in the bosonic Hamiltonian appears when the bosonic transformations are applied to the spin model Hamiltonian. In contrast to the second quartic term, the hard core repulsion component (({\mathcal {U}} \rightarrow \infty)) has a stronger influence. Therefore, it is adequate to consider how hard core repulsion affects the excitation spectrum of the bosonic model hamiltonian in Eq. (15). The off-diagonal components of the bosonic self-energy matrix should disappear due to the intrasite interaction nature of hard core bosons. Dyson’s equation may be used as follows to determine how hard core repulsion affects the Green’s function of bosons: $$\begin{aligned} \textbf{G}(\textbf{k},i\omega {n})=\textbf{G}^{(0)}(\textbf{k},i\omega {n})+ \textbf{G}^{(0)}(\textbf{k},i\omega {n})\mathbf{\Sigma }(\textbf{k},i\omega {n})\textbf{G}(\textbf{k},i\omega _{n}), \end{aligned}$$ (16) such that the diagonal normal self-energy 3(\times)3 matrix resulting from hard core repulsion between bosons is described by (\mathbf{\Sigma }(\textbf{k},i\omega {n})). Additionally, the equation (\textbf{G}^{(0)}(\textbf{k},i\omega {n})(\textbf{G}(\textbf{k},i\omega _{n}))) suggests that the bosonic Green’s function is noninteracting (interacting). The interacting Green’s function matrix elements at the low energy limit are provided by Eq. (16), which is based on Dyson’s equation. $$\begin{aligned} G_{aa}(\textbf{k},\omega )= & \frac{Z^{h}{\textbf{k}}\|u{12}(\textbf{k})\|^{2}}{i\omega {n}-\Omega {3}(\textbf{k})}+ \frac{Z^{a}{\textbf{k}}\|u{22}(\textbf{k})\|^{2}}{i\omega {n}-\Omega {1}(\textbf{k})}+ \frac{Z^{b}{\textbf{k}}\|u{32}(\textbf{k})\|^{2}}{i\omega {n}-\Omega {2}(\textbf{k})},\nonumber \ G_{bb}(\textbf{k},\omega )= & \frac{Z^{h}{\textbf{k}}\|u{13}(\textbf{k})\|^{2}}{i\omega {n}-\Omega {3}(\textbf{k})}+ \frac{Z^{a}{\textbf{k}}\|u{23}(\textbf{k})\|^{2}}{i\omega {n}-\Omega {1}(\textbf{k})}+ \frac{Z^{b}{\textbf{k}}\|u{33}(\textbf{k})\|^{2}}{i\omega {n}-\Omega {2}(\textbf{k})},\nonumber \ G_{hh}(\textbf{k},\omega )= & \frac{Z^{a}{\textbf{k}}\|u{21}(\textbf{k})\|^{2}}{i\omega {n}-\Omega {1}(\textbf{k})}+ \frac{Z^{b}{\textbf{k}}\|u{31}(\textbf{k})\|^{2}}{i\omega {n}-\Omega {2}(\textbf{k})},\nonumber \ G_{ha}(\textbf{k},i\omega {n})= & \frac{Z^{b}{\textbf{k}}u_{21}(\textbf{k})u^{}{22}(\textbf{k})}{i\omega {n}-\Omega {2}(\textbf{k})}+ \frac{Z^{h}{\textbf{k}}u_{31}(\textbf{k})u^{}{32}(\textbf{k})}{i\omega {n}-\Omega {3}(\textbf{k})},\nonumber \ G{ah}(\textbf{k},i\omega {n})= & \frac{Z^{b}{\textbf{k}}u^{}{21}(\textbf{k})u{22}(\textbf{k})}{i\omega {n}-\Omega {2}(\textbf{k})}+ \frac{Z^{h}_{\textbf{k}}u^{}{31}(\textbf{k})u{32}(\textbf{k})}{i\omega {n}-\Omega {3}(\textbf{k})},\nonumber \ G_{hb}(\textbf{k},i\omega {n})= & \frac{Z^{b}{\textbf{k}}u_{21}(\textbf{k})u^{}{23}(\textbf{k})}{i\omega {n}-\Omega {2}(\textbf{k})}+ \frac{Z^{h}{\textbf{k}}u_{31}(\textbf{k})u^{}{33}(\textbf{k})}{i\omega {n}-\Omega {3}(\textbf{k})},\nonumber \ G{bh}(\textbf{k},i\omega {n})= & \frac{Z^{b}{\textbf{k}}u^{}{21}(\textbf{k})u{23}(\textbf{k})}{i\omega {n}-\Omega {2}(\textbf{k})}+ \frac{Z^{h}_{\textbf{k}}u^{}{31}(\textbf{k})u{33}(\textbf{k})}{i\omega {n}-\Omega {3}(\textbf{k})},\nonumber \ G_{ab}(\textbf{k},i\omega {n})= & \frac{Z^{a}{\textbf{k}}u_{12}(\textbf{k})u^{}{13}(\textbf{k})}{i\omega {n}-\Omega {1}(\textbf{k})}+\frac{ Z^{b}{\textbf{k}}u_{22}(\textbf{k})u^{}{23}(\textbf{k})}{i\omega {n}-\Omega {2}(\textbf{k})}+ \frac{Z^{h}{\textbf{k}}u_{32}(\textbf{k})u^{}{33}(\textbf{k})}{i\omega {n}-\Omega {3}(\textbf{k})},\nonumber \ G{ba}(\textbf{k},i\omega {n})= & \frac{Z^{a}{\textbf{k}}u^{}{12}(\textbf{k})u{13}(\textbf{k})}{i\omega {n}-\Omega {1}(\textbf{k})}+\frac{Z^{b}{\textbf{k}}u^{}{22}(\textbf{k})u{23} (\textbf{k})}{i\omega {n}-\Omega {2}(\textbf{k})}+ \frac{Z^{h}_{\textbf{k}}u^{}{32}(\textbf{k})u_{33}(\textbf{k})}{i\omega {n}-\Omega {3}(\textbf{k})}, \end{aligned}$$ (17) with (\Omega {1}(\textbf{k})=Z^{a}{\textbf{k}}(\omega {1}(\textbf{k})+\Sigma {aa}(\textbf{k},0))), (\Omega {2}(\textbf{k})=Z^{b}{\textbf{k}}(\omega {2}(\textbf{k})+\Sigma {bb}(\textbf{k},0))) and (\Omega {3}(\textbf{k})=Z^{h}{\textbf{k}}(\omega {3}(\textbf{k})+\Sigma {hh}(\textbf{k},0))) introduce the renormalized excitation spectrum of interacting bosonic gas. Additionally, the renormalized weight of a single particle is provided by $$\begin{aligned} (Z^{a}{\textbf{k}})^{-1}= & 1-(\frac{\partial Re\Big (\Sigma ^{Ret}{aa}(\textbf{k},\omega )\Big )}{\partial \omega }) {\omega =0}, (Z^{b}{\textbf{k}})^{-1}=1-(\frac{\partial Re\Big (\Sigma ^{Ret}{bb}(\textbf{k},\omega )\Big )}{\partial \omega })\nonumber \ (Z^{h}{\textbf{k}})^{-1}= & 1-(\frac{\partial Re\Big (\Sigma ^{Ret}{hh}(\textbf{k},\omega )\Big )}{\partial \omega }) {\omega =0}\cdot \end{aligned}$$ (18) Brueckner’s method may be used to determine the bosonic self-energy matrix elements77,79,80. Using this method, the ladder diagram summation method is used to generate the self-energy matrix elements for low temperature and the low density limit of bosonic gas. This method’s validity is maintained when the interaction intensity (\mathcal {U}) is extremely big and the interacting Hamiltonian is short ranged, i.e., (\mathcal {H}{int}) in Eq. (15). Where (p{i}\equiv (\textbf{p},(p_{0})){i}), the scattering amplitude (\Gamma (p{1},p_{2};p_{3},p_{4})) of hard core bosons is introduced. We have taken into account the bosonic diagonal elements. We use the Green’s function to create (\Gamma (K\equiv p_{1}+p_{2})). This results from the boson-to-boson interaction being strongly intrasite. The scattering amplitude for each sublattice A,B,H is obtained by utilizing the Feynman’s rules in momentum space for Matsubara’s representation and after taking limit (\mathcal {U}\longrightarrow \infty). $$\begin{aligned} \Gamma {aaaa}(\textbf{K},i\omega {n})= & -\Big (\frac{k_{B}T}{N}\sum {\textbf{Q},m} G^{(0)}{aa}(\textbf{Q},iQ_{m}) G^{(0)}{aa}(\textbf{K}-\textbf{Q},i\omega {n}-iQ_{m}) \Big )^{-1},\nonumber \ \Gamma {bbbb}(\textbf{K},i\omega {n})= & -\Big (\frac{k_{B}T}{N}\sum {\textbf{Q},m} G^{(0)}{bb}(\textbf{Q},iQ_{m}) G^{(0)}{bb}(\textbf{K}-\textbf{Q},i\omega {n}-iQ_{m}) \Big )^{-1},\nonumber \ \Gamma {hhhh}(\textbf{K},i\omega {n})= & -\Big (\frac{k_{B}T}{N}\sum {\textbf{Q},m} G^{(0)}{hh}(\textbf{Q},iQ_{m}) G^{(0)}{hh}(\textbf{K}-\textbf{Q},i\omega {n}-iQ_{m}) \Big )^{-1}. \end{aligned}$$ (19) The wave vector (\textbf{Q}) is a member of the Lieb lattice’s first Brillouin zone, and N is the number of unit cells in the aforementioned relations. Feynman’s principles may be used to conduct the summation over internal Matsubara frequencies (Q_{m}=2m\pi k_{B}T) in the Eq. (19)71. To derive the explicit equations for scattering amplitudes, $$\begin{aligned} \Gamma {aaaa}(\textbf{K},i\omega {n})= & -\Big (\frac{1}{N} \sum {\textbf{Q}}[ \|u{12}(\textbf{Q})u_{12}(\textbf{K}-\textbf{Q})\|^{2}\frac{n_{B}(\omega {1}(\textbf{Q}))-n{B} (-\omega {1}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {1}(\textbf{Q})-\omega {1}(\textbf{K}-\textbf{Q})}\+ & \|u_{12}(\textbf{Q})u_{22}(\textbf{K}-\textbf{Q})\|^{2}\frac{n_{B}(\omega {1}(\textbf{Q}))-n{B} (-\omega {2}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {1}(\textbf{Q})-\omega {2}(\textbf{K}-\textbf{Q})}\ + & \|u_{12}(\textbf{Q})u_{32}(\textbf{K}-\textbf{Q})\|^{2}\frac{n_{B}(\omega {1}(\textbf{Q}))-n{B} (-\omega {3}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {1}(\textbf{Q})-\omega {1}(\textbf{K}-\textbf{Q})}\ + & \|u_{22}(\textbf{Q})u_{12}(\textbf{K}-\textbf{Q})\|^{2}\frac{n_{B}(\omega {2}(\textbf{Q}))-n{B} (-\omega {1}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {2}(\textbf{Q})-\omega {1}(\textbf{K}-\textbf{Q})}\ + & \|u_{22}(\textbf{Q})u_{22}(\textbf{K}-\textbf{Q})\|^{2}\frac{n_{B}(\omega {2}(\textbf{Q}))-n{B} (-\omega {2}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {2}(\textbf{Q})-\omega {2}(\textbf{K}-\textbf{Q})}\ + & \|u_{22}(\textbf{Q})u_{32}(\textbf{K}-\textbf{Q})\|^{2}\frac{n_{B}(\omega {2}(\textbf{Q}))-n{B} (-\omega {3}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {2}(\textbf{Q})-\omega {3}(\textbf{K}-\textbf{Q})}\ + & \|u_{32}(\textbf{Q})u_{12}(\textbf{K}-\textbf{Q})\|^{2}\frac{n_{B}(\omega {3}(\textbf{Q}))-n{B} (-\omega {1}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {3}(\textbf{Q})-\omega {1}(\textbf{K}-\textbf{Q})}\ + & \|u_{32}(\textbf{Q})u_{22}(\textbf{K}-\textbf{Q})\|^{2}\frac{n_{B}(\omega {3}(\textbf{Q}))-n{B} (-\omega {2}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {3}(\textbf{Q})-\omega {2}(\textbf{K}-\textbf{Q})}\ + & \|u_{32}(\textbf{Q})u_{32}(\textbf{K}-\textbf{Q})\|^{2}\frac{n_{B}(\omega {3}(\textbf{Q}))-n{B} (-\omega {3}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {3}(\textbf{Q})-\omega {3}(\textbf{K}-\textbf{Q})}] \Big )^{-1}, \end{aligned}$$ $$\begin{aligned} \Gamma {bbbb}(\textbf{K},i\omega {n})= & -\Big (\frac{1}{N} \sum {\textbf{Q}}[ \|u{13}(\textbf{Q})u_{13}(\textbf{K}-\textbf{Q})\|^{2} \frac{n_{B}(\omega {1}(\textbf{Q}))-n{B} (-\omega {1}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {1}(\textbf{Q})-\omega {1}(\textbf{K}-\textbf{Q})}\+ & \|u_{13}(\textbf{Q})u_{23}(\textbf{K}-\textbf{Q})\|^{2} \frac{n_{B}(\omega {1}(\textbf{Q}))-n{B} (-\omega {2}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {1}(\textbf{Q})-\omega {2}(\textbf{K}-\textbf{Q})}\ + & \|u_{13}(\textbf{Q})u_{33} (\textbf{K}-\textbf{Q})\|^{2}\frac{n_{B}(\omega {1}(\textbf{Q}))-n{B} (-\omega {3}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {1}(\textbf{Q})-\omega {1}(\textbf{K}-\textbf{Q})}\ + & \|u_{23}(\textbf{Q})u_{13}(\textbf{K}-\textbf{Q})\|^{2}\frac{n_{B}(\omega {2}(\textbf{Q}))-n{B} (-\omega {1}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {2}(\textbf{Q})-\omega {1}(\textbf{K}-\textbf{Q})}\ + & \|u_{23}(\textbf{Q})u_{23}(\textbf{K}-\textbf{Q})\|^{2} \frac{n_{B}(\omega {2}(\textbf{Q}))-n{B} (-\omega {2}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {2}(\textbf{Q})-\omega {2}(\textbf{K}-\textbf{Q})}\ + & \|u_{23}(\textbf{Q})u_{33}(\textbf{K}-\textbf{Q})\|^{2} \frac{n_{B}(\omega {2}(\textbf{Q}))-n{B} (-\omega {3}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {2}(\textbf{Q})-\omega {3}(\textbf{K}-\textbf{Q})}\ + & \|u_{33}(\textbf{Q})u_{13}(\textbf{K}-\textbf{Q})\|^{2} \frac{n_{B}(\omega {3}(\textbf{Q}))-n{B} (-\omega {1}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {3}(\textbf{Q})-\omega {1}(\textbf{K}-\textbf{Q})}\ + & \|u_{33}(\textbf{Q})u_{23}(\textbf{K}-\textbf{Q})\|^{2} \frac{n_{B}(\omega {3}(\textbf{Q}))-n{B} (-\omega {2}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {3}(\textbf{Q})-\omega {2}(\textbf{K}-\textbf{Q})}\+ & \|u_{33}(\textbf{Q})u_{33}(\textbf{K}-\textbf{Q})\|^{2}\frac{n_{B}(\omega {3}(\textbf{Q}))-n{B} (-\omega {3}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {3}(\textbf{Q})-\omega {3}(\textbf{K}-\textbf{Q})}] \Big )^{-1}, \end{aligned}$$ $$\begin{aligned} \Gamma {hhhh}(\textbf{K},i\omega {n})= & -\Big (\frac{1}{N} \sum {\textbf{Q}}[ \|u{21}(\textbf{Q})u_{21}(\textbf{K}-\textbf{Q})\|^{2} \frac{n_{B}(\omega {2}(\textbf{Q}))-n{B} (-\omega {2}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {2}(\textbf{Q})-\omega {2}(\textbf{K}-\textbf{Q})}\nonumber \+ & \|u_{21}(\textbf{Q})u_{31}(\textbf{K}-\textbf{Q})\|^{2} \frac{n_{B}(\omega {2}(\textbf{Q}))-n{B} (-\omega {3}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {2}(\textbf{Q})-\omega {3}(\textbf{K}-\textbf{Q})}\nonumber \+ & \|u_{31}(\textbf{Q})u_{21}(\textbf{K}-\textbf{Q})\|^{2} \frac{n_{B}(\omega {3}(\textbf{Q}))-n{B} (-\omega {2}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {3}(\textbf{Q})-\omega {2}(\textbf{K}-\textbf{Q})}\nonumber \+ & \|u_{31}(\textbf{Q})u_{31}(\textbf{K}-\textbf{Q})\|^{2}\frac{n_{B} (\omega {3}(\textbf{Q}))-n{B} (-\omega {3}(\textbf{K}-\textbf{Q}))}{i\omega {n}-\omega {3}(\textbf{Q})-\omega {3}(\textbf{K}-\textbf{Q})} \Big )^{-1}, \end{aligned}$$ (20) in which the Bose-Einstein distribution function is implied by (n_{B}(x)=1/(e^{x/(k_{B}T)}-1)). Hard core bosonic particle self-energy matrix elements have been determined by using the vertex-function equations in Eq. (19) as $$\begin{aligned} \Sigma {aa}({\textbf {k}},i\omega {n})= & -2\frac{k_{B}T}{N}\sum {p{m},\textbf{p}} \Gamma {aaaa}(p,k;p,k)G^{(0)}{aa}(\textbf{p},ip_{m}), \nonumber \ \Sigma {bb}({\textbf {k}},i\omega {n})= & -\frac{k_{B}T}{N}\sum {p{m},\textbf{p}} \Gamma {bbbb}(p,k;p,k)G^{(0)}{bb}(\textbf{p},ip_{m}), \nonumber \ \Sigma {hh}({\textbf {k}},i\omega {n})= & -\frac{k_{B}T}{N}\sum {p{m},\textbf{p}} \Gamma {hhhh}(p,k;p,k)G^{(0)}{hh}(\textbf{p},ip_{m}). \end{aligned}$$ (21) The following relations for hard core self-energy matrix elements result from the summation over internal Matsubara’s frequencies, or (p_{m}). $$\begin{aligned} \Sigma {aa}(\textbf{k},i\omega {n})= & \frac{1}{N}\sum {\textbf{p}}\Big (\|u{12}(\textbf{p})\|^{2}n_{B}(\omega {1}(\textbf{p})) \Gamma {aaaa}(\textbf{p}+\textbf{k},\omega {1}(\textbf{p})+i\omega {n})\nonumber \ + & \|u_{22}(\textbf{p})\|^{2}n_{B}(\omega {2}(\textbf{p}))\Gamma {aaaa}(\textbf{p}+\textbf{k},\omega {2}(\textbf{p})+i\omega {n})\nonumber \+ & \|u_{32}(\textbf{p})\|^{2}n_{B}(\omega {3}(\textbf{p}))\Gamma {aaaa}(\textbf{p}+\textbf{k},\omega {3}(\textbf{p})+i\omega {n})\Big ),\nonumber \ \Sigma {bb}(\textbf{k},i\omega {n})= & \frac{1}{N}\sum {\textbf{p}}\Big (\|u{13}(\textbf{p})\|^{2}n_{B}(\omega {1}(\textbf{p})) \Gamma {bbbb}(\textbf{p}+\textbf{k},\omega {1}(\textbf{p})+i\omega {n})\nonumber \ + & \|u_{23}(\textbf{p})\|^{2}n_{B}(\omega {2}(\textbf{p}))\Gamma {bbbb}(\textbf{p}+\textbf{k},\omega {2}(\textbf{p})+i\omega {n})\nonumber \ + & \|u_{33}(\textbf{p})\|^{2}n_{B}(\omega {3}(\textbf{p}))\Gamma {bbbb}(\textbf{p}+\textbf{k},\omega {3}(\textbf{p})+i\omega {n})\Big ),\nonumber \ \Sigma {hh}(\textbf{k},i\omega {n})= & \frac{1}{N}\sum {\textbf{p}}\Big (\|u{21}(\textbf{p})\|^{2}n_{B}(\omega {2}(\textbf{p})) \Gamma {hhhh}(\textbf{p}+\textbf{k},\omega {2}(\textbf{p})+i\omega {n})\nonumber \ + & \|u_{31}(\textbf{p})\|^{2}n_{B}(\omega {3}(\textbf{p}))\Gamma {hhhh}(\textbf{p}+\textbf{k},\omega {3}(\textbf{p})+i\omega {n}). \end{aligned}$$ (22) The scattering amplitudes in Eqs. (19, 20) are derived using the ladder diagram summation in Brueckner’s formalism , summing over Matsubara frequencies and wave vectors to account for two-particle interactions in the low-density limit. The self-energies in Eqs. (21, 22) are obtained by integrating the scattering amplitudes with the non-interacting Green’s functions, as detailed above, ensuring consistency with the hard-core boson model. Analytical continuation ((i\omega _{n}\longrightarrow \omega +i0^{+})) of relations in Eq. (22) quickly yields delayed forms of self-energy matrix elements, which are the basis for calculating the system’s physical attributes. Theoretical calculation of dynamical and static spin conductivity Spin conductivity in the Heisenberg model describes the transport of spin degrees of freedom, specifically the (z)-component of magnetization, carried by quantum excitations such as magnons or spinons. This is distinct from charge transport and is a well-established concept in low-dimensional quantum magnets59,60,61,63,64,65,79. In the field-induced spin-polarized phase ((B > B_c)), the strong magnetic field aligns spins, and the DM interaction enables spin flips, facilitating spin transport that we quantify using the Kubo formula. One may add the Zeeman interaction to the original spin model Hamiltonian to get the spin conductivity of localized electrons on the Lieb lattice. The spin operators of the confined electron system are coupled with the inhomogenous time-dependent fictitious magnetic field by this additional Zeeman term. This Zeeman term is shown as the model Hamiltonian that follows. $$\begin{aligned} \mathcal {H}(t)=H-g\mu {B}\sum {i}S^{z}(i)B^{z}(i,t), \end{aligned}$$ (23) where the imaginary inhomogenous and time-dependent magnetic field is implied by (B^{z}(i,t)). According to Eq. (1), H represents the initial antiferromagnetic Heisenberg model Hamiltonian on Lieb lattice. The transport of the z component of magnetization ((j^{z}(x_{i})\equiv j_{i\rightarrow i+x}^{z})) on Lieb lattice from site i to site j along x is obtained using the continuity equation to determine the spin current density operator form, as shown in Fig.1. One example of such an equation is $$\begin{aligned} \sum {i}\Big (\frac{\partial }{\partial x{i}}j^{z}(x_{i})+\frac{\partial }{\partial t}S^{z}_{i}\Big )=0, \end{aligned}$$ (24) Thus, the lattice divergence of the local spin current density at site i is described by (\frac{\partial }{\partial x_{i}}j^{z}(x_{i})). The identity is used to determine the spin current for the z component of spin and along the x axis. $$\begin{aligned} \sum {i}\frac{\partial }{\partial x{i}}\Big (x_{i}j^{z}(x_{i})\Big )=0. \end{aligned}$$ (25) Eq. (24) rewrites Eq. (25) as $$\begin{aligned} \sum {i}j^{z}(x{i})-\sum {i}x{i}\frac{\partial }{\partial t}S^{z}_{i}=0. \end{aligned}$$ (26) The spin current operator in the x direction for localized electrons on the Lieb lattice is stated in terms of spin operators by computing the second term of Eq. (26) using the formula below: $$\begin{aligned} j_{x}^{z}(q=0)=\sum {i}j^{z}(x{i})=\sum {l,\Delta }\Big ( (iJ/2-D/2)S^{+}{l,H}S^{-}{l+\Delta ,A} -(iJ/2+D/2)S^{-}{l,H}S^{+}_{l+\Delta ,A}\Big ) \Big ). \end{aligned}$$ (27) Additionally, using Fourier transformations of hard core bosonic operators, the spin current density operator may be obtained in Eq. (27) as $$\begin{aligned} & j^{z}{x}(q=0)=\sum \textbf{k} \Big (\phi (\textbf{k})a^{\dag }\textbf{k}h\textbf{k}+h.c.\Big ),\nonumber \ & \phi (\textbf{k})=(-D+iJ)e^{-ik_{x}/2}cos(k_{x}/2) \end{aligned}$$ (28) According to the Kubo formula for charge transport, the spin conductivity in two-dimensional objects has already been examined using linear response theory81. Here, the magnetization current is connected to the wavelength and frequency dependent magnetic field via the longitudinal spin conductivity coefficient along the x direction, (\sigma ^{tot}_{xx}(\omega )), which is based on $$\begin{aligned} j_{m,x}(q,\omega )=g\mu {B}j^{z}{x}(q,\omega )=\sigma ^{tot}{xx}(q,\omega )iq{x}B^{z}(q,\omega ). \end{aligned}$$ (29) The following formula represents the regular component of dynamical spin conductivity82,83: $$\begin{aligned} \sigma {xx}(\omega )=\frac{Im(\Pi {xx}(\omega ))}{\omega }. \end{aligned}$$ (30) According to the Kubo formula, the longitudinal retarded current-current correlation function is represented by (\Pi _{xx}(\omega ))78,84. It is easy to derive this correlation function between two current operators using the Matsubara’s formulation as $$\begin{aligned} \Pi {xx}(\omega )=\frac{1}{\hbar N}lim{i\omega {n}\longrightarrow \omega +i0^{+}}\int ^{1/k{B}T}{0}d\tau e^{i\omega {n}\tau }\langle T(j^{z}{x}(q=0,\tau )j^{z}{x}(-q=0,0))\rangle , \end{aligned}$$ (31) In this case, (\omega {n}=2n\pi k{B}T) denotes the bosonic Matsubara. The current-current correlation function may be expressed in terms of bosonic Green’s function elements by substituting the operator form of spin current into Eq. (31) based on Wick’s theorem. $$\begin{aligned} \Pi {xx}(\omega )= & \frac{1}{\hbar N}lim{i\omega {n}\longrightarrow \omega +i0^{+}}\int ^{1/k{B}T}{0}d\tau e^{i\omega {n}\tau } \Big [\phi ^{2}(\textbf{k})G_{ha}(\textbf{k},\tau )G_{ha}(\textbf{k},-\tau )+\phi ^{2}(\textbf{k})G_{ah}(\textbf{k},\tau )G_{ah}(\textbf{k},-\tau )\nonumber \+ & \|\phi (\textbf{k})\|^{2}G_{aa}(\textbf{k},\tau )G_{hh}(\textbf{k},-\tau )+\|\phi (\textbf{k})\|^{2}G_{hh}(\textbf{k},\tau )G_{aa}(\textbf{k},-\tau )\Big ], \end{aligned}$$ (32) Using the Matsubara notation’s description of the Fourier transform of Green’s function, which is (G_{aa}(\textbf{k},\tau )=k_{B}T\sum {m}e^{-i\omega {m}\tau } G_{aa}(\textbf{k},i\omega {m})), using the Fourier transforms of Green’s functions in Eq. (17), yields the following outcome for (\Pi {xx}(\omega )). $$\begin{aligned} \Pi {xx}(\omega )= & \frac{1}{4}\sum \textbf{k}\Big [\phi ^{2}(\textbf{k})Z^{b}{\textbf{k}}u{21}(\textbf{k})u^{}{22}(\textbf{k}) Z^{h}{\textbf{k}}u_{31}(\textbf{k})u^{}{32}(\textbf{k})+\phi ^{2}(\textbf{k})Z^{b}_{\textbf{k}}u^{}{21}(\textbf{k})u_{22}(\textbf{k}) Z^{h}{\textbf{k}}u^{}{31}(\textbf{k})u_{32}(\textbf{k})\nonumber \+ & Z^{h}{\textbf{k}}\|\phi (\textbf{k})\|^{2}Z^{b}{\textbf{k}}\|u_{31}(\textbf{k})\|^{2} \|u_{12}(\textbf{k})\|^{2}\Big ] \Big (\frac{n_{B}(\Omega {2}(\textbf{k}))-n{B}(\Omega {3}(\textbf{k}))}{\omega +\Omega {2}(\textbf{k})-\Omega {3}(\textbf{k})+i0^{+}}+\frac{n{B}(\Omega {3}(\textbf{k}))- n{B}(\Omega {2}(\textbf{k}))}{\omega +\Omega {3}(\textbf{k})-\Omega {2}(\textbf{k})+i0^{+}}\Big )\nonumber \+ & Z^{h}\textbf{k}Z^{b}{\textbf{k}}\|\phi (\textbf{k})\|^{2}\|u{21}(\textbf{k})\|^{2}\|u_{12}(\textbf{k})\|^{2} \Big (\frac{n_{B}(\Omega {1}(\textbf{k}))-n{B}(\Omega {3}(\textbf{k}))}{\omega +\Omega {1}(\textbf{k})-\Omega {3}(\textbf{k})+i0^{+}}+\frac{n{B}(\Omega {3}(\textbf{k}))- n{B}(\Omega {1}(\textbf{k}))}{\omega +\Omega {3}(\textbf{k})-\Omega {1}(\textbf{k})+i0^{+}}\Big )\nonumber \+ & \|\phi (\textbf{k})\|^{2}Z^{a}{\textbf{k}}Z^{b}{\textbf{k}}\Big [\|u{21}(\textbf{k})\|^{2}\|u_{32}(\textbf{k})\|^{2} +\|u_{22}(\textbf{k})\|^{2}\|u_{31}(\textbf{k})\|^{2}\Big ] \Big (\frac{n_{B}(\Omega {1}(\textbf{k}))-n{B}(\Omega {2}(\textbf{k}))}{\omega +\Omega {1}(\textbf{k})-\Omega {2}(\textbf{k})+i0^{+}}\nonumber \ + & \frac{n{B}(\Omega {2}(\textbf{k}))- n{B}(\Omega {1}(\textbf{k}))}{\omega +\Omega {2}(\textbf{k})-\Omega _{1}(\textbf{k})+i0^{+}}\Big ) \end{aligned}$$ (33) Eqs. (33) can be substituted with Eq. (30) to yield the dynamical spin conductivity expression. Additionally, as a measure of dissipative transport, we present the static spin conductivity ((\sigma {d.c}=lim{\omega \longrightarrow 0}\sigma {xx}(\omega ))). The ultimate solution for static spin conductivity ((\sigma {d.c}(T))) follows the following statement following additional algebraic computations. $$\begin{aligned} & \sigma {d.c}(T)=lim{\omega \longrightarrow 0}\frac{Im\Pi {xx}(i\omega \longrightarrow \omega +i0^{+})}{\omega }\nonumber \ = & -\frac{\pi }{4k{B}T}\sum \textbf{k}Z\textbf{k}^{2}\Big [\phi ^{2}(\textbf{k})Z^{b}{\textbf{k}}u{21}(\textbf{k})u^{}{22}(\textbf{k}) Z^{h}{\textbf{k}}u_{31}(\textbf{k})u^{}{32}(\textbf{k})+\phi ^{2}(\textbf{k})Z^{b}_{\textbf{k}}u^{}{21}(\textbf{k})u_{22}(\textbf{k}) Z^{h}{\textbf{k}}u^{}{31}(\textbf{k})u_{32}(\textbf{k})\nonumber \ + & Z^{h}{\textbf{k}}\|\phi (\textbf{k})\|^{2}Z^{b}{\textbf{k}}\|u_{31}(\textbf{k})\|^{2} \|u_{12}(\textbf{k})\|^{2}\Big ] \Big (\frac{e^{\frac{\Omega {3}(\textbf{k})}{k{B}T}}}{(e^{\frac{\Omega {3}(\textbf{k})}{k{B}T}}-1)^2}+ \frac{e^{\frac{\Omega {2}(\textbf{k})}{k{B}T}}}{(e^{\frac{\Omega {2}(\textbf{k})}{k{B}T}}-1)^2}\Big )\nonumber \ + & \Big [Z^{h}\textbf{k}Z^{b}{\textbf{k}}\|\phi (\textbf{k})\|^{2}\|u_{21}(\textbf{k})\|^{2}\|u_{12}(\textbf{k})\|^{2}\Big ] \Big (\frac{e^{\frac{\Omega {3}(\textbf{k})}{k{B}T}}}{(e^{\frac{\Omega {3}(\textbf{k})}{k{B}T}}-1)^2}+ \frac{e^{\frac{\Omega {1}(\textbf{k})}{k{B}T}}}{(e^{\frac{\Omega {1}(\textbf{k})}{k{B}T}}-1)^2}\Big )\nonumber \+ & \|\phi (\textbf{k})\|^{2}Z^{a}{\textbf{k}}Z^{b}{\textbf{k}}\Big [\|u_{21}(\textbf{k})\|^{2}\|u_{32}(\textbf{k})\|^{2} +\|u_{22}(\textbf{k})\|^{2}\|u_{31}(\textbf{k})\|^{2}\Big ]\Big (\frac{e^{\frac{\Omega {2}(\textbf{k})}{k{B}T}}}{(e^{\frac{\Omega {2}(\textbf{k})}{k{B}T}}-1)^2}+ \frac{e^{\frac{\Omega {1}(\textbf{k})}{k{B}T}}}{(e^{\frac{\Omega {1}(\textbf{k})}{k{B}T}}-1)^2}\Big ) \end{aligned}$$ (34) The spin current operator (Eq. (27)) is derived using the continuity equation for the (z)-component of magnetization, consistent with linear response theory79,80. The correlation function (\Pi _{xx}(\omega )) (Eqs. (31–33)) is computed using the Kubo formula, and the static conductivity (Eq. (34)) follows from the zero-frequency limit, capturing the transport of spin excitations in the Heisenberg model. Results and discussions This section presents numerical results for the spin conductivity of the (S=1/2) Heisenberg model on the Lieb lattice in the field-induced spin-polarized phase. We analyze the temperature dependence of static spin conductivity, the frequency dependence of dynamic spin conductivity, and the effects of the Dzyaloshinskii-Moriya interaction strength and magnetic field. In the limit of (B/J\longrightarrow \infty), the ground state of the original spin model Hamiltonian is a field induced spin-polarized state, and there is a finite energy gap to the lowest excited state. The energy gap narrows as the magnetic field weakens and finally disappears at the critical magnetic field ((B_c)). In the setting of Brueckner’s formalism above threshold field (B_{c}), when the density of bosonic gas is tiny, the effects of hard core contact on the excitation spectrum have been determined using the Green’s function technique. In order to prevent duplicate occupation of bosons at each lattice site and maintain the SU(2) algebra of the spin model Hamiltonian, the original spin model was represented by a bosonic model in the presence of hard core repulsion. The hard-core boson constraint, enforced by an infinite on-site repulsion (\mathcal {U} \rightarrow \infty) (Eq. (15)), prevents multiple bosons from occupying the same lattice site, preserving the SU(2) spin algebra of the original Heisenberg model73 Using Brueckner’s formalism above the threshold field (B_{c}), where the density of bosonic gas is modest, we have applied the Green’s function technique to determine the influence of interaction on the diagonal component of the bosonic Hamiltonian. With the replacements (\omega {j=1,2,3}(\textbf{k})\longrightarrow \Omega {j=1,2,3}(\textbf{k})) in the corresponding equations, a self-consistent solution of Eqs. (11, 13, 18, 20, 22) should yield the single particle excitation. An initial guess is made for (Z^{a}\textbf{k},\Sigma {aa}(\textbf{k},0),Z^{b}\textbf{k},\Sigma {bb}(\textbf{k},0),Z^{h}\textbf{k},\Sigma {hh}(\textbf{k},0)), and corrected excitation energy is found using Eq. (18). Until convergence is achieved, this process is repeated. Eqs. (33, 34) allow us to compute the dynamical and static spin susceptibilities using the final values for the excitation spectrum. We present the numerical results for spin structure factors in the field-induced spin polarized domain, where a limited energy gap between the ground state and the first excited state is part of the hamiltonian spin model’s energy spectrum. Thus, the system maintains its gapped spin polarized phase as long as the excitation spectrum (\Omega {\textbf{Q}{0}=(\pi ,\pi )}) has non-zero values. We have investigated the Lieb lattice’s spin conductivities along the x direction. Over a set normalized temperature (k_{B}T/J=0.05), the energy gap ((E_{g}=\Omega {2}(\pi ,\pi ))) in terms of magnetic field (g\mu {B}B/J) has been displayed over a range of Dzyaloshinskii-Moriya interaction strength D/J values. Figure2 makes it clear that when the magnetic field gets closer to the normalized critical value (g\mu {B}B{c}/J), the energy gap disappears. The gap disappears for all values of D/J at the crucial point (g\mu {B}B{c}/J), which marks the change from the gapped spin liquid phase to the gapless magnetic ordering phase. The critical field value rises with D/J, as seen in Fig.2. In magnetic fields greater than the critical field (g\mu {B}B{c}/J), the ground state and the lowest excited state have different energies. Spinon modes are the name given to such an excitation spectrum. For magnetic fields below (g\mu {B}B{c}/J) for each D/J, a gapless magnetic ordering state forms and the energy gap disappears as the magnetic field decreases. This graphic shows that the anisotropy parameter increases the magnetic field area where the excitation spectrum becomes gapless. at other words, when D/J decreases, the field-induced spin-polarized phase forms at a lower magnetic field. The dynamical spin conductivity ((\sigma (\omega ))) results are shown in Fig. 3 as a function of normalized frequency ((\omega /J)) for various magnetic fields at fixed (D/J=0.2). The normalized temperature has been set at (k_{B}T/J=0.05). Since our method is expected to work in the field-induced spin polarized state, the value of the normalized magnetic field is limited to (g\mu {B}B/J>g\mu {B}B_{c}/J). A number of noteworthy features are included: each curve has nine peaks in the frequency range (0<\omega /J<3.5). According to Fig. 3, the height of the peaks in dynamical spin conductivity also rises as the magnetic field decreases. An increasing magnetic field widens the energy gap, thereby reducing the transition rate of bosons from the ground state to the first excited state. As a result, at a certain frequency, conductivity decreases with magnetic field. Furthermore, Fig. 3 shows that the frequency location of peaks is unaffected by the strength of the magnetic field. The transition rate of bosons between energy bands provides an explanation for the several peaks in dynamical spin conductivity curves. The bosonic model is really used to analyze the frequency behavior of conductivities since the original spin model Hamiltonian has been transferred to it. The energy needed for boson transitions between energy levels is implied by the dynamical spin conductivity peak’s frequency location. Therefore, the interband transition of bosons between energy levels is shown by the peaks in each dynamical spin conductivity curve. Figure 4 illustrates how the dynamical spin conductivity (\sigma (\omega )) behaves for varying DM interaction strengths for (g\mu {B}B/J=6.0) as a function of normalized frequency (\omega /J). It is assumed that the normalized temperatures are (k{B}T/J=0.05). (0<\omega /J<3.5) is the frequency range where the conductivity obtains a finite non-zero value. As D/J increases, the conductivity’s peak positions shift to higher frequencies. Furthermore, as Fig. 4 illustrates, the conductivity value increases with the intensity of the DM interaction at a certain frequency. At frequency (\omega /J<1.0), the dynamical conductivity is zero for all Dzyaloshinskii-Moriya interaction strength values. Small peaks in the dynamical spin conductivity for (\omega /J < 1.0) Fig.4 result from low-energy transitions involving the flat band ((\Omega _1)), which has a high density of states, facilitating contributions to the conductivity at low frequencies. The two-dimensional Lieb lattice’s static spin conductivity for various magnetic field levels at fixed (D/J=0.2) is displayed in Fig. 5. Every spin conductivity curve has a peak, and as the magnetic field increases, the peak’s location increases in temperature. Furthermore, as Fig. 5 illustrates, the height of the peak diminishes with magnetic field. These findings stem from the observation that an increase in magnetic field causes the energy gap in the model’s excitation spectrum to widen, which in turn causes the static spin conductivity to drop at a given temperature. The density of the current-carrying bosons increases when the temperature rises below the typical temperature of peak for each magnetic field. Thus, as seen in Fig. 5, we observe an increase in spin conductivity at low temperatures. The bosons experience mutual scattering effects as the temperature rises, which lowers the spin conductivity. The bosons experience mutual scattering effects as the temperature rises, which lowers the spin conductivity. Therefore, the conflict between the two phenomena,the rise in boson density within the physical bounds and the boson scattering at higher temperatures, is the cause of each curve’s temperature dependency. We see a temperature range where the spin conductivity disappears for every magnetic field value. As seen in Fig. 5, the diameter of this temperature band rises with magnetic field. We also looked at how the anisotropy parameter, or D, affected the Lieb lattice’s spin conductivity’s temperature dependency. For a range of DM interaction strength values, including (D/J=0.2,0.4,0.6,0.8) for the magnetic field (g\mu {B}B/J=7) above the critical magnetic field, we display (\sigma) vs normalized temperature in Fig. 6. This graphic illustrates how, throughout the whole temperature range, the spin conductivities rise as _D increases. This makes sense since increasing the intensity of the DM interaction lowers the energy gap and raises the static conductivity as a result. Additionally, given the selected anisotropy values, the spin conductivity exhibits a non-zero value at temperatures (k_{B}T/J>0.4). It is evident that the peak location in static spin conductivity is unaffected by the intensity of the Dzyaloshinskii-Moriya interaction. At this temperature, the spin conductivity peak is located around the normalized value (k_{B}T/J=1.75). The conductivity decreases when the DM integration strength decreases at a certain temperature. The spin conductivity’s magnetic field dependency for different normalized anisotropy values D/J for fixed (k_{B}T/t=0.05) is shown in Fig. 7. Figure 7 shows that when the magnetic field decreases, (\sigma {d.c}) increases quickly, approaching the threshold magnetic field, which is dependent on the anisotropy parameter. The enhancement of the energy gap with the magnetic field, which reduces the density of bosons, explains the decreasing behavior of the conductivity on the magnetic field as seen in Fig. 9. Additionally, for all levels of anisotropy, the spin conductivity very nearly approaches zero for magnetic fields greater than 8. In addition, at fixed value of magnetic field, the static conductivity increases with _D which arises from the reduction of energy gap with anisotropy parameter D. In Fig. 8 we present the magnetic field dependence of the spin conductivity for various normalized temperature (k_{B}T/J) at fixed (D/J=0.4). The increase of the magnetic field causes to reduce (\sigma {d.c}), for each temperature (k{B}T/J). Here, it is evident that for any temperature value, the static conductivity decreases with magnetic field. We deduce that whereas the static conductivity has a substantial temperature dependency at the critical magnetic field, this dependence is modest for large magnetic fields (above about 8). Furthermore, for a given magnetic field, the static spin conductivity increases with temperature. This finding may be explained by the fact that rising temperatures accelerate the transition of bosons from their ground state to their excited state, which in turn causes conductivity to increase. Finally we have plotted the dependence of static spin conductivity on Dzyaloshinskii-Moriya interaction strength for different values of magnetic field in Fig. 9. The temperature is considered to be (k_{b}T/J=0.05). The static conductivity curves is clearly independent of D/J for (D/J<0.4), however the conductivity shows a monotonically increasing behavior at (D/J>0.4) according to Fig. 9. The increasing behavior of spin conductivity in terms of D/J is justified based on the reduction of energy gap in in excitation spectrum with D/J so that the transition rate of bosons increases with D/J. In addition at fixed D/J, the static spin conductivity decreases with magnetic field since the enhancement of magnetic field leads to increase energy gap in excitation spectrum of bosonic gas. Here we add few comments regarding the asymmetric properties of peaks in dynamical spin conductivities in this study. The slight asymmetry in the dynamical spin conductivity peaks (Figs.3 and 4) originates from the DM interaction’s complex hopping terms (Eq. (9)), which introduce non-Hermitian effects in the bosonic Hamiltonian, combined with interference between the three excitation branches of the Lieb lattice. It is worthwhile to add few discussions regarding influence of the flat band on spin transport. The Lieb lattice’s flat band, characterized by (\omega _1(\textbf{k}) = g \mu _B B - J), leads to a high density of states, enhancing the transition rates of bosonic excitations. This contributes to the multiple peaks in the dynamical spin conductivity (Fig.3), as the flat band facilitates interband transitions, distinguishing the Lieb lattice from other structures like the square lattice. Compared to the square lattice, the Lieb lattice’s spin conductivity exhibits multiple peaks (Fig.3) due to its three-sublattice structure and flat band. For a square lattice, the simpler two-band dispersion results in a single dominant peak in the dynamical spin conductivity, highlighting the Lieb lattice’s unique topology in enhancing spin transport. Some crude explanation for studying the properties of the model in the gapless regime would be suitable. Below the critical magnetic field (B_c), the energy gap closes due to a quantum phase transition to a gapless magnetic ordering phase, driven by quantum fluctuations and DM-induced spin rotations67,68. Future studies could employ quantum Monte Carlo or DMRG to investigate spin correlations and transport in this regime, complementing our analysis of the gapped phase. Conclusions This study investigates spin conductivity in a two-dimensional Heisenberg model on a Lieb lattice, focusing on the effects of a perpendicular magnetic field and spin-orbit interactions, particularly the Dzyaloshinskii-Moriya interaction (DMI). By transforming the spin Hamiltonian into a bosonic gas model and using the Green’s function method, the research reveals that increasing DMI shifts the dynamic spin conductivity peak to higher frequencies and enhances its intensity, indicating a stronger influence on spin dynamics than the magnetic field, which does not affect peak position. The static transverse structure factor decreases with increasing magnetic field strength across various DMI values, suggesting suppression of spin correlations. Additionally, static spin conductivity shows a temperature-dependent peak within a limited range for fixed magnetic fields, highlighting thermal effects on spin transport. These findings of spin transport on a Lieb lattice broaden our understanding of spin dynamics in 2D materials governed by the Heisenberg model, with possible implications for optimizing spintronic devices through spin-orbit coupling and magnetic field manipulation Fig. 2 Energy gap ((E_{g})) versus magnetic field ((g\mu {B}B/J)) and different Dzyaloshinskii-Moriya strength values _D/J by setting (k_{B}T/J=0.05). Full size image Fig. 3 Dynamical spin conductivity ((\sigma (\omega ))) versus frequency ((\omega /J)) for different values of magnetic field (g\mu {B}B/J) and (k{B}T/J=0.05). The Dzyaloshinskii Moriya interaction strength has been fixed at (D/J=0.5). Full size image Fig. 4 Dynamical spin conductivity ((\sigma (\omega ))) versus frequency ((\omega /J)) for different values of Dzyaloshinskii-Moriya interaction strength D/J and (k_{B}T/J=0.05). The magnetic field has been fixed at (g\mu _{B}B/J=6). Full size image Fig. 5 Static spin conductivity ((\sigma {d.c})) versus temperature ((k{B}T/J)) for different values of magnetic field, namely (g\mu _{B}B/J=6.5,7.0,7.5,8.0) at fixed Dzyaloshinskii-Moriya interaction strength (D/J=0.4). Full size image Fig. 6 Static spin conductivity ((\sigma {d.c})) of localized electrons on Lieb lattice versus normalized temperature((k{B}T/J)) for different values of Dzyaloshinskii-Moriya interaction strength D/J at fixed magnetic field (g\mu _{B}B/J=7.0). Full size image Fig. 7 Static spin conductivity ((\sigma {d.c})) versus normalized magnetic field ((g\mu {B}B/J)) for different values of Dzyaloshinskii-Moriya interaction strength D/J at fixed temperature (k_{B}T/J=0.05.). Full size image Fig. 8 Static spin conductivity ((\sigma {d.c})) versus normalized magnetic field ((g\mu {B}B/J)) for different values of normalized temperature (k_{B}T/J) at fixed (D/J=0.4). 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Study of magnetization and specific heat of Heisenberg model on Lieb nanolattice: Anisotropic effects. Physica E142, 115269 (2022). ArticleCASGoogle Scholar Azizi, F. & Rezania, H. Spin structure factors of Heisenberg antiferromagnet on two-dimensional Lieb nanolattice. J. Supercond. Novel Magn.35(6), 1597–1612 (2022). ArticleCASGoogle Scholar Azizi, F. & Rezania, H. Thermal transport properties of Heisenberg antiferromagnet on honeycomb lattice: The effects of anisotropy. Physica E135, 114985 (2022). ArticleCASGoogle Scholar Azizi, F. & Rezania, H. Thermal conductivity of localized electrons on magnetic ordered monolayer graphene. ECS J. Solid State Sci. Technol.10(8), 081012 (2021). ArticleCASGoogle Scholar Rezania, H. & Azizi, F. Dynamical and static spin conductivities in the Heisenberg model on honeycomb lattice: The effects of magnetic long range ordering. Chin. J. Phys.65, 593–603 (2020). ArticleMathSciNetCASGoogle Scholar Azizi, F., & Rezania, H. “Spin transport properties of anisotropic Heisenberg antiferromagnet on honeycomb lattice in the presence of magnetic field.” Euro. Phys. J. B93, 1–11 (2020). Azizi, F. & Rezania, H. Magnetic structure factors Heisenberg model for magnetic ordered graphene like structure. Chin. J. Phys.61, 358–367 (2019). ArticleMathSciNetGoogle Scholar Azizi, F. & Rezania, H. Dynamical and static spin structure factors of Heisenberg antiferromagnet on honeycomb lattice in the presence of Dzyaloshinskii-Moriya interaction. Physica E114, 113603 (2019). ArticleGoogle Scholar Côté, R. & MacDonald, A. H. Spin-ordering and magnon collective modes for two-dimensional electron lattices in strong magnetic fields. Phys. Rev. B53(15), 10019 (1996). ArticleADSGoogle Scholar Chen, K.-J. & Zhang, W. Nematic ferromagnetism on the Lieb lattice. Chin. Phys. Lett.31(11), 110303 (2014). ArticleADSGoogle Scholar Fouet, J.-B., Tchernyshyov, O. & Mila, F. Field-induced gap in ordered Heisenberg antiferromagnets. Phys. Rev. B70(17), 174427 (2004). ArticleADSGoogle Scholar Zhuo, F., Kang, J., Cheng, Z. & Manchon, A. Topological band engineering and tunable thermal Hall effect in trimerized Lieb lattice ferromagnets. Phys. Rev. B109(5), 054412 (2024). ArticleADSCASGoogle Scholar Auerbach, A. Interacting electrons and quantum magnetism. Springer Science & Business Media, (2012). Mahan, G. D. Many-particle physics. Springer Science & Business Media, (2013). Yildirim, T., Harris, A. B., Aharony, A., & Entin-Wohlman, O. “Anisotropic spin Hamiltonians due to spin-orbit and Coulomb exchange interactions.” Phys. Rev. B, 52(14), 10239 (1995). Volosniev, A. G., et al. “Engineering the dynamics of effective spin-chain models for strongly interacting atomic gases.” Phys. Rev. A, 91(2), 023620 (2015). Da Silva, W. M. & Montenegro-Filho, R. R. Role of density-dependent magnon hopping and magnon-magnon repulsion in ferrimagnetic spin-(1/2, S) chains in a magnetic field. Phys. Rev. B103(5), 054432 (2021). ArticleADSGoogle Scholar Rezania, H. Thermodynamics of anisotropic antiferromagnetic Heisenberg chain in the presence of longitudinal magnetic field. Physica E101, 239–244 (2018). ArticleADSCASGoogle Scholar Rezania, H. Thermodynamic properties of anisotropic spin ladder in a longitudinal magnetic field. J. Magn. Magn. Mater.388, 68–75 (2015). ArticleADSCASGoogle Scholar Abrikosov, A. A., Gorkov, L. P., & Dzyaloshinski, I. E. Meth. Quant. Field Theo. stat. Phys.. Courier Corporation, (2012). Sentef, M., Kollar, M., & Kampf, A. P. “Spin transport in Heisenberg antiferromagnets in two and three dimensions.” Phys. Rev. B,75(21), 214403 (2007). Rezania, H., Langari, A. & Thalmeier, P. Green’s function approach to quantum criticality in the anisotropic Kondo necklace model. Phys. Rev. B77(9), 094438 (2008). ArticleADSGoogle Scholar Fetter, A. L. & Walecka, J. D. Quant. Theo. Many-part. Syst.. Courier Corporation, (2012). Scalapino, D. J., White, S. R. & Zhang, S. Insulator, metal, or superconductor: The criteria. Phys. Rev. B47(13), 7995 (1993). ArticleADSCASGoogle Scholar Chandra, P., Coleman, P., & Larkin, A. I. “A quantum fluids approach to frustrated Heisenberg models.” J. Phys.: Condensed Matter2, no. 39, p. 7933 (1990). Shastry, B. S., & Sutherland, Bill. “Twisted boundary conditions and effective mass in Heisenberg-Ising and Hubbard rings.” Phys. Rev. Lett.65(2), 243 (1990). Alvarez, J. V., & Gros, C. “Low-temperature transport in Heisenberg chains.” Phys. Rev. Lett.88(7), 077203 (2002). Download references Author information Authors and Affiliations Department of Physics, Jundi-Shapur University of Technology, Dezful, Iran Farshad Azizi Department of Physics, Razi University, Kermanshah, Iran Hamed Rezania Authors 1. Farshad AziziView author publications Search author on:PubMedGoogle Scholar 2. Hamed RezaniaView author publications Search author on:PubMedGoogle Scholar Contributions The authors of this article, Farshad Azizi and Hamed Rezania, have equally contributed to all stages of the research, writing, and editing of this manuscript. Both authors were involved in the study design, data collection, analysis of results, and the final drafting of the text, and they share joint responsibility for the content of the article. Corresponding author Correspondence to Farshad Azizi. Additional information Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. 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Copy shareable link to clipboard Provided by the Springer Nature SharedIt content-sharing initiative Keywords Spin conductivity Spin-orbit Lieb lattice Subjects Condensed-matter physics Electronic properties and materials Magnetic properties and materials Spintronics Download PDF Sections Figures References Abstract Introduction Theoretical formalism Theoretical calculation of dynamical and static spin conductivity Results and discussions Conclusions Data availability References Author information Additional information Rights and permissions About this article Advertisement Fig. 1 View in articleFull size image Fig. 2 View in articleFull size image Fig. 3 View in articleFull size image Fig. 4 View in articleFull size image Fig. 5 View in articleFull size image Fig. 6 View in articleFull size image Fig. 7 View in articleFull size image Fig. 8 View in articleFull size image Fig. 9 View in articleFull size image Broholm, C., et al. Quantum spin liquids. 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Chen, L., et al. “Magnetic anisotropy in ferromagnetic CrI(3).” _Physical Review B,101(13), 134418 (2020). Jafari, R., Kargarian, M., Langari, A., & Siahatgar, M. “Phase diagram and entanglement of the Ising model with Dzyaloshinskii-Moriya interaction.” Phys. Revi. B78, (21), 214414 (2008). Zakeri, K., et al. “Asymmetric spin-wave dispersion on Fe(110): Direct evidence of the Dzyaloshinskii-Moriya interaction.” Phys. Rev. Lett.104(13), 137203 (2010). Zakeri, K., et al. “Chirality-inverted Dzyaloshinskii-Moriya interaction.” Phys. Rev. B108(10), L100403 (2023). Michels, A., et al. “Microstructural-defect-induced Dzyaloshinskii-Moriya interaction.” Phys. Rev. B99(1), 014416 (2019). Azizi, F. & Rezania, H. Study of magnetization and specific heat of Heisenberg model on Lieb nanolattice: Anisotropic effects. Physica E142, 115269 (2022). ArticleCASGoogle Scholar Azizi, F. & Rezania, H. Spin structure factors of Heisenberg antiferromagnet on two-dimensional Lieb nanolattice. J. Supercond. Novel Magn.35(6), 1597–1612 (2022). ArticleCASGoogle Scholar Azizi, F. & Rezania, H. Thermal transport properties of Heisenberg antiferromagnet on honeycomb lattice: The effects of anisotropy. Physica E135, 114985 (2022). ArticleCASGoogle Scholar Azizi, F. & Rezania, H. Thermal conductivity of localized electrons on magnetic ordered monolayer graphene. ECS J. Solid State Sci. Technol.10(8), 081012 (2021). ArticleCASGoogle Scholar Rezania, H. & Azizi, F. Dynamical and static spin conductivities in the Heisenberg model on honeycomb lattice: The effects of magnetic long range ordering. Chin. J. Phys.65, 593–603 (2020). ArticleMathSciNetCASGoogle Scholar Azizi, F., & Rezania, H. “Spin transport properties of anisotropic Heisenberg antiferromagnet on honeycomb lattice in the presence of magnetic field.” Euro. Phys. J. B93, 1–11 (2020). Azizi, F. & Rezania, H. Magnetic structure factors Heisenberg model for magnetic ordered graphene like structure. Chin. J. Phys.61, 358–367 (2019). ArticleMathSciNetGoogle Scholar Azizi, F. & Rezania, H. Dynamical and static spin structure factors of Heisenberg antiferromagnet on honeycomb lattice in the presence of Dzyaloshinskii-Moriya interaction. Physica E114, 113603 (2019). ArticleGoogle Scholar Côté, R. & MacDonald, A. H. Spin-ordering and magnon collective modes for two-dimensional electron lattices in strong magnetic fields. Phys. Rev. B53(15), 10019 (1996). ArticleADSGoogle Scholar Chen, K.-J. & Zhang, W. Nematic ferromagnetism on the Lieb lattice. Chin. Phys. Lett.31(11), 110303 (2014). ArticleADSGoogle Scholar Fouet, J.-B., Tchernyshyov, O. & Mila, F. Field-induced gap in ordered Heisenberg antiferromagnets. Phys. Rev. B70(17), 174427 (2004). ArticleADSGoogle Scholar Zhuo, F., Kang, J., Cheng, Z. & Manchon, A. Topological band engineering and tunable thermal Hall effect in trimerized Lieb lattice ferromagnets. Phys. Rev. B109(5), 054412 (2024). ArticleADSCASGoogle Scholar Auerbach, A. Interacting electrons and quantum magnetism. Springer Science & Business Media, (2012). Mahan, G. D. Many-particle physics. Springer Science & Business Media, (2013). Yildirim, T., Harris, A. B., Aharony, A., & Entin-Wohlman, O. “Anisotropic spin Hamiltonians due to spin-orbit and Coulomb exchange interactions.” Phys. Rev. B, 52(14), 10239 (1995). Volosniev, A. G., et al. “Engineering the dynamics of effective spin-chain models for strongly interacting atomic gases.” Phys. Rev. A, 91(2), 023620 (2015). Da Silva, W. M. & Montenegro-Filho, R. R. Role of density-dependent magnon hopping and magnon-magnon repulsion in ferrimagnetic spin-(1/2, S) chains in a magnetic field. Phys. Rev. B103(5), 054432 (2021). ArticleADSGoogle Scholar Rezania, H. Thermodynamics of anisotropic antiferromagnetic Heisenberg chain in the presence of longitudinal magnetic field. Physica E101, 239–244 (2018). ArticleADSCASGoogle Scholar Rezania, H. Thermodynamic properties of anisotropic spin ladder in a longitudinal magnetic field. J. Magn. Magn. Mater.388, 68–75 (2015). ArticleADSCASGoogle Scholar Abrikosov, A. A., Gorkov, L. P., & Dzyaloshinski, I. E. Meth. Quant. Field Theo. stat. Phys.. Courier Corporation, (2012). Sentef, M., Kollar, M., & Kampf, A. P. “Spin transport in Heisenberg antiferromagnets in two and three dimensions.” Phys. Rev. B,75(21), 214403 (2007). Rezania, H., Langari, A. & Thalmeier, P. Green’s function approach to quantum criticality in the anisotropic Kondo necklace model. Phys. Rev. B77(9), 094438 (2008). ArticleADSGoogle Scholar Fetter, A. L. & Walecka, J. D. Quant. Theo. Many-part. Syst.. Courier Corporation, (2012). Scalapino, D. J., White, S. R. & Zhang, S. Insulator, metal, or superconductor: The criteria. Phys. Rev. B47(13), 7995 (1993). ArticleADSCASGoogle Scholar Chandra, P., Coleman, P., & Larkin, A. I. “A quantum fluids approach to frustrated Heisenberg models.” J. Phys.: Condensed Matter2, no. 39, p. 7933 (1990). Shastry, B. S., & Sutherland, Bill. “Twisted boundary conditions and effective mass in Heisenberg-Ising and Hubbard rings.” Phys. Rev. Lett.65(2), 243 (1990). Alvarez, J. V., & Gros, C. “Low-temperature transport in Heisenberg chains.” Phys. Rev. Lett.88(7), 077203 (2002). 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5810	http://www.erbion.com/index_files/PHYS52/Chapter%2019.pdf	Chapter 19 p The first law of thermodynamics 19.1 Thermodynamic systems 19.2 Work done during volume changes g g 19.3 Paths between thermodynamic states 19 4 Internal energy and first law of 19.4 Internal energy and first law of thermodynamics 19 5 Kinds of thermodynamic processes 19.5 Kinds of thermodynamic processes 19.6 Internal energy of an ideal gas 19 7 Heat capacities of an ideal gas 19.7 Heat capacities of an ideal gas 19.8 Adiabatic processes for an ideal gas Thermodynamics Thermodynamics • The study of relationships involving heat, The study of relationships involving heat, mechanical work, and other aspects of energy and energy transfer • Name few example of thermodynamics processes • First law of thermodynamics deals with conversion of energy in a thermodynamic system • Name an example of mechanical energy transfer • Name an example of heat transfer • First law will combine two forms of energy Thermodynamic systems Thermodynamic systems • Thermodynamic systems: Any collection of objects that is convenient to regard as a unit and that may that is convenient to regard as a unit, and that may have potential to exchange energy with its surroundings g • Thermodynamic processes: any process during which there are changes in the state of the thermodynamic system • State of a thermodynamic system is defined by its state variables such as T, P, V • Thermodynamics does not depend on details of the structure of matter so both microscopic and macroscopic treatment lead to the same result. Sign convention for exchange of energy between a thermodynamic system and environment a thermodynamic system and environment Thermodynamics: work done by the system is + Mechanics: work done by the forces acting on the body is + Sign of work in thermodynamics and mechanics are opposite Sign of the work done during a volume change (whoever does the work has the positive W) (whoever does the work has the positive W) A Contraction: W(molecules)<0 Expansion: p W(molecules)>0 Work done in a volume change F A F pA dW Fdx pAdx Ad dV = = = V Adx dV dW pdV = = 2 1 ( ) V V W p V dV = ∫ 1 2 Always is initial state is final If is a variable (not a constant) V V p we need to know its functionality with to solve the integral. V Name an example of a PV relationship Work done in constant p 2 1 ressure: ( ) W p V V = − Calculating work using PV diagrams Work is the area under the curve that represents pressure change as a function of volume change Area under the curve i i i W p V pV = Δ = ∑ • Verify sign of the work during an expansion and contraction process with the convention set before Example: Isothermal expansion of an ideal gas • An ideal gas undergoes an isothermal g g (constant-temperature) expansion at T, during which its volume changes from V1 g g 1 to V2. How much work does the gas do? • How much is the volume expansion of an How much is the volume expansion of an ideal gas if the work is 2nRT? Paths between thermodynamic states: work states: work • Calculate the work done by the system to get from P1V1 to P2V2 on the paths 132, 142,12. D th k d d th th? • Does the work depend on the path? Paths between thermodynamic states: heat states: heat Gradual expansion and moving the piston up Rapid removal of the wall and free expansion Q1 W1 Q2=0 W2=0 • How much is the amount of the work done by the system in a and b for one mole of an ideal gas W2 0 ΔT=0 in a and b for one mole of an ideal gas Stop and think Stop and think • What do you think about “the • What do you think about the amount of work contained in a system”? Can it be defined clearly? clearly? • Can you name a system that the work done by it does not depend on the path? on the path? Internal energy and the first law of thermodynamics: a generalization thermodynamics: a generalization of the conservation of energy Internal energ U of a s stem is s m of the kinetic energies of all of its • Internal energy, U, of a system is sum of the kinetic energies of all of its constituents plus sum of the potential energy of the interaction among these particles. • When heat added to a system its internal energy increases ΔU=Q When heat added to a system, its internal energy increases ΔU Q • When work is done by a system its internal energy decreases by the amount of work by ΔU=-W • Total change in internal energy of a system is ΔU=Q-W Total change in internal energy of a system is ΔU Q W How we determine the internal f ? energy of a system? • We can’t determine absolute value of the internal energy of a system (similar to the potential energy). • We choose an state of a system as reference We choose an state of a system as reference and give an arbitrary value to its internal energy. • Then we can measure change in internal energy i th h t h d k d b / using the heat exchange and work done by/on the system ΔU=Q-W • While dQ and dW both depend on the path, it e dQ a d d bot depe d o t e pat , t turns out that ΔU does not depend on the path. ΔU only depends on the state of a system which is a function of its state variables (p V T) which is a function of its state variables (p, V, T) Change in internal energy of human body in 24 hours human body in 24 hours Work done during a change of sate Summary Summary: The first law of h d i thermodynamics Summary: Internal energy Example Example • You intake 900 calorie and run up a cliff You intake 900 calorie and run up a cliff. How high do you have to climb to burn all calories consumed? (your mass is 60 kg) calories consumed? (your mass is 60 kg) (ans. 6410 m) • Evaluate your answer • Evaluate your answer. A cyclic process A cyclic process • State of a system starts and y ends at a. The total work is -500 J. Why is the work negative? – Why is the work negative? – Find the change in internal energy and the heat added to the system in this process the system in this process. • For pV diagrams total work is positive for clockwise processes and negative for counterclockwise. Can you contradict it? contradict it? Comparing thermodynamic processes Kinds of thermodynamic processes Kinds of thermodynamic processes • Adiabatic: no heat transfer Adiabatic: no heat transfer • Isobaric: constant pressure I h i t t l • Isochoric: constant volume • Isothermal: constant temperature Adiabatic processes and the first law Adiabatic processes and the first law • No heat transfer: Q=0 can be achieved in – Isolated systems – Fast processes • First Law: ΔU=-W First Law: ΔU W • In adiabatic expansion system does positive work and hence looses energy. W>0 and ΔU<0 I di b ti i t d ti k • In adiabatic compression system does negative work and hence gains energy. W<0 and ΔU>0 • Example: compression stroke in an internal combustion, p p expansion of the fuel during the power stoke (temperature rise or drop happens during the process) Isochoric processes and the first law Isochoric processes and the first law • No change of volume: W=0 is zero No change of volume: W 0 is zero • First Law: ΔU=Q; • All the energy added to the system remains in it All the energy added to the system remains in it and cusses temperature change • Example: heating gas in a cylinder (don’t try it at Example: heating gas in a cylinder (don t try it at home) • Although there are other kinds of work that do t oug t e e a e ot e ds o o t at do not involve volume change but “isochoric” is used for the systems that no work is done at all. Isobaric processes and the first law Isobaric processes and the first law • No change of pressure: none of the ΔU Q No change of pressure: none of the ΔU, Q, or W is zero • First Law: ΔU=Q W; • First Law: ΔU=Q-W; • Work is calculated easily W=p(V2-V1) Isothermal processes and the first law Isothermal processes and the first law • No change of temperature: process has to No change of temperature: process has to occur slowly and none of the ΔU, Q, or W is zero is zero • First Law: ΔU=Q-W; I i l ΔT 0 ΔU 0 • In special cases ΔT=0 means ΔU=0 (Example: ideal gas) but generally ΔU is t not zero. Thermodynamic processes on PV di diagrams Internal energy of an ideal gas Internal energy of an ideal gas • Example: adiabatic expansion of a gas If U depended on V or p then gas. If U depended on V or p then we would have observed a change in T. • Experiment shows no change in T • Experiment shows no change in T for expansion of low density gasses. I t l f id l • Internal energy of an ideal gas depends only on its temperature, not on its pressure or volume. • For a non-ideal gas adiabatic expansion is accompanied with drop in temperature. p p • Does internal energy of a solid depend on its volume? Heat capacities of an ideal gas Heat capacities of an ideal gas • Cp molar heat capacity p p y under constant pressure (usually for gasses) C l h t it • Cv molar heat capacity under constant volume (for solids or liquids) • For a given system (ideal gas) why these should be different? Which one is different? Which one is greater? Cp=Cv+R molar heat capacities of an ideal gas Usually Cv=.4Cp For water Cv<Cp between 0 and 40C More about heat capacities More about heat capacities • Ratio of heat capacities γ=C /C Ratio of heat capacities γ=Cp/Cv • For an ideal gas internal energy change only depends on T so ΔU=nC ΔT no matter what depends on T so ΔU=nCvΔT no matter what the volume change is. E l 19 6 Fi d th h i i t l • Example 19.6: Find the change in internal energy of a dorm room containing 2400 l f i h it i l d f 23 90C t moles of air when it is cooled from 23.90C to 11.60C at a constant pressure of 1.00 atm. T t th i id l ith 1 4 Treat the air as an ideal gas with γ=1.4. Adiabatic Process for an ideal gas • Q=0 and ΔU=-W<0 • An adiabatic curve on PV An adiabatic curve on PV diagram is always steeper than an isotherm. • Adiabatic process has to be Adiabatic process has to be fast for not letting heat exchange to happen. • If we use ideal gas law to g deduce any results the process has to be slow to allow equilibrium condition to b i t i d be maintained. • In reality it is hard to meet both of these conditions simultaneously So whatever simultaneously. So whatever we discuss here is an approximation. Adiabatic Process for an ideal gas • For an adiabatic process in an ideal gas: • TVγ-1=constant so T V γ-1= T V γ-1 T1V1 γ-1= T2V2 γ-1 • PVγ=constant so P1V1 γ= P2V2 γ P1V1 P2V2 • W=-ΔU=nCv (T1-T2) • Use PV=nRT Use PV nRT • W=Cv (P1V1-P2V2) /R • W= (P1V1-P2V2) /(γ-1) ( 1 1 2 2) (γ ) Summary Summary Summary Summary Summary Example: 19 7 Example: 19.7 • The compression ratio of a diesel engine is 15 to 1 a diesel engine is 15 to 1. If the initial pressure is 1.01x105pa and the initial t t i 270C fi d temperature is 270C, find the final pressure and the temperature after temperature after compression. Air is a mixture of mostly diatomic d it th t oxygen and nitrogen that is an ideal gas with γ=1.40. (T2=886K, γ ( 2 , p2=44.8x105Pa=44 atm) Work done in an adiabatic process Work done in an adiabatic process • In the previous example how much work is In the previous example how much work is done by the gas do during the compression if the initial volume of the cylinder is 1.00 L=1.00x10-3m3? Assume that CV for air is 20.8 J/mol.K and γ=1.40
5811	https://www.cnblogs.com/1024th/p/10623541.html	Loading [MathJax]/jax/output/CommonHTML/fonts/TeX/fontdata.js 排列组合的一些公式及推导(非常详细易懂) 绪论：加法原理、乘法原理# 分类计数原理：做一件事，有n类办法，在第1类办法中有m1种不同的方法，在第2类办法中有m2种不同的方法，…，在第n类办法中有mn种不同的方法，那么完成这件事共有N=m1+m2+…+mn种不同的方法。分步计数原理：完成一件事，需要分成n个步骤，做第1步有m1种不同的方法，做第2步有m2种不同的方法，…，做第n步有mn种不同的方法,那么完成这件事共有N=m1×m2×⋯×mn种不同的方法。排列问题# 排列数# 从n个不同元素种取出m(m≤n)个元素的所有不同排列的个数，叫做从n个不同元素种取出m个元素的排列数，用符号Amn表示。排列数公式# Amn=n(n−1)(n−2)⋯(n−m+1)=n!(n−m)!,n,m∈N∗,并且m≤n （规定0!=1）推导：把n个不同的元素任选m个排序，按计数原理分步进行：取第一个：有n种取法；取第二个：有(n−1)种取法；取第三个：有(n−2)种取法； …… 取第m个：有(n−m+1)种取法；排列数性质# Amn=nAm−1n−1 可理解为“某特定位置”先安排，再安排其余位置。 Amn=mAm−1n−1+Amn−1 可理解为：含特定元素的排列有mAm−1n−1，不含特定元素的排列为Amn−1。组合问题# 组合数# 从n个不同元素种取出m(m≤n)个元素的所有不同组合的个数，叫做从n个不同元素种取出m个元素的组合数，用符号Cmn表示。组合数公式# Cmn=AmnAmm=n(n−1)(n−2)⋯(n−m+1)m!=n!m!(n−m)!,n,m∈N∗,并且m≤n C0n=Cnn=1 证明：利用排列和组合之间的关系以及排列的公式来推导证明。将部分排列问题Amn分解为两个步骤：第一步，就是从n个球中抽m个出来，先不排序，此即组合数问题Cmn；第二步，则是把这m个被抽出来的球排序，即全排列Amm。根据乘法原理，Amn=CmnAmm，那么 Cmn=AmnAmm=n(n−1)(n−2)⋯(n−m+1)m!=n!m!(n−m)! 组合数的性质# Cmn=Cn−mn 可以理解为：将原本的每个组合都反转，把原来没选的选上，原来选了的去掉，这样就变成从n个元素种取出n−m个元素，显然方案数是相等的。递推公式Cmn=Cmn−1+Cm−1n−1 可理解为：含特定元素的组合有Cm−1n−1，不含特定元素的排列为Cmn−1。还不懂？看下面。 Example 从1，2，3，4，5（n=5）中取出2（m=2）个元素的组合（Cmn）： 12 13 14 15 23 24 25 34 35 45 显然，这些组合中要么含有元素“1”，要么不含。其中含有“1”的是：12 13 14 15 把里面的“1”都挖掉：2 3 4 5 而上面这个等价于从2，3，4，5（n−1）中取出1（m−1）个元素的组合。其中不含“1”的是：23 24 25 34 35 45 上面等价于从2，3，4，5（n−1）中取出2（m）个元素的组合。而总方案数等于上面两种情况方案数之和，即Cmn=Cmn−1+Cm−1n−1。组合数求和公式# C0n+C1n+C2n+⋯+Cnn=2n 我们感性认知一下，上面这个式子的左边表示什么呢？把从n个球中抽出0个球的组合数（值为1）、抽出1个球的组合数、抽出2个球的组合数、……、抽出n个球的组合数相加。换句话说，就是从n个球中随便抽出一些不定个数球，问一共有多少种组合。对于第1个球，可以选，也可以不选，有2种情况。对于第2个球，可以选，也可以不选，有2种情况。对于任意一个球，可以选，也可以不选，有2种情况。根据乘法原理，一共2×2×⋯×2⏟n个2相乘=2n种组合。想要严谨的证明？数学归纳法：当n=1时，C01+C11=2=21成立。假设n=k(k∈N∗)时等式成立，即 k∑i=0Cik=2n 成立，当n=k+1时， C0k+1+C1k+1+C2k+1+⋯+Ckk+1+Ck+1k+1=C0k+1+(C0k+C1k)+(C1k+C2k)+⋯+(Ck−1k+Ckk)+Ck+1k+1=(C0k+C1k+C2k+⋯+Ckk)+(C0k+C1k+C2k+⋯+Ckk)=2×2k=2k+1 等式也成立。 3. 由1、2得，等式对n∈N∗都成立。也可偷懒地用二项式定理证明： (a+b)n=n∑k=0Cknan−kbk 令a=b=1，就得到了 n∑i=0Cin=2n 类似的公式（由Cmn=Cn−mn推导）： C0n+C2n+C4n+⋯=C1n+C3n+C5n+⋯=2n−1 杨辉三角# 这个神奇的图形和组合数、二项式定理密切相关。（图片来自百度百科）杨辉三角可以帮助你更好地理解和记忆组合数的性质：第n行的m个数可表示为 Cm−1n−1，即为从n−1个不同元素中取m−1个元素的组合数。第n行的数字有n项。每行数字左右对称（第n行的第m个数和第n−m+1个数相等，Cmn=Cn−mn），由1开始逐渐变大。每个数等于它上方两数之和（第n+1行的第i个数等于第n行的第i−1个数和第i个数之和，即Cin+1=Cin+Ci−1n）。 (a+b)n的展开式中的各项系数依次对应杨辉三角的第n+1行中的每一项（二项式定理）。以下来自维基百科（我只是随便贴这）二项式系数二项式系数可排列成帕斯卡三角形。在数学上，二项式系数是二项式定理中各项的系数。一般而言，二项式系数由两个非负整数n和k为参数决定，写作，定义为的多项式展开式中，项的系数，因此一定是非负整数。如果将二项式系数写成一行，再依照顺序由上往下排列，则构成帕斯卡三角形。 (nk)(1+x)nxk(n0),(n1),…,(nn)n=0,1,2,… 二项式系数常见于各数学领域中，尤其是组合数学。事实上，可以被理解为从n个相异元素中取出k个元素的方法数，所以大多读作「n取k」。二项式系数的定义可以推广至n是复数的情况，而且仍然被称为二项式系数。二项式系数亦有不同的符号表达方式，包括：C(n,k)、nCk、nCk、、[注3]，其中的C代表组合（combinations）或选择（choices）。很多计算机使用含有C的变种记号，使得算式只占一行的空间，相同理由也发生在置换数，例如写作P(n,k)。 CknCnkPnk 定义及概念对于非负整数n和k，二项式系数定义为的多项式展开式中，项的系数，即 (nk)(1+x)nxk (1+x)n=n∑k=0(nk)xk=(n0)+(n1)x+⋯+(nn)xn 事实上，若x、y为交换环上的元素，则 (x+y)n=∑nk=0(nk)xn−kyk 此数的另一出处在组合数学，表达了从n物中，不计较次序取k物有多少方式，亦即从一n元素集合中所能组成k元素子集的数量。计算二项式系数除展开二项式或点算组合数量之外，尚有多种方式计算的值。 (nk) 递归公式以下递归公式可计算二项式系数： (nk)=(n−1k−1)+(n−1k)∀n,k∈N 其中特别指定： (n0)=1∀n∈N∪{0},(0k)=0∀k∈N. 此公式可由计算(1 + X ) n −1 (1 + X )中的X k项，或点算集合{1, 2, ..., n }的k个元素组合中包含n与不包含n的数量得出。显然，如果k > n，则。而且对所有n，，故此上述递归公式可于此等情况下中断。递归公式可用作建构帕斯卡三角形。 \tbinom nk=0\tbinom nn=1 帕斯卡三角形(杨辉三角) 有关二项式系数的恒等式关系式阶乘公式能联系相邻的二项式系数，例如在k是正整数时，对任意n有： (n+1k)=(nk)+(nk−1) (nk)=nk(n−1k−1) (n−1k)−(n−1k−1)=n−2kn(nk) 两个组合数相乘可作变换： (ni)(im)=(nm)(n−mi−m) n∑r=0(nr)=2n k∑r=0(n+r−1r)=(n+kk) n−k∑r=0(−1)r(n+1)k+r+1(n−kr)=(nk)−1 n∑r=0(dndr)=1dd∑r=1(1+e2πrid)dn n∑i=m(a+ii)=(a+n+1n)−(a+mm−1) (a+mm−1)+(a+mm)+(a+m+1m+1)+...+(a+nn)=(a+n+1n) Fn=∞∑i=0(nii) Fn−1+Fn=∞∑i=0(n−1−ii)+∞∑i=0(n−ii)=1+∞∑i=1(n−ii−1)+∞∑i=1(n−ii)=1+∞∑i=1(n+1−ii)=∞∑i=0(n+1−ii)=Fn+1 主条目：朱世杰恒等式 n∑i=m(ia)=(n+1a+1)−(ma+1) (ma+1)+(ma)+(m+1a)...+(na)=(n+1a+1) 二阶求和公式 n∑r=0(nr)2=(2nn) n∑i=0(r1+n−1−ir1−1)(r2+i−1r2−1)=(r1+r2+n−1r1+r2−1) (1−x)−r1(1−x)−r2=(1−x)−r1−r2 (1−x)−r1(1−x)−r2=(∞∑n=0(r1+n−1r1−1)xn)(∞∑n=0(r2+n−1r2−1)xn)=∞∑n=0(n∑i=0(r1+n−1−ir1−1)(r2+i−1r2−1))xn (1−x)−r1−r2=∞∑n=0(r1+r2+n−1r1+r2−1)xn 主条目：范德蒙恒等式 k∑i=0(ni)(mk−i)=(n+mk) 三阶求和公式主条目：李善兰恒等式 (n+kk)2=k∑j=0(kj)2(n+2k−j2k) 分类: 数学好文要顶关注我收藏该文微信分享樱花赞粉丝 - 151 关注 - 5 +加关注 81 0 « 上一篇：相同字符串问题_题解 » 下一篇：简单证明圆锥体积为三分之一圆柱 posted @ 2019-03-29 19:47 樱花赞阅读(419457) 评论(17) 收藏举报刷新页面返回顶部登录后才能查看或发表评论，立即登录或者逛逛博客园首页【推荐】注册飞算 JavaAI 开发助手，立得京东e卡！分享体验再领30元【推荐】100%开源！大型工业跨平台软件C++源码提供，建模，组态！【推荐】AI 的力量，开发者的翅膀：欢迎使用 AI 原生开发工具 TRAE 【推荐】2025 HarmonyOS 鸿蒙创新赛正式启动，百万大奖等你挑战编辑推荐： · WPF 引用 ASP.NET Core 的 AOT 版本 · 通过 Canvas 将后端发来的一帧帧图片渲染成“视频”的实现过程 · 当加密ID需要变成Guid：为什么我选择了AES-CBC而非GCM？ · 基于 epoll 的协程调度器——零基础深入浅出 C++20 协程 · 下划线字段在golang结构体中的应用阅读排行： · WPF 引用 ASP.NET Core 的 AOT 版本 · .NET周刊【8月第1期 2025-08-03】 · C# 13 与 .NET 9 跨平台开发实战（第一章：开发环境搭建与.NET概述-下篇） · 别再被VO、BO、PO、DTO、DO绕晕！今天用一段代码把它们讲透 · 深入理解 C# 异步编程：同步、Task.Wait () 与 await 的本质区别及实践指南公告昵称：樱花赞园龄： 6年7个月粉丝： 151 关注： 5 +加关注 \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| \| \| \| --- \| < \| 2025年8月 \| > \| \| \| \| \| \| \| \| \| 日 \| 一 \| 二 \| 三 \| 四 \| 五 \| 六 \| \| 27 \| 28 \| 29 \| 30 \| 31 \| 1 \| 2 \| \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| \| 10 \| 11 \| 12 \| 13 \| 14 \| 15 \| 16 \| \| 17 \| 18 \| 19 \| 20 \| 21 \| 22 \| 23 \| \| 24 \| 25 \| 26 \| 27 \| 28 \| 29 \| 30 \| \| 31 \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 字节旗下的 AI IDE ✕ 绪论：加法原理、乘法原理排列问题排列数排列数公式排列数性质组合问题组合数组合数公式组合数的性质组合数求和公式杨辉三角
5812	https://math.stackexchange.com/questions/3398062/show-that-there-are-two-a-i-and-a-j-with-i%E2%89%A0j-such-that-either-a-ia-j	combinatorics - Show that there are two, $a_i$ and $a_j$ with $i≠j,$ such that either $a_i+a_j$ or $a_i- a_j$ is divisible by $100.$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Show that there are two, a i a i and a j a j with i≠j,i≠j, such that either a i+a j a i+a j or a i−a j a i−a j is divisible by 100.100. [duplicate] Ask Question Asked 5 years, 11 months ago Modified5 years, 11 months ago Viewed 1k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. This question already has answers here: Prove:that in any set of 1009 positive integers exits two numbers a i a i and a j a j such that a i−a j a i−a j or a i+a j a i+a j is divisible by 2014 (2 answers) Closed 5 years ago. Suppose (a 1,a 2,…,a 52(a 1,a 2,…,a 52) are integers, not necessarily distinct. Show that there are two, a i a i and a j a j with i≠j i≠j, such that either a i+a j a i+a j or a i−a j a i−a j is divisible by 100 100. Show that this is not necessarily true for integers (a 1,a 2,…,a 51).(a 1,a 2,…,a 51). If we divide a i a i by 100 100, we get 99 possible remainders- hence we can get 52 distinct remainders on dividing each of 52 distinct a i a i's by 100. Can't proceed from here. combinatorics Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Oct 17, 2019 at 20:21 TapiTapi asked Oct 17, 2019 at 20:04 TapiTapi 1,753 15 15 silver badges 36 36 bronze badges 2 1 What have you tried?Arthur –Arthur 2019-10-17 20:08:26 +00:00 Commented Oct 17, 2019 at 20:08 1 Assume at least two of them are non distinct, then their subtraction would be 0 and 0 is divisible by anything. Therefore, assume all of them are distinct.Someone –Someone 2019-10-17 20:08:52 +00:00 Commented Oct 17, 2019 at 20:08 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. For each index i i, set b i=a i(mod 100)b i=a i(mod 100) so that we obtain a new list of 52 52 numbers (b 1,b 2,…b 52)(b 1,b 2,…b 52) where 0≤b i≤99 0≤b i≤99. If there exist indices i i and j j for which b i=b j b i=b j, we're done since b i−b j≡0(mod 100)b i−b j≡0(mod 100) (i.e. their difference is divisible by 100 100). So we can consider the case in which all of the b′i s b i′s are different. If all of the b′i s b i′s take on different values, there must exist two indices i i and j j such that b i+b j≡0(mod 100)b i+b j≡0(mod 100). This is true by the Pigeonhole Principle because we can consider the 100 100 possible values that the b i b i's can take on in pairs as follows: {{1,99},{2,98},{3,97},⋯{49,51}}.{{1,99},{2,98},{3,97},⋯{49,51}}. We're guaranteed that at least two integers will lie in the same pair (and by construction, the sum of these two integers will be divisible by 100 100). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Oct 17, 2019 at 20:13 Ekesh KumarEkesh Kumar 3,474 1 1 gold badge 12 12 silver badges 23 23 bronze badges 1 You only really need the final paragraph. If two of the b i b i are equal, they will still be in the same pigeonhole. Whether you have to add or subtract to get to 0 mod 100 0 mod 100 will of course change, but still.Arthur –Arthur 2019-10-17 20:17:48 +00:00 Commented Oct 17, 2019 at 20:17 Add a comment\| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 4Prove:that in any set of 1009 positive integers exits two numbers a i a i and a j a j such that a i−a j a i−a j or a i+a j a i+a j is divisible by 2014 Related 4Prove:that in any set of 1009 positive integers exits two numbers a i a i and a j a j such that a i−a j a i−a j or a i+a j a i+a j is divisible by 2014 0Prove that there are i i and j j, with 1≤i≤j≤n 1≤i≤j≤n, such that a i+a i+1+⋯+a j a i+a i+1+⋯+a j is divisible by n n 1In a sequence of 34 34 odd integers a 1,a 2,⋯,a 34 a 1,a 2,⋯,a 34 between 1 1 and 100 100 there exist i≠j i≠j such that a i∣a j a i∣a j. 17Is ∏1≤i<j≤n a i−a j i−j∏1≤i<j≤n a i−a j i−j, with distinct integers a i a i, an integer? 1How many different 2 n 2 n-block A i A i can we construct such that \|A i∩A j\|=n\|A i∩A j\|=n? 1Family of sets {A i}n i=1{A i}i=1 n such that A i⊈A j A i⊈A j and \|A i\|≠\|A j\|\|A i\|≠\|A j\| for all distinct i i,j j. 1Among 9 9 tuples (a i,b i)∈Z 2(a i,b i)∈Z 2 there are 3 3 with 3∣a i+a j+a k,b i+b j+b k 3∣a i+a j+a k,b i+b j+b k. 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5813	https://www.storyofmathematics.com/glossary/lateral/	JUMP TO TOPIC Lateral\|Definition & Meaning Definition In solid geometry (three-dimensional objects), lateral refers to the side faces of objects (not the bottom or top face). For example, a cone has a circular base and a curved lateral surface. Similarly, a square pyramid has four triangular lateral faces and one square base. What Is a Lateral? A lateral, in mathematical terms, is simply a side or surface that’s perpendicular to a base. This idea is used in the study of geometry to describe shapes and their dimensions. Figure 1 – Lateral Surface is the slanted surface of any shape For instance, in two–dimensional geometry, a lateral is a side that is perpendicular to a base and connected to it by two perpendicular lines. In three-dimensional geometry, a lateral is a surface that’s perpendicular to the base and connected to it by perpendicular lines. We use the concept of laterals to describe a variety of shapes. For example, cones are three-dimensional shapes with a circular base and a single lateral surface that extends from the base to a point called the vertex. The lateral surface of a cone is curved and perpendicular to the base at all points. Similarly, in shapes like cylinders, pyramids, and prisms, the lateral surfaces are perpendicular to the base and connected to it by perpendicular lines. Significance of Lateral Surface The lateral surface is a type of surface that runs perpendicular to the main axis of an object. It’s one of three primary surfaces found on solid objects, with the others being the top and bottom surfaces. The lateral surface can often be thought of as the side surface of an object, but it can also refer to the outer surface of a cylinder or the circumference of a circular object. Understanding the properties of lateral surfaces is crucial for many industries, including engineering, design, and manufacturing. When it comes to engineering, the lateral surface of an object is frequently exposed to outside forces and stresses, such as friction, wind resistance, and impact. As a consequence, it’s essential to evaluate the resilience and durability of lateral surfaces when designing and fabricating products. For instance, a product with a sturdy lateral surface will be better equipped to handle external forces and maintain its shape. In contrast, a product with a weak lateral surface may deform or break under pressure. The lateral surface also plays a role in the design, specifically the appearance and style of an object. Designers might choose to add features or designs to the lateral surface to make a product look better or to set it apart from similar products. For example, a designer may add a unique pattern to the lateral surface to make a product stand out. In the manufacturing industry, the lateral surface of an object can impact the production process as well. For instance, when machining a solid object, the lateral surface may need to be smooth and uniform to guarantee consistent and accurate measurements. Sometimes, the lateral surface may need to be treated with a special coating or finish to improve its strength or appearance. Lateral surfaces also play a crucial role in the function of an object. For instance, on a cylindrical object, the lateral surface may act as a bearing surface, allowing the object to rotate smoothly. In these cases, the lateral surface must be smooth and free of defects to ensure proper function. Additionally, the lateral surface may serve as a surface for attaching other components or for attaching to a support structure. Lateral Surface Area of a Cone When it comes to geometry, there is a three-dimensional body that has a round base and a sharp-ended vertex, known as a cone. The lateral surface area of a cone directs to the area of the curved surface that extends from the circular base to the vertex. It is the portion of the cone that lies between the base and the slant height. The slanted or lateral area of a cone can be easily calculated using the formula πrs, where r is the radius of the circular base and s is the slant height of the cone. A = πrs The slant height is described as the length between the vertex point and the point on the circumference of the base. In other words, it is the length of the line segment that links the apex of the cone to the boundary of the round base. So, to put it all together, the equation to calculate the lateral surface extent of a perfectly symmetrical cone is π multiplied by the radius of the base multiplied by the slant height. The lateral surface area of a cone comes out to be an important factor in specifying the general surface area of the cone. By using the formula πrs, it is achievable to calculate the exact lateral surface area of any cone, no matter the size or shape. Lateral Surface Area of a Pyramid A pyramid is a 3D geometric shape with a base and sides that meet at a point called the apex. The lateral surface area of a pyramid refers to the area of the pyramid’s sides, excluding its base. In other words, it’s the combined area of all the triangular faces of a pyramid. Figure 2 – Lateral Surface Area of a Pyramid Now, here’s the formula for calculating the lateral surface area of a pyramid: Lateral Surface Area = Perimeter of Base Slant Height / 2 Where “Perimeter of Base” refers to the total length of the sides of the pyramid’s base, and “Slant Height” refers to the length of the side from the apex of the pyramid to the base. Let’s put this formula into practice! Suppose we have a square-based pyramid with a base that measures 4 units long on each side and a slant height of 5 units. The perimeter of the base would be 4 4 = 16 units, and the lateral surface area would be (16 5) / 2 = 40 square units. Lateral Surface of a Cuboid In the case of a 3D cuboid shape, it is the combined area of all its faces, excluding the base. It’s like the total area of the four sides of the cuboid. The procedure to calculate the lateral surface extent of a cuboid is pretty simple. All you need to know is the length of any side of the cube, and you can easily find the lateral surface area by using the formula: L.S Area = 4 x (Side Length)^2 It’s important to recall that the lateral surface area is not the same as the total surface area of a cuboid. The total surface area of a cube includes both the lateral surface area and the area of the base. To discover the total surface area, you just need to add the total area of the base to the lateral area. Lateral Surface Area of a Prism The lateral surface area of a prism can be considered as the combined area of all its rectangular sides, omitting the top and bottom faces. In simpler phrases, it’s the total area of the four sides of the prism. Figure 3 – Lateral Surface Area of a Prism Calculating the lateral surface area of a prism is a straightforward process. The formula is just 2 times the product of the length and width of the rectangular sides. This can be written as: L.S Area = 2 x (Length x Width) It’s worth noting that the lateral surface area is not the same as the total surface area of the prism. The surface area is obtained by summing up the lateral area, the top area, and the bottom area. Practical Uses of Lateral Surfaces The idea of lateral surface area has a number of practical uses in various fields and industries. Here’s a closer look at a few: Architecture: Architects use lateral surface area to figure out the quantity of material they need to cover the sides of their structures and buildings. Engineering: In the field of engineering, the lateral surface area is used to work out the heat transfer and insulation requirements for equipment like pipelines and tanks. Biology: Biologists use the surface-to-volume ratio, which is linked to the lateral surface area. It is employed to specify the quantity of oxygen, cells require and to calculate heat loss in living organisms. Chemistry: Chemists utilize the surface-to-volume ratio in their calculations of chemical reaction rates. It is especially used in the case of catalysts and reaction vessels. Environmental Science: Environmental scientists employ lateral surface area to find out how much pollutants can be absorbed by wetlands and other natural systems. Agriculture: Farmers utilize the lateral surface area to estimate the size and volume of equipment, such as irrigation systems and storage tanks. Solved Examples Involving Lateral Surface Areas of Different Shapes Example 1 A cone-shaped grain silo has a wide area diameter equal to 20 m and a total elevation of 30 m. If the grain silo needs to be coated with a material to protect it from moisture, what is the lateral surface area that needs to be coated? Solution Given the diameter, we will determine the radius first. The radius of the base can be calculated as r = d/2 r = 20/2 r = 10 meters. Using the formula of The lateral surface area of the cone: A = πrs A = 3.14 x 10 x √(10^2 + 30^2) A = 628.3 square meters. Example 2 A company is building a cylindrical oil storage tank to store crude oil. The tank has a diameter of 50 meters and a height of 60 meters. The company management requires us to keep in view that the tank is protected from the elements. Solution To begin, we need to calculate the lateral surface area of the tank. We can start by finding the radius of the tank, which is, 50/2 = 25 meters Next, we can use the formula for the lateral surface area of a cylinder, which is 2πrh = 2 x 3.14 x 25 x 60 = 15140 m$^2$ Next, the company wants to determine the amount of material required to cover the lateral surface area. First, we need to determine the total area that each roll can cover. This can be calculated as, 5 x 20 = 100 m$^2$ Next, we can divide the lateral surface area of the tank by the total area that each roll can cover to determine the number of rolls required. In this case, 15140 / 100 = 151.4 This means that the company will need to purchase 152 rolls of material to cover the lateral surface area of the tank. All images are created using GeoGebra. Kite Definition < Glossary Index > Least Common Denominator Definition Back to Top of Page © 2025 SOM – STORY OF MATHEMATICS
5814	https://www.youtube.com/playlist?list=PLJ47IhBihbBwFpZBOJyR58G4VMXMigsLJ	Right triangle trig - YouTube Back Skip navigation Search Search with your voice Sign in Home HomeShorts ShortsSubscriptions SubscriptionsYou YouHistory History Play all Right triangle trig by Linda Clark • Playlist•11 videos•74 views Play all PLAY ALL Right triangle trig 11 videos 74 views Last updated on Apr 15, 2024 Save playlist Shuffle play Share Show more Linda Clark Linda Clark Subscribe Play all Right triangle trig by Linda Clark Playlist•11 videos•74 views Play all 1 4:30 4:30 Now playing Distance formula Linda Clark Linda Clark • 93 views • 1 year ago • 2 3:01 3:01 Now playing Pythagorean Theorem word problem - ladder (find leg) Linda Clark Linda Clark • 683 views • 1 year ago • 3 2:50 2:50 Now playing Pythagorean Theorem word problem - runners (hypotenuse) Linda Clark Linda Clark • 969 views • 1 year ago • 4 6:21 6:21 Now playing Pythagorean Theorem (formula equations) Linda Clark Linda Clark • 901 views • 1 year ago • 5 4:47 4:47 Now playing Finding all 6 trig values (using a right triangle) Linda Clark Linda Clark • 125 views • 1 year ago • 6 4:17 4:17 Now playing Right triangle word problem (water slide) Linda Clark Linda Clark • 408 views • 1 year ago • 7 4:47 4:47 Now playing Right triangle word problem (Tower) Linda Clark Linda Clark • 167 views • 1 year ago • 8 4:53 4:53 Now playing Special right triangles (30-60-90) Linda Clark Linda Clark • 167 views • 1 year ago • 9 5:51 5:51 Now playing Special right triangles (45-45-90) Linda Clark Linda Clark • 282 views • 1 year ago • 10 5:30 5:30 Now playing Solving right triangles (find angle using sides) Linda Clark Linda Clark • 344 views • 1 year ago • 11 6:15 6:15 Now playing Solving a right triangle (trig functions) Linda Clark Linda Clark • 275 views • 1 year ago • Search Info Shopping Tap to unmute 2x If playback doesn't begin shortly, try restarting your device. • You're signed out Videos you watch may be added to the TV's watch history and influence TV recommendations. To avoid this, cancel and sign in to YouTube on your computer. Cancel Confirm Share - [x] Include playlist An error occurred while retrieving sharing information. Please try again later. Watch later Share Copy link 0:00 / •Watch full video Live • • NaN / NaN [](
5815	https://www.gauthmath.com/solution/1801513413829637/What-is-the-units-digit-of-8127-A1-B-2-C-4-D-6-E-8	Solved: What is the units digit of 8^(127) ? (A) 1 (B) 2 (C) 4 (D) 6 (E) 8 [Math] Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Math Arithmetic Questions Question What is the units digit of 8^(127) ? (A) 1 (B) 2 (C) 4 (D) 6 (E) 8 Gauth AI Solution 100%(2 rated) Answer (B) 2. Explanation The units digit of powers of 8 cycle in the pattern 8, 4, 2, 6. Divide the exponent 127 by 4: 127 ÷ 4 = 31 remainder 3. The remainder is 3, so the units digit is the 3rd in the cycle, which is 2. Helpful Not Helpful Explain Simplify this solution Gauth AI Pro Back-to-School 3 Day Free Trial Limited offer! Enjoy unlimited answers for free. Join Gauth PLUS for $0 Previous questionNext question Related What is the units digit of 8127 ? A1 B 2 C 4 D 6 E 8 , 100% (2 rated) What is the units digit of 8127 ? A1 B 2 C 4 D6 E B 100% (3 rated) What is the units digit of 8127 ? A1 B2 C 4 D E、 100% (4 rated) Which of the following lists only contains shapes that fall under the category of parallelogram? A square, rectangle, triangle B trapezoid, square, rectangle C quadrilateral, square, rectangle D rhombus, rectangle, square 100% (3 rated) Solve the following inequality algebraically. 5x-5/x+2 ≤ 4 What is the solution? -2,13 Type your answer in interval notation. Simplify your answer. Use integers or fractions for any numbers in the expression. 100% (4 rated) How may different arrangements are there of the letters in The number of possible arrangements is MISSISSIPPI？ 100% (2 rated) Rearrange each set of digits to make l example: 5 9 0 3 9,530 digits largest number you can make with the digits b. 5177 _ 。. 298 0_ d. 4302 _ c. 6 0 4 2_ e. 1600 _ d. 7291_ Rearrange each set of digits to make the smallest number possible. example: 5 9 1 3- 1,359 digits smallest number you can make with the digits r. 8127 g. 4 9 3 2 _ h. 5 1 16 _6798 i. _ j. 6 3 5 6 k.5 9 93_ 100% (1 rated) In a right triangle, if one acute angle is 45 ° , what is the measure of the other acute angle? 60 ° 90 ° 30 ° 45 ° 100% (1 rated) The product of eight and seven when multiplied by F is less than the product of four and seven plus ten. a. 8+7F<4+7+10 b. 87F>47+10 C. 87F ≤ 47+10 d. 87F<47+10 100% (5 rated) Rearrange each set of digits to make the smallest numbar possible example: 5 9 1 3 - 1,359 digits smallest number you can make with the digits r. 8127 o、 49 3 2 _ 5116 6798 _ 6356_ k. 5 9 93_ 100% (1 rated) Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App
5816	https://artofproblemsolving.com/wiki/index.php/Angle_bisector?srsltid=AfmBOoppp9-Z6aji5H7HonbONBEDUE1TgwRmbHUdSwaND1cP1Z3hZ3lP	Art of Problem Solving Angle bisector - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Angle bisector Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Angle bisector This is an AoPSWiki Word of the Week for June 6-12 Contents 1 Angle Bisector 2 Features of Angle Bisectors 3 Proofs 3.1 Angle Bisector Locus Theorem 3.2 Angle Bisector Theorem 3.3 Internal and External Angle Bisectors Are Perpendicular 4 See also Angle Bisector For an angle, the (internal) angle bisector of is the line from such that the angle between this line and is congruent to the angle between this line and : An angle also has an external angle bisector, which bisects external angle : The external angle is defined by and the two angle bisectors are perpendicular to each other. Features of Angle Bisectors The distances from a point on an angle bisector to both of its sides are equal. The angle bisectors are the locus of points which are equidistant from the two sides of the angle. A reflection about either angle bisector maps the two sides of the angle to each other. In a triangle, the Angle Bisector Theorem gives the ratio in which the angle bisector cuts the opposite side. In a triangle, the internal angle bisectors (which are cevians) all intersect at the incenter of the triangle. The internal angle bisector of one angle and the external angle bisectors of the other two angles all intersect at an excenter of the triangle. A bisector of an angle can be constructed using a compass and straightedge. Triangle with incenterI, incircle (blue), angle bisectors (orange), and external angle bisectors (green) Proofs Angle Bisector Locus Theorem Theorem: A point lies on the internal angle bisector of if and only if it is equidistant from the sides and . Proof: Let point lie on the angle bisector of . Drop perpendiculars from to lines and , meeting them at points and respectively. Since and both triangles and are right triangles with a shared hypotenuse and equal angles, they are congruent by AAS. Hence, Conversely, if is equidistant from and , then , and lies on the angle bisector of . Angle Bisector Theorem Theorem: In triangle , if the internal angle bisector of intersects at point , then Proof: Let , and draw the angle bisector . Construct a line through parallel to , and let it intersect the extension of at point . Since (by alternate interior angles), triangles and are similar by AA. Therefore, Internal and External Angle Bisectors Are Perpendicular Theorem: The internal and external angle bisectors of any angle are perpendicular. Proof: Let the internal and external angle bisectors divide and its supplement into two equal parts. The internal angle is , and the external angle is . So the internal bisector makes an angle of with one side, and the external bisector makes an angle of with the same side. Adding them: so the two bisectors are perpendicular. See also Cevian Geometry Stewart's Theorem Retrieved from " Category: Geometry Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
5817	https://blog.stackademic.com/what-is-the-off-by-one-error-oboe-07d30f426f15	Sitemap Open in app Sign in Sign in · Stackademic is a learning hub for programmers, devs, coders, and engineers. Our goal is to democratize free coding education for the world. What is the Off-By-One Error (OBOE)? Tharun Kumar Reddy Polu 5 min readSep 3, 2024 Uncovering the Common Bug That Trips Up Programmers and the Best Practices You Can Use to Avoid the Error. My Frustrating Journey with Off-By-One Errors I can still remember hours of frustration staring at my computer screen, trying to work out why my code wasn’t working. I was still a relatively new programmer, just starting to feel confident that I was finally mastering the basics. But every time I ran my program, either it crashed or it gave out results that absolutely made no sense. The problem was one bug, a single bug that kept slipping through my fingers: the off-by-one error. This was the enemy of my loops, array indices, and conditionals, turning what in my mind was a logically tight piece of code into some type of unruly mess. I’d add a couple of lines of code, thinking this finally fixed the issue, and then just sit in frustration as the bug persisted. It wasn’t until I grasped the mechanics of the off-by-one error that I could overcome it once and for all. In the world of programming, there are some errors that are so common they are almost an important aspect for programmers to look for. One such error is what is referred to as the off-by-one error (OBOE)-a stealthy, yet potentially disastrous bug that can cause your code to malfunction. This error generally occurs when you deal with loops, arrays, or basically any operation involving some sort of counting. In this post, let’s delve deep into what an off-by-one error is, why it occurs, and how you can prevent it from breaking your code. What is an Off-By-One Error? An off-by-one error occurs when a loop or iterative process runs one time too many, or one time too few. Usually, this results because the boundary conditions in a loop or an index inside an array were inadequately specified. Essentially, it is when you want to work your way through a list of 10 items but accidentally make your loop run 11 times. Common Examples of Off-By-One Errors Looping Through Arrays: When you forget that array/strings(group of characters) indices in most programming languages start at 0, you might accidentally go out of bounds by iterating or looping through one extra time. Range Specification: Using <= instead of < or >= instead of > in your loop conditions can cause the loop to run one additional time than intended. Fencepost Problems: This is when you are trying to find the number of things within something that has discrete sections, like posts on a fence, but forget you have one more post than what you actually need. Why Do Off-By-One Errors Occur? Off-by-one bugs are everywhere because they involve a deep misunderstanding of how the raw process of counting works in programming. It is very easy to forget if you are looping or describing indices, whether the first index should be 0 or 1. In addition, some languages also make the matter worse due to zero-based indexing as used in Python, JavaScript, and C, among others, wherein the first element in a sequence has an index of 0, not 1. Best Practices to Prevent Off-By-One Errors The good news is that off-by-one errors are preventable with careful attention and here are some of the best practices: 1. Understand the Indexing of Your Programming Language Know whether your programming language uses zero-based or one-based indexing. This understanding will help you set up your loops correctly. 2. Be Mindful of Loop Boundaries When writing loops, double-check your boundary conditions such as >, <, >=, <=. Ask yourself whether the loop should include the last element or stop just before handling the last element. 3. Use Built-in Functions Many languages offer built-in functions to handle iterations more safely, such as Python’s range() or Java’s for-each loop. These functions can reduce the chances of introducing an off-by-one error. 4. Test Boundary Conditions Make sure to test your code with boundary values or edge cases. For example, if you’re iterating over a list, test with an empty list, a list with one element, and a list with several elements to ensure your loop behaves as expected. 5. Add Comments to Your Code Adding comments to your loops and boundary conditions can help you and others understand the logic behind the loop, making it easier to spot potential off-by-one errors if there are any. Get Tharun Kumar Reddy Polu’s stories in your inbox Join Medium for free to get updates from this writer. However, The off-by-one error is a classic way to get into programming trouble, but it doesn’t have to be a disaster. By being aware of how these errors arise and taking consideration during coding, this can be avoided and your code will run smoothly. Because, after all, the devil is in the details — especially when it comes to counting in loops and arrays. Next time you are writing any loop, take the time to check your boundaries and save yourself from the silent nemesis of off-by-one errors. Happy Programming! Connect with me through linktr.ee to know more! Stackademic 🎓 Thank you for reading until the end. Before you go: Please consider clapping and following the writer! 👏 Follow us X \| LinkedIn \| YouTube \| Discord Visit our other platforms: In Plain English \| CoFeed \| Differ More content at Stackademic.com Data Structures Algorithms Coding Errors Best Practices ## Published in Stackademic 57K followers ·Last published 1 day ago Stackademic is a learning hub for programmers, devs, coders, and engineers. Our goal is to democratize free coding education for the world. ## Written by Tharun Kumar Reddy Polu 910 followers ·12 following A software engineer documenting the personal tech experiences and exploring What, Why and How behind software engineering and large scale distributed systems. No responses yet Write a response What are your thoughts? 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5818	https://www.cazoommaths.com/free-maths-worksheet/ratio-using-bar-models-free-worksheet/	Menu Ratio - Using Bar Models WORKSHEET Ratio - Using Bar Models WORKSHEET DESCRIPTION Bar models are used throughout this worksheet as a tool for students to divide amounts into given ratios. The worksheet is scaffolded by including drawn bar models for sections A to D. The scaffolding is decreased as labels are removed and students have to label more of the bar models themselves. In section B, learners will divide amounts into given ratios. Section C asks students to use one known amount to find the second amount and in section D, the difference is known. Finally, section E is made up of three questions and students must use an appropriate method to solve each one including constructing their own bar models. RELATED TO Ratio - Using Bar Models WORKSHEET Ratio Problems Year groups: 8, 9 Recipes (C) Year groups: 8, 9 Dividing Amounts into Ratios Year groups: 7, 8, 9 Finding Fractions of Amounts Year groups: 7, 8 ABOUT Visual Maths Resources Ltd is a company registered in England and Wales with a company no. 10607102 with a registered office at 71-75 Shelton Street, WC2H 9JQ. Cazoom Maths is the trading name of Visual Maths Resources Ltd. CONNECT GET STARTED MEMBERS AREA
5819	https://stackoverflow.com/questions/42732225/fast-integer-sqrt-upper-bound-approximation	Skip to main content Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Fast integer sqrt upper bound approximation Ask Question Asked Modified 4 years, 8 months ago Viewed 2k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. This is a question, regarding my homework, specifically on NASM. I am writing an algorithm to find the least whole factor of a number. (Greater than 1) In pseudo-code it can be summed up as: ``` if(n%2==0) return 2; for(i=3; i <= n/2; i+=2) if(n%i==0) return i; return n; ``` The program is just slightly slower than the requirement for large numbers. (n > 1 000 000 000) The most obvious (to me) improvement would be replacing n/2 with sqrt(n). However I am not supposed to know how to use floating-point numbers and finding the integer sqrt by Newton's method seems overkill. (Because I don't actually need to know the exact value and although I haven't tested it, I would imagine finding the isqrt recursively/iteratively would be quite slow) So I was wondering, whether there is a fast algorithm for some function, such that sqrt(n) < f(n) < n/2. By "fast" I mean preferably constant time and by f(n) < n/2 I mean significantly less for big n. Some of the options I am considering are: Check, for i <= min(sqrt(2^32), n/2), where sqrt(2^32) = 2^16 is a constant. Replace i <= n/2 with i <= (2^p), where p = ⌈log_2(n)/2⌉ or something. (p is half the position of most significant bit of n) nasm approximation sqrt square-root upperbound Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Mar 11, 2017 at 17:34 RuRo asked Mar 11, 2017 at 6:43 RuRoRuRo 37355 silver badges2222 bronze badges Add a comment \| 3 Answers 3 Reset to default This answer is useful 3 Save this answer. Show activity on this post. There is an iterative process to find a squareroot: ``` def approximate_sqrt(number, depth, start_val): for i in range(depth): start_val=(start_val+number/start_val)/2 return start_val ``` The better the initial guess (start_val), the faster it converges to a reasonable solution. ``` If start_val>sqrt(number) then every iterative value>sqrt(number) ``` and so it provides an upper bound (similarly for start_val < sqrt(number)). You can reduce the iteration depth to 1 or 2 if your initial guess is pretty close. So for iteratively guessing an upper bound for prime number candidates for example you can call ``` sqrt_appr=approximate_sqrt(i, 1, sqrt_appr+1) ``` for the next prime number candidate with the previous estimation for the square root of sqrt_appr and get upper bounds with an error of about 10E-6. (Although every time I checked how close the approximation was, which was for intervals of 3 million numbers, I set sqrt_appr=sqrt(number)+1 to reset the process.) Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Dec 6, 2020 at 12:27 M-Chen-3 2,05455 gold badges1616 silver badges3939 bronze badges answered Dec 5, 2020 at 15:58 physics_studentphysics_student 3944 bronze badges Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. Replace the i with ii: ``` if (n % 2 == 0) return 2; for (int i = 3; i i <= n; i += 2) if (n % i == 0) return i; return n ``` Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Mar 11, 2017 at 12:15 answered Mar 11, 2017 at 10:02 Angela PretoriusAngela Pretorius 24311 silver badge77 bronze badges 2 I think you meant ii <= n. I would certainly try this, when I get home, but wouldn't performing around 50000 extra multiplication operations be slower than the log_2 option I mentioned, even though it is saving some iteration time. RuRo – RuRo 03/11/2017 10:38:23 Commented Mar 11, 2017 at 10:38 Okay, I may be doing something wrong, but the algorithm got slower when calculating ii in the loop. It gets around 5 seconds in the worst case scenario (large prime). My old version was around 2 seconds for the same input. The algorithm I chose in the end, gets sub 0.025s. Anyways, thank you for your effort. RuRo – RuRo 03/11/2017 17:33:14 Commented Mar 11, 2017 at 17:33 Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. In the end I settled on the ⌈log_2(n)/2⌉ version. Since sqrt(n) = 2^(log_2(n)/2). So for anyone in need this is my solution. The sqrt(n) upper bound approximation is O(1). Whole algorithm is O(sqrt(n)) (I think). ``` mov esi,ebx ;ebx = N shr esi,1 ;esi = N/2 cmovnc esi,2 ;if no remainder RETURN 2 jnc RETURN mov edi,2 ;edi is max counter bsr eax,ebx ;eax is most significant set bit of ebx (log_2(eax)) shr eax,1 ;eax/=2 mov cl,al shl edi,cl ;max counter is 2^(log_2(N)/2 + 1) >= sqrt(N) mov esi,1 ;default answer is 1 mov ecx,3 ;start counter is 3 START: mov edx,0 mov eax,ebx div ecx ;divide N by counter test edx,edx ;if no remainder RETURN counter cmovz esi,ecx jz RETURN add ecx,2 ;else counter += 2 cmp ecx,edi jb START ;if counter <= max counter try agian with new divisor RETURN: ;the answer is in (esi) ``` P.S. If you are wondering why I don't just check ii <= N. It is actually significantly slower than this version. Just adding a mul inside the loop shouldn't slow it down that much, so I suspect it breaks the branch prediction algorithm each iteration. The cmp ecx,edi is comparing a counter against a constant, so it might be predicted right most of the times, where the cmp 'ecxecx',ebx would be comparing a square of the counter, which may not be so obvious to the processor. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Mar 11, 2017 at 17:46 answered Mar 11, 2017 at 17:29 RuRoRuRo 37355 silver badges2222 bronze badges Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions nasm approximation sqrt square-root upperbound See similar questions with these tags. 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5820	https://www.quora.com/What-is-the-14-22-and-26-Fibonacci-using-the-binet-formula	What is the 14, 22, and 26 Fibonacci using the binet formula? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Binet's Formula Fibonacci Series Formulas (functions) Mathematical Equations Mathematical Sciences Fibonacci Numbers Mathematical Formulas 4 What is the 14, 22, and 26 Fibonacci using the binet formula? All related (33) Sort Recommended Sol Warda Author has 5.8K answers and 2.5M answer views ·3y I shall give you the Binet’s formula and I expect you to use it accurately to find the Fibonacci numbers that you want! Here is the Binet’s formula: n=14;F=((1 + sqrt(5))^n - (1 - sqrt(5))^n) / (2^nsqrt(5)) Note: Do you think you can? If not, why not? Hint: Here are the first few Fibonacci numbers: 1, 1, 2, 3, 5, 8……….and so on. Do you see the pattern? Upvote · 9 2 Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Download 999 207 Related questions More answers below What is the 25th Fibonacci using Binet's formula? What is the 46th Fibonacci number using Binet's formula? What is the formula for the nth Fibonacci number? Prove it. What are the 51st Fibonacci numbers using Binet's formula? Can you help me find the nth Fibonacci number using the Binet formula? Klaus Hauptfleisch B.A. from University of Bonn (Graduated 1985) · Author has 8.4K answers and 2.7M answer views ·2y For larger Fibonacci numbers the approximation formula (according to the Binet formula) is good enough: [math]f_n \approx \frac{ \phi^n}{ \sqrt 5} \approx \frac{1.61803398874989^n}{ \sqrt 5}[/math]. When applying it to [math]f_{14}[/math] my Windows calculator spits out a very close 377.0005305032164566221929875403… The Binet formula itself is: [math]f_n= \frac{ \phi^n- \psi^n}{ \sqrt 5}= \frac{1}{ \sqrt 5} \biggl [ \bigl ( \frac{1+ \sqrt 5}{2} \bigr )^n-\bigl ( \frac{1+ \sqrt 5}{2} \bigr )^n \biggr ][/math] The latter fraction or [math]\psi[/math] (Psi), the negative inverse of [math]\phi[/math] tends towards 0 and can be ignored with larger Fibonacci numbers. Upvote · 9 1 Unnikrishnan Menon Integrated my Brain By Parts! · Upvoted by Shubhankar Datta , Master of Science Mathematics, Jadavpur University (2022) · Author has 521 answers and 4.7M answer views ·7y Related How do you derive the formula for the nth Fibonacci number? [math]1,1,2,3,5,8,13,21,.....[/math] That’s the Fibonacci Series. Other than the first 2 terms, every subsequent term is the sum of the previous 2 terms that come before it. It’s easy to see the pattern. In other words, [math]y_{n+2}=y_{n+1}+y_{n}\tag{1}[/math] Also since we are starting off our series with the first 2 terms as 1, we can say that [math]y_0=y_1=1[/math] This is a pretty cool application of Z-transforms and Difference Equatio Continue Reading [math]1,1,2,3,5,8,13,21,.....[/math] That’s the Fibonacci Series. Other than the first 2 terms, every subsequent term is the sum of the previous 2 terms that come before it. It’s easy to see the pattern. In other words, [math]y_{n+2}=y_{n+1}+y_{n}\tag{1}[/math] Also since we are starting off our series with the first 2 terms as 1, we can say that [math]y_0=y_1=1[/math] This is a pretty cool application of Z-transforms and Difference Equations :) I’ll take the Z-Transform of both sides of equation 1 [math]\begin{equation}\begin{split}\sum_{n=0}^{\infty}y_{n+2}z^{-n}=\sum_{n=0}^{\infty}y_{n+1}z^{-n}+\sum_{n=0}^{\infty}y_{n}z^{-n}\end{split}\end{equation}\tag{}[/math] Now on, I’ll write the Z-transform of [math]y_n[/math] as [math]Y(z)[/math]. Just so that it doesn’t get too messy. I’ll use the Left-Shift property of Z-transforms to break down the Z-transforms of [math]y_{n+2}[/math] and [math]y_{n+1}[/math]. Then we’ll have… [math]\begin{equation}\begin{split}z^2Y(z)-z^2\underbrace{y_0}{1}-z\underbrace{y_1}{1}=zY(z)-z\underbrace{y_0}_{1}+Y(z)\end{split}\end{equation}\tag{}[/math] Let’s solve for [math]Y(z)[/math] [math]\begin{equation}\begin{split}Y(z)=\dfrac{z^2}{z^2-z-1}=Z(y_n)\end{split}\end{equation}\tag{}[/math] That means… [math]\begin{equation}\begin{split}y_n&=Z^{-1}\left[\dfrac{z^2}{z^2-z-1}\right]\&=Z^{-1}\left[z\left(\dfrac{z}{z^2-z-1}\right)\right]\end{split}\end{equation}\tag{}[/math] I wanna factor that denominator. [math]\begin{equation}\begin{split}z^2-z-1&=0\\implies z&=\dfrac{1\pm\sqrt{5}}{2}\end{split}\end{equation}\tag{}[/math] Let’s say for now, [math]a=\dfrac{1+\sqrt{5}}{2}[/math] and [math]b=\dfrac{1-\sqrt{5}}{2}[/math] I’m doing this just to keep things tidy. [math]\begin{align}\implies y_n&=Z^{-1}\left[z\left(\dfrac{z}{(z-a)(z-b)}\right)\right]\&=Z^{-1}\left[z\left(\dfrac{... Upvote · 999 124 9 5 Assistant Bot · 1y The Binet formula for the Fibonacci sequence is given by: [math]F(n) = \frac{\phi^n - \psi^n}{\sqrt{5}}[/math] where: [math]\phi = \frac{1 + \sqrt{5}}{2}[/math] (approximately 1.6180339887) is the golden ratio, [math]\psi = \frac{1 - \sqrt{5}}{2}[/math] (approximately -0.6180339887). Now, we can use this formula to calculate the 14th, 22nd, and 26th Fibonacci numbers. 14th Fibonacci Number [math]F(14) = \frac{\phi^{14} - \psi^{14}}{\sqrt{5}} \approx \frac{377 - (-0.0006)}{\sqrt{5}} \approx \frac{377}{2.236} \approx 169[/math] 22nd Fibonacci Number [math]F(22) = \frac{\phi^{22} - \psi^{22}}{\sqrt{5}} \approx \frac{17711 - (-0.00002)}{\sqrt{5}} \approx \f[/math] Continue Reading The Binet formula for the Fibonacci sequence is given by: [math]F(n) = \frac{\phi^n - \psi^n}{\sqrt{5}}[/math] where: [math]\phi = \frac{1 + \sqrt{5}}{2}[/math] (approximately 1.6180339887) is the golden ratio, [math]\psi = \frac{1 - \sqrt{5}}{2}[/math] (approximately -0.6180339887). Now, we can use this formula to calculate the 14th, 22nd, and 26th Fibonacci numbers. 14th Fibonacci Number [math]F(14) = \frac{\phi^{14} - \psi^{14}}{\sqrt{5}} \approx \frac{377 - (-0.0006)}{\sqrt{5}} \approx \frac{377}{2.236} \approx 169[/math] 22nd Fibonacci Number [math]F(22) = \frac{\phi^{22} - \psi^{22}}{\sqrt{5}} \approx \frac{17711 - (-0.00002)}{\sqrt{5}} \approx \frac{17711}{2.236} \approx 7901[/math] 26th Fibonacci Number [math]F(26) = \frac{\phi^{26} - \psi^{26}}{\sqrt{5}} \approx \frac{196418 - 0}{\sqrt{5}} \approx \frac{196418}{2.236} \approx 8784[/math] Summary of Fibonacci Numbers F(14): 377 F(22): 17711 F(26): 196418 These values are rounded to the nearest integer, which is how Fibonacci numbers are typically expressed. Upvote · Related questions What is the 25th Fibonacci using Binet's formula? What is the 46th Fibonacci number using Binet's formula? What is the formula for the nth Fibonacci number? Prove it. What are the 51st Fibonacci numbers using Binet's formula? Can you help me find the nth Fibonacci number using the Binet formula? What is the Fibonacci number of n=14 using the binet formula? What is the formula to calculate the [math]n^\text{th}[/math] Fibonacci number? What is the Binet formula to determine the 50th Fibonacci number? What is the 100th term using Binet's formula? What is the nth Fibonacci number of n=14, n=22, n+20, n-25, and n-50 using the binet formula? What is the 18th term in the Fibonacci sequence applying the binets formula? What is the 90th term in the Fibonacci sequence using Binet’s formula? How do I get the 25th term using the binet formula? What is the following using the Fibonacci number formula X12? What are the 32nd Fibonacci numbers using Binet's simplified formula? Related questions What is the 25th Fibonacci using Binet's formula? What is the 46th Fibonacci number using Binet's formula? What is the formula for the nth Fibonacci number? Prove it. What are the 51st Fibonacci numbers using Binet's formula? Can you help me find the nth Fibonacci number using the Binet formula? What is the Fibonacci number of n=14 using the binet formula? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
5821	https://2012books.lardbucket.org/books/finance-banking-and-money-v1.1/s24-01-aggregate-output-and-keynesian.html	This is “Aggregate Output and Keynesian Cross Diagrams”, section 21.1 from the book Finance, Banking, and Money (v. 1.1). For details on it (including licensing), click here. This book is licensed under a Creative Commons by-nc-sa 3.0 license. See the license for more details, but that basically means you can share this book as long as you credit the author (but see below), don't make money from it, and do make it available to everyone else under the same terms. This content was accessible as of December 29, 2012, and it was downloaded then by Andy Schmitz in an effort to preserve the availability of this book. Normally, the author and publisher would be credited here. However, the publisher has asked for the customary Creative Commons attribution to the original publisher, authors, title, and book URI to be removed. Additionally, per the publisher's request, their name has been removed in some passages. More information is available on this project's attribution page. For more information on the source of this book, or why it is available for free, please see the project's home page. You can browse or download additional books there. To download a .zip file containing this book to use offline, simply click here. Has this book helped you? Consider passing it on: Help Creative Commons Creative Commons supports free culture from music to education. Their licenses helped make this book available to you. Help a Public School DonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators. Previous Section Table of Contents Next Section 21.1 Aggregate Output and Keynesian Cross Diagrams Learning Objectives What does this equation mean: Y = Yad = C + I + G + NX? Why is this equation important? What is the equation for C and why is it important? What is the Keynesian cross diagram and what does it help us to do? Developed in 1937 by economist and Keynes disciple John Hicks, the IS-LM model is still used today to model aggregate output (gross domestic product [GDP], gross national product [GNP], etc.) and interest rates in the short run. It begins with John Maynard Keynes’s recognition that where: Y = aggregate output (supplied) Yad = aggregate demand C = consumer expenditure I = investment (on new physical capital like computers and factories, and planned inventory) G = government spending NX = net exports (exports minus imports) Keynes further explained that C = a + (mpc × Yd) where: Yd = disposable income, all that income above a a = autonomous consumer expenditure (food, clothing, shelter, and other necessaries) mpc = marginal propensity to consume (change in consumer expenditure from an extra dollar of income or “disposable income;” it is a constant bounded by 0 and 1) Practice calculating C in Exercise 1. Exercises Calculate consumer expenditure using the formula C = a + (mpc x Yd). \| Autonomous Consumer Expenditure \| Marginal Propensity to Consume \| Disposable Income \| Answer: C \| --- --- \| \| 200 \| 0.5 \| 0 \| 200 \| \| 400 \| 0.5 \| 0 \| 400 \| \| 200 \| 0.5 \| 200 \| 300 \| \| 200 \| 0.5 \| 300 \| 350 \| \| 300 \| 0.5 \| 300 \| 450 \| \| 300 \| 0.75 \| 300 \| 525 \| \| 300 \| 0.25 \| 300 \| 375 \| \| 300 \| 0.01 \| 300 \| 303 \| \| 300 \| 1 \| 300 \| 600 \| \| 100 \| 0.5 \| 1000 \| 600 \| \| 100 \| 0.75 \| 1000 \| 850 \| You can plot a consumption functionA mathematical equation thought to express the level of consumer spending.by drawing a graph, as in Figure 21.1 "A consumption function", with consumer expenditure on the vertical axis and disposable income on the horizontal. (Autonomous consumer expenditure a will be the intercept and mpc × Yd will be the slope.) Figure 21.1 A consumption function Investment is composed of so-called fixed investment on equipment and structures and planned inventory investment in raw materials, parts, or finished goods. For the present, we will ignore G and NX and, following Keynes, changes in the price level. (Remember, we are talking about the short term here. Remember, too, that Keynes wrote in the context of the gold standard, not an inflationary free floating regime, so he was not concerned with price level changes.) The simple model that results, called a Keynesian cross diagram, looks like the diagram in Figure 21.2 "A Keynesian cross diagram". Figure 21.2 A Keynesian cross diagram The 45-degree line simply represents the equilibrium Y = Yad. The other line, the aggregate demand function, is the consumption function line plus planned investment spending I. Equilibrium is reached via inventories (part of I). If Y > Yad, inventory levels will be higher than firms want, so they’ll cut production. If Y < Yad, inventories will shrink below desired levels and firms will increase production. We can now predict changes in aggregate output given changes in the level of I and C and the marginal propensity to consume (the slope of the C component of Yad). Suppose I increases. Due to the upward slope of Yad, aggregate output will increase more than the increase in I. This is called the expenditure multiplier and it is summed up by the following equation: So if a is 200 billion, I is 400 billion, and mpc is .5, Y will be If I increases to 600 billion, Y = 800 × 2 = $1,600 billion. If the marginal propensity to consume were to increase to .75, Y would increase to Y = 800 × 1/.25 = 800 × 4 = $3,200 billion because Yad would have a much steeper slope. A decline in mpc to .25, by contrast, would flatten Yad and lead to a lower equilibrium: Practice calculating aggregate output in Exercise 2. 2. Calculate aggregate output with the formula: Y = (a + I) × 1/(1 − mpc) \| Autonomous Spending \| Marginal Propensity to Consume \| Investment \| Answer: Aggregate Output \| --- --- \| \| 200 \| 0.5 \| 500 \| 1400 \| \| 300 \| 0.5 \| 500 \| 1600 \| \| 400 \| 0.5 \| 500 \| 1800 \| \| 500 \| 0.5 \| 500 \| 2000 \| \| 500 \| 0.6 \| 500 \| 2500 \| \| 200 \| 0.7 \| 500 \| 2333.33 \| \| 200 \| 0.8 \| 500 \| 3500 \| \| 200 \| 0.4 \| 500 \| 1166.67 \| \| 200 \| 0.3 \| 500 \| 1000 \| \| 200 \| 0.5 \| 600 \| 1600 \| \| 200 \| 0.5 \| 700 \| 1800 \| \| 200 \| 0.5 \| 800 \| 2000 \| \| 200 \| 0.5 \| 400 \| 1200 \| \| 200 \| 0.5 \| 300 \| 1000 \| \| 200 \| 0.5 \| 200 \| 800 \| Stop and Think Box During the Great Depression, investment (I) fell from $232 billion to $38 billion (in 2000 USD). What happened to aggregate output? How do you know? Aggregate output fell by more than $232 billion − $38 billion = $194 billion. We know that because investment fell and the marginal propensity to consume was > 0, so the fall was more than $194 billion, as expressed by the equation Y = (a + I) × 1/(1 − mpc). To make the model more realistic, we can easily add NX to the equation. An increase in exports over imports will increase aggregate output Y by the increase in NX times the expenditure multiplier. Likewise, an increase in imports over exports (a decrease in NX) will decrease Y by the decrease in NX times the multiplier. Government spending (G) also increases Y. We must realize, however, that some government spending comes from taxes, which consumers view as a reduction in income. With taxation, the consumption function becomes the following: T means taxes. The effect of G is always larger than that of T because G expands by the multiplier, which is always > 1, while T is multiplied by MPC, which never exceeds 1. So increasing G, even if it is totally funded by T, will increase Y. (Remember, this is a short-run analysis.) Nevertheless, Keynes argued that, to help a country out of recession, government should cut taxes because that will cause Yd to rise, ceteris paribus. Or, in more extreme cases, it should borrow and spend (rather than tax and spend) so that it can increase G without increasing T and thus decreasing C. Stop and Think Box As noted in Chapter 11 "The Economics of Financial Regulation", many governments, including that of the United States, responded to the Great Depression by increasing tariffs in what was called a beggar-thy-neighbor policy. Today we know that such policies beggared everyone. What were policymakers thinking? They were thinking that tariffs would decrease imports and thereby increase NX (exports minus imports) and Y. That would make their trading partner’s NX decrease, thus beggaring them by decreasing their Y. It was a simple idea on paper, but in reality it was dead wrong. For starters, other countries retaliated with tariffs of their own. But even if they did not, it was a losing strategy because by making neighbors (trading partners) poorer, the policy limited their ability to import (i.e., decreased the first country’s exports) and thus led to no significant long-term change in NX. Figure 21.3 "The determinants of aggregate demand"sums up the discussion of aggregate demand. Figure 21.3 The determinants of aggregate demand Key Takeaways The equation Y = Yad = C + I + G + NX tells us that aggregate output (or aggregate income) is equal to aggregate demand, which in turn is equal to consumer expenditure plus investment (planned, physical stuff) plus government spending plus net exports (exports – imports). It is important because it allows economists to model aggregate output (to discern why, for example, GDP changes). In a taxless Eden, like the Gulf Cooperation Council countries, consumer expenditure equals autonomous consumer expenditure (spending on necessaries) (a) plus the marginal propensity to consume (mpc) times disposable income (Yd), income above a. In the rest of the world, C = a + mpc × (Yd − T), where T = taxes. C, particularly the marginal propensity to consume variable, is important because it gives the aggregate demand curve in a Keynesian cross diagram its upward slope. A Keynesian cross diagram is a graph with aggregate demand (Yad) on the vertical axis and aggregate output (Y) on the horizontal. It consists of a 45-degree line where Y = Yad and a Yad curve, which plots C + I + G + NX with the slope given by the expenditure multiplier, which is the reciprocal of 1 minus the marginal propensity to consume: Y = (a + I + NX + G) × 1/(1 − mpc). The diagram helps us to see that aggregate output is directly related to a, I, exports, G, and mpc and indirectly related to T and imports. 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5822	https://www2.math.uconn.edu/~stein/virtual/Notes/Calculus/inequalities.pdf	Analyzing the Sign of a Function c ⃝Alan H. Stein The University of Connecticut at Waterbury stein@math.uconn.edu In Calculus, we see that many problems involving change can be dealt with by analyzing the sign of a derivative. This makes the ability to analyze the sign of a function (remember, among other things, a derivative is itself a function) extremely important (no pun intended). Indeed, hardly a day will go by in which we will not have to do such an analysis several times. Analyzing the sign of a function is equivalent to solving an inequality. For example x is in the solution set of the inequality f(x) < g(x) if and only f(x) −g(x) is negative. Thus, any method that one can use to solve an inequality can be used to analyze the sign of a function. The method described here is probably the easiest method to use in a large proportion of cases. We will describe the method for the case where the function involved is a rational function, i.e. a quotient of polynomial functions, but the basic idea can be used with more general functions. Laying the Groundwork with Linear Functions. First, consider linear functions, that is, functions of the form f(x) = ax+b for constants a, b. Actually, since ax+b = a(x+b/a), we really only have to consider functions of the form f(x) = x + c or f(x) = x −c. The key fact is that it’s very easy to determine the sign of f(x) = x+c on any interval not containing −c, and it’s very easy to determine the sign of f(x) = x −c on any interval not containing c. Consider the following examples. (1) Determine the sign of x −3 on the interval (5, ∞) = {x : x > 5}. Analysis: If x > 5, then x −3 > 5 −3, so x −3 > 2 and thus x −3 is positive on (5, ∞). (2) Determine the sign of x −3 on the interval (5, 10) = {x : 5 < x < 10}. Analysis: If 5 < x < 10, then 5 −3 < x −3 < 10 −3, so 2 < x −3 < 7 and thus x −3 is positive on (5, 10). (3) Determine the sign of x −3 on the interval (3, ∞) = {x : x > 3}. Analysis: If x > 3, then x −3 > 3 −3, so x −3 > 0 and thus x −3 is positive on (3, ∞). (4) Determine the sign of x −3 on the interval (−∞, 3) = {x : x < 3}. Analysis: If x < 3, then x −3 < 3 −3, so x −3 < 0 and thus x −3 is negative on (−∞, 3). (5) Determine the sign of x + 3 on the interval (−∞, −3) = {x : x < −3}. Analysis: If x < −3, then x + 3 < −3 + 3, so x + 3 < 0 and thus x + 3 is negative on (−∞, −3). (6) Determine the sign of x + 3 on the interval (5, 10) = {x : 5 < x < 10}. Analysis: If 5 < x < 10, then 5 + 3 < x + 3 < 10 + 3, so 8 < x < 13 and thus x + 3 is positive on (5, 10). 1 2 Look for the general idea: To analyze the sign of x ± c on an interval (α, β), take the inequality α < x < β, add or substract c to all sides to get α ± c < x ± c < β ± c and draw the obvious conclusion about the sign of x ± c. This suggests the following general strategy for analyzing the sign of more general func-tions. Strategy. (1) Factor the function as much as possible. If the function involves a quotient, factor both the numerator and denominator as much as possible. In the best of all possible worlds, each factor will be a linear function. (2) Find all the zeroes of each of the factors of both the numerator and the denominator. (3) Mark oﬀthose points on a number line. These points partition the real line into a set of intervals. (4) Determine the sign of the function on each of those intervals. This may be done by examining the sign of each of the factors of the function. If the factors are linear functions, then the method described earlier will work. Example. Consider the function f(x) = 5(x −3)(x + 8) x(x −1)2 . Note that this function is com-pletely factored. If the function you are dealing with is not factored, you will need to factor it completely. One can tell at sight that x −3 = 0 when x is 3, that x + 8 = 0 when x is −8, that x = 0 when x is 0 (that didn’t take much thought, did it?), and that x −1 = 0 when x is 1. So set up and mark oﬀa number line as shown below. Now take a look at each interval, one at a time. (1) (3, ∞) If x > 3, each factor is clearly positive, so f is positive on this interval. (2) (1, 3) If 1 < x < 3, then x −3 is negative, since x < 3, but every other factor is positive, so f is negative on this interval. (3) (0, 1) If 0 < x < 1, then x−3 is negative. Although x−1 is negative, (x−1)2 is positive, since it’s a square of a non-zero number, and every other factor is positive, so f is negative on this interval. (4) (−8, 0) If −8 < x < 0, then x −3 is negative, x + 8 is positive (since x > 8), but x is negative while (x −1)2 is positive, so f is positive on this interval. (5) (−∞, −8) If x < −8, then x−3 is negative, x+8 is negative (since x < −8), and x is negative, but (x −1)2 is positive, so f is negative on this interval. 3 We may visualize all this as follows. If we wanted to express this result in standard mathematical notation, as we always should want to, we might say that f is positive on (−8, 0) ∪(3, ∞), and f is negative on (−∞, −8) ∪(0, 1) ∪(1, 3). Alternatively, we might say that f is positive on {x : −8 < x < 0 or x > 3} and f is negative on {x : x < −8 or 0 < x < 1 or 1 < x < 3}. The Theoretical Basis for the Method. The Zero Theorem tells us that if a function is continuous on a closed interval but has diﬀerent signs at the endpoints of that interval it must be zero somewhere in between. It follows that a function must maintain the same sign at all points of an interval which does not contain any zeroes or discontinuities. If one factors the numerator and denominator of a function completely, the zeroes of the function will occur at the points where factors of the numerator are zero, while the discontinuities will generally occur where factors of the denominator are zero.
5823	https://emedicine.medscape.com/article/239927-workup	For You News & Perspective Tools & Reference Edition English Medscape Editions About You Professional Information Newsletters & Alerts Your Watch List Formulary Plan Manager Log Out Log In EN Medscape Editions English Deutsch EspaÃ±ol FranÃ§ais PortuguÃªs UK X Univadis from Medscape About You Professional Information Newsletters & Alerts Your Watch List Formulary Plan Manager Log Out Register Log In close Please confirm that you would like to log out of Medscape. If you log out, you will be required to enter your username and password the next time you visit. Log out Cancel processing.... Tools & Reference>Nephrology IgA Nephropathy Workup Updated: Apr 18, 2025 Author: Sohail Abdul Salim, MD, FASN, FACP; Chief Editor: Vecihi Batuman, MD, FASN more...;) Share Print Feedback Close Facebook Twitter LinkedIn WhatsApp Email Sections IgA Nephropathy Sections IgA Nephropathy Overview Practice Essentials Pathophysiology Etiology Epidemiology Prognosis Show All Presentation History Physical Show All DDx Workup Approach Considerations Histologic Findings Show All Treatment Approach Considerations Medical Care Diet Tonsillectomy Kidney Transplantation Show All Medication Medication Summary Complement Inhibitors Corticosteroids Angiotensin-converting enzyme inhibitors Angiotensin II receptor antagonists Omega-3 polyunsaturated fatty acids Immunosuppressive agents Endothelin Antagonists Show All Media Gallery;) References;) Workup Approach Considerations The first step in confirming the diagnosis of IgA nephropathy is a careful urinalysis of a first-void urine sample performed by an experienced urine analyst. Direct examination of the urine sediment is required to identify red blood cells (RBCs) and RBC casts, both of which indicate glomerular injury. Proteinuria testing can be accomplished quantitatively by a 24-hour measurement of urinary protein or semiquantitatively by measuring a urine protein/creatinine ratio. A normal ratio should be less than approximately 0.1. Also, adults older than 50 years with proteinuria should have urine protein electrophoresis performed to exclude monoclonal light chains as a cause of proteinuria. Assess kidney function in patients with proteinuria or hematuria by a 24-hour creatinine clearance test. Alternatively, the glomerular filtration rate (GFR) can be estimated using the Modification of Diet in Renal Disease (MDRD) formula or CKD-EPI. Although the serum IgA level is elevated in up to half of patients, this finding is insensitive, nonspecific, and of no clinical utility Kidney biopsy is necessary for confirming the diagnosis of IgA nephropathy. In addition, biopsy findings are used to determine the MEST-C score (mesangial [M] and endo-capillary [E] hypercellularity, segmental sclerosis [S], interstitial fibrosis/tubular atrophy [T], and crescents [C]), for predicting risk of progression (see Overview/Prognosis). In IgA nephropathy, proteinuria rarely occurs without microscopic hematuria. Mild proteinuria is common. Nephrotic-range proteinuria is uncommon, occurring in only 5% of patients with IgA nephropathy, and is more commonly seen in children and adolescents. Nephrotic-range proteinuria can be seen early in the disease course as well as in patients with advanced disease Patients with heavy proteinuria and nephrotic syndrome are likely to have IgA deposition with diffuse proliferative glomerular lesions or minimal-change light microscopic findings Acute kidney injury Acute kidney injury, with edema, hypertension, and oliguria, occurs in fewer than 5% of patients. It can develop from either of the following two distinct mechanisms: Acute severe immune injury can manifest as necrotizing glomerulonephritis and crescent formation Alternatively, only mild glomerular injury is observed with gross hematuria, and kidney injury is presumably due to tubular occlusion by RBCs; this is reversible, and kidney function recovers with supportive measures Next: Histologic Findings Light microscopy On kidney biopsy, the most common light microscopy findings are focal or, more often, diffuse mesangial proliferation and extracellular matrix expansion (as seen in the image below). Morphology can range from normal to moderate or severe intracapillary or extracapillary proliferative lesions. Some patients with little or no change by light microscopy have IgA deposits on immunofluorescence. Light microscopy of a glomerulus from a patient with immunoglobulin A nephropathy showing increased mesangial matrix and cellularity. View Media Gallery) Occasionally, patients have focal glomerular sclerosis indistinguishable from focal segmental glomerulosclerosis on light microscopy. A number of other findings can be observed in advanced disease, including interstitial fibrosis, tubular atrophy, and vascular sclerosis. A few patients have segmental necrotizing lesions with crescent formation due to extensive disruption of the capillaries. These findings can be helpful prognostic tools in patients with IgA nephropathy (see Overview/Prognosis). Electron microscopy Electron microscopy shows mesangial hypercellularity and increased mesangial matrix. The important finding is electron-dense deposits of IgA in the mesangium, such as those in the image below, but deposits in the subendothelial and subepithelial region of the glomerular capillary wall are found in a minority of patients, especially those with more severe disease. Subendothelial deposits are often present when there is endocapillary hypercellularity. Electron microscopy showing large dark mesangial deposits. View Media Gallery) Immunofluorescence Immunofluorescence findings are the pathologic hallmark of this disease. IgA is deposited in a diffuse granular pattern in the mesangium (as seen in the image below) and occasionally in the capillary wall. The deposits are predominantly polymeric IgA of the IgA1 subclass; in addition, IgG is found in 43% of cases, and IgM in 54%. C3 is often present. Polyclonal deposits, often with more lambda than kappa light chains, are seen. Presence of C4d indicates a worse prognosis. Immunofluorescence microscopy demonstrating large mesangial immunoglobulin A (IgA) deposits diagnostic of IgA nephropathy. View Media Gallery) Previous Treatment & Management References Stamellou E, Seikrit C, Tang SCW, Boor P, TesaÅ V, Floege J, et al. IgA nephropathy. Nat Rev Dis Primers. 2023 Nov 30. 9 (1):67. [QxMD MEDLINE Link]. Rout P, Limaiem F, Hashmi MF. IgA Nephropathy (Berger Disease). 2025 Jan. 2:16002. 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Manno C, Torres DD, Rossini M, et al. Randomized controlled clinical trial of corticosteroids plus ACE-inhibitors with long-term follow-up in proteinuric IgA nephropathy. Nephrol Dial Transplant. 2009 Jul 23. [QxMD MEDLINE Link]. Heerspink HJL, Radhakrishnan J, Alpers CE, Barratt J, Bieler S, Diva U, et al. Sparsentan in patients with IgA nephropathy: a prespecified interim analysis from a randomised, double-blind, active-controlled clinical trial. Lancet. 2023 May 13. 401 (10388):1584-1594. [QxMD MEDLINE Link]. Rovin BH, et al; DUPRO steering committee and PROTECT Investigators. Efficacy and safety of sparsentan versus irbesartan in patients with IgA nephropathy (PROTECT): 2-year results from a randomised, active-controlled, phase 3 trial. Lancet. 2023 Dec 2. 402 (10417):2077-2090. [QxMD MEDLINE Link]. Heerspink HJL, Jardine M, Kohan DE, Lafayette RA, Levin A, Liew A, et al. Atrasentan in Patients with IgA Nephropathy. N Engl J Med. 2025 Feb 6. 392 (6):544-554. [QxMD MEDLINE Link]. Fabhalta (iptacopan) [package insert]. August 2024 East Hanover, NJ: Novartis. August 2024. Available at [Full Text]. Shoji T, Nakanishi I, Suzuki A, et al. Early treatment with corticosteroids ameliorates proteinuria, proliferative lesions, and mesangial phenotypic modulation in adult diffuse proliferative IgA nephropathy. Am J Kidney Dis. 35(2):194-201. [QxMD MEDLINE Link]. Floege J, Feehally J. Treatment of IgA nephropathy and Henoch-SchÃ¶nlein nephritis. Nat Rev Nephrol. 2013 Jun. 9 (6):320-7. [QxMD MEDLINE Link]. Pozzi C, Andrulli S, Del Vecchio L, et al. Corticosteroid effectiveness in IgA nephropathy: long-term results of a randomized, controlled trial. J Am Soc Nephrol. 2004 Jan. 15(1):157-63. [QxMD MEDLINE Link]. Lv J, Zhang H, Chen Y, Li G, Jiang L, Singh AK, et al. Combination therapy of prednisone and ACE inhibitor versus ACE-inhibitor therapy alone in patients with IgA nephropathy: a randomized controlled trial. Am J Kidney Dis. 2009 Jan. 53 (1):26-32. [QxMD MEDLINE Link]. Tesar V, Troyanov S, Bellur S, Verhave JC, Cook HT, Feehally J, et al. Corticosteroids in IgA Nephropathy: A Retrospective Analysis from the VALIGA Study. J Am Soc Nephrol. 2015 Sep. 26 (9):2248-58. [QxMD MEDLINE Link].[Full Text]. Lv J, et al; TESTING Study Group. Effect of Oral Methylprednisolone on Clinical Outcomes in Patients With IgA Nephropathy: The TESTING Randomized Clinical Trial. JAMA. 2017 Aug 1. 318 (5):432-442. [QxMD MEDLINE Link]. Tang SC, Tang AW, Wong SS, et al. Long-term study of mycophenolate mofetil treatment in IgA nephropathy. Kidney Int. 2010 Mar. 77(6):543-9. [QxMD MEDLINE Link]. Rauen T, Eitner F, Fitzner C, Sommerer C, Zeier M, Otte B, et al. Intensive Supportive Care plus Immunosuppression in IgA Nephropathy. N Engl J Med. 2015 Dec 3. 373 (23):2225-36. [QxMD MEDLINE Link].[Full Text]. Vecchio M, Bonerba B, Palmer SC, Craig JC, Ruospo M, Samuels JA, et al. Immunosuppressive agents for treating IgA nephropathy. Cochrane Database Syst Rev. 2015 Aug 3. CD003965. [QxMD MEDLINE Link]. FellstrÃ¶m BC, Barratt J, Cook H, Coppo R, Feehally J, de Fijter JW, et al. Targeted-release budesonide versus placebo in patients with IgA nephropathy (NEFIGAN): a double-blind, randomised, placebo-controlled phase 2b trial. Lancet. 2017 May 27. 389 (10084):2117-2127. [QxMD MEDLINE Link]. Lafayette R, Kristensen J, Stone A, Floege J, TesaÅ V, Trimarchi H, et al. Efficacy and safety of a targeted-release formulation of budesonide in patients with primary IgA nephropathy (NefIgArd): 2-year results from a randomised phase 3 trial. Lancet. 2023 Sep 9. 402 (10405):859-870. [QxMD MEDLINE Link]. Brooks A. FDA Awards Full Approval to Budesonide (Tarpeyo) in Treatment of IgA Nephropathy. HCP Live. Available at December 20, 2023; Accessed: February 26, 2024. Wheeler DC, Toto RD, StefÃ¡nsson BV, Jongs N, Chertow GM, Greene T, et al. A pre-specified analysis of the DAPA-CKD trial demonstrates the effects of dapagliflozin on major adverse kidney events in patients with IgA nephropathy. Kidney Int. 2021 Jul. 100 (1):215-224. [QxMD MEDLINE Link].[Full Text]. Dillon JJ. Fish oil therapy for IgA nephropathy: efficacy and interstudy variability. J Am Soc Nephrol. 1997 Nov. 8(11):1739-44. [QxMD MEDLINE Link]. Liu LJ, Lv JC, Shi SF, et al. Oral calcitriol for reduction of proteinuria in patients with IgA nephropathy: a randomized controlled trial. Am J Kidney Dis. 2012 Jan. 59(1):67-74. [QxMD MEDLINE Link]. Lafayette RA, Canetta PA, Rovin BH, Appel GB, Novak J, Nath KA, et al. A Randomized, Controlled Trial of Rituximab in IgA Nephropathy with Proteinuria and Renal Dysfunction. J Am Soc Nephrol. 2017 Apr. 28 (4):1306-1313. [QxMD MEDLINE Link]. Zand L, Canetta P, Lafayette R, Aslam N, Jan N, Sethi S, et al. An Open-Label Pilot Study of Adrenocorticotrophic Hormone in the Treatment of IgA Nephropathy at High Risk of Progression. Kidney Int Rep. 2020 Jan. 5 (1):58-65. [QxMD MEDLINE Link].[Full Text]. Liu LJ, Yang YZ, Shi SF, Bao YF, Yang C, Zhu SN, et al. Effects of Hydroxychloroquine on Proteinuria in IgA Nephropathy: A Randomized Controlled Trial. Am J Kidney Dis. 2019 Jul. 74 (1):15-22. [QxMD MEDLINE Link].[Full Text]. Roccatello D, Ferro M, Cesano G, Rossi D, Berutti S, Salomone M, et al. Steroid and cyclophosphamide in IgA nephropathy. Nephrol Dial Transplant. 2000 Jun. 15 (6):833-5. [QxMD MEDLINE Link]. Klahr S, Levey AS, Beck GJ, et al. The effects of dietary protein restriction and blood-pressure control on the progression of chronic renal disease. Modification of Diet in Renal Disease Study Group. N Engl J Med. 1994 Mar 31. 330(13):877-84. [QxMD MEDLINE Link]. Levey AS, Adler S, Caggiula AW, et al. Effects of dietary protein restriction on the progression of advanced renal disease in the Modification of Diet in Renal Disease Study. Am J Kidney Dis. 1996 May. 27(5):652-63. [QxMD MEDLINE Link]. Vergano L, Loiacono E, Albera R, Coppo R, Camilla R, Peruzzi L, et al. Can tonsillectomy modify the innate and adaptive immunity pathways involved in IgA nephropathy?. J Nephrol. 2014 Apr 23. [QxMD MEDLINE Link]. KovÃ¡cs T, Vas T, KÃ¶vesdy CP, Degrell P, Nagy G, RÃ©kÃ¡si Z, et al. Effect of tonsillectomy and its timing on renal outcomes in Caucasian IgA nephropathy patients. Int Urol Nephrol. 2014 Sep 3. [QxMD MEDLINE Link]. Maeda I, Hayashi T, Sato KK, Shibata MO, Hamada M, Kishida M, et al. Tonsillectomy has beneficial effects on remission and progression of IgA nephropathy independent of steroid therapy. Nephrol Dial Transplant. 2012 Jul. 27(7):2806-13. [QxMD MEDLINE Link]. Watanabe H, Goto S, Kondo D, Takata T, Yamazaki H, Hosojima M, et al. Comparison of methods of steroid administration combined with tonsillectomy for IgA nephropathy patients. Clin Exp Nephrol. 2016 May 23. [QxMD MEDLINE Link]. Yu L, Jiang L, Zhou XJ, et al. Common genetic variants in the chromogranin a promoter are associated with renal injury in IgA nephropathy patients with malignant hypertension. Ren Fail. 2010 Jan. 32(1):41-6. [QxMD MEDLINE Link]. Clayton P, McDonald S, Chadban S. Steroids and recurrent IgA nephropathy after kidney transplantation. Am J Transplant. 2011 Aug. 11 (8):1645-9. [QxMD MEDLINE Link].[Full Text]. Baek CH, Lee JG, Park JH, Kim H, Yang WS, Kim YS, et al. The clinical outcomes of second kidney transplantation in IgA nephropathy: a multicenter retrospective study. Clin Nephrol. 2016 Aug. 86 (2):87-93. [QxMD MEDLINE Link]. Media Gallery Light microscopy of a glomerulus from a patient with immunoglobulin A nephropathy showing increased mesangial matrix and cellularity. Electron microscopy showing large dark mesangial deposits. Immunofluorescence microscopy demonstrating large mesangial immunoglobulin A (IgA) deposits diagnostic of IgA nephropathy. of 3 Tables Back to List Contributor Information and Disclosures Author Sohail Abdul Salim, MD, FASN, FACP Consultant Physician, Central Nephrology; Affiliate Faculty, Department of Internal Medicine, Division of Nephrology, University of Mississippi Medical CenterSohail Abdul Salim, MD, FASN, FACP is a member of the following medical societies: American College of Physicians, American Society of Nephrology, Mississippi State Medical Association, Renal Physicians AssociationDisclosure: Nothing to disclose. Coauthor(s) Tibor Fulop, MD, PhD, FACP, FASN Professor of Medicine, Department of Medicine, Division of Nephrology, Medical University of South Carolina College of Medicine; Attending Physician, Medical Services, Ralph H Johnson VA Medical CenterTibor Fulop, MD, PhD, FACP, FASN is a member of the following medical societies: American Academy of Urgent Care Medicine, American College of Physicians, American Society of Hypertension, American Society of Nephrology, International Society for Apheresis, International Society for Hemodialysis, Magyar Orvosi Kamara (Hungarian Chamber of Medicine)Disclosure: Nothing to disclose. Luis A Juncos, MD, FASN, FAHA Chief of Nephrology, Central Arkansas Veterans Healthcare System, John L McClellan Memorial Veterans Hospital; Professor of Medicine, University of Arkansas for Medical Sciences College of Medicine Luis A Juncos, MD, FASN, FAHA is a member of the following medical societies: American College of Sports Medicine, American Federation for Clinical Research, American Federation for Medical Research, American Heart Association, American Physiological Society, American Society of Nephrology, Inter-American Society of Hypertension, International Society for Peritoneal Dialysis, International Society of Nephrology, Minnesota Medical Association, National Kidney Foundation, Society of Critical Care Medicine, Southern Society for Clinical Investigation, Zumbro Valley Medical SocietyDisclosure: Nothing to disclose. Wisit Cheungpasitporn, MD, FACP, FASN Nephrologist, Division of Nephrology and Hypertension, Mayo Clinic Wisit Cheungpasitporn, MD, FACP, FASN is a member of the following medical societies: American College of Physicians, American Society of TransplantationDisclosure: Nothing to disclose. Specialty Editor Board Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug ReferenceDisclosure: Received salary from Medscape for employment. Christie P Thomas, MBBS, FRCP, FASN, FAHA Professor, Department of Internal Medicine, Division of Nephrology, Departments of Pediatrics and Obstetrics and Gynecology, Medical Director, Kidney and Kidney/Pancreas Transplant Program, University of Iowa Hospitals and ClinicsChristie P Thomas, MBBS, FRCP, FASN, FAHA is a member of the following medical societies: American College of Physicians, American Heart Association, American Society of Nephrology, Royal College of PhysiciansDisclosure: Nothing to disclose. Chief Editor Vecihi Batuman, MD, FASN Professor of Medicine, Section of Nephrology-Hypertension, Deming Department of Medicine, Tulane University School of MedicineVecihi Batuman, MD, FASN is a member of the following medical societies: American College of Physicians, American Society of Hypertension, American Society of Nephrology, Southern Society for Clinical InvestigationDisclosure: Nothing to disclose. Additional Contributors James H Sondheimer, MD, FACP, FASN Professor of Medicine, Nephrology and Hypertension, Department of Medicine, Wayne State University School of MedicineJames H Sondheimer, MD, FACP, FASN is a member of the following medical societies: American College of Physicians, American Society of NephrologyDisclosure: Nothing to disclose. Mona Brake, MD Assistant Professor, Department of Internal Medicine, Kansas University School of MedicineMona Brake, MD is a member of the following medical societies: American College of Physicians, American Society of NephrologyDisclosure: Nothing to disclose. Acknowledgements Douglas Somers, MD Assistant Professor, Department of Internal Medicine, Division of Nephrology, University of Iowa Medical Center Douglas Somers, MD is a member of the following medical societies: American Society of Nephrology Disclosure: Nothing to disclose. Acknowledgments The authors thank Dr. Tim Timmerman, pathologist, for his invaluable help with the pathology slides. Close;) What would you like to print? What would you like to print? Print this section Print the entire contents of Print the entire contents of article Sections IgA Nephropathy Overview Practice Essentials Pathophysiology Etiology Epidemiology Prognosis Show All Presentation History Physical Show All DDx Workup Approach Considerations Histologic Findings Show All Treatment Approach Considerations Medical Care Diet Tonsillectomy Kidney Transplantation Show All Medication Medication Summary Complement Inhibitors Corticosteroids Angiotensin-converting enzyme inhibitors Angiotensin II receptor antagonists Omega-3 polyunsaturated fatty acids Immunosuppressive agents Endothelin Antagonists Show All Media Gallery;) References;) encoded search term (IgA Nephropathy) and IgA Nephropathy What to Read Next on Medscape Related Conditions and Diseases IgA Nephropathy Pediatric IgA Nephropathy IgA Nephropathy: Recognition and Management Rapid Review Quiz: IgA Nephropathy Fast Five Quiz: IgA Nephropathy Causes Fast Five Quiz: IgA Nephropathy Overview New Therapies Update in the Management of IgA Nephropathy News & Perspective Advice on Developing Opportunities in IgA Nephropathy The Impact of Health Disparities in IgA Nephropathy IgA Nephropathy: Reflections on a Powerful Lesson From a Patient Q&A: Are You Sure the Diagnosis Is IgA Nephropathy? Transforming IgA Nephropathy Care: New Drugs Expand Treatment Options Immunoglobulin A Nephropathy (IgAN): 5 Differential Diagnoses to Know Drug Interaction Checker Pill Identifier Calculators Formulary 2001 /viewarticle/1002158 Q&A: Are You Sure the Diagnosis Is IgA Nephropathy? 2003 /viewarticle/1002763 The Basis of Success: Recognizing IgAN and Promoting Earlier Diagnosis to Facilitate Treatment Initiation 0.5 CME Credits You are being redirected to Medscape Education Yes, take me there 0.5 CME The Basis of Success: Recognizing IgAN and Promoting Earlier Diagnosis to Facilitate Treatment Initiation 2001 /viewarticle/1000378 Experts And Viewpoints Experts And Viewpoints Q&A: Immunosuppressive Therapy in Immunoglobulin A Nephropathy 2002 239927-overview Diseases & Conditions Diseases & Conditions IgA Nephropathy
5824	https://brilliant.org/wiki/rule-of-product/	Rule of Product Sign up with Facebook or Sign up manually Already have an account? Log in here. Mei Li, Karleigh Moore, Daniel Lim, and Arron Kau Pawan Kumar Abhiram Rao Calvin Lin Jimin Khim contributed The rule of product states that if there are n ways of doing something, and m ways of doing another thing after that, then there are n×m ways to perform both of these actions. In other words, when choosing an option for n and an option for m, there are n×m different ways to do both actions. Contents Basic Examples Intermediate Examples Problem Solving See Also Basic Examples Lily is trying to decide what to wear. She has shirts in the following colors: red, purple, and blue, and she has pants in the following colors: black and white. How many different outfits can Lily choose from (assuming she selects one shirt and one pair of pants)? We know from the definition of the rule of product that if there are n options for doing one thing (like choosing a shirt), and m options for doing another thing (like choosing a pair of pants), then there are n×m total combinations we can choose from. In this case, there are 3 options for choosing a shirt, and there are 2 options for choosing pants. Thus, there are 3×2=6 total options. Here is a table where each row represents a possible outfit. \| \| \| --- \| \| Shirt \| Pants \| \| Red \| Black \| \| Blue \| Black \| \| Purple \| Black \| \| Red \| White \| \| Blue \| White \| \| Purple \| White \| As expected, there are 6 possible combinations. □ In the example above, there were two things to choose: a shirt and a pair of pants. However, the rule of product can extend to however many things to choose from. For example, if there are n choices for a shirt, m choices for a pair of pants, x choices for a pair of shoes, and y choices for a hat, the rule of product states that there are n×m×x×y total possible combinations. Unknown 175 145000 15000 142500 You go to check out three books at the library, and you want one history book, one science book, and one fantasy book. The library has 50 history books, 95 fantasy novels, and 30 books about science. How many combinations of books do you have to choose from? The correct answer is: 142500 There are 8 daily newspapers and 5 weekly magazines published in Chicago. If Colin wants to subscribe to exactly one daily newspaper and one weekly magazine, how many different choices does he have? Colin has 8×5=40 choices. □ Intermediate Examples Calvin wants to go to Milwaukee. He can choose from 3 bus services or 2 train services to head from home to downtown Chicago. From there, he can choose from 2 bus services or 3 train services to head to Milwaukee. How many ways are there for him to get to Milwaukee? Since Calvin can either take a bus or a train downtown , he has 3+2=5 ways to head downtown (Rule of sum). After which, he can either take a bus or a train to Milwaukee, and hence he has another 2+3=5 ways to head to Milwaukee (Rule of sum). Thus in total, he has 5×5=25 ways to head from home to Milwaukee (Rule of product). □ Six friends Andy, Bandy, Candy, Dandy, Endy, and Fandy want to sit in a row at the cinema. If there are only six seats available, how many ways can we seat these friends? For the first seat, we have a choice of any of the 6 friends. After seating the first person, for the second seat, we have a choice of any of the remaining 5 friends. After seating the second person, for the third seat, we have a choice of any of the remaining 4 friends. After seating the third person, for the fourth seat, we have a choice of any of the remaining 3 friends. After seating the fourth person, for the fifth seat, we have a choice of any of the remaining 2 friends. After seating the fifth person, for the sixth seat, we have a choice of only 1 of the remaining friends. Hence, by the rule of product, there are 6×5×4×3×2×1=720 ways to seat these 6 people. More generally, this problem is known as a permutation. There are n!=n×(n−1)×(n−2)×⋯×1 ways to seat n people in a row. □ My toy piano keyboard has 7 distinct white notes: letters A-G in English alphabet. I'm going to create a melody by playing three random notes. I am not allowed to repeat any notes and the melody cannot be ended with E, F or G. How many different melodies can I play? Examples: C G A is permitted. A F A isn't permitted because of repetition. A B E is not permitted because of last note rule. The correct answer is: 120 How many positive divisors does 2000=2453 have? Any positive divisor of 2000 must have the form 2a5b, where a and b are integers satisfying 0≤a≤4,0≤b≤3. There are 5 possibilities for a and 4 possibilities for b, hence there are 5×4=20 (rule of product) positive divisors of 2000 in all. □ Problem Solving Problem solving See Also Rule of Sum Permutations Cite as: Rule of Product. Brilliant.org. Retrieved 02:45, September 28, 2025, from
5825	https://mathcentral.quora.com/In-modular-arithmetic-does-2-3-mod-5-mean-that-2-and-3-are-identical	In modular arithmetic, does -2≡3 (mod 5) mean that -2 and 3 are identical? - Math Central - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Math Central Mathematics is the "Language of the universe" Follow · 125.5K 125.5K In modular arithmetic, does -2≡3 (mod 5) mean that -2 and 3 are identical? See parent question Answer Request Follow 1 Answer Sort Recommended Mohammad Afzaal Butt B.Sc in Mathematics&Physics, Islamia College Gujranwala (Graduated 1977) · 4y · −2≡3(mod 5)−2≡3(mod 5) means that -2 and 3 leaves the same remainder when divided by 5, or difference of -2 and 3 is divisible by 5. 421 views · View upvotes · · 9 1 View 3 other answers on parent question Related questions What is the general integral Z(x,y) of the PDE x(y+Z) Zx + Z (Z - y) Zy = y(y-Z)? Can you find the general integral of the PDE (2xy-1) zₓ + (z-2x^2) zᵧ = 2 (x-yz)? How do I solve the partial differential equation (D²-2DD'-15D'²) z=12xy? How do you graph the equation y=-4x^2? What are the intercepts and then use them to graph the equation 2y=-18+9x? How do we prove that ∫1 0 ln(1+x 2)x√x 2−1 d x=−i(a r c s i n h 2(1))?∫0 1 ln⁡(1+x 2)x x 2−1 d x=−i(a r c s i n h 2⁡(1))? How do we evaluate the integral ∫π 0 x cos x 1+sin 2 x d x?∫0 π x cos⁡x 1+sin 2⁡x d x? How do we evaluate the integral ∫2 π 0 x 2 sin x 8+sin 2 x d x?∫0 2 π x 2 sin⁡x 8+sin 2⁡x d x? How do we evaluate the integral ∫∞0 x 4(x 4−x 2+1)4 d x?∫0∞x 4(x 4−x 2+1)4 d x? A can company charges a $3 flat rate in addition to $1.50 per mile the domain of this function is The range of this function is? Agatha put $775 into a simple interest bearing account at a rate of 2.4% for 8 years. Her friend Dominic invested the same amount into an account that is compounded quarterly at the same rate for 6 years? Jeff has $28.75. He purchased three cookies that cost $1.50 each, five newspapers that each cost $0.50, five flowers for $1.25 each, and used the remainder of the cash on a pair of sunglasses. How much were the sunglasses? When a factory operates from 6 AM to 6 PM, its total fuel consumption varies according to the formula f(t) = 0.91 - 0.110.4 + 12, where t is the time in hours after 6 AM and f() is the number of barre? Cooper designs a photo album that he is going to give his parents as a gift. He created 15 pages of the album in June. He designed 1/10 of the album in August. He finishes the remaining 3/5 just in time for his parent's anniversary in November? The following scatterplot relates the life expectancy of animals to their heart rate. Ignoring humans (which are labeled "Man" in the scatterplot), which two conclusions can be made from the scatterplot? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
5826	https://www.wikihow.com/Calculate-Relative-Frequency	Skip to Content Courses Hot Guides Tech Help Pro Expert Videos About wikiHow Pro Upgrade QUIZZES All Quizzes Hot Love Quizzes Personality Quizzes Trivia Quizzes Taylor Swift Quizzes EDIT Edit this Article EXPLORE Tech Help ProAbout UsRandom ArticleQuizzes Request a New ArticleCommunity DashboardTrendingForums Arts and EntertainmentArtworkBooksMovies Computers and ElectronicsComputersPhone SkillsTechnology Hacks HealthMen's HealthMental HealthWomen's Health RelationshipsDatingLoveRelationship Issues Hobbies and CraftsCraftsDrawingGames Education & CommunicationCommunication SkillsPersonal DevelopmentStudying Personal Care and StyleFashionHair CarePersonal Hygiene Quizzes Love Quizzes Personality Quizzes Fun Games ForumsArts and EntertainmentFinance and BusinessHome and Garden Relationship QuizzesCars & Other VehiclesFood and EntertainingPersonal Care and Style Sports and FitnessComputers and ElectronicsHealthPets and Animals TravelEducation & CommunicationHobbies and CraftsPhilosophy and Religion Work WorldFamily LifeHolidays and TraditionsRelationships LOG IN ### Log in wikiHow Account No account yet? Create an account RANDOM Home Random Browse Articles TrendingNew Quizzes & Games All QuizzesHot Love Quizzes Personality Quizzes Fun Games Dating Simulator Learn Something New Forums Courses Happiness Hub Explore More Support wikiHow About wikiHow Log in / Sign up Terms of Use wikiHow is where trusted research and expert knowledge come together. Learn why people trust wikiHow Categories Education and Communications Studying Mathematics Probability and Statistics How to Calculate Relative Frequency Download Article Preparing, calculating, and reporting your data Reviewed by Joseph Meyer Last Updated: March 19, 2025 Fact Checked Download Article Preparing the Data \| Calculating Relative Frequency Results \| Reporting Relative Frequency Data \| Calculator, Practice Problems, and Answers \| Q&A \| Tips X This article was reviewed by Joseph Meyer. Joseph Meyer is a High School Math Teacher based in Pittsburgh, Pennsylvania. He is an educator at City Charter High School, where he has been teaching for over 7 years. Joseph is also the founder of Sandbox Math, an online learning community dedicated to helping students succeed in Algebra. His site is set apart by its focus on fostering genuine comprehension through step-by-step understanding (instead of just getting the correct final answer), enabling learners to identify and overcome misunderstandings and confidently take on any test they face. He received his MA in Physics from Case Western Reserve University and his BA in Physics from Baldwin Wallace University. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 133,267 times. Absolute frequency is a simple concept to grasp: it refers to the number of times a particular value appears in a specific data set (a collection of objects or values). However, relative frequency can be a little trickier. It refers to the proportion of times a particular value appears in a specific data set. In other words, relative frequency is, in essence, how many times a given event occurs divided by the total number of outcomes. If you organize your data, calculating and presenting relative frequency can become a simple task. Steps Part 1 Part 1 of 3: Preparing the Data Download Article 1 Collect your data. Unless you are just completing a math homework assignment, calculating relative frequency generally implies that you have some form of data. Conduct your experiment or study and collect the data. Decide how precisely you wish to report your results. X Research source For example, suppose you are collecting data on the ages of people who attend a particular movie. You could decide to collect and report the exact age of everyone who attends. But this is likely to give you 60 or 70 different results, being every number from about 10 through 70 or 80. You may instead wish to collect data in groups, like “Under 20,” “20-29,” “30-39,” “40-49,” “50-59,” and “60 plus.” This would be a more manageable set of six data groups. As another example, a doctor might collect body temperatures of patients on a given day. In this case, just collecting whole numbers, like 97, 98, 99, might not be precise enough. It might be necessary to report data in decimals in this case. 2. 2 Sort the data. After you complete your study or experiment, you are likely to have a collection of data values that could look like 1, 2, 5, 4, 6, 4, 3, 7, 1, 5, 6, 5, 3, 4, 5, 1. In this form, the data appear almost meaningless and difficult to use. It is more helpful to sort the data in order from lowest to highest. This would result in the list 1,1,1,2,3,3,4,4,4,5,5,5,5,6,6,7. X Research source When you are sorting and rewriting your collection of data, be careful to include every point correctly. Count the data set to make sure you do not leave off any values. Advertisement 3. 3 Use a data table. You can summarize the results of your data collection by creating a simple data frequency table. This is a chart with three columns that you will use for your relative frequency calculations. Label the columns as follows: X Research source . This column will be filled with each value that appears in your data set. Do not repeat items. For example, if the value 4 appears several times in the list, just put under the column once. , or . In statistics, the variable is conventionally used to represent the count of a particular value. You may also write , which is read as “n of x,” and means the count of each x-value. A final alternative is , which means the “frequency of x.” In this column, you will put the number of times that the value appears. For example, if the number 4 appears three times, you will place a 3 next to the number 4. Relative Frequency or . This final column is where you will record the relative frequency of each data item or grouping. The label , which is read “P of x,” could mean the probability of x or the percentage of x. The calculation of relative frequency appears below. This column will be used after you complete that calculation for each value of x. Advertisement Part 2 Part 2 of 3: Calculating Relative Frequency Results Download Article 1 Count your full data set. Relative frequency is a measure of the number of times a particular value results, as a fraction of the full set. In order to calculate relative frequency, you need to know how many data points you have in your full data set. The will become the denominator in the fraction that you use for calculating. X Research source In the sample data set provided above, counting each item results in 16 total data points. 2. 2 Count each result. You need to determine the number of times that each data point appears in your results. You may want to calculate the relative frequency of one particular item, or you may be summarizing the overall data for the full data set. X Research source For example, in the data set provided above, consider the value . This value appears three times in the list. 3. 3 Divide each result by the total size of the set. This is the final calculation to determine the relative frequency of each item. You can set it up as a fraction or use a calculator or spreadsheet to perform the division. X Research source Continuing with the example above, because the value appears three times, and the full set contains 16 items, you can determine that the relative frequency of the value is 3/16. This is equal to a decimal result of 0.1875. Advertisement Part 3 Part 3 of 3: Reporting Relative Frequency Data Download Article 1 Present your results in a frequency table. The frequency table that you began above can be used to present the results in a format that is easy to review. As you perform each of the calculations, fill in the results in the corresponding places in the table. It is common to round your answers to two decimal places, although you will need to decide this for yourself based on the needs of your study. Because of rounding the end result may total something close to , but not exactly 1.0. X Research source For example, using the data set above, the relative frequency table would appear as follows: x : n(x) : P(x) 1 : 3 : 0.19 2 : 1 : 0.06 3 : 2 : 0.13 4 : 3 : 0.19 5 : 4 : 0.25 6 : 2 : 0.13 7 : 1 : 0.06 total : 16 : 1.01 2. 2 Report items that do not appear. It may be just as meaningful to report items whose frequency is 0 as to report those items that do appear in your data set. Look at the kind of data you are collecting, and if you notice any gaps in your sorted data, you may need to report them as 0s. For example, the sample data set you have been working with includes all values from 1 to 7. But suppose that the number 3 never appeared. That could be important, and you would report the relative frequency of the value 3 as 0. 3. 3 Show your results as percentages. You may wish to turn your decimal results into percentages. This is a common practice, as relative frequency is often used as a predictor of the percentage of times that some value will occur. To convert a decimal number to a percentage, simply shift the decimal point two spaces to the right, and add a percent symbol. X Research source For example, the decimal result of 0.13 is equal to 13%. The decimal result of 0.06 is equal to 6%. (Don’t just skip over the 0.) Advertisement Calculator, Practice Problems, and Answers Sample Relative Frequency Calculator Sample Calculating Relative Frequency Practice Problems Sample Calculating Relative Frequency Practice Answers Community Q&A Search Add New Question Question What is frequency of the event? Donagan Top Answerer It's a measurement of how often the event occurs in a given time period. Thanks! We're glad this was helpful. Thank you for your feedback. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow Yes No Not Helpful 4 Helpful 7 Question How can you calculate frequency from relative frequency? Community Answer The word "frequency" alone is not very clear. In statistics, there are absolute frequency (the number of times a data point appears), relative frequency (usually presented as a percentage), or cumulative frequency. Cumulative frequency begins at 0 and adds up the frequencies as you move through your list. If you are just asked for "frequency," from the relative frequency, it probably means the absolute frequency. Take your relative frequency, and multiply it by the total number of items in the full data set, and you will have the absolute frequency. Thanks! We're glad this was helpful. Thank you for your feedback. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow Yes No Not Helpful 2 Helpful 10 Ask a Question 200 characters left Include your email address to get a message when this question is answered. Submit Advertisement Tips Physically speaking, the relative frequency tells you the presence or occurrence of a particular event in a set of events. Thanks Helpful 0 Not Helpful 0 If you add up the relative frequencies of all items in a data set, you should get a sum of 1. If you round off your values, the sum may not be exactly 1.0. Thanks Helpful 0 Not Helpful 0 If your data set is too large for simple counting, you may need to use a software package like MS-Excel or MATLAB to avoid mistakes. Thanks Helpful 0 Not Helpful 0 Submit a Tip All tip submissions are carefully reviewed before being published Name Please provide your name and last initial Submit Thanks for submitting a tip for review! Advertisement You Might Also Like How to Calculate Cumulative FrequencyHow to Make a Ratio How to Find the Mode of a Set of NumbersHow to Understand ProbabilityHow to Calculate ProbabilityHow to Calculate PrecisionHow to Convert Percents, Fractions, and DecimalsHow to Find the Five Number Summary in StatisticsHow to Convert to PercentageHow to Calculate OddsHow to Calculate Relative ErrorFinding Class Width for a Frequency Distribution TableHow to Calculate PercentagesHow to Determine the Empirical Formula of a Chemical Substance Advertisement References ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ About This Article Reviewed by: Joseph Meyer Math Teacher This article was reviewed by Joseph Meyer. Joseph Meyer is a High School Math Teacher based in Pittsburgh, Pennsylvania. He is an educator at City Charter High School, where he has been teaching for over 7 years. Joseph is also the founder of Sandbox Math, an online learning community dedicated to helping students succeed in Algebra. His site is set apart by its focus on fostering genuine comprehension through step-by-step understanding (instead of just getting the correct final answer), enabling learners to identify and overcome misunderstandings and confidently take on any test they face. He received his MA in Physics from Case Western Reserve University and his BA in Physics from Baldwin Wallace University. This article has been viewed 133,267 times. 38 votes - 62% Co-authors: 3 Updated: March 19, 2025 Views: 133,267 Categories: Probability and Statistics Article SummaryX To stop face sweating, try applying an astringent containing tannic acid, like witch hazel, to your face twice a day using a cotton ball. Additionally, apply an antiperspirant spray to your scalp, temples, and upper forehead to temporarily block your sweat glands. Alternatively, try using a dry shampoo to manage scalp sweating by holding it 8 inches from your head, then spraying it in 2 inch sections of your hair at a time. After that, massage the dry shampoo into your scalp for even distribution. For more tips, like how to show your results as percentages, read on! Did this summary help you? In other languages Spanish Russian Portuguese Italian German French Indonesian Print Send fan mail to authors Thanks to all authors for creating a page that has been read 133,267 times. Did this article help you? Advertisement Cookies make wikiHow better. By continuing to use our site, you agree to our cookie policy. Reviewed by: Joseph Meyer Math Teacher 38 votes - 62% Click a star to vote % of people told us that this article helped them. 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5827	https://www.coresta.org/sites/default/files/abstracts/2024_STPOST36_Martin.pdf	Figure 5. Overlaid quantifier (m/z 95) and qualifier (m/z 97) ion chromatograms of a calibration standard (left), HTP sample (middle), and a mid-level spiked HTP sample. Peaks marked with an asterisk indicate the common fragment ions shared between 2-bromoethanol and 2-bromoethanol-d4. 9.60 9.62 9.64 9.66 9.68 9.70 9.72 9.74 9.76 9.78 9.80 9.82 9.84 9.86 9.88 9.90 31397-19-2 [2-bromoethanol height] 3 x10 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 95.0 97.0 2-Bromoethanol 9.60 9.62 9.64 9.66 9.68 9.70 9.72 9.74 9.76 9.78 9.80 9.82 9.84 9.86 9.88 9.90 MD R10C P1 0000-999 A1101 [2-bromoethanol height] 3 x10 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 95.0 97.0 2-Bromoethanol 9.60 9.62 9.64 9.66 9.68 9.70 9.72 9.74 9.76 9.78 9.80 9.82 9.84 9.86 9.88 9.90 MD R10C P2 0000-999 A1102-1 [2-bromoethanol height] 3 x10 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 95.0 97.0 2-Bromoethanol Ethylene oxide 2.91 2.92 2.93 2.94 2.95 2.96 2.97 2.98 2.99 3 3.01 3.02 3.03 3.04 3.05 3.06 3.07 3.08 3.09 3.1 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.2 3.21 3.22 3.23 3.24 3.25 4 x10 1.75 2 2.25 2.5 2.75 3 3.25 3.5 3.75 4 4.25 4.5 4.75 5 5.25 5.5 5.75 6 6.25 6.5 6.75 7 7.25 7.5 7.75 8 8.25 8.5 8.75 S4 Ethylene oxide 0124-005 A Ethylene oxide 0124-005 A Spiked Ethylene oxide Ethylene oxide Abstract HTP aerosol was collected into a glass impinger containing methanol behind a glass fiber filter pad. An aliquot of the methanol solution was derivatized with concentrated hydrobromic acid and a separate aliquot was analyzed as is for the remaining VOCs. These extracts were each analyzed by GC-MS. The ions chosen to monitor as quantitation ions were based on the 2-bromoethanol mass spectrum2. Early testing showed promise using m/z 95 and m/z 97 as these generated the highest responses. Since d6-benzene was already being used as an internal standard, it was the first internal standard tested for 2-bromoethanol. However, the relative recoveries were variable. 2-Bromoethanol-d4 was evaluated and found to be a more suitable internal standard. A method was successfully validated for the determination of ethylene oxide in HTP aerosols by utilizing HBr derivatization to form 2-bromoethanol. We were able to use a single GC configuration and aerosol collection method for ENDS and HTP for 8 VOCs, with minor alterations to the sample preparation procedure and GC method parameters. Despite HTP aerosols having more complex matrices than ENDS aerosols, we were able to validate this method with comparable linearity, range, LOD/LOQ, accuracy, and precision to the ENDS aerosol method. This presentation will describe the challenges of expanding a validated method for the analysis of volatile organic compounds (VOCs) in Electronic Nicotine Delivery Systems (ENDS), to evaluate VOCs in heated tobacco product (HTP) aerosols. In 2023, we successfully validated a method for VOCs in ENDS e-liquid and aerosol for 1,3-butadiene, vinyl chloride, ethylene oxide, isoprene, propylene oxide, acrylonitrile, benzene, and toluene. This method was based upon CRM No 70 and ISO 21330/ISO 23923, with expected modifications to the analysis (different column, ion source, and calibration range). In 2024, we applied the new method to VOCs in HTP aerosol. While levels of VOCs in ENDS aerosol were typically near, or below, the limit of quantitation (LOQ), all analytes, except vinyl chloride, were detected in the HTP aerosol samples at or above the LOQ. The chromatograms for the HTP aerosol were more complex than for the ENDS aerosol and changes to the GC method were required to resolve some interferences from the analytes of interest. However, the biggest challenge was a large interference coeluting with ethylene oxide. Additional investigation determined the interfering compound to be acetaldehyde, which is present at a level 1000-times that of ethylene oxide in HTP aerosols. To aid in ethylene oxide selectivity, a derivatization method was adopted and incorporated into the method for HTP aerosols. The validation of the new ethylene oxide method was conducted according to FDA/ICH Guidelines, and included selectivity, linearity and range, LOD/LOQ, accuracy, precision, stability, and robustness. The extract LOQ was 100 ng/mL, equivalent to 1 µg/collection, recoveries ranged from 91.5 to 102.8%, and method precision was 2.9%. • Limited number of HTP products were evaluated • Although the objective of this work was to use an existing aerosol collection method and GC configuration, it may be possible to achieve lower LOQs and improve peak resolution using different collection methods and a different column. • We may be able to apply this derivatization procedure to other analytes (e.g., propylene oxide) as well as mainstream cigarette smoke. Other GC columns have been shown to provide acceptable resolution of ethylene oxide from acetaldehyde, however, this typically results in higher LOQs and coelution with other large peaks. A drawback to using 2-bromoethanol-d4 as the internal standard is that common masses in both spectra led to less than ideal resolution of the internal standard peak from the analyte peak (Figure 5). In order to combat this, m/z 95 was chosen as the quantifier ion to increase the overall analyte response and the internal standard concentration was decreased from 2 µg/mL to 1 µg/mL. Optimization of a GC-MS Method for the Determination of Ethylene Oxide in Heated Tobacco Product (HTP) Aerosols Collins, J.M. and Martin, A.M.; McKinney Specialty Labs, Richmond, VA, USA Poster # STPOST36, CORESTA Congress, October 13 – 17, 2024, Edinburgh, Scotland, UK Pre-validation Summary Methodology Validation Results Validation Results Continued Conclusions Limitations and Future Work References Figure 4. TIC overlay of an HTP volatiles extract injected as is (green), spiked with 2-bromoethanol-d4 and ethylene oxide (blue), and spiked and derivatized (black) Calibration Range 0.1-3.0 µg/mL LOD 0.12 µg/collection LOQ 1 µg/collection Accuracy 91.5-102.8% Instrument Precision 1.0-3.3% Intermediate Accuracy and Precision(3-day) (by area) 97.9 ± 6.7% Intermediate Accuracy and Precision(3-day) (by height) 98.3 ± 2.8% Sample Stability Derivatized extracts are stable for 4 days when stored under ambient conditions Table 1. Validation Summary For example, interferences were observed for acrylonitrile in HTP aerosol, but these were resolved by altering the oven temperature ramp. A large interference was observed eluting just before ethylene oxide (Figure 2) that could not be resolved chromatographically. Attempts to resolve the peaks by mass were unsuccessful as there was not a unique mass between them. Further investigation revealed this large interference to be acetaldehyde, which may be present at much higher levels in HTP aerosols. The goal of this work was to expand the scope of a validated method for VOCs in ENDS aerosol to include the analysis of VOCs in HTP aerosol. Because of the increased complexity of the HTP matrix (Figure 1), the chromatography had to be carefully evaluated. In most cases, the ENDS method was suitable for the analysis of VOCs in HTP, however, there were a few challenges. Feasibility and selectivity were determined by analysis of a 2-bromoethanol standard. A derivatized and underivatized aliquot of a stock solution containing ethylene oxide was also analyzed. The derivatized extract indicated the presence of 2-bromoethanol at the confirmed retention time. The underivatized extract confirmed the presence of ethylene oxide and its disappearance in the derivatized extract. Selectivity was further evaluated through the comparison of a derivatized and underivatized HTP aerosol extract, shown in Figure 4. As a result of this interference, we shifted our focus from creating a single HTP aerosol method for VOCs, to validating ethylene oxide separately. The plan was to use an aliquot of the same sample collection and the same instrument (GC column) to optimize sample throughput. In order to improve selectivity, a derivatization procedure was tested for feasibility1. Ethylene oxide was derivatized to form 2-bromoethanol using hydrobromic acid (HBr). Reaction shown in Figure 3. Figure 3. Reaction of ethylene oxide with hydrobromic acid to form 2-bromoethanol Figure 2. Overlaid ion chromatograms (m/z 44) of a calibration standard for ethylene oxide at 500 ng/mL (black), HTP aerosol extract (blue) and the same HTP aerosol extract spiked with ethylene oxide (green) 1. LIU X.; JOZA P.; MASTERS A.; RICKERT W. ST71 Determination of ethylene oxide in mainstream cigarette smoke using hydrobromic acid derivatization and gas chromatography-mass spectrometry. CORESTA Congress, Quebec City, October 12-16, 2014. Available on-line: Accessed: 30Aug24. 2. Simmons, M. Ethylene Oxide (OSHA Method 1010, version 2.0), 2014. Available on-line: Accessed: 30Aug24. Quantitation was evaluated using both area and height. While all validation parameters were acceptable for both, quantitation by area required significantly more manual integrations to 2-bromoethanol peaks in both standards and samples than quantitation by height. Additionally, quantitation by height resulted in improved method accuracy and precision. Since the method used as the starting point for this scope expansion was for the determination of ethylene oxide in mainstream cigarette smoke, the amount of HBr used for derivatization was also evaluated. A composite HTP aerosol sample was spiked at the upper end of the calibration range and aliquots were taken for derivatization using different amounts of HBr in order to determine the appropriate amount of derivatizing reagent required. Ethylene oxide 2-Bromoethanol Figure 1. Total ion chromatogram for an HTP aerosol VOC extract. 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 5 x10 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 + TIC SIM AM284_240205-036.D (Run 4 P1) 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 5 x10 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 4.4 + TIC SIM AM284_240405_003.D (A-spiked/underivatized-) + TIC SIM AM284_240405_002.D (A-unspiked/underivatized-) + TIC SIM AM284_240405_009.D (A-100µL-A) 9.54 9.56 9.58 9.6 9.62 9.64 9.66 9.68 9.7 9.72 9.74 9.76 9.78 9.8 9.82 9.84 9.86 9.88 9.9 9.92 9.94 9.96 9.98 4 x10 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2-Bromoethanol 2-Bromoethanol-d4 2024_STPOST36_Martin.pdf Congress2024 - Document not peer-reviewed by CORESTA
5828	https://math.stackexchange.com/questions/56235/h%C3%B6lders-inequality-with-three-functions	functional analysis - Hölder's inequality with three functions - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Hölder's inequality with three functions Ask Question Asked 14 years, 1 month ago Modified4 years, 11 months ago Viewed 14k times This question shows research effort; it is useful and clear 33 Save this question. Show activity on this post. Let p,q,r∈(1,∞)p,q,r∈(1,∞) with 1/p+1/q+1/r=1 1/p+1/q+1/r=1. Prove that for every functions f∈L p(R)f∈L p(R), g∈L q(R)g∈L q(R),and h∈L r(R)h∈L r(R) ∫R\|f g h\|≤∥f∥p⋅∥g∥q⋅∥h∥r.∫R\|f g h\|≤‖f‖p⋅‖g‖q⋅‖h‖r. I was going to use Hölder's inequality by letting 1/p+1/q=1/(p q/p+q)1/p+1/q=1/(p q/p+q) and WLOG let p<q p<q so that L q(R)⊆L p(R)L q(R)⊆L p(R), but I cannot use this inclusion because R R does not have finite measure. Would you please help me if you have any other method to approach this problem? functional-analysis measure-theory inequality lp-spaces Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jul 13, 2019 at 1:12 Eric Wofsey 343k 28 28 gold badges 485 485 silver badges 701 701 bronze badges asked Aug 8, 2011 at 4:22 Yeonjoo YooYeonjoo Yoo 975 1 1 gold badge 10 10 silver badges 16 16 bronze badges Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 40 Save this answer. Show activity on this post. The rough idea is to show a series of inequalities: ∫\|f g h\|≤∥f g∥p′∥h∥r≤∥f∥p∥g∥q∥h∥r∫\|f g h\|≤‖f g‖p′‖h‖r≤‖f‖p‖g‖q‖h‖r where p′=p q p+q p′=p q p+q or 1 p′=1 p+1 q 1 p′=1 p+1 q or 1=1 p/p′+1 q/p′1=1 p/p′+1 q/p′. First we show that ∥f g∥p′≤∥f∥p∥g∥q‖f g‖p′≤‖f‖p‖g‖q. This follows from computing ∥f g∥p′=(∫\|f g\|p′)1 p′≤(∥f p′∥p/p′∥g p′∥q/p′)1 p′=∥f∥p∥g∥q,‖f g‖p′=(∫\|f g\|p′)1 p′≤(‖f p′‖p/p′‖g p′‖q/p′)1 p′=‖f‖p‖g‖q, where the middle inequality comes from Holder's inequality. (Holder's inequality applies because \|f\|∈L p(R)\|f\|∈L p(R) implies \|f\|p′∈L p/p′(R)\|f\|p′∈L p/p′(R), and p′p+p′q=1 p′p+p′q=1.) As a result, \|f g\|∈L p′(R)\|f g\|∈L p′(R). Apply Holder's inequality again to get the very first inequality up above. Hope this will help you. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Oct 20, 2020 at 2:29 xion3582 529 6 6 silver badges 14 14 bronze badges answered Aug 8, 2011 at 5:10 user14242user14242 3,030 1 1 gold badge 29 29 silver badges 34 34 bronze badges 4 3 Could please tell me how did you get the first inequality on the left hand side i.e. (∫\|f g\|p′)1 p′≤(∥f p′∥p/p′∥g p′∥q/p′)1 p′(∫\|f g\|p′)1 p′≤(‖f p′‖p/p′‖g p′‖q/p′)1 p′Noob101 –Noob101 2017-05-09 02:16:43 +00:00 Commented May 9, 2017 at 2:16 2 This proof is elegant.Sam Wong –Sam Wong 2018-03-30 02:57:35 +00:00 Commented Mar 30, 2018 at 2:57 @SuryakantShrivastava t.b. just used the Holder's inequality to get the inequality you asked about, as he said.Sam Wong –Sam Wong 2018-03-30 02:59:00 +00:00 Commented Mar 30, 2018 at 2:59 @SuryakantShrivastava Maybe you also need to note that the function f(x)=x 1 p′f(x)=x 1 p′ is strictly increasing on [0,∞)[0,∞), and the integral of non-negative measurable function is also non-nagetive.Sam Wong –Sam Wong 2018-03-30 03:02:19 +00:00 Commented Mar 30, 2018 at 3:02 Add a comment\| This answer is useful 33 Save this answer. Show activity on this post. We can use a generalized AM-GM inequality to deduce that if 1/p+1/q+1/r=1 1/p+1/q+1/r=1, then a b c≤a p p+b q q+c r r a b c≤a p p+b q q+c r r for nonnegative a,b,c a,b,c. Let a=\|f(x)\|/∥f∥p,b=\|g(x)\|/∥g∥q,c=\|h(x)\|/∥h∥r a=\|f(x)\|/‖f‖p,b=\|g(x)\|/‖g‖q,c=\|h(x)\|/‖h‖r, and then integrate both sides of the inequality over R R to obtain ∥f g h∥1∥f∥p∥g∥q∥h∥r≤1 p+1 q+1 r=1.‖f g h‖1‖f‖p‖g‖q‖h‖r≤1 p+1 q+1 r=1. Multiply out and you have Holder's inequality for three functions. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 8, 2011 at 5:27 anonanon 156k 14 14 gold badges 249 249 silver badges 422 422 bronze badges 2 1 Why is f g h∈L 1 f g h∈L 1?Dragonite –Dragonite 2017-04-28 13:41:37 +00:00 Commented Apr 28, 2017 at 13:41 @Dragonite, by the very first inequality and then integrating, you can get the integrability of f g h f g h.Fardad Pouran –Fardad Pouran 2018-04-20 16:38:57 +00:00 Commented Apr 20, 2018 at 16:38 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions functional-analysis measure-theory inequality lp-spaces See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 11 p+1 q+1 r=1 1 p+1 q+1 r=1 then Holder's inequality 7IMO 2001 2001 problem 2 2 5How prove this inequality ∑m i=1∑m j=1\|A i\|⋅\|A i∩A j\|≥1 m n(∑m i=1\|A i\|)3∑i=1 m∑j=1 m\|A i\|⋅\|A i∩A j\|≥1 m n(∑i=1 m\|A i\|)3 Related 3Two proofs concerning Hölder's inequality 0How to use Hölder's inequality to show L q L q is a subspace of L p L p? 2Hölder inequality conditions for L p L p spaces? 0Every L q L q function as a product of L p L p and L r L r functions 6An inequality by using general Hölder's inequality 1Contradiction with inclusion of L p[0,1],L q[0,1]L p[0,1],L q[0,1]? if 1/p+1/q=1 1/p+1/q=1 1Applying Hölder's inequality Hot Network Questions What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel? 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5829	https://www.nku.edu/~longa/classes/biomath/www.biology.arizona.edu/biomath/tutorials/Functions/Properties.html	BioMath: Functions ----------------------------------------------------------------------------------------------------------------------------------- The Biology Project>Biomath>Functions>Properties Functions --------- Properties of Functions This section will introduce a few terms to qualify functions. Understanding these properties comes in handy when you are sketching graphs of functions or looking at what happens when a given quantity is maximized or minimized. Increasing and Decreasing Functions Imagine you are observing how a certain property of a biological system is changing over time. Mathematically, you would be looking at whether or not the function modeling the property increases or decreases. To determine whether a function is increasing or decreasing we must “read” the graph of the function from left to right. You can think of an increasing function as one that rises from left to right, and a decreasing function as on that falls from left to right. Most functions do not increase/decrease for all x in the domain, but rather, they increase on some intervals and decrease on others. We say that f increases on some interval, I if for x 1 and x 2 in I, On the other hand, we say that f decreases on I if for x 1 and x 2 in I, To determine if the graph of a function increases or decreases over a specified interval, we read the graph left to right and determine whether it has an upward or downward trend as pictured below, If you imagine a ball rolling on the graph, the ball will be rolling downhill on portions of the graph that decrease, and it will be rolling uphill on portions of the graph that increase. Odd and Even Functions Some functions have the special property of being odd or even. Odd and even functions have special graphical symmetries. Definition A function f(x) is called odd if, for all x in the domain of f. How do we know if a function is odd? An example of an odd function is f(x) = x. We can demonstrate that f(x) = x is odd by showing that f(−x) = −f(x) by substituting -x for x as, f(−x) = −x = −f(x). If a function is not odd, it might (but not necessarily) be even. Definition A function f(x) is called even if, f(-x) = f(x), for all x in the domain of f. An example of an even function is f(x) = x 2. We can demonstrate that f(x) = x 2 is even by showing that f(-x) = f(x) as, f(-x) = (- x 2) = (−1)2 x 2 = x 2 = f(x). As stated above, the graphs of odd and even functions each have special symmetries. The graph of an odd function is symmetric about the origin. A graph is symmetric about the origin if the point (-x, -y) lies on the graph whenever (x, y) does. The graph of an even function, on the other hand, is symmetric about the y-axis. A graph is symmetric about the y-axis if the point (-x, y) lies on the graph whenever (x, y) does. Not all functions are classified as odd or even; a function that is not even or odd is called neither. An example of a function that is neither even nor odd is f(x) = x + 1. We can demonstrate this by finding f(−x) as, f(−x) = −x + 1. Notice that f(−x) is not equal to −f(x), because −f(x) = −(x + 1) = −x−1; and f(−x) is not equal to f(x) , because f(x) =x + 1. The fact that even and odd functions have certain graphical symmetries is useful to know. In particular, if you know some function f(x) is odd (even) and know the value of f(a), then you also know that the value of f(−a) is −f(a) (f(a)). Similarly, if you know some function f(x) is even and know the value of f(a), then you also know that the value of f(−a) is −f(a). This becomes important when you study integral calculus because the integral of an odd or even function over a symmetric interval can be greatly simplified. Maxima and Minima of Functions A function may achieve values that can be classified as maxima and/or minima. Collectively, maxima and minima are referred to as extrema. Graphically, you can think of maxima and minima as the functional values of peaks and valleys. Extrema may be classified as global or local. As the names imply, a global maximum is the maximum value a function achieves, while the global minimum is the minimum value a function achieves. If a function is bounded and has domain consisting of the entire real line, the global maximum is the highest peak and the global minimum is the lowest valley. Both the maximum and minimum of a function must be a finite number. Therefore, if a function increases without bound, that function has no global maximum. Likewise, if a function decreases without bound, there is no global minimum. However, these unbounded functions can have local maxima and minima, as shown in the figure below, The extrema depicted in the above graph are local because the function increases and decreases without bound as indicated by the arrows on the graph. Concave Up and Concave Down Functions A function may be classified as concave up or concave down on an interval I based its shape. Formally, we define the terms concave up and concave down using calculus concepts. However, we can understand these terms by looking at the following graphs: It may be helpful to think of concave up functions as being able to hold water; concave down functions would spill water. A function need not be concave up or down for all x in the domain; a function may change concavity. Points at which concavity changes are called points of inflection. We will discuss these concepts in depth when we study differential calculus. > In the next section we will learn how to add, subtract, multiply, and divide functions. Operations The Biology Project>Biomath>Functions>Properties The Biology ProjectDepartment of Biochemistry and Molecular Biophysics The University of Arizona January 2006 Contact the Development Team All contents copyright © 2006. All rights reserved.
5830	https://www.sciencedirect.com/topics/engineering/kg-dry-air	Skip to Main content My account Sign in Kg Dry Air In subject area:Engineering Kilogram of dry air (kg dry air) refers to the mass of dry air associated with one cubic meter of humid air, which is essential for calculating the density and specific volume of humid air. AI generated definition based on: Engineering Principles of Unit Operations in Food Processing, 2021 How useful is this definition? Add to Mendeley Discover other topics Chapters and Articles You might find these chapters and articles relevant to this topic. Drying 2002, Chemical Engineering (Fifth Edition), Volume 2J.F. RICHARDSON, ... J.R. BACKHURST Example 16.3 A granular material containing 40 per cent moisture is fed to a countercurrent rotary dryer at a temperature of 295 K and is withdrawn at 305 K, containing 5 per cent moisture. The air supplied, which contains 0.006 kg water vapour/kg dry air, enters at 385 K and leaves at 310 K. The dryer handles 0.125 kg/s wet stock. Assuming that radiation losses amount to 20 kJ/kg dry air used, determine the mass flowrate of dry air supplied to the dryer and the humidity of the exit air. The latent heat of water vapour at 295 K = 2449 kJ/kg, specific heat capacity of dried material = 0.88 kJ/kg K, the specific heat capacity of dry air = 1.00 kJ/kg K, and the specific heat capacity of water vapour = 2.01 kJ/kg K. Solution This example involves a heat balance over the system. 273 K will be chosen as the datum temperature, and it will be assumed that the flowrate of dry air = G kg/s. : Heat in: (a) : G kg/s dry air enter with 0.006G kg/s water vapour and hence the heat content of this stream (b) : Wet solid 0.125 kg/s enter containing 0.40 kg water/kg wet solid, assuming the moisture is expressed on a wet basis. Thus: and: : Heat out: (a) : Air Heat in exit air = (1.00 G) + (0.006 G × 2.01) = 37.45G kW. Mass flowrate of dry solids = 0.075 kg/s containing 0.05 kg water/kg wet solids. Hence: and:Assuming evaporation takes place at 295 K, then:and: (b) : Dried solids The dried solids contain 0.0036 kg/s water and hence heat content of this stream is: (c) : Losses These amount to 20 kJ/kg dry air or 20m kW. : Heat balance and:and: View chapterExplore book Read full chapter URL: Book2002, Chemical Engineering (Fifth Edition), Volume 2J.F. RICHARDSON, ... J.R. BACKHURST Chapter Dehydration 2009, Food Process Engineering and TechnologyZeki Berk 22.2Thermodynamics of Moist air (psychrometry) 22.2.1Basic principles In most food dehydration processes, air is the drying medium. It is therefore appropriate to review basic concepts pertaining to the thermodynamic behavior of air–water vapor mixtures (moist air), before discussing drying kinetics and drying processes. Although dry air is, in itself, a mixture of gases (nitrogen, oxygen, carbon dioxide etc.), we shall consider moist air as consisting of two components only: dry air and water vapor. Gibbs’ phase rule (Josiah Willard Gibbs, American mathematician-physicist, 1839–1903) establishes the number of possible ‘degrees of freedom’ in a system in equilibrium, as follows: (22.1) where: : F=number of degrees of freedom (number of possible variables) : C=number of components : P=number of phases. For mono-phase (homogeneous) moist air C=2 and P=1. It follows that the number of independent variables is 3, e.g. temperature, pressure and moisture content. Psychrometry is most commonly applied to air at atmospheric pressure. In this case, the pressure is not a ‘variable’. It follows that the state of homogeneous moist air at atmospheric pressure can be unequivocally defined by two variables: temperature and moisture content (humidity). Consequently, it is customary to represent the state of moist air graphically on a system of two axes: temperature as the abscissa and humidity as the ordinate. The resulting graph is known as the ‘Psychrometric chart’ (see Figure A.2). The state of two-phase moist air (e.g. moist air containing mist) at atmospheric pressure can be unequivocally defined by one variable only. For processes at variable pressure one more degree of freedom has to be added. 22.2.2Humidity The moisture content of air (humidity, absolute humidity) is expressed as the mass ratio of water vapor to dry air (kg water vapor per kg dry air) and is dimensionless. At atmospheric pressure, the air–moisture mixture can be treated as an ideal mixture of gases and Dalton's law (Section 13.2) can be applied: (22.2) where: : P=total pressure (usually atmospheric) : pw and pa=partial pressures of water vapor and dry air, respectively. Assuming perfect gas behavior, the mass of water vapor per unit volume of mixture is: (22.3) Similarly, the mass of dry air (average molecular weight=29) per unit volume of mixture is: (22.4) It follows that the absolute humidity of the air, H, is: (22.5) Note: The approximation P−pw≈P, is permissible only at ambient temperatures and in relatively dry air where pw is much smaller than the total pressure P. 22.2.3Saturation, relative humidity (RH) Moisture-saturated air is air in equilibrium with pure liquid water at a given temperature. Consequently, the partial vapor pressure of water vapor in moisture-saturated air is equal to the vapor pressure of liquid water at the same temperature: (22.6) The saturation humidity Hs is the maximum quantity of water vapor that air can contain at a given temperature, without phase separation. The relative humidity (φ or RH) is the ratio (as percentage) of the partial pressure of water vapor in air, to the vapor pressure of liquid water at the same temperature. (22.7) ‘Percent saturation’ (S), often confounded with relative humidity, is: (22.8) The saturation and constant percent saturation lines are indicated on the psychrometric chart. 22.2.4Adiabatic saturation, wet-bulb temperature If a mass of air is brought into contact with water under adiabatic conditions (no heat transfer with the exterior), the humidity of the air increases until saturation is reached.Since there is no external source of heat, water is evaporated using heat from the air itself. Consequently, the air is cooled at the same time that it is humidified. The pro-cess described is called ‘adiabatic saturation’ and the temperature reached at saturation is called the ‘temperature of adiabatic saturation’. It can be demonstrated that: 1. : The temperature of adiabatic saturation is a sole function of the initial conditions (temperature and humidity) of the air. It follows that the temperature of adiabatic saturation is a thermodynamic property of moist air. 2. : The process of adiabatic saturation follows a straight line on the psychrometric chart. If the bulb of a mercury thermometer is enveloped in wet tissue and air is flown over the bulb, the air (and the thermometer) will be eventually cooled to the adiabatic saturation temperature. For this reason, the wet bulb temperature is often confounded with the adiabatic saturation temperature of the air. One of the ways to describe unequivocally the state of moist air is to specify its dry bulb (normal) and wet-bulb temperatures. It should be remembered, however, that the ‘wet-bulb temperature’ is an empirical value (not a property) depending on the kinetics of the measurement, while the adiabatic saturation temperature is a thermodynamic property. 22.2.5Dew point ‘Dew point’ is yet another property of moist air. If a mass of moist air is cooled at constant humidity, saturation is reached and liquid water is formed in the form of dew or mist. The temperature at which this occurs is called the ‘dew point’ of the air. Dew point is one of the phenomena on which certain hygrometers (instruments for measuring humidity in air) are based. The air is contacted with a mirror that is gradually cooled. The temperature at which the mirror becomes foggy is the dew point of the air. Based on the relation between the water activity of a food sample and the relative humidity of the air in equilibrium contact with the food, the ‘foggy mirror’ phenomenon is one of the methods for measuring water activity in the laboratory. Example 22.1 A mass of air at 30°C has a percent saturation of 60%. Find the absolute humidity, wet-bulb temperature and dew point of that air. Solution: All the properties are obtainable from the psychrometric chart in Appendices (Figure A.2) : The absolute humidity is H=0.016 kg/kg : The adiabatic saturation (wet-bulb) temperature is 23.6°C : The dew point is 21.2°C. Example 22.2 The air in the room used for packaging instant coffee must have a percent saturation of 20% or less, in order to prevent moisture sorption and caking of the product. The dehumidified air is ‘produced’ by cooling air at 30°C and 80% percent saturation beyond saturation, separating the condensed water and heating the air back to 30°C. Show the process on the psychrometric chart and find the temperature to which the air has to be cooled. Solution: The process is shown on the general form of the psychrometric chart (Figure 22.1). Line AB represents cooling at constant humidity until saturation is reached. The section BC represents condensation at 100% saturation. The line CD represents heating of the dehumidified air. Point C shows the temperature to which air has to be cooled to perform the desired dehumidification. The result is 5°C. View chapterExplore book Read full chapter URL: Book2009, Food Process Engineering and TechnologyZeki Berk Chapter Humidification and Water Cooling 2001, Chemical EngineeringJ.R. BACKHURST, J.H. HARKER PROBLEM 13.3 Wet material, containing 70% moisture on a wet basis, is to be dried at the rate of 0.15 kg/s in a counter-current dryer to give a product containing 5% moisture (both on a wet basis). The drying medium consists of air heated to 373 K and containing water vapour with a partial pressure of 1.0 kN/m2. The air leaves the dryer at 313 K and 70% saturated. Calculate how much air will be required to remove the moisture. The vapour pressure of water at 313 K may be taken as 7.4 kN/m2. Solution The feed is 0.15 kg/s wet material containing 0.70 kg water/kg feed. Thus water in feed = (0.15 × 0.70) = 0.105 kg/s and dry solids = (0.15 − 0.105) = 0.045 kg/s. The product contains 0.05 kg water/kg product. Thus, if w kg/s is the amount of water in the product, then: and: water to be removed = (0.105 − 0.00237) = 0.1026 kg/s. The inlet air is at 373 K and the partial pressure of the water vapour is 1 kN/m2. Assuming a total pressure of 101.3 kN/m2, the humidity is: (equation 13.1) The outlet air is at 313 K and is 70% saturated. Thus, as in Example 13.1, Volume 1: The increase in humidity is (0.0335 − 0.0062) = 0.0273 kg/kg dry air and this must correspond to the water removed, 0.1026 kg/s. Thus if G kg/s is the mass flowrate of dry air, then: In the inlet air, this is associated with 0.0062 kg water vapour, or: Hence, the mass of moist air required at the inlet conditions= (3.76 + 0.0233) = View chapterExplore book Read full chapter URL: Book2001, Chemical EngineeringJ.R. BACKHURST, J.H. HARKER Chapter Fundamental Relationships of Heat and Mass Transfer in Solar Seawater Desalination Systems 2017, Solar Energy Desalination TechnologyHongfei Zheng 3.1Physical Properties and State Parameters of Moist Air 3.1.1Physical Properties of Moist Air The seawater desalination systems, especially the ones based on distillation either through multistage flash distillation or atmospheric distillation, are affected by air due to air leakage. Air is a noncondensable gas, which has a significant influence on water evaporation and condensation. In reality, the air in the system is the combination of air and water vapor, or moist air in other words. Therefore it is important to understand the properties of moist air for improving the efficiency of the desalination system. Moist air refers to the air with water vapor, and dry air is the air without water vapor. Water vapor changes state to gas or solid under certain conditions. Dry air is a mixture of gases, which consists of nitrogen (N2, 78.084%), oxygen (O2, 20.948%), argon (Ar, 0.934%), carbon dioxide (CO2, 0.0314%), neon (Ne, 0.0018%), and helium (He, 0.00052%) by volume fraction and various other components. The scientific test presents that the combination of dry air is relatively stable under normal circumstances. Hence it is reasonable to consider the dry air as a whole when investigating the physical properties of moist air. Water vapor takes a small part in moist air at a normal pressure but changes with temperature and pressure. Table 3.1 demonstrates the variation of water vapor content in saturated humid air. Table 3.1. Physical properties of saturated air at different temperature (reference temperature Treference = 0°C) . \| Properties of water and water vapor \| Physical properties of 1 m3 saturated air at temperature T \| --- \| \| T (°C) \| pw (kPa) \| hw (kJ/kg) \| hfg (kJ/kg) \| hvap (kJ/kg) \| mv \| Hv,a \| Hair \| Ha,s \| \| 100 \| 101.3 \| 418.7 \| 2255.2 \| 2674.0 \| ∞ \| ∞ \| 131.4 \| ∞ \| \| 95 \| 84.5 \| 397.6 \| 2268.1 \| 2666.0 \| 3.774 \| 10,060 \| 124.7 \| 10,185 \| \| 90 \| 70.1 \| 376.6 \| 2281.5 \| 2658.1 \| 1.684 \| 4476.5 \| 118.0 \| 4594.5 \| \| 85 \| 57.8 \| 355.6 \| 2294.5 \| 2650.1 \| 1.007 \| 2668.5 \| 111.3 \| 2779.9 \| \| 80 \| 47.3 \| 334.6 \| 2307.0 \| 2641.7 \| 0.634 \| 1675.3 \| 104.4 \| 1779.7 \| \| 75 \| 38.5 \| 313.8 \| 2319.6 \| 2633.4 \| 0.443 \| 1168.2 \| 97.9 \| 1266.1 \| \| 70 \| 31.1 \| 292.8 \| 2332.2 \| 2625.0 \| 0.321 \| 843.5 \| 91.4 \| 934.9 \| \| 65 \| 25.0 \| 271.9 \| 2344.3 \| 2616.3 \| 0.237 \| 621.3 \| 84.5 \| 705.8 \| \| 60 \| 19.9 \| 251.0 \| 2357.0 \| 2607.9 \| 0.177 \| 461.9 \| 78.2 \| 540.1 \| \| 55 \| 15.7 \| 230.1 \| 2369.0 \| 2599.1 \| 0.133 \| 346.0 \| 71.8 \| 417.8 \| \| 50 \| 12.3 \| 209.2 \| 2381.1 \| 2590.3 \| 0.100 \| 259.8 \| 65.3 \| 325.1 \| \| 45 \| 9.58 \| 188.3 \| 2394.0 \| 2581.1 \| 0.076 \| 195.1 \| 58.7 \| 253.8 \| \| 40 \| 7.38 \| 167.4 \| 2405.0 \| 2572.3 \| 0.057 \| 146.2 \| 52.2 \| 198.4 \| \| 35 \| 5.62 \| 146.4 \| 2417.1 \| 2563.5 \| 0.042 \| 108.9 \| 45.7 \| 154.6 \| \| 30 \| 4.25 \| 125.5 \| 2428.9 \| 2554.3 \| 0.032 \| 80.8 \| 39.2 \| 120.0 \| \| 25 \| 3.16 \| 104.6 \| 2440.5 \| 2545.1 \| 0.023 \| 59.3 \| 32.6 \| 92.0 \| \| 20 \| 2.34 \| 83.6 \| 2452.7 \| 2536.3 \| 26.1 \| Note: Dry air is at 25°C, 101.3 kPa. In thermodynamics, dry air can be assumed as an ideal gas at standard ambient temperature and pressure (some seawater desalination system belongs to this area). When the temperature is low, the water vapor in moist air can be also assumed as an ideal gas approximately because the small portion of water vapor is sometimes overheated so that it has low partial pressure and huge specific volume. Meanwhile, the moist air consisting of air and water vapor follows the principle of ideal air approximately. Hence the relationship of state parameters can be expressed by the ideal gas equation of state: (3.1) (3.2) where p is the pressure of moist air in Pa; v is the specific volume of moist air in m3/kg; R is the gas constant, which is determined by gas properties in J/(kg K); V is the total volume in m3; T is the thermodynamic temperature of gas in K; and m is the total mass of moist air in kg. When the total mass of moist air is in kmol, the ideal gas equation of state is expressed by pVm = R0T, where Vm is the volume of 1 kmol gas in m3/kmol; R0 is the universal gas constant in J/kmol K. But at high temperature and high pressures, the moist air should be calculated as real gas. The common gas equations of state are Van der Waals equation of state and Virial equations of state, which are: (3.3) and (3.4) where a(m6 Pa/kg2) and b(m3/kg) are Van der Waals constants. Table 3.2 demonstrates some common values of them. Table 3.2. Constants of Van der Waals equation of state. \| Substance \| \| \| --- \| Nitrogen N2 \| 0.136788 \| 0.03864 \| \| Oxygen O2 \| 0.138105 \| 0.03184 \| \| Argon Ar \| 0.136687 \| 0.03227 \| \| Carbon dioxide CO2 \| 0.365377 \| 0.04280 \| \| Water vapor H2O \| 0.552626 \| 0.03042 \| If water vapor occupies a large proportion in moist air, the physical properties of moist air can be obtained by searching the tables and diagrams of water vapor directly, or calculating the related parameters and obtaining by weighted sum of relative dry air and water vapor content. The physical properties of dry air and dry steam vapor are demonstrated in Appendix Tables 1 and 2. 3.1.2State Parameters of Moist Air The state parameters of moist air are critical to determine the state and physical properties of moist air. The calculation of moist air is commonly used. Hence it is necessary to discuss the state parameters in the calculation of seawater desalination in industry. 1. : Pressure Daltons Law states that the total pressure p of moist air is the sum of partial pressures of dry air pa and water vapor pv: p = pa + pv. : At a certain temperature, the more water vapor in air, the wetter the air, and the larger the partial pressure of water vapor. The extra water vapor will dissolve out when it exceeds a certain value. Therefore when the air contains a maximum amount of moisture that it can hold at a particular temperature, the air is in a state of saturation and is called “saturated air”. The corresponding partial pressure of water vapor is referred as “partial pressure water vapor” by pv,s. It can be found in a common saturated steam table, as shown in Appendix Table 3 The internal pressure of a seawater desalination system is tested by pressure gauge or vacuum gauge. The pressure gauge is used to test the operating pressures under positive pressures, and the vacuum gauge is for the working pressures under negative pressures or vacuum conditions. The operating pressure is not the actual or absolute pressure in the system but the difference with local barometric pressure. The relationship between them is: Absolute pressure = local barometric pressure + operating pressure (under positive pressure) Absolute pressure = local barometric pressure − operating pressure (under negative pressure) Only the absolute pressure is the state parameter of moist air. When the operating pressure is not indicated, the pressure should be considered as absolute pressure. The local barometric pressure can be obtained by barometer. 2. : Temperature The temperature of moist air represents its hotness and coldness. The common international criteria are absolute temperature scale (or Kelvin scale) in K, represented by T; Celsius temperature scale in °C represented by t; and Fahrenheit temperature scale in °F, represented by t. The relationship between the three scales is: (3.5) (3.6) 3. : Density and specific volume Density is the mass of a unit of volume for moist air and referred by ρ: (3.7) where m is the mass of moist air in kg; V is the volume occupied by moist air in m3. Specific volume is the volume occupied by a unit of mass of moist air and represented by v: (3.8) : Because the content of water vapor changes during the transmission of moist air, the benchmarks of moist air are different at initial and final states. However, the quantity of dry air is stable, resulting in applying 1 kg dry air as the standard in the calculation of moist air, which is represented by the subscript of a. Therefore it is important to notice the units and meanings in calculations of moist air. 4. : Absolute humidity and relative humidity Humidity is the amount of water vapor in the air, which can be signified by absolute humidity and relative humidity. Absolute is the total mass of water vapor present in a given volume of air, which is the density of water vapor ρv (3.9) where Rv is the gas constant of water vapor. Relative humidity (RH) is the ratio of the absolute humidity ρv and the maximum absolute humidity (absolute humidity of saturated air) ρv·s of moist air at the same temperature: (3.10) According to Eq. (3.9), (3.11) where pv·s is the maximum partial pressure of water vapor in moist air. Therefore the RH is also the partial pressure of water vapor to the equilibrium vapor pressure of water at the same temperature. The smaller the RH, the drier the moist air and vice versa. When the RH φ = 100%, the moist air is saturated and cannot hold more moisture. 5. : Moisture content (humidity ratio) Moisture content d is the mass of water vapor present in a unit of mass for dry air, which can be expressed as: (3.12) where mv and ma are the mass of water vapor and mass of dry air in moist air; the unit kg(a) refers to per kilogram of dry air. Then the following equation can be driven based on the ideal gas equation of state (3.13) where pb is barometric pressure. 6. : Dew point The water vapor is overheated in unsaturated air. Dew point td is the temperature to which moist air must be cooled to make the water vapor from overheating to saturation and maintain a constant partial pressure of water vapor pv in moist air at the same time. For an air mixture with constant moisture content d, when the temperature decreases to td, the moist air is saturated (φ = 100%), and water will leave the air and condense into “dew” if the temperature continues to fall. Therefore dew point is the critical point of water condensation. In addition, dew point depends on the moisture content in moist air only, which means td is constant when the moisture content is stable. In seawater desalination, the condensing temperature must be lower than the dew point if condensing fresh water from moist air. The lower the condensing temperature compared with the dew point, the fresher water will be condensed. 7. : Enthalpy of moist air Moist air is a mixture of dry air and water vapor. The enthalpy of moist air is the total of the enthalpy of the dry air and the enthalpy of the water vapor: (3.14) Because only the quantity of dry air is constant in the thermodynamic process of moist air, specific enthalpy is for a unit mass of dry air, which can be expressed as: (3.15) As the enthalpy of dry air is 0 at 0°C, the enthalpy of dry air at temperature t is: (3.16) where Cp,a is the specific heat of air at constant pressure and Cp,a = 1.005 kJ/(kg K). The enthalpy of water vapor can be calculated by: (3.17) where hc is the enthalpy of vaporization at 0°C, which is hc = 2501 kJ/kg; cp,a is the specific heat of air at constant pressure and equals cp,v = 1.86 kJ/(kg K). The enthalpy of moist air can be expressed as: (3.18) View chapterExplore book Read full chapter URL: Book2017, Solar Energy Desalination TechnologyHongfei Zheng Chapter Solid State Fermentation and Its Applications 2007, Bioprocessing for Value-Added Products from Renewable ResourcesLiping Wang, Shang-Tian Yang NOMENCLATURE \| a, b, c, d, and e \| constants \| --- \| \| Cs \| concentration of support in bed (kg dry support m−3 support) \| \| Cpg \| dry gas heat capacity (J kg−1 K−1) \| \| Cpg \| heat capacity of dry solid (J kg−1 K−1) \| \| Cpv \| water vapor heat capacity (J kg−1 K−1) \| \| Cpw \| heat capacity of liquid water (J kg−1 K−1) \| \| CWG \| concentration of water in air (kg water m−3 air) \| \| CX \| concentration of biomass on support (kg dry biomass m−3 support) \| \| dp \| particle diameter; m \| \| dave \| particle average diameter; m \| \| deqv \| particle equivalent diameter; m \| \| Dc \| reactor body diameter; m \| \| Di \| air inlet diameter; m \| \| F″a \| superficial aeration rate (kg dry air m−2 s−1) \| \| ha \| enthalpy of (moist) air (J [kg dry air]−1) \| \| ha \| heat transfer coefficient for heat transfer between the solid and gas phases (J s−1 m−3 K) \| \| H \| Bed height; m \| \| ΔHo \| reaction enthalpy (J [kg O2]−1) \| \| K \| constant in Charlton correlation \| \| K’a \| mass transfer coefficient for water transfer between the solid and gas phases (kg s−1 m−1) \| \| r’”o \| oxygen production biomass (kg O2 m−3 reactor s−1) \| \| R’”w \| water production biomass (kg H2O m−3 reactor s−1) \| \| S \| volumetric concentration of total dry solid (kg m−3) \| \| T \| time (s) \| \| Tg \| temperature of gas phase (K) \| \| Ts \| temperature of solid phase (K) \| \| Ums \| reactor superficial minimum spouting velocity; m/s \| \| V \| volume of spouting bed in Charlton correlation; cm3 \| \| xwx \| water content of biomass (kg water [kg dry biomass]−1) \| \| Yq \| heat yield from growth (J kg−1) \| \| YWB \| stoichiometric coefficient relating water production to growth (kg kg−1) \| \| z \| axial position in bed (m) \| \| ρg \| gas phase density (kg m−3) \| \| ρp \| particle density; kg/m3 \| \| ρf \| spouting gas density; kg/m3 \| \| φg \| water content of gas phase (kg kg−1 dry air) \| \| φs \| water content of solid phase (kg kg−1 dry solid) \| \| φS \| solid phase Water content for equilibrium with the gas phase at Tg (kg kg−1) \| \| λ \| enthalpy of evaporation of water (J kg−1) \| \| ɛ \| void fraction (m3 air m−3 reactor) \| View chapterExplore book Read full chapter URL: Book2007, Bioprocessing for Value-Added Products from Renewable ResourcesLiping Wang, Shang-Tian Yang Chapter Humidification and Water Cooling 2001, Chemical EngineeringJ.R. BACKHURST, J.H. HARKER PROBLEM 13.7 A rotary dryer is fed with sand at the rate of 1 kg/s. The feed is 50% wet and the sand is discharged with 3% moisture. The entering air is at 380 K and has an absolute humidity of 0.007 kg/kg. The wet sand enters at 294 K and leaves at 309 K and the air leaves at 310 K. Calculate the mass flowrate of air passing through the dryer and the humidity of the air leaving the dryer. Allow for a radiation loss of 25 kJ/kg dry air. Latent heat of water at 294 K = 2450 kJ/kg. Specific heat capacity of sand = 0.88 kJ/kg K. Specific heat capacity of dry air = 0.99 kJ/kg k. Specific heat capacity of vapour = 2.01 kg K. Solution The feed rate of wet sand is 1 kg/s and it contains 50% moisture or (1.0 × 50/100) = 0.50 kg/s water. If the dried sand contains w kg/s water, then: and: the water evaporated = (0.50 − 0.0155) = 0.4845 kg/s. Assuming a flowrate of G kg/s dry air, then a heat balance may be made based on a datum temperature of 294 K. Inlet streams: (i) : Sand: this enters at 294 K and hence the enthalpy = 0. (ii) : Air: G kg/s of dry air is associated with 0.007 kg/kg moisture. ∴ enthalpy = (G × 0.99) + (0.007G × 2.01) = 86.4G kW and: the total heat into the system = 86.4G kW. Outlet streams: (i) : Sand: 0.50 kg/s dry sand contains 0.0155 kg/s water and leaves the unit at 309 K. ∴ enthalpy = (0.5 × 0.88) + (0.0155 × 4.18) = 7.6 kW (ii) : Air: the air leaving contains 0.07 G kg/s water from the inlet air plus the water evaporated. It will be assumed that evaporation takes place at 294 K. Thus:enthalpy of dry air = G × 0.99(310 − 294) = 15.8m kWenthalpy of water from inlet air = 0.007G × 2.01(310 − 294) = 0.23G kWenthalpy in the evaporated water = 0.4845[2430 + 2.01(310 − 294)] = 1192.9 kW, a total of (16.03G + 1192.9) kW (iii) : Radiation losses = 25 kJ/kg dry air or 25G kW and the total heat out = (41.03G + 1200.5) kW. Mass balance: Thus the flow of dry air through the dryer =and the flow of inlet air = (26.5 × 1.007) = As in Problem 13.5, water leaving with the air is: (26.5 × 0.007) + 0.4845 = 0.67 kg/s and humidity of the outlet air = (0.67/26.5) = . View chapterExplore book Read full chapter URL: Book2001, Chemical EngineeringJ.R. BACKHURST, J.H. HARKER Chapter Solar drying system 2022, Technologies for Solar Thermal EnergyMd Zillur Rahman, Md Hasanuzzaman Mass balance for drying air Conservation of drying air mass represents that the product loses the moisture, while the drying air absorbs it. It is stated as (Lamrani, Khouya, & Draoui, 2019) (8.17) where is the mass of drying air (kg), is the mass flow rate of drying air (kg/s), is the absolute humidity of the air inside the drying chamber (kg water/kg dry air), is the absolute humidity of the air at the inlet of the drying chamber (kg water/kg dry air), is the product mass (dry basis) (kg), is the moisture content of the product (kg water/kg dry product), and is the time (s). View chapterExplore book Read full chapter URL: Book2022, Technologies for Solar Thermal EnergyMd Zillur Rahman, Md Hasanuzzaman Chapter Ventilation and Air-Conditioning in Powertrain Test Facilities 2012, Engine Testing (Fourth Edition)A.J. Martyr, M.A. Plint Fundamentals of Psychometry The study of the properties of moist air is known as psychometry. It is treated in many standard texts and only a very brief summary will be given here. Air-conditioning processes are represented on the psychometric chart in Figure 6.11. This relates the following properties of moist air: • : The moisture content or specific humidity, ω kg moisture/kg dry air. Note that even under fairly extreme conditions (saturated air at 30 °C) the moisture content does not exceed 3% by weight. • : The percentage saturation or relative humidity, φ. This is the ratio of the mass of water vapor present to the mass that would be present if the air were saturated at the same conditions of temperature and pressure. The mass of vapor under saturated conditions is very sensitive to temperature. A consequence of this relationship is the possibility of drying air by cooling. As the temperature is lowered the percentage saturation increases until at the dew point temperature the air is fully saturated and any further cooling results in the deposition of moisture. \| \| \| \| \| \| \| --- --- --- \| \| Temperature (°C) \| 10 \| 15 \| 20 \| 25 \| 30 \| \| Moisture content ω (kg/kg) \| 0.0076 \| 0.0106 \| 0.0147 \| 0.0201 \| 0.0273 \| • : The wet- and dry-bulb temperatures. The simplest method of measuring relative humidity is by means of a wet- and dry-bulb thermometer. If unsaturated air flows past a thermometer having a wetted sleeve of cotton around the bulb the temperature registered will be less than the actual temperature of the air, as registered by the dry-bulb thermometer, owing to evaporation from the wetted sleeve. The difference between the wet- and dry-bulb temperatures is a measure of the relative humidity. Under saturated conditions the temperatures are identical, and the depression of the wet-bulb reading increases with increasing dryness. Wet- and dry-bulb temperatures are shown in a psychometric chart. • : The specific enthalpy of the air, relative to an arbitrary zero corresponding to dry air at 0 °C. We have seen earlier that on this basis the specific enthalpy of dry air, (6.6a) : The specific enthalpy of moist air must include both the sensible heat and the latent heat of evaporation of the moisture content. The specific enthalpy of moist air is: (6.6b) : the last two terms representing the sum of the sensible and latent heats of the moisture. Taking the example of saturated air at 300 °C: : The first two terms represent the sensible heat of air plus moisture, and it is apparent that ignoring the sensible heat of the latter, as is usual in air cooling calculations, introduces no serious error. The third term, however, representing the latent heat of the moisture content, is much larger than the sensible heat terms. This accounts for the heavy cooling load associated with the process of drying air by cooling: condensation of the moisture in the air is accompanied by a massive release of latent heat. View chapterExplore book Read full chapter URL: Book2012, Engine Testing (Fourth Edition)A.J. Martyr, M.A. Plint Chapter Heat pumps in the drying industry 2020, Low-Temperature Energy Systems with Applications of Renewable EnergyAndriy Redko, ... Ronald DiPippo 4.1.1Case A: Basic open system using ambient air The simplest unit is an open, flow-through arrangement where ambient air is introduced to the drying chamber, passes over the moist product, evaporates the moisture, and is discharged to the surroundings. A heater is built in to boost the ambient air temperature as needed. See Fig. 4.2A. The figure also shows a recuperated system (B) that will be described shortly. Before moving on to more complex systems, we present a graphical method to analyze the basic case. Consider the adiabatic drying process which takes place in a dryer operating on the basic scheme (Fig. 4.2A), i.e., the air is heated in an air heater and then makes a single pass through the drying chamber. The processes may be visualized in a psychrometric chart (PC) (Fig. 4.3) that presents the properties of moist air in coordinates of moisture content (ordinate) versus the dry-bulb temperature (abscissa) [14,15]. The curved lines that fan out and upward from the lower left represent lines of constant relative humidity, with the highest one being 100%. In this figure, point 1 is determined from the properties of the ambient air. Assume that the ambient air temperature, t1 = 20 °C and moisture content is 0.10 kg/kg dry air; from the PC one can read that the relative humidity is ∼70%. The process of heating the air prior to the dryer is represented by line 1 → 2′, where point 2′ is determined by the assumed temperature t2 = 49 °C and a constant moisture content of 0.10 kg/kg dry air. Process 2′→2″ is an adiabatic process of cooling and humidifying the air in the dryer that follows a line of constant enthalpy to the outlet temperature of the air, in this example, ∼28.3 °C. It can be seen that the outlet relative humidity is ∼78% and the absolute moisture is 0.19 kg/kg dry air. In an ideal adiabatic saturation process, point 2″ would move to the 100% relative humidity line. In practice, the dry- and wet-bulb temperatures of the incoming air can be measured using a “sling psychrometer,” and the PC can then be used to determine the relative humidity, absolute humidity and specific enthalpy. View chapterExplore book Read full chapter URL: Book2020, Low-Temperature Energy Systems with Applications of Renewable EnergyAndriy Redko, ... Ronald DiPippo Chapter Adsorption 2002, Chemical Engineering (Fifth Edition), Volume 2J.F. RICHARDSON, ... J.R. BACKHURST Example 17.5 An adsorption unit is to be designed to dry air using silica gel. A moving-bed design is considered in which silica gel moves down a cylindrical column in plug flow while air flows up the column. Air enters the unit at the rate of 0.129 kg of dry air/m2s and with a humidity of 0.00267 kg water/kg dry air. It leaves essentially bone dry. There is equilibrium between air and gel at the entrance to and the exit from the adsorption zone. Experiments were carried out to find the relative resistances of the external gas film and pellet diffusion. Referred to a driving force expressed as mass ratios then: (a) : for the gas film, the coefficient kgaz = 31.48G′0.55 kg/m3 s where G’ is the mass flowrate of dry air per unit cross-section of bed. (b) : for pellet diffusion, solid-film coefficent kpaz = 0.964 kg/m3s where az is the external area of adsorbent per unit volume of bed (a) : Using the transfer-unit concept, calculate the minimum length of packing which will reduce the moisture content of the air to 0.0001 kg water/kg dry air. At what rate should the gel travel through the bed? The properties of the gel and the condition of the air are as given in Example 17.4. (b) : After operating for some time, the gel jams and the unit continues operating as a fixed bed. How long after jamming will it be before the moisture content of the effluent rises to half the inlet value? Solution (a) : The bed must be long enough to contain the adsorption zone. From equation 17.112 the number of transfer units may be written as: ƒ dyr/(yr — yr) and the height of a transfer unit in appropriate units is G′/k0gaz, where k°g is the overall mass transfer coefficient. The integral may be evaluated graphically from a plot of 1/(yr - yr\) against yr over the concentration range of the adsorption zone. A mass balance over a part of the bed gives an operating line: This line, together with the equilibrium line, is similar to that shown in Figure 17.27. Corresponding values of y and y\ may be measured and the integral evaluated. The pinches between the operating and equilibrium lines which occur at each end of the zone prevent the end concentration from being used as the limits of the integration. If the lower limit of yr = 0.0001 and the upper limit 0.0024, then from a graphical construction:The height of a transfer unit is:(from equation 17.112) k0g may be evaluated from the film coefficients, as discussed in Chapter 12. When the zone is fully developed, each part will move at the same constant velocity. If f'(C) is the mean slope of the isotherm over the range of concentrations of interest, then, in appropriate units: The inter-pellet air velocity = 0.233 m/s. The velocity uc with which the adsorption wave moves through the column may be obtained from equation 17.79. Hence: and:The rate at which clean adsorbent must be added and spent adsorbent removed in order to maintain steady state may also be found from an overall balance: (b) : When the gel stops moving, the bed behaves as a fixed-bed already at its breakpoint. The concentration of water in the effluent begins to rise. The time ta for the adsorption zone to move its own length za is given by: The time taken for a point at a distance z’ into the zone to emerge is given by:where:The results for graphical integration are tabulated below.where m is the slope of the operating , as shown in Figure 17.22.Hence:and:The length of the adsorption zone = (10.95 × 0.0169) = 0.185 m Hence the minimum length of bed to contain the adsorption zone is . In practice, a somewhat greater length would be used to allow for variations in the length of the zone that might result from fluctuations in operating conditions. The data are summarised as follows: \| yr \| yr\ \| \| \| \| \| t (h) \| --- --- --- \| 0.0001 \| 0.00005 \| 20,000 \| 0 \| 0 \| 0.038 \| 0 \| \| 0.0002 \| 0.00010 \| 10,000 \| 1.50 \| 0.137 \| 0.075 \| 0.8 \| \| 0.0006 \| 0.00032 \| 3570 \| 4.00 \| 0.362 \| 0.225 \| 3.1 \| \| 0.0010 \| 0.00062 \| 2630 \| 5.18 \| 0.473 \| 0.374 \| 4.1 \| \| 0.0014 \| 0.00100 \| 2500 \| 6.13 \| 0.560 \| 0.525 \| 4.5 \| \| 0.0018 \| 0.00133 \| 3700 \| 7.38 \| 0.674 \| 0.674 \| 5.8 \| \| 0.0022 \| 0.00204 \| 6250 \| 9.33 \| 0.852 \| 0.825 \| 7.1 \| \| 0.0024 \| 0.00230 \| 10,000 \| 10.95 \| 1.000 \| 0.899 \| 7.7 \| By interpolation, yr/yro =0.5 when . View chapterExplore book Read full chapter URL: Book2002, Chemical Engineering (Fifth Edition), Volume 2J.F. RICHARDSON, ... J.R. BACKHURST Related terms: Relative Humidity Energy Engineering Battery (Electrochemical Energy Engineering) Partial Pressure Dry Bulb Temperature Flue Gas Humidity Ratio Moist Air Wet-Bulb Temperature Mass Flowrate View all Topics
5831	https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality?srsltid=AfmBOoqbvZT74iaKpeNp-pqMUibk5wfMr51bVN6euG-4JEWjxrx9-BFK	Art of Problem Solving AM-GM Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki AM-GM Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search AM-GM Inequality In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same. In symbols, the inequality states that for any real numbers , with equality if and only if . The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests. Contents 1 Proofs 2 Generalizations 2.1 Weighted AM-GM Inequality 2.2 Mean Inequality Chain 2.3 Power Mean Inequality 3 Problems 3.1 Introductory 3.2 Intermediate 3.3 Olympiad 4 See Also Proofs Main article: Proofs of AM-GM All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for . Generalizations The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM. Weighted AM-GM Inequality The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article. Mean Inequality Chain Main article: Mean Inequality Chain The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain. Power Mean Inequality Main article: Power Mean Inequality The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality. Problems Introductory For nonnegative real numbers , demonstrate that if then . (Solution) Find the maximum of for all positive . (Solution) Intermediate Find the minimum value of for . (Source) Olympiad Let , , and be positive real numbers. Prove that (Source) See Also Proofs of AM-GM Mean Inequality Chain Power Mean Inequality Cauchy-Schwarz Inequality Inequality Retrieved from " Categories: Algebra Inequalities Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
5832	https://www.sapere.it/sapere/strumenti/studiafacile/fisica/L-elettromagnetismo/L-induzione-elettromagnetica-e-le-equazioni-di-Maxwell/La-legge-di-Lenz-e-l-autoinduzione.html	L'induzione elettromagnetica e le equazioni di Maxwell La legge di Lenz e l'autoinduzione Avvicinando un magnete a un circuito, la variazione di flusso del campo magnetico produce nel circuito una corrente indotta. Questa corrente genera a sua volta un campo magnetico, il cui effetto sul circuito è, come dimostrato sperimentalmente, quello di opporsi al campo magnetico esterno. In altre parole, il verso della f.e.m. indotta è tale da opporsi alla causa che l'ha prodotta. Questo significa che se, per esempio, la f.e.m. è stata generata da un aumento del flusso del campo magnetico concatenato col circuito, essa tende a far circolare una corrente di verso tale da produrre un flusso di verso opposto a quello inducente. Quanto detto è riassunto dalla legge di Lenz, che stabilisce che la corrente indotta circola sempre con verso tale da opporsi alle variazioni di flusso che l'hanno generata. Quindi la legge di Lenz permette di prevedere quale sarà il verso della corrente indotta in un circuito da un campo magnetico variabile. La legge di Faraday-Neumann, integrata con la legge di Lenz (legge di Faraday-Neumann-Lenz), stabilisce che la f.e.m. indotta in un circuito sia esprimibile attraverso la relazione: dove ΔΦ è la variazione di flusso nel tempo (Δt) e il segno meno tiene conto della legge di Lenz. L'autoinduzione Poiché si ha una "forza controelettromotrice" ogni volta che interviene una variazione di flusso di un campo magnetico, se consideriamo un solenoide nel quale si faccia variare − per esempio, aumentare − l'intensità della corrente, si produrrà un campo magnetico variabile. Man mano che l'intensità della corrente aumenta, aumenta anche il flusso del campo magnetico generato dalla corrente stessa, quindi sul solenoide si produrrà una corrente indotta, il cui effetto è quello di opporsi all'aumento della corrente inducente. Questo fenomeno prende il nome di autoinduzione e la f.e.m. che si genera prende il nome di f.e.m. autoindotta. In sintesi, quando il flusso di campo magnetico concatenato con un circuito varia per effetto della variazione dell'intensità della corrente del circuito stesso, la f.e.m. è detta di autoinduzione. Lo schema della figura 19.3 riassume i fenomeni principali connessi ai magnetismi. Quanto detto è riassunto dalla legge di Lenz, che stabilisce che la corrente indotta circola sempre con verso tale da opporsi alle variazioni di flusso che l'hanno generata. Quindi la legge di Lenz permette di prevedere quale sarà il verso della corrente indotta in un circuito da un campo magnetico variabile. La legge di Faraday-Neumann, integrata con la legge di Lenz (legge di Faraday-Neumann-Lenz), stabilisce che la f.e.m. indotta in un circuito sia esprimibile attraverso la relazione: dove ΔΦ è la variazione di flusso nel tempo (Δt) e il segno meno tiene conto della legge di Lenz. L'autoinduzione Poiché si ha una "forza controelettromotrice" ogni volta che interviene una variazione di flusso di un campo magnetico, se consideriamo un solenoide nel quale si faccia variare − per esempio, aumentare − l'intensità della corrente, si produrrà un campo magnetico variabile. Man mano che l'intensità della corrente aumenta, aumenta anche il flusso del campo magnetico generato dalla corrente stessa, quindi sul solenoide si produrrà una corrente indotta, il cui effetto è quello di opporsi all'aumento della corrente inducente. Questo fenomeno prende il nome di autoinduzione e la f.e.m. che si genera prende il nome di f.e.m. autoindotta. In sintesi, quando il flusso di campo magnetico concatenato con un circuito varia per effetto della variazione dell'intensità della corrente del circuito stesso, la f.e.m. è detta di autoinduzione. Lo schema della figura 19.3 riassume i fenomeni principali connessi ai magnetismi. La legge di Faraday-Neumann, integrata con la legge di Lenz (legge di Faraday-Neumann-Lenz), stabilisce che la f.e.m. indotta in un circuito sia esprimibile attraverso la relazione: dove ΔΦ è la variazione di flusso nel tempo (Δt) e il segno meno tiene conto della legge di Lenz. L'autoinduzione Poiché si ha una "forza controelettromotrice" ogni volta che interviene una variazione di flusso di un campo magnetico, se consideriamo un solenoide nel quale si faccia variare − per esempio, aumentare − l'intensità della corrente, si produrrà un campo magnetico variabile. Man mano che l'intensità della corrente aumenta, aumenta anche il flusso del campo magnetico generato dalla corrente stessa, quindi sul solenoide si produrrà una corrente indotta, il cui effetto è quello di opporsi all'aumento della corrente inducente. Questo fenomeno prende il nome di autoinduzione e la f.e.m. che si genera prende il nome di f.e.m. autoindotta. In sintesi, quando il flusso di campo magnetico concatenato con un circuito varia per effetto della variazione dell'intensità della corrente del circuito stesso, la f.e.m. è detta di autoinduzione. Lo schema della figura 19.3 riassume i fenomeni principali connessi ai magnetismi. dove ΔΦ è la variazione di flusso nel tempo (Δt) e il segno meno tiene conto della legge di Lenz. L'autoinduzione Poiché si ha una "forza controelettromotrice" ogni volta che interviene una variazione di flusso di un campo magnetico, se consideriamo un solenoide nel quale si faccia variare − per esempio, aumentare − l'intensità della corrente, si produrrà un campo magnetico variabile. Man mano che l'intensità della corrente aumenta, aumenta anche il flusso del campo magnetico generato dalla corrente stessa, quindi sul solenoide si produrrà una corrente indotta, il cui effetto è quello di opporsi all'aumento della corrente inducente. Questo fenomeno prende il nome di autoinduzione e la f.e.m. che si genera prende il nome di f.e.m. autoindotta. In sintesi, quando il flusso di campo magnetico concatenato con un circuito varia per effetto della variazione dell'intensità della corrente del circuito stesso, la f.e.m. è detta di autoinduzione. Lo schema della figura 19.3 riassume i fenomeni principali connessi ai magnetismi. dove ΔΦ è la variazione di flusso nel tempo (Δt) e il segno meno tiene conto della legge di Lenz. L'autoinduzione Poiché si ha una "forza controelettromotrice" ogni volta che interviene una variazione di flusso di un campo magnetico, se consideriamo un solenoide nel quale si faccia variare − per esempio, aumentare − l'intensità della corrente, si produrrà un campo magnetico variabile. Man mano che l'intensità della corrente aumenta, aumenta anche il flusso del campo magnetico generato dalla corrente stessa, quindi sul solenoide si produrrà una corrente indotta, il cui effetto è quello di opporsi all'aumento della corrente inducente. Questo fenomeno prende il nome di autoinduzione e la f.e.m. che si genera prende il nome di f.e.m. autoindotta. In sintesi, quando il flusso di campo magnetico concatenato con un circuito varia per effetto della variazione dell'intensità della corrente del circuito stesso, la f.e.m. è detta di autoinduzione. Lo schema della figura 19.3 riassume i fenomeni principali connessi ai magnetismi. dove ΔΦ è la variazione di flusso nel tempo (Δt) e il segno meno tiene conto della legge di Lenz. L'autoinduzione Poiché si ha una "forza controelettromotrice" ogni volta che interviene una variazione di flusso di un campo magnetico, se consideriamo un solenoide nel quale si faccia variare − per esempio, aumentare − l'intensità della corrente, si produrrà un campo magnetico variabile. Man mano che l'intensità della corrente aumenta, aumenta anche il flusso del campo magnetico generato dalla corrente stessa, quindi sul solenoide si produrrà una corrente indotta, il cui effetto è quello di opporsi all'aumento della corrente inducente. Questo fenomeno prende il nome di autoinduzione e la f.e.m. che si genera prende il nome di f.e.m. autoindotta. In sintesi, quando il flusso di campo magnetico concatenato con un circuito varia per effetto della variazione dell'intensità della corrente del circuito stesso, la f.e.m. è detta di autoinduzione. Lo schema della figura 19.3 riassume i fenomeni principali connessi ai magnetismi. L'autoinduzione Poiché si ha una "forza controelettromotrice" ogni volta che interviene una variazione di flusso di un campo magnetico, se consideriamo un solenoide nel quale si faccia variare − per esempio, aumentare − l'intensità della corrente, si produrrà un campo magnetico variabile. Man mano che l'intensità della corrente aumenta, aumenta anche il flusso del campo magnetico generato dalla corrente stessa, quindi sul solenoide si produrrà una corrente indotta, il cui effetto è quello di opporsi all'aumento della corrente inducente. Questo fenomeno prende il nome di autoinduzione e la f.e.m. che si genera prende il nome di f.e.m. autoindotta. In sintesi, quando il flusso di campo magnetico concatenato con un circuito varia per effetto della variazione dell'intensità della corrente del circuito stesso, la f.e.m. è detta di autoinduzione. Lo schema della figura 19.3 riassume i fenomeni principali connessi ai magnetismi. Poiché si ha una "forza controelettromotrice" ogni volta che interviene una variazione di flusso di un campo magnetico, se consideriamo un solenoide nel quale si faccia variare − per esempio, aumentare − l'intensità della corrente, si produrrà un campo magnetico variabile. Man mano che l'intensità della corrente aumenta, aumenta anche il flusso del campo magnetico generato dalla corrente stessa, quindi sul solenoide si produrrà una corrente indotta, il cui effetto è quello di opporsi all'aumento della corrente inducente. Questo fenomeno prende il nome di autoinduzione e la f.e.m. che si genera prende il nome di f.e.m. autoindotta. In sintesi, quando il flusso di campo magnetico concatenato con un circuito varia per effetto della variazione dell'intensità della corrente del circuito stesso, la f.e.m. è detta di autoinduzione. Lo schema della figura 19.3 riassume i fenomeni principali connessi ai magnetismi. Media correlati Figura 19.3 Schema dei fenomeni connessi al magnetismo. Enciclopedia di Sapere Non conosci qualche termine? Cerca nel dizionario di lingua italiana Seguici su © 2024 Copyright Media Data Factory S.R.L. I contenuti sono di proprietà di Media Data Factory S.R.L, è vietata la riproduzione. Media Data Factory S.R.L. sede legale in via Guercino 9 20154 Milano - PI/CF 09595010969 I contenuti sono di proprietà di Media Data Factory S.R.L, è vietata la riproduzione. Media Data Factory S.R.L. sede legale in via Guercino 9 20154 Milano - PI/CF 09595010969
5833	https://www.youtube.com/watch?v=sj-mC03aYIQ	Graphing Inequalities: Standard Form MATHguide 11700 subscribers 17 likes Description 2184 views Posted: 14 Jun 2014 This MATHguide [ ] video will demonstrate how to graph linear inequalities that are in standard form. The text lesson is at 4 comments Transcript: hello there this is mathguy.com my name is Mark carradimos today we are going to be graphing linear inequalities now these linear inequalities are in a different form than our last video this is in these are in that I should say uh standard form what is standard form standard form is when you've got both variables the X and the Y both on one side of the equation or an inequality in this case and we've got the constant term the non-variable on the other side of the equation all right so I've explained how to do this with equations now we're going to do it with inequalities uh slightly different of course with inequalities but it's fairly the same very much the same uh all right where do we start we start with making a table sometimes we call this T table I call this a little baby table because it's small all right we put a zero in for x and we're going to calculate what the Y is now remember when we put in 0 for x six times I mean actually before I actually get into the computation actually physically picture putting in zero there right picture putting a zero right here so it'd be six times zero is zero and that was zero in other words this cancels zero minus y is just going to be minus y so this whole term lands up canceling the only thing you have here is a minus y now for the purposes of what we're solving here I'm actually going to be solving an equation I want to know what the value is for y so I'm not even looking at the inequality right now I'm looking at an equality so the last step for this particular problem is to figure out how to get y alone so we're going to picture the coefficient really is negative one here so we're going to divide both sides by negative one so we get y equals negative 1. so I'm going to put that value right here into the table all right I'm just going to get rid of this junk so I'm going to put y equals negative 1 right next to that value so I'm just going to back up a little bit get that chunk out of there Put negative one right there okay so now let's do the same thing but we'll just put in 0 for y now again let's actually physically put in 0 here now we're going to put in a zero picture putting in the value 0 right there so we have 6X minus nothing so 6X minus nothing is just 6x so it's like this thing disappears so we're left with 6 x equals one yep I'm just ignoring the inequality right now I want to know what the value is for X I don't care about shading at the moment just want to figure out what the corresponding x value is so of course what I'm going to do is divide by 6. so if you divide it by 6 well I get a fraction I get 1 6 or I think that's 0.17 but whatever it's a very small value okay so I'm going to put in 1 6 into the table right over here all right it's going to back up get this junk out of the way I got 1 6. let's put that right there all right so what do I do I graph them so keep in mind that these are separate points this is one point this is another point so I'm going to graph each one of those points separately all right so let's graph the first one 0 negative one let's graph 1 6 0 so that's one-sixth it's just tiny bit to the right right it's a very small decimal value right here up zero so it's going to be right here just really close to the origin not the origin but very close to it all right now I'm going to graph this when we graph keep in mind that we do have an inequality here so there's going to be some shading notice that there is equality so that means I'm going to be using a solid line all right so anytime there's a equal sign we use a solid line and that's what I'm going to do right now put that in Black so I'm going to put a solid line here almost missed those points all right so we've got our line now we got to figure out which side of the line to shade again you could shade I'm sorry you could test any value you want off of this line I always like using the origin if I can it's the easiest point there is to use because calculating with zero is easy now this line almost goes through this point if the line goes through the origin you can't use the origin but this line doesn't go through the origin so I'm going to use it all right so let's use it let's put in 0 for X 0 for y so I guess what I could do is go off over here on the side and show my work so I get 6 times x minus y is less than or equal to one okay so I'm going to put that value 0. 00 for X 0 for y six times zero zero zero times negative one or I should say just the minus zero if you're looking at it that way in other words I get zero is less than or equal to one is zero less than or equal to one yep that is definitely true that is the case so the origin is to the left side of this line and the origin works so therefore I'm going to be shading everything on the left side because the left side works all right so I'm going to get my black tool out here and I'm going to shade everything to the left of that line okay so everything over here nothing on the right side gets shaded okay let's graph the next one because we're done with number one all right number two same thing we start off by making a baby table or very small T table so our x value our y value we're going to put in 0 for X 0 for y and do some calculating so all right we're going to put in 0 for X so imagine putting in zero there that's going to cancel out this term so we're going to divide both sides by three we're going to get y equals again I'm just looking at this like it's an equal sign I'm temporarily suspending the less than I just want to know what the value is here so I'll worry about the shading in a moment so I get y equals three just doing the math I get y equals three we back up so I put y equals 3 in the table all right likewise let's do the same thing now for y I'm going to go I'm going to put in 0 for y so let's see I'm going to get uh let's do this in white to be consistent so I'm going to get uh oh well I guess I don't have to do that three times zero is zero this term cancels out so I get 2x equals 9. again I'm just thinking that like an equal sign at the moment so I divide both sides by two I get x equals 4.5 right that's half of 9. or nine halves if you prefer but since I'm going to be graphing this in a number line I'd like to use a decimal all right so I get 4.5 so I'm going to put that in there get rid of this junk 4.5 let's put that right there 4.5 all right now we graph the two points so let's graph the first one zero three so zero on the x-axis 3 up Let's see we got 4.5 x technically I should be using an open circle and I'll explain why in a second okay 4.5 to the right zero up pretty open circle all right now the reason why I'm using open circles and that closed circles like on the last one here I use close circles you see those closed circles there I'm using Open Circles because there is no equality there's no equal sign if there's no equal sign there's really no line all we really have is the Border the Border exists I mean I have to show that there's some type of Border but we don't the border is actually not part of the problem we're going to be shading some side of the line but we're not going to include the line like over here we included the line from problem number one because there was an equal sign okay if the Z equal sign you include the Border all right no equal sign no border so I'm going to shade now one side now to determine which side to shade you're going to have to use a test Point yep I know I'm getting boring I'm using the same points over and over and but I'm going to use 0 0 as my test point so I go back to my equation and I put in let's see 2 times x put 2 times x plus 3 times Y is less than nine all right again I'm using 0 for X 0 for y oh 2 times 0 is 0. 3 times 0 0. zero plus zero zero is zero less than nine yep zero is definitely less than nine so it turns out that this point is the correct side it works the equation works when I when I put in this test point so what I want to do is now shade the side that works and we found out that the origin does work so we shade below the line for this particular problem okay so that's how we do it sometimes we use dashed lines sometimes we use solid lines depending on what the problem looks like all right make sure you go back to mathguide.com we have thorough lessons interactive quizzes and a bunch of other videos thanks have a great day
5834	https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_9?srsltid=AfmBOorzCWPj_5H1eyZlrM2hYjHbbKmfqXjbmpJK0tokCagjg0MyxILy	Page Toolbox Search 2021 AIME II Problems/Problem 9 Contents Problem Find the number of ordered pairs such that and are positive integers in the set and the greatest common divisor of and is not . Solution 1 This solution refers to the Remarks section. By the Euclidean Algorithm, we have We are given that Multiplying both sides by gives which implies that must have more factors of than does. We construct the following table for the first positive integers: To count the ordered pairs we perform casework on the number of factors of that has: Together, the answer is ~Lcz ~MRENTHUSIASM Solution 2 Consider any ordered pair such that . There must exist some odd number such that and . Let be the order of modulo . Note that . From this, we can say that and are both multiples of , but is not. Thus, we have and . Substituting the latter equation into the inequality before gives . Since and are integers, this implies . The rest of the solution now proceeds as in Solution 1. ~Sedro Remarks Claim 1 (GCD Property) If and are positive integers such that then As and are relatively prime (have no prime divisors in common), this property is intuitive. ~MRENTHUSIASM Claim 2 (Olympiad Number Theory Lemma) If and are positive integers such that then There are two proofs to this claim, as shown below. ~MRENTHUSIASM Claim 2 Proof 1 (Euclidean Algorithm) If then from which the claim is clearly true. Otherwise, let without the loss of generality. For all integers and such that the Euclidean Algorithm states that We apply this result repeatedly to reduce the larger number: Continuing, we have from which the proof is complete. ~MRENTHUSIASM Claim 2 Proof 2 (Bézout's Identity) Let It follows that and By Bézout's Identity, there exist integers and such that so from which We know that Next, we notice that Since is a common divisor of and we conclude that from which the proof is complete. ~MRENTHUSIASM Video Solution ~MathProblemSolvingSkills.com Video Solution by Interstigation ~Interstigation See Also \| \| \| \| --- \| 2021 AIME II (Problems • Answer Key • Resources) \| \| \| \| Preceded by Problem 8 \| Followed by Problem 10 \| \| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 \| \| \| \| All AIME Problems and Solutions \| \| \| These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Something appears to not have loaded correctly. Click to refresh.
5835	https://www.mytutor.co.uk/answers/7268/GCSE/Maths/Factorise-x-2x-8/	+44 (0) 203 773 6024 Factorise x² + 2x - 8 Since an x² is produced, we know that the factorised version will be (x+a)(x+b). To find a and b we need to consider the relationship between 2 and -8. It is going to be factorised into two brackets, due to the x², so we need to break the 2 down into 2 numbers that will add to give 2 and multiply together to give -8. The 8 is negative so one of these numbers must also be negative. These 2 numbers must be 4 and -2. Then substitude these numbers in to be a and b so the equation factorised is (x+4)(x-2) Since an x² is produced, we know that the factorised version will be (x+a)(x+b). To find a and b we need to consider the relationship between 2 and -8. It is going to be factorised into two brackets, due to the x², so we need to break the 2 down into 2 numbers that will add to give 2 and multiply together to give -8. The 8 is negative so one of these numbers must also be negative. These 2 numbers must be 4 and -2. Then substitude these numbers in to be a and b so the equation factorised is (x+4)(x-2) 32697 Views Need help with Maths? One-to-one online tuition can be a great way to brush up on your Maths knowledge. Have a Free Meeting with one of our hand picked tutors from the UK's top universities Download MyTutor's free revision handbook? This handbook will help you plan your study time, beat procrastination, memorise the info and get your notes in order. 8 study hacks, 3 revision templates, 6 revision techniques, 10 exam and self-care tips. Free weekly group tutorials Join MyTutor Squads for free (and fun) help with Maths, Coding & Study Skills. Related Maths GCSE answers Solve the following simultaneous equations: 3a + 2b = 36 equation ( 1), and 5a + 4b = 64 equation (2) Write x^2+4x-12 in the form (x+a)^2+b where 'a' and 'b' are constants to be determined. I know the formula, but I don't understand it. Solve the following simultaneous equations: 2x-3y=16 x+2y=-6 We're here to help Company Information Popular Requests CLICK CEOP Internet Safety Payment Security Cyber Essentials MyTutor is part of the IXL family of brands: IXL Comprehensive K-12 personalized learning Rosetta Stone Immersive learning for 25 languages Wyzant Trusted tutors for 300 subjects Education.com 35,000 worksheets, games, and lesson plans Vocabulary.com Adaptive learning for English vocabulary Emmersion Fast and accurate language certification Thesaurus.com Essential reference for synonyms and antonyms Dictionary.com Comprehensive resource for word definitions and usage SpanishDictionary.com Spanish-English dictionary, translator, and learning resources FrenchDictionary.com French-English dictionary, translator, and learning Ingles.com Diccionario ingles-espanol, traductor y sitio de apremdizaje ABCya Fun educational games for kids © 2025 by IXL Learning
5836	https://www.sparkl.me/learn/cambridge-igcse/mathematics-us-0444-advanced/the-perpendicular-bisector-of-a-chord-passes-through-the-center/revision-notes/2639	Past Papers Log In Past Papers Courses Collegeboard AP IB DP Cambridge IGCSE AS & A Level IB MYP 1-3 IB MYP 4-5 All Topics mathematics-us-0444-advanced \| cambridge-igcse Trigonometry 1.1.1 Apply the Pythagorean Theorem to find missing sides 1.1.2 Use trigonometric ratios (sine, cosine, tangent) to solve right-angled triangles 1.2.1 Know exact values for sine, cosine, and tangent of 0°, 30°, 45°, 60°, 90° 1.3.1 Solve problems involving bearings 1.3.2 Understand and use the concepts of angle of elevation and angle of depression 1.4.1 Extend sine and cosine values to angles between 0° and 360° 1.4.2 Use the relationship between sine and cosine of complementary angles 1.5.1 Graph and understand the properties of sine, cosine, and tangent functions 1.6.1 Apply the Sine Rule for solving triangles (ASA, SSA cases) 1.7.1 Apply the Cosine Rule for solving triangles (SAS, SSS cases) 1.8.1 Use the formula for the area of a triangle (1/2 ab sinC) Transformations and Vectors 2.1.1 Describe and perform stretches 2.1.2 Describe and perform translations using column vectors 2.1.3 Describe and perform reflections in a given axis or line 2.1.4 Describe and perform rotations about a point 2.1.5 Describe and perform enlargements (dilations) with scale factors 2.2.1 Find the inverse of a given transformation 2.3.1 Find the single transformation equivalent to a sequence of transformations 2.4.1 Understand directed line segment notation and component form of vectors 2.4.2 Use appropriate symbols for vectors and their magnitudes 2.5.1 Find the components of a vector by subtracting coordinates of an initial point from a terminal point 2.5.2 Use position vectors 2.5.3 Calculate the magnitude of a vector 2.5.4 Understand vector subtraction as v - w = v + (-w) 2.5.5 Multiply a vector by a scalar 2.5.6 Add and subtract vectors algebraically (component form) and geometrically (parallelogram rule) Statistics 3.1.1 Draw a straight line of best fit by eye through the mean on a scatter diagram 3.2.1 Read and interpret data from graphs and tables 3.3.1 Understand discrete and continuous data 3.4.1 Calculate mean, mode, median, and range from discrete data 3.4.2 Calculate mean, modal class, median, and range from grouped and continuous data 3.5.1 Construct and interpret histograms with frequency density on the vertical axis 3.5.2 Interpret histograms with unequal class intervals 3.6.1 Create and interpret cumulative frequency tables and curves 3.6.2 Determine median, quartiles, percentiles, and interquartile range 3.7.1 Use statistics (median, mean, interquartile range) to compare different data sets 3.8.1 Understand and describe correlation (positive, negative, or zero) using scatter diagrams 3.9.1 Construct and interpret compound bar charts, dot plots, line graphs, pie charts, scatter diagrams Geometry 4.1.1 Angle properties of triangles, quadrilaterals, and polygons 4.1.2 Interior and exterior angles of a polygon 4.2.1 Make formal geometric constructions using a compass and straight edge 4.2.2 Copy and bisect a segment or an angle 4.2.3 Construct perpendicular lines and perpendicular bisectors 4.2.4 Construct equilateral triangles, squares, and hexagons inscribed in circles 4.2.5 Construct inscribed and circumscribed circles of a triangle 4.2.6 Construct tangent lines from a point outside a circle 4.3.1 Tangent perpendicular to radius at the point of contact 4.3.2 Tangents from a point 4.3.3 Angle in a semicircle 4.3.4 Angles at the center and at the circumference on the same arc 4.3.5 Cyclic quadrilateral properties 4.3.6 Equal chords are equidistant from the center 4.3.7 The perpendicular bisector of a chord passes through the center 4.3.8 Tangents from an external point are equal in length 4.4.1 Calculate lengths of similar figures using scale factors 4.4.2 Use area and volume scale factors for similar figures and solids 4.5.1 Recognize and use congruence in solving geometric problems 4.6.1 Know definitions of acute, obtuse, right angle, reflex, equilateral, isosceles, congruent, similar, 4.6.2 Know definitions of pentagon, hexagon, octagon, rectangle, square, kite, rhombus, parallelogram, tra 4.7.1 Understand definitions of angle, circle, perpendicular line, parallel line, and line segment 4.8.1 Line and rotational symmetry in 2D and 3D 4.8.2 Recognize symmetry properties of prisms and pyramids 4.9.1 Angles around a point 4.9.2 Angles on a straight line and intersecting straight lines 4.9.3 Vertically opposite angles 4.9.4 Alternate and corresponding angles on parallel lines Functions 5.1.1 Use function notation 5.1.2 Understand domain and range 5.1.3 Use mapping diagrams 5.2.1 Graph an equation in two variables as the set of all its solutions 5.2.2 Construct tables of values and graphs for functions of the form ax^n where n = -2, -1, 0, 1, 2, 3 5.2.3 Solve equations approximately using graphical methods 5.3.1 Write a function that describes a relationship between two quantities 5.4.1 Compare properties of two functions represented in different ways (algebraically, graphically, numer 5.5.1 Recognize linear, quadratic, cubic, reciprocal, exponential, and trigonometric functions from their 5.5.2 Interpret key features of function graphs (intercepts, increasing/decreasing behavior, maxima/minima 5.6.1 Relate the domain of a function to its graph and quantitative relationships 5.7.1 Calculate and interpret the average rate of change of a function over a specified interval 5.7.2 Estimate rate of change using a graph 5.8.1 Understand and compare the behavior of linear, quadratic, and exponential functions 5.8.2 Observe that exponential growth eventually exceeds polynomial growth 5.8.3 Use properties of exponents to interpret expressions for exponential functions 5.9.1 Construct linear and exponential functions given graphs, descriptions, or two input-output pairs 5.10.1 Simplify expressions for composite functions such as f(g(x)) 5.11.1 Find and interpret inverse functions 5.12.1 Describe changes to the graph of y = f(x) when transformed as y = f(x) + k, y = kf(x), and y = f(x + 5.13.1 Graph the solutions to a linear inequality in two variables as a half-plane 5.13.2 Graph the solution set of a system of inequalities as the intersection of corresponding half-planes Number 6.1.1 Converting between units (e.g., currency, time, length) 6.1.2 Understanding units in problems 6.1.3 Choosing appropriate levels of accuracy 6.2.1 Units of time: seconds, minutes, hours, days, months, years 6.2.2 24-hour and 12-hour clock formats 6.3.1 Understanding and solving problems related to motion 6.3.2 Using the speed formula correctly 6.4.1 Natural numbers, integers, prime numbers, square numbers 6.4.2 Rational and irrational numbers, real numbers 6.4.3 Sum or product of two rational numbers is rational 6.4.4 Sum of a rational and an irrational number is irrational 6.4.5 Product of a non-zero rational and an irrational number is irrational 6.4.6 Symbols: =, ≠, ⩽, ⩾, <, > 6.5.1 Four operations and parentheses 6.5.2 Applies to integers, fractions, and decimals 6.6.1 Greatest Common Factor (GCF) 6.6.2 Least Common Multiple (LCM) 6.7.1 Understanding and solving problems using ratios 6.7.2 Proportions in real-world contexts 6.8.1 Convert between decimals, fractions, ratios, and percentages 6.8.2 Ordering different forms by magnitude 6.9.1 Calculating percentage increases and decreases 6.9.2 Applications like interest and profit 6.9.3 Includes reverse percentages 6.9.4 Includes both simple and compound interest 6.9.5 Includes percentiles 6.10.1 Scientific notation (Standard Form) 6.10.2 Basic exponent rules 6.10.3 Positive, negative, zero, and fractional exponents 6.11.1 Calculation and simplification of square root and cube root expressions 6.11.2 Simplify radical expressions Geometrical Measurement 7.1.1 Understand and convert between metric units (mm, cm, m, km) 7.1.2 Understand and convert between area and volume units (mm², cm², m², ha, km², mm³, cm³, ml, l, m³) 7.2.1 Calculate perimeter and area of rectangles and triangles 7.2.2 Calculate area of compound shapes derived from rectangles and triangles 7.2.3 Calculate area of trapezoids and parallelograms 7.3.1 Calculate circumference and area of circles 7.3.2 Calculate arc length and area of a sector (sector angles in degrees only) 7.4.1 Calculate surface area and volume of prisms and pyramids (including cuboids, cylinders, and cones) 7.4.2 Calculate surface area and volume of spheres 7.5.1 Calculate areas and volumes of compound shapes 7.6.1 Describe objects using geometric shapes and their properties 7.7.1 Identify the shapes of 2D cross sections of 3D objects 7.7.2 Identify 3D objects generated by rotations of 2D shapes 7.8.1 Apply concepts of density based on area and volume in modeling situations 7.9.1 Apply geometric methods to solve design problems Algebra 8.1.1 Writing, showing, and interpreting inequalities on the real number line 8.1.2 Create and solve linear inequalities 8.2.1 Create expressions and solve linear equations, including fractional expressions 8.2.2 Explain algebraic steps of a solution 8.2.3 Interpret solutions in a given context 8.3.1 Rules of exponents, including negative and fractional exponents 8.3.2 Basic exponent calculations 8.4.1 Rearrange and evaluate formulae, including algebraic manipulation to prove identities 8.4.2 Make a variable the subject of an equation 8.5.1 Create and solve systems of linear equations algebraically and graphically 8.6.1 Identify terms, factors, and coefficients 8.7.1 Expansion of parentheses, including the square of a binomial 8.7.2 Simplify expressions 8.8.1 Simplification using factorization 8.8.2 Addition or subtraction of fractions with linear denominators 8.8.3 Multiplication or division and simplification of two fractions 8.9.1 Create and solve quadratic equations by inspection, factorization, quadratic formula, and completing 8.9.2 Write quadratic expressions in the form (x-a)^2 + b and state the minimum value 8.10.1 Solve simple rational and radical equations and identify extraneous solutions 8.11.1 Continuation of sequences of numbers or patterns 8.11.2 Recognize patterns and generalize to algebraic statements 8.11.3 Determine the nth term of a sequence 8.11.4 Derive and use the formula for the sum of a finite geometric series 8.12.1 Express direct and inverse variation in algebraic terms 8.12.2 Use variation formulas to find unknown quantities 8.13.1 Factorization using common factors, difference of squares, trinomials, and four-term expressions Probability 9.1.1 Understand probability P(A) as a fraction, decimal, or percentage 9.1.2 Interpret probability values and their significance 9.1.3 Calculate probability using the rule P(A) = 1 – P(A') 9.2.1 Use relative frequency as an estimate of probability 9.3.1 Calculate the expected number of occurrences in probability experiments 9.4.1 Apply the addition rule P(A or B) = P(A) + P(B) – P(A and B) 9.4.2 Apply the multiplication rule P(A and B) = P(A) × P(B) 9.5.1 Understand and determine if two events are independent 9.6.1 Use possibility diagrams to list all outcomes 9.7.1 Construct and interpret tree diagrams for successive selections with or without replacement Coordinate Geometry 10.1.1 Plot points and read coordinates in the Cartesian plane 10.2.1 Calculate the distance between two points using the distance formula 10.3.1 Find the midpoint of a line segment 10.4.1 Find the point on a directed line segment that partitions the segment in a given ratio 10.5.1 Calculate the slope (gradient) of a line segment 10.6.1 Interpret and obtain the equation of a straight line in the form y = mx + b 10.6.2 Interpret and obtain the equation of a straight line in the form ax + by = d (where a, b, and d are 10.6.3 Find the equation of a straight line given two points 10.7.1 Understand and find the slope of parallel lines 10.7.2 Find the equation of a line parallel to a given line that passes through a given point 10.8.1 Understand the relationship between slopes of perpendicular lines 10.8.2 Find the equation of a line perpendicular to a given line that passes through a given point The perpendicular bisector of a chord passes through the center Topic 2/3 Revision Notes Flashcards Past Paper Analysis Questions Videos Your Flashcards are Ready! 15 Flashcards in this deck. or How would you like to practise? Choose Difficulty Level. Choose Easy, Medium or Hard to match questions to your skill level. Choose Learning Method. Choose Easy, Medium or Hard to match questions to your skill level. 3 Still Learning I know 12 Previous Next The Perpendicular Bisector of a Chord Passes Through the Center Introduction Understanding the properties of circles is fundamental in geometry, particularly for students preparing for the Cambridge IGCSE Mathematics - US - 0444 - Advanced syllabus. One such pivotal concept is that the perpendicular bisector of a chord always passes through the center of the circle. This principle not only reinforces the intrinsic symmetry of circles but also serves as a foundational tool in solving various geometric problems and proofs. Key Concepts Defining the Perpendicular Bisector A perpendicular bisector of a line segment is a line that divides the segment into two equal parts at a right angle (90 degrees). When applied to a chord of a circle, it reveals significant insights about the circle's geometry. Understanding Chords in a Circle A chord is a line segment with both endpoints lying on the circumference of a circle. Unlike the diameter, which passes through the center, a chord can be of any length shorter than the diameter. The properties of chords are essential in understanding the symmetry and other geometric features of circles. The Perpendicular Bisector of a Chord When a perpendicular bisector is drawn to a chord, it not only divides the chord into two equal parts but also extends to pass through the center of the circle. This means that the perpendicular bisector is a radial line emanating from the center, ensuring that the bisected chord is equidistant from the center at all points. Proof: The Perpendicular Bisector Passes Through the Center To understand why the perpendicular bisector of a chord passes through the center, consider the following proof: Let’s assume we have a circle with center C and a chord AB. Draw the perpendicular bisector CD of the chord AB, where D is the midpoint of AB. Since CD is perpendicular to AB at D, triangles ADC and BDC are congruent by the Side-Angle-Side (SAS) criterion: AD = BD (as D is the midpoint) CD = CD (common side) ∠ADC = ∠BDC = 90° By congruence, AC = BC, indicating that C is equidistant from A and B, reinforcing that C lies on the perpendicular bisector. Since C is the center of the circle and lies on the perpendicular bisector, the bisector must pass through the center. Applications of the Perpendicular Bisector Understanding that the perpendicular bisector of a chord passes through the center can be applied in various geometric constructions and proofs, such as: Constructing perpendicular diameters. Determining the center of a circle given a chord. Solving problems related to tangents and secants intersecting at specific points. Equidistant Property of the Perpendicular Bisector Every point on the perpendicular bisector of a chord is equidistant from the endpoints of the chord. This property is instrumental in various geometric proofs and constructions, ensuring symmetry and balance within the circle. Connecting Perpendicular Bisectors and Diameters A diameter is a special type of chord that passes through the center of the circle. Since the perpendicular bisector of any chord passes through the center, it follows that for a diameter, the perpendicular bisector coincides with itself, affirming that the diameter is its own perpendicular bisector. Graphical Representation Consider the following figure illustrating the concept: $$ \begin{align} & \text{Let } C \text{ be the center of the circle.} \ & \text{Let } AB \text{ be a chord with midpoint } D. \ & \text{Draw the perpendicular bisector } CD \text{ intersecting } AB \text{ at } D. \ & \text{Then, } CD \text{ passes through } C. \end{align} $$ Real-World Examples The principle that the perpendicular bisector of a chord passes through the center can be observed in various real-world scenarios, such as: Designing circular tracks where equal divisions are necessary for fairness. Urban planning where circular layouts require symmetry for infrastructure development. Engineering applications involving rotational symmetry and balance. Common Misconceptions A common misunderstanding is believing that any perpendicular line to a chord will pass through the center. However, it is specifically the perpendicular bisector of the chord that guarantees passage through the center. Merely being perpendicular without bisecting does not ensure this property. Mathematical Formulation Mathematically, if a chord AB has a midpoint D, then the perpendicular bisector CD satisfies the following conditions: AD = DB ∠ADC = ∠BDC = 90° C lies on CD, the perpendicular bisector. Thus, confirming that CD passes through the center C. Using Coordinates to Prove the Concept Consider a circle centered at the origin (0,0) with radius r. Let's take a chord AB with endpoints at A(x₁,y₁) and B(x₂,y₂). The midpoint D of the chord is: $$ D\left(\frac{x₁ + x₂}{2}, \frac{y₁ + y₂}{2}\right) $$ The slope of chord AB is: $$ m_{AB} = \frac{y₂ - y₁}{x₂ - x₁} $$ The slope of the perpendicular bisector is the negative reciprocal: $$ m_{CD} = -\frac{1}{m_{AB}} = -\frac{x₂ - x₁}{y₂ - y₁} $$ The equation of the perpendicular bisector passing through D and C(0,0) can be derived, showing that it indeed passes through the center. Conclusion of Key Concepts The perpendicular bisector of a chord is a fundamental concept in the study of circles, providing essential insights into the symmetrical properties of circles and serving as a critical tool in geometric constructions and proofs. Mastery of this concept is crucial for excelling in the Cambridge IGCSE Mathematics syllabus. Advanced Concepts Theorem: Perpendicular Bisector and Central Symmetry Building upon the basic understanding, the theorem stating that the perpendicular bisector of any chord passes through the center reinforces the intrinsic symmetry of the circle. This theorem is vital in advanced geometric proofs and constructions, especially when dealing with multiple chords and their relationships. Mathematical Derivation Using Coordinate Geometry To delve deeper, let's consider the circle with center at C(h,k) and a chord AB with endpoints A(x₁,y₁) and B(x₂,y₂). 1. Midpoint of AB: $$ D\left(\frac{x₁ + x₂}{2}, \frac{y₁ + y₂}{2}\right) $$ 2. Slope of AB: $$ m_{AB} = \frac{y₂ - y₁}{x₂ - x₁} $$ 3. Slope of Perpendicular Bisector: $$ m_{CD} = -\frac{1}{m_{AB}} = -\frac{x₂ - x₁}{y₂ - y₁} $$ 4. Equation of Perpendicular Bisector: Using point-slope form: $$ y - y_D = m_{CD}(x - x_D) $$ 5. Substituting Center Coordinates: Since the perpendicular bisector passes through the center C(h,k), substituting h and k verifies the theorem. Proving the Theorem Using Vector Geometry Alternatively, vector geometry offers a robust framework to prove the theorem. Let vectors represent points in the plane, and use vector operations to demonstrate that the perpendicular bisector must intersect the center. Represent points A and B as position vectors a and b, respectively. The midpoint D is given by: $$ \mathbf{d} = \frac{\mathbf{a} + \mathbf{b}}{2} $$ The direction vector of chord AB is b - a. The direction vector of the perpendicular bisector is orthogonal to b - a, implying: $$ (\mathbf{b} - \mathbf{a}) \cdot (\mathbf{r} - \mathbf{d}) = 0 $$ where r is any point on the perpendicular bisector. Since C is the center, its position vector satisfies the above equation, confirming that C lies on the perpendicular bisector. Exploring the Converse: Lines Through the Center While the perpendicular bisector of a chord necessarily passes through the center, the converse also holds true: any line passing through the center of a circle is the perpendicular bisector of every chord it intersects. This reciprocity underscores the central role of the center in defining the properties of chords and their bisectors. Application in Circle Theorems The concept extends to several circle theorems, such as: Alternate Segment Theorem: Relies on perpendicular bisectors to define angles related to tangents. Cyclic Quadrilaterals: Use perpendicular bisectors to establish the cyclic nature of quadrilaterals inscribed in circles. Intersecting Chords Theorem: Involves bisectors to determine relationships between intersecting chords. Intersection of Multiple Perpendicular Bisectors In geometry, if multiple chords are given, their respective perpendicular bisectors will all intersect at the unique center of the circle. This property is pivotal in geometric constructions and in determining circle centers when only chords are provided. Implications in Tangent Properties The perpendicular bisector property is intricately linked to tangent properties. Specifically, the radius drawn to the point of tangency is perpendicular to the tangent line. Understanding the perpendicular bisector aids in establishing these perpendicularities, crucial for solving tangent-related problems. Algorithm for Finding the Center Using Perpendicular Bisectors To find the center of a circle given two chords: Identify two chords in the circle. Construct the perpendicular bisector for each chord. The intersection point of these bisectors is the center of the circle. This method is widely used in geometric constructions and proofs, ensuring precision in locating the circle's center. Advanced Problem-Solving Examples Let's explore some complex problems that utilize the perpendicular bisector property: Example 1: Determining the Center from Two Chords Given two chords AB and CD in a circle, with midpoints M and N, and perpendicular bisectors OM and PN intersecting at point O, prove that O is the center of the circle. Using the perpendicular bisector property, since both OM and PN pass through the center by definition, their intersection point O must be the center. Example 2: Proving Tangent Perpendicularity Prove that the radius drawn to the point of tangency is perpendicular to the tangent line. Given a tangent line T at point P on the circle, and radius CP: Since T is tangent at P, any chord through P will have its perpendicular bisector passing through C. Therefore, the angle between CP and T must be 90°, establishing perpendicularity. Interdisciplinary Connections The principle that the perpendicular bisector of a chord passes through the center finds applications beyond pure mathematics: Engineering: Designing circular gears and ensuring symmetry. Physics: Understanding rotational dynamics and circular motion. Computer Graphics: Creating algorithms for rendering circular shapes with precision. Exploring Through Technology: Dynamic Geometry Software Tools like GeoGebra allow students to visualize and interact with the perpendicular bisector of chords dynamically. By manipulating chords and observing the formation of bisectors converging at the center, learners can gain an intuitive understanding of the concept. Challenging Proofs Involving Perpendicular Bisectors Advanced geometric proofs often require leveraging the perpendicular bisector property to establish relationships between various elements within a circle, such as angles, arcs, and other chords. Mastery of this property accelerates the ability to navigate complex proofs with confidence. Integration with Trigonometry Incorporating trigonometric principles with the perpendicular bisector property can solve intricate problems involving circular segments, sector areas, and arc lengths, enhancing problem-solving versatility. Advanced Theorems and Concepts Several advanced theorems rest upon the foundation of the perpendicular bisector property, including: Circumcircle Theorem: Determines the circumcircle of a triangle by finding the intersection of perpendicular bisectors of its sides. Power of a Point Theorem: Explores relationships involving tangents and intersecting chords, utilizing perpendicular bisectors. Simson Line Theorem: Investigates the collinearity of points related to a given point and its projections onto the sides of a triangle, employing perpendicular bisector properties. Case Study: Application in Architectural Design In architectural design, circular elements require precise calculations to ensure structural integrity and aesthetic symmetry. The perpendicular bisector property aids architects in determining precise center points, facilitating balanced and harmonious designs. Exploring Non-Euclidean Geometry While the perpendicular bisector property holds in Euclidean geometry, exploring its implications in non-Euclidean geometries can provide deeper insights into the versatility and limitations of geometric principles across different frameworks. Mathematical Software and Simulations Utilizing mathematical software for simulations can offer enhanced visualization and experimentation with the perpendicular bisector of chords, fostering a deeper comprehension through interactive learning. Enhancing Problem-Solving Skills By engaging with advanced problems that incorporate the perpendicular bisector property, students can sharpen their logical reasoning and analytical skills, preparing them for higher-level mathematics and related disciplines. Research and Further Reading For those interested in exploring further, academic papers and geometry textbooks delve into the intricacies of circle theorems and their applications, offering comprehensive perspectives and advanced methodologies. Conclusion of Advanced Concepts The perpendicular bisector of a chord serves as a cornerstone in the study of circles, bridging fundamental geometric principles with advanced applications and interdisciplinary connections. Mastery of this concept equips students with the tools necessary for sophisticated problem-solving and a deeper appreciation of geometric symmetry. Comparison Table \| \| \| \| --- \| Aspect \| Perpendicular Bisector of a Chord \| Diameter \| \| Definition \| A line that cuts a chord into two equal parts at a right angle. \| A chord that passes through the center of the circle. \| \| Passes Through Center \| Yes, by definition. \| Always, as it is a diameter. \| \| Perpendicularity \| Perpendicular to the chord. \| Perpendicular bisector of any chord it intersects. \| \| Length \| Varies, depends on the chord. \| Maximum length, equal to twice the radius. \| \| Uniqueness \| Each chord has a unique perpendicular bisector. \| There are infinitely many diameters through the center. \| \| Uses in Proofs \| Establishing center points, symmetry, and congruence. \| Determining maximum chords, symmetry axes. \| Summary and Key Takeaways The perpendicular bisector of a chord in a circle consistently passes through the circle's center. This property is fundamental in geometric proofs, constructions, and various applications across disciplines. Advanced studies integrate this concept with coordinate and vector geometry, enhancing problem-solving capabilities. Understanding this principle reinforces the inherent symmetry and balance within circular geometry. Coming Soon! Examiner Tip Tips Mnemonic: "Perpendicular Bisector Points to the Center" can help you remember that drawing a perpendicular bisector of a chord will lead you directly to the circle's center. Actionable Advice: When solving problems, always start by accurately finding the midpoint of the chord before drawing the perpendicular bisector. Utilizing coordinate geometry can simplify the process and enhance precision. Exam Success: Practice constructing perpendicular bisectors using both geometric tools and algebraic methods to build versatility in problem-solving for your exams. Did You Know Did You Know The concept of perpendicular bisectors is not only pivotal in geometry but also plays a significant role in astronomy. For instance, astronomers use the perpendicular bisector method to determine the centers of celestial bodies by analyzing the chords formed by observed stellar positions. Additionally, in engineering, this principle ensures the balanced design of circular structures, such as domes and rotors, by maintaining symmetry and structural integrity. Common Mistakes Common Mistakes Mistake 1: Assuming any perpendicular line to a chord passes through the center. Incorrect Approach: Drawing a perpendicular line without ensuring it bisects the chord. Correct Approach: Always verify that the perpendicular line divides the chord into two equal parts. Mistake 2: Misidentifying the midpoint of the chord, leading to an inaccurate bisector. Incorrect Approach: Estimating the midpoint without precise measurement. Correct Approach: Use geometric tools or coordinate calculations to accurately find the midpoint before drawing the bisector. FAQ What is a perpendicular bisector in the context of a circle? A perpendicular bisector in a circle is a line that cuts a chord into two equal parts at a right angle, ensuring that it passes through the circle's center. How does the perpendicular bisector help in finding the center of a circle? By constructing the perpendicular bisector of a chord, the line will inevitably pass through the circle's center, allowing you to determine the exact location of the center. Is the perpendicular bisector of a chord unique? Yes, each chord has one unique perpendicular bisector that passes through the circle's center. Can the perpendicular bisector of a chord be a diameter? Yes, when the chord is a diameter itself, the perpendicular bisector coincides with the diameter, reinforcing that the diameter is its own perpendicular bisector. What are common applications of the perpendicular bisector property? This property is used in geometric constructions, proofs, engineering designs, and various real-world applications requiring symmetry and precise center determination. Trigonometry 1.1.1 Apply the Pythagorean Theorem to find missing sides 1.1.2 Use trigonometric ratios (sine, cosine, tangent) to solve right-angled triangles 1.2.1 Know exact values for sine, cosine, and tangent of 0°, 30°, 45°, 60°, 90° 1.3.1 Solve problems involving bearings 1.3.2 Understand and use the concepts of angle of elevation and angle of depression 1.4.1 Extend sine and cosine values to angles between 0° and 360° 1.4.2 Use the relationship between sine and cosine of complementary angles 1.5.1 Graph and understand the properties of sine, cosine, and tangent functions 1.6.1 Apply the Sine Rule for solving triangles (ASA, SSA cases) 1.7.1 Apply the Cosine Rule for solving triangles (SAS, SSS cases) 1.8.1 Use the formula for the area of a triangle (1/2 ab sinC) Transformations and Vectors 2.1.1 Describe and perform stretches 2.1.2 Describe and perform translations using column vectors 2.1.3 Describe and perform reflections in a given axis or line 2.1.4 Describe and perform rotations about a point 2.1.5 Describe and perform enlargements (dilations) with scale factors 2.2.1 Find the inverse of a given transformation 2.3.1 Find the single transformation equivalent to a sequence of transformations 2.4.1 Understand directed line segment notation and component form of vectors 2.4.2 Use appropriate symbols for vectors and their magnitudes 2.5.1 Find the components of a vector by subtracting coordinates of an initial point from a terminal point 2.5.2 Use position vectors 2.5.3 Calculate the magnitude of a vector 2.5.4 Understand vector subtraction as v - w = v + (-w) 2.5.5 Multiply a vector by a scalar 2.5.6 Add and subtract vectors algebraically (component form) and geometrically (parallelogram rule) Statistics 3.1.1 Draw a straight line of best fit by eye through the mean on a scatter diagram 3.2.1 Read and interpret data from graphs and tables 3.3.1 Understand discrete and continuous data 3.4.1 Calculate mean, mode, median, and range from discrete data 3.4.2 Calculate mean, modal class, median, and range from grouped and continuous data 3.5.1 Construct and interpret histograms with frequency density on the vertical axis 3.5.2 Interpret histograms with unequal class intervals 3.6.1 Create and interpret cumulative frequency tables and curves 3.6.2 Determine median, quartiles, percentiles, and interquartile range 3.7.1 Use statistics (median, mean, interquartile range) to compare different data sets 3.8.1 Understand and describe correlation (positive, negative, or zero) using scatter diagrams 3.9.1 Construct and interpret compound bar charts, dot plots, line graphs, pie charts, scatter diagrams Geometry 4.1.1 Angle properties of triangles, quadrilaterals, and polygons 4.1.2 Interior and exterior angles of a polygon 4.2.1 Make formal geometric constructions using a compass and straight edge 4.2.2 Copy and bisect a segment or an angle 4.2.3 Construct perpendicular lines and perpendicular bisectors 4.2.4 Construct equilateral triangles, squares, and hexagons inscribed in circles 4.2.5 Construct inscribed and circumscribed circles of a triangle 4.2.6 Construct tangent lines from a point outside a circle 4.3.1 Tangent perpendicular to radius at the point of contact 4.3.2 Tangents from a point 4.3.3 Angle in a semicircle 4.3.4 Angles at the center and at the circumference on the same arc 4.3.5 Cyclic quadrilateral properties 4.3.6 Equal chords are equidistant from the center 4.3.7 The perpendicular bisector of a chord passes through the center 4.3.8 Tangents from an external point are equal in length 4.4.1 Calculate lengths of similar figures using scale factors 4.4.2 Use area and volume scale factors for similar figures and solids 4.5.1 Recognize and use congruence in solving geometric problems 4.6.1 Know definitions of acute, obtuse, right angle, reflex, equilateral, isosceles, congruent, similar, 4.6.2 Know definitions of pentagon, hexagon, octagon, rectangle, square, kite, rhombus, parallelogram, tra 4.7.1 Understand definitions of angle, circle, perpendicular line, parallel line, and line segment 4.8.1 Line and rotational symmetry in 2D and 3D 4.8.2 Recognize symmetry properties of prisms and pyramids 4.9.1 Angles around a point 4.9.2 Angles on a straight line and intersecting straight lines 4.9.3 Vertically opposite angles 4.9.4 Alternate and corresponding angles on parallel lines Functions 5.1.1 Use function notation 5.1.2 Understand domain and range 5.1.3 Use mapping diagrams 5.2.1 Graph an equation in two variables as the set of all its solutions 5.2.2 Construct tables of values and graphs for functions of the form ax^n where n = -2, -1, 0, 1, 2, 3 5.2.3 Solve equations approximately using graphical methods 5.3.1 Write a function that describes a relationship between two quantities 5.4.1 Compare properties of two functions represented in different ways (algebraically, graphically, numer 5.5.1 Recognize linear, quadratic, cubic, reciprocal, exponential, and trigonometric functions from their 5.5.2 Interpret key features of function graphs (intercepts, increasing/decreasing behavior, maxima/minima 5.6.1 Relate the domain of a function to its graph and quantitative relationships 5.7.1 Calculate and interpret the average rate of change of a function over a specified interval 5.7.2 Estimate rate of change using a graph 5.8.1 Understand and compare the behavior of linear, quadratic, and exponential functions 5.8.2 Observe that exponential growth eventually exceeds polynomial growth 5.8.3 Use properties of exponents to interpret expressions for exponential functions 5.9.1 Construct linear and exponential functions given graphs, descriptions, or two input-output pairs 5.10.1 Simplify expressions for composite functions such as f(g(x)) 5.11.1 Find and interpret inverse functions 5.12.1 Describe changes to the graph of y = f(x) when transformed as y = f(x) + k, y = kf(x), and y = f(x + 5.13.1 Graph the solutions to a linear inequality in two variables as a half-plane 5.13.2 Graph the solution set of a system of inequalities as the intersection of corresponding half-planes Number 6.1.1 Converting between units (e.g., currency, time, length) 6.1.2 Understanding units in problems 6.1.3 Choosing appropriate levels of accuracy 6.2.1 Units of time: seconds, minutes, hours, days, months, years 6.2.2 24-hour and 12-hour clock formats 6.3.1 Understanding and solving problems related to motion 6.3.2 Using the speed formula correctly 6.4.1 Natural numbers, integers, prime numbers, square numbers 6.4.2 Rational and irrational numbers, real numbers 6.4.3 Sum or product of two rational numbers is rational 6.4.4 Sum of a rational and an irrational number is irrational 6.4.5 Product of a non-zero rational and an irrational number is irrational 6.4.6 Symbols: =, ≠, ⩽, ⩾, <, > 6.5.1 Four operations and parentheses 6.5.2 Applies to integers, fractions, and decimals 6.6.1 Greatest Common Factor (GCF) 6.6.2 Least Common Multiple (LCM) 6.7.1 Understanding and solving problems using ratios 6.7.2 Proportions in real-world contexts 6.8.1 Convert between decimals, fractions, ratios, and percentages 6.8.2 Ordering different forms by magnitude 6.9.1 Calculating percentage increases and decreases 6.9.2 Applications like interest and profit 6.9.3 Includes reverse percentages 6.9.4 Includes both simple and compound interest 6.9.5 Includes percentiles 6.10.1 Scientific notation (Standard Form) 6.10.2 Basic exponent rules 6.10.3 Positive, negative, zero, and fractional exponents 6.11.1 Calculation and simplification of square root and cube root expressions 6.11.2 Simplify radical expressions Geometrical Measurement 7.1.1 Understand and convert between metric units (mm, cm, m, km) 7.1.2 Understand and convert between area and volume units (mm², cm², m², ha, km², mm³, cm³, ml, l, m³) 7.2.1 Calculate perimeter and area of rectangles and triangles 7.2.2 Calculate area of compound shapes derived from rectangles and triangles 7.2.3 Calculate area of trapezoids and parallelograms 7.3.1 Calculate circumference and area of circles 7.3.2 Calculate arc length and area of a sector (sector angles in degrees only) 7.4.1 Calculate surface area and volume of prisms and pyramids (including cuboids, cylinders, and cones) 7.4.2 Calculate surface area and volume of spheres 7.5.1 Calculate areas and volumes of compound shapes 7.6.1 Describe objects using geometric shapes and their properties 7.7.1 Identify the shapes of 2D cross sections of 3D objects 7.7.2 Identify 3D objects generated by rotations of 2D shapes 7.8.1 Apply concepts of density based on area and volume in modeling situations 7.9.1 Apply geometric methods to solve design problems Algebra 8.1.1 Writing, showing, and interpreting inequalities on the real number line 8.1.2 Create and solve linear inequalities 8.2.1 Create expressions and solve linear equations, including fractional expressions 8.2.2 Explain algebraic steps of a solution 8.2.3 Interpret solutions in a given context 8.3.1 Rules of exponents, including negative and fractional exponents 8.3.2 Basic exponent calculations 8.4.1 Rearrange and evaluate formulae, including algebraic manipulation to prove identities 8.4.2 Make a variable the subject of an equation 8.5.1 Create and solve systems of linear equations algebraically and graphically 8.6.1 Identify terms, factors, and coefficients 8.7.1 Expansion of parentheses, including the square of a binomial 8.7.2 Simplify expressions 8.8.1 Simplification using factorization 8.8.2 Addition or subtraction of fractions with linear denominators 8.8.3 Multiplication or division and simplification of two fractions 8.9.1 Create and solve quadratic equations by inspection, factorization, quadratic formula, and completing 8.9.2 Write quadratic expressions in the form (x-a)^2 + b and state the minimum value 8.10.1 Solve simple rational and radical equations and identify extraneous solutions 8.11.1 Continuation of sequences of numbers or patterns 8.11.2 Recognize patterns and generalize to algebraic statements 8.11.3 Determine the nth term of a sequence 8.11.4 Derive and use the formula for the sum of a finite geometric series 8.12.1 Express direct and inverse variation in algebraic terms 8.12.2 Use variation formulas to find unknown quantities 8.13.1 Factorization using common factors, difference of squares, trinomials, and four-term expressions Probability 9.1.1 Understand probability P(A) as a fraction, decimal, or percentage 9.1.2 Interpret probability values and their significance 9.1.3 Calculate probability using the rule P(A) = 1 – P(A') 9.2.1 Use relative frequency as an estimate of probability 9.3.1 Calculate the expected number of occurrences in probability experiments 9.4.1 Apply the addition rule P(A or B) = P(A) + P(B) – P(A and B) 9.4.2 Apply the multiplication rule P(A and B) = P(A) × P(B) 9.5.1 Understand and determine if two events are independent 9.6.1 Use possibility diagrams to list all outcomes 9.7.1 Construct and interpret tree diagrams for successive selections with or without replacement Coordinate Geometry 10.1.1 Plot points and read coordinates in the Cartesian plane 10.2.1 Calculate the distance between two points using the distance formula 10.3.1 Find the midpoint of a line segment 10.4.1 Find the point on a directed line segment that partitions the segment in a given ratio 10.5.1 Calculate the slope (gradient) of a line segment 10.6.1 Interpret and obtain the equation of a straight line in the form y = mx + b 10.6.2 Interpret and obtain the equation of a straight line in the form ax + by = d (where a, b, and d are 10.6.3 Find the equation of a straight line given two points 10.7.1 Understand and find the slope of parallel lines 10.7.2 Find the equation of a line parallel to a given line that passes through a given point 10.8.1 Understand the relationship between slopes of perpendicular lines 10.8.2 Find the equation of a line perpendicular to a given line that passes through a given point Get PDF How would you like 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5837	https://brainly.com/question/63135683	[FREE] Factor the polynomial. $27 a^3-64 b^3$ A. $(3 a-4 b)(9 a^2+16 b^2)$ B. $(3 a-4 b)(9 a^2+12 a b+16 - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +81,9k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +28,1k Ace exams faster, with practice that adapts to you Practice Worksheets +7,5k Guided help for every grade, topic or textbook Complete See more / Mathematics Factor the polynomial. 27 a 3−64 b 3 A. (3 a−4 b)(9 a 2+16 b 2) B. (3 a−4 b)(9 a 2+12 ab+16 b 2) C. (3 a+4 b 2)(9 a 2−12 ab+16 b 2) D. (27 a−4 b)(a 2+12 ab+16 b 2) 1 See answer Explain with Learning Companion NEW Asked by jovi123 • 07/20/2025 0:01 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Something went wrong! Please refresh to get your Expert Answer. Community Answer This answer helped 1183294037 people 1183M 0.0 0 Upload your school material for a more relevant answer Recognize the polynomial as a difference of cubes: 27 a 3−64 b 3=(3 a)3−(4 b)3. Apply the difference of cubes factorization formula: x 3−y 3=(x−y)(x 2+x y+y 2). Substitute x=3 a and y=4 b into the formula and simplify. The factored form of the polynomial is: (3 a−4 b)(9 a 2+12 ab+16 b 2). Explanation Recognizing the Problem Type We are asked to factor the polynomial 27 a 3−64 b 3. This expression is a difference of cubes, which can be factored using a specific formula. Applying the Difference of Cubes Formula The difference of cubes factorization formula is:x 3−y 3=(x−y)(x 2+x y+y 2) In our case, we have 27 a 3−64 b 3, which can be written as (3 a)3−(4 b)3. So, x=3 a and y=4 b. Substituting and Simplifying Substituting x=3 a and y=4 b into the formula, we get: (3 a)3−(4 b)3=(3 a−4 b)((3 a)2+(3 a)(4 b)+(4 b)2) Now, we simplify the expression: (3 a−4 b)(9 a 2+12 ab+16 b 2) Final Answer Comparing our result with the given options, we find that the correct factorization is: (3 a-4 b)\left(9 a^2+12 a b+16 b^2\right) Examples Factoring polynomials like 27 a 3−64 b 3 is useful in simplifying complex algebraic expressions and solving equations. For example, in engineering, you might encounter such expressions when analyzing the volume of a structure or the behavior of a system. By factoring, you can break down the problem into simpler components, making it easier to understand and solve. This skill is also fundamental in calculus and other advanced mathematical fields, where simplification is key to finding solutions. Answered by GinnyAnswer •8M answers•1.2B people helped Thanks 0 0.0 (0 votes) jovi123 has a question! Can you help? Add your answer Advertisement ### Free Mathematics solutions and answers Community Answer (iii) [tex][ \left[ \left( \frac{27 a^3 + 54 a^2 b + 36 a b^2 + 8 b^3}{9 a^2 - 4 b^2} \right) \div \left( \frac{9 a^2 + 12 a b + 4 b^2}{27 a^3 - 8 b^3} \right) - \left( 9 a^2 + 6 a b + 4 b^2 \right) \right] ][/tex] Community Answer [Simplify the expression: [ \frac{9 a^2 + 4 b^2}{9 a^2 - 4 b^2} + \frac{3 a}{2 b - 3 a} + \frac{2 b}{3 a + 2 b} ]]( "Simplify the expression: [ \frac{9 a^2 + 4 b^2}{9 a^2 - 4 b^2} + \frac{3 a}{2 b - 3 a} + \frac{2 b}{3 a + 2 b} ]") Community Answer What is the product? [tex][ \left(-6 a^3 b+2 a b^2\right)\left(5 a^2-2 a b^2-b\right) ][/tex] A. tex[/tex] B. tex[/tex] C. tex[/tex] D. tex[/tex] Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? New questions in Mathematics Yesterday, Selma read 75 pages of her book. If she reads at a pace of 2 pages per minute today, which table shows only viable solutions for the total number of pages she has read, y, after x minutes have elapsed? A. \begin{table} \captionsetup{labelformat=empty} \caption{Selma's Reading} \| Minutes of Reading (x) \| Total Pages Read (y) \| :---: \| \| 2 \| 79 \| \| 14 \| 101 \| \| 39 \| 153 \| \| 55 \| 185 \| \end{table} B. \| Minutes of Reading (x) \| Total Pages Read (y) \| :---: \| \| -16 \| 43 \| \| 6 \| 87 \| \| 27 \| 129 \| \| 52 \| 179 \| C. \begin{table} \captionsetup{labelformat=empty} \caption{Selma's Reading} \| Minutes of Reading (x) \| Total Pages Read ′(y) \| :---: \| \| 0 \| 0 \| \| 19 \| 113 \| \| 32 \| 139 \| \| 47 \| 169 \| \end{table} D. \begin{table} \captionsetup{labelformat=empty} \caption{Selma's Reading} \| Minutes of Reading (x) \| Total Pages Read (y) \| :---: \| \| 0 \| 75 \| \| 11 \| 97 \| \| 20.25 \| 115.5 \| \end{table} What values are needed to make each expression a perfect square trinomial? x 2+2 x+□x 2−20 x+◯x 2+5 x+□ Given two square matrices A and B of the same size such that (A−B)(A+B)=A 2−B 2. Which of the following statement is correct?A. B A=A B B. A B=−2 I n where I n is the identity matrix of size n C. None of the given optionsD. A B=O where O is the null The sea floor is 104 feet below sea level. Katy is 28 feet below sea level. She is moving downward at a rate of 4 feet per minute. Colin used the following calculation to determine how long it will take Katy to reach the sea floor. Is he correct? Explain. 104−28÷4=97 minutes Write h(x)=x 2−4 x−3 in vertex form and then identify the transformations of its graph. The function h written in vertex form is h(x)=(x−□)2+□ . To graph the function h, shift the graph of f(x)=x 2 right □ units and down □ units. 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5838	https://www.indeed.com/career-advice/career-development/how-to-calculate-total-cost	How To Calculate Total Cost (Plus Definition and Example) \| Indeed.com Skip to main content Home Company reviews Find salaries Sign in Sign in Employers / Post Job 1 new update Home Company reviews Find salaries Employers Create your resume Resume services Change country 🇺🇸 United States Help Start of main content Career Guide Finding a job Resumes & cover letters Resumes & cover letters articles Resume samples Cover letter samples Interviewing Pay & salary Career development Career development articles Starting a new job Career paths News Career development How To Calculate Total Cost (Plus Definition and Example) How To Calculate Total Cost (Plus Definition and Example) Written by Indeed Editorial Team Updated July 24, 2025 Knowing how much an activity or function costs in a business or organization can be an important part of managing expenses and earning consistent profits. This simple but powerful formula combines fixed costs with variable costs to provide a total dollar amount spent on a venture or a portion of it. Businesses and individuals can then use that information to interpret, analyze and maximize profits.In this article, we discuss what total cost is, describe ways you can use this calculation, provide steps for its calculation and include an example of total cost for a business. Related jobs on Indeed Cost accountant jobs Part-time jobs Full-time jobs Remote jobs View more jobs on Indeed What is total cost? Total cost is the combined sum of fixed and variable costs. Businesses usually calculate this figure per sales unit and then multiply it by the actual number of items produced. These items can range from tangible physical goods, such as cars or housewares, to less physical sales units, such as services or customer contacts.Here's the formula for total cost:Total cost = (average fixed cost per unit + average variable cost per unit) x number of units produced The total cost formula is flexible enough to account for a wide variety of product types, which can make it useful in most industries and at most levels. Business owners, managers andfinancial analystscan use this information to make decisions about prices, purchasing and other factors that affect revenue. They can also use the figure as a basis to further analyze parts of a business that might be more or less profitable.Related:Total Cost: Definition and Formula Uses for total cost in business Total cost, the sum of average fixed costs and variable costs in a business, can be useful for tracking profits. External stakeholders often want this information because it can be a factor in supply chain negotiations. Leaders also often use this metric to guide pricing decisions because it can help find a price point that consumers are willing to pay while still maximizing profit.Many businesses find that total cost is most useful when applied to relatively simple products or services. Sometimes the two components of total cost, fixed and variable costs, encompass multiple complex factors. This is especially true when variable costs change relatively frequently or to a large degree. For example, if some materials are much more expensive during winter than in the summer, total cost might be a less reliable metric. Knowing this formula can help business owners and leadersthink criticallyabout their decisions.Related:How To Calculate Average Total Cost How to calculate total cost If you've analyzed your product line and determined that total cost is the right formula for your purposes, you can follow these calculation steps: 1. Decide which product or product line to analyze It's useful to calculate total cost for multiple products or services you offer. Other times, you might need the total cost for one particular kind of sales unit. Knowing the specific information you want before gathering the details to place in the formula ensures you reach the most appropriate result for your purposes.Related:How To Calculate Total Revenue 2. Calculate your average fixed cost per item Fixed costs are the expenses for a product or service that doesn't change. Examples include overhead costs, like rent and utilities, salaried employee payroll and insurance. Determine the dollar amount of these kinds of expenses that contribute to each item you included in your total cost calculation.Related:Total Fixed Cost: What It Is and How To Find It 3. Calculate your average variable cost per item Variable costs are the changing expenses included in producing an item. For example, direct labor wages, materials and transportation charges can all change depending on market circumstances and are therefore variable costs. Since these costs change, average the variable cost per unit over a period. Be mindful of the period you choose, since variable costs can change day-to-day and month-to-month. For example, your utilities are probably different in the winter than in the summer. Averaging your variable costs during an intermediary season or over a longer period could offset this variability.Related:Variable Cost: Definition, Examples and Formulas 4. Measure the number of units produced You calculate and communicate total cost based on individual units, measure the number of units produced over a period. Be sure the period you use to count units is the same as the time for which you measured fixed and variable costs. If your production process is complex, consider returning to your original goals for calculating total cost to ensure that you're counting the appropriate specific items.Related:Management for Productivity: What It Is and Tips for Using It 5. Complete the formula After you've determined your fixed cost per item, variable cost per item and number of units produced, you can insert those numbers into the formula. You can use this information to determine whether your costs align with your expectations and if you can change them. For example, if the total cost of producing a product is higher than your expectation, you may change production processes to make it less expensive to produce.Related:Production Efficiency Formula: What It Is and Who Uses It Search jobs and companies hiring now Job title, keywords or company Location Search Example of how to calculate total cost Here's an example of using this process to calculate total cost:A tropical fish shop owner calculates the total cost for one variety of fish they carry. To isolate the fixed and variable costs of keeping these fish, they place them all in separate tanks. For six months, the shop owner tracks the overhead for those specific tanks and records them as fixed expenses. They also track variable expenses, such as the cost of powering those tanks, and average them for those six months. Finally, they tabulate the number of those fish they sold over the same period.Using this process, the tropical fish shop owner determines that the average fixed cost for keeping one of these fish over six months is $20. Theaverage variable costis $10 over the same period. Finally, they sold 300 of these fish during the same six months. When they insert these numbers into the total cost formula, they get this calculation: $9,000 = ($20 + $10) X 300 fish The information on this site is provided as a courtesy and for informational purposes only. Indeed is not a career or legal advisor and does not guarantee job interviews or offers Cost accountant Share: Related Articles How To Calculate Fixed Cost in 3 Steps (With Examples) How To Calculate Variable Cost Full Cost Accounting: Definition, How To Calculate and Examples Explore more articles How To Write an Effective Mission Statement (With Examples) Teamwork vs. Individual Work: Definitions and 8 Differences How To Become a Bartender With No Experience (Plus Tips) 11 Careers in Basketball (Plus Primary Duties and Salaries) Is Being Discharged from a Job the Same as Laid Off or Fired? 14 Mentoring Goals (And How To Set Them) Top 51 Nurse Practitioner Interview Questions (Plus Answers) 8 Steps to Take When Relocating for a Job 70+ Resume Objective Examples (With Tips and How-To Guide) What Is Followership? 14 Qualities of Good Followers Listing Receptionist Duties on Your Resume (With Examples) How To Ask for Feedback at Work (With Examples) Hiring Lab Career advice Browse jobs Browse companies Salaries Indeed Events Work at Indeed Countries About Help ESG at Indeed © 2025 Indeed Your Privacy Choices Accessibility at Indeed Privacy Center and Ad Choices Terms
5839	https://www.sciencedirect.com/topics/physics-and-astronomy/equipartition-theorem	Equipartition Theorem - an overview \| ScienceDirect Topics Skip to Main content Journals & Books Equipartition Theorem In subject area:Physics and Astronomy The equipartition theorem is defined as a principle stating that, for a harmonic potential, the potential energy can be expressed as U = (1/2)k_BT, where k_B is the Boltzmann constant and T is the temperature, allowing for the estimation of trapping stiffness using time averages in ergodic systems. AI generated definition based on:Encyclopedia of Condensed Matter Physics (Second Edition), 2024 How useful is this definition? Press Enter to select rating, 1 out of 3 stars Press Enter to select rating, 2 out of 3 stars Press Enter to select rating, 3 out of 3 stars About this page Add to MendeleySet alert Also in subject area s: Earth and Planetary Sciences Mathematics Discover other topics 1. On this page On this page Definition Chapters and Articles Related Terms Recommended Publications Featured Authors On this page Definition Chapters and Articles Related Terms Recommended Publications Featured Authors Chapters and Articles You might find these chapters and articles relevant to this topic. Review article Statistical mechanics of systems with negative temperature 2021, Physics ReportsMarco Baldovin, ... Angelo Vulpiani 3.2.1 Equipartition theorem A useful result of statistical mechanics is the so-called “Equipartition theorem”, which states that, given a Hamiltonian system H(Q,P), the following relation holds: (48)x i∂H∂x i=k B T, where x i is a certain canonical coordinate and 〈⋅〉 denotes a microcanonical average. If H(Q,P) depends quadratically on the momenta, Eq.(48) implies that energy is equally distributed, on average, among the different kinetic degrees of freedom, hence the name of the theorem. In deriving such result (see e.g.Ref.) one makes use of Gibbs’ definition of entropy; consequently, Eq.(48) is exactly verified only if one interprets T as T G. See also Ref. on this point. View article Read full article URL: Journal2021, Physics ReportsMarco Baldovin, ... Angelo Vulpiani Chapter Optical tweezers: Theory and practice 2024, Encyclopedia of Condensed Matter Physics (Second Edition)Giuseppe Pesce Equipartition theorem The potential associated with an optical trap is to a very good first approximation harmonic, i.e., (16)U x=1 2 κ x x−x eq 2, where κ x is the trap stiffness and x eq is the equilibrium position. Once this hypothesis has been verified, it is possible to use the equipartition theorem, which states that (17)U x=1 2 κ x x−x eq 2=1 2 k B T, where instead of the ensemble average one can also employ a time average because of the ergodicity of the system. As in the determination of the potential described above, the experimental data points need to sample the probability distribution so that they do not need to be acquired at high frequency rate or with a fixed time step as long as a sufficient number of independent points is acquired. It is the easiest and simplest method nevertheless the estimation of trapping stiffness is also prone to systematic errors due to the presence of uncorrelated noise (Florin et al., 1998) and it is limited to harmonic potentials. View chapterExplore book Read full chapter URL: Reference work 2024, Encyclopedia of Condensed Matter Physics (Second Edition)Giuseppe Pesce Chapter Ideal Monatomic Gases 2011, Introductory Statistical ThermodynamicsNils Dalarsson, ... Leonardo Golubović 11.4.6 Equipartition Theorem The probability for a particle to have its velocity vector within the velocity space element d v x d v y d v z is simply obtained from the equation (11.33) by replacing the expression 4 π v 2 d v therein by d v x d v y d v z, i.e., (11.58)4 π v 2 d v→d v x d v y d v z, yielding, (11.59)d ω(v x,v y,v z)=(m 2 π k T)3/2 d v x d v y d v z exp(−m v 2 2 k T), or (11.60)d ω(v x,v y,v z)=v(v x)v(v y)v(v z)d v x d v y d v z, where (11.61)v(v j)=(m 2 π k T)1/2 exp(−m v j 2 2 k T),j=x,y,z. The one-dimensional probability distribution functionsv(v j) satisfy the normalization condition (11.62)∫−∞∞v(v j)d v j=1,j=x,y,z. The normalization condition (11.62) is easily verified as (11.63)∫−∞∞v(v j)d v j=(m 2 π k T)1/2∫−∞∞exp(−m v j 2 2 k T)d v j =(m 2 π k T)1/2(2 k T m)1/2∫−∞∞e−w 2 d w =1 π∫−∞∞e−w 2 d w=1, where we introduced a new variable w as follows: (11.64)v j=2 k T m w⇒d v j=2 k T m d w. Using now the probability distribution (11.61), we can calculate (11.65)=(m 2 π k T)1/2∫−∞∞v j 2 exp(−m v j 2 2 k T)d v j. Introducing a new variable w defined by (11.64), we get (11.66)=(m 2 π k T)1/2(2 k T m)3/2∫−∞∞w 2 e−w 2 d w, or (11.67)=1 π(m 2 k T)1/2(2 k T m)3/2 2 I 2, where the integral I 2 was calculated in (11.25) as (11.68)I 2=∫0∞w 2 e−w 2 d w=1 2∫0∞w 2 e−w 2 d w=x 4. Substituting (11.68) into (11.67), we obtain (11.69)=2 k T m 1 π 2 π 4=k T m,j=x,y,z. Using the result (11.69), we can calculate the average kinetic energy of a single particle (11.56) using the following definition (11.70)=1 2 m(++). Substituting (11.69) into (11.70), we obtain (11.71)=1 2 m(k T m+k T m+k T m)=3 2 k T. The internal energy of the system can then be calculated as follows: (11.72)U=N=N(k T 2+k T 2+k T 2)=3 2 k T. From the result (11.72), we see that each of the three translational degrees of freedom makes the same contribution, equal to 1 2 NkT, to the internal energy of the system. This observation is a special case of the so-called equipartition theorem in classical thermodynamics. According to the equipartition theorem, each term in a nonrelativistic expression E = E(x, y, z, p x, p y, p z) for the energy of a constituent particle that is quadratic in a dynamical variable contributes the amount of 1 2 NkT to the internal energy of the system. For example, for N classical simple harmonic oscillators with classical energy (11.73)e V(x,p x)=p x 2 2 m+1 2 m ω 2 x 2, there are, by (11.73), two such quadratic terms. By the equipartition theorem, we thus expect that the internal energy of any system with N classical harmonic oscillators is given by U=2×1 2 N k T=N k T This result for the vibrational degrees of freedom in the classical limit, will be derived from quantum statistical physics in Chapter 12, “Ideal Diatomic Gases.” In Section 11.6.2, “Harmonic and Anharmonic Oscillators,” we will derive this result from the classical statistical physics of oscillators in thermal equilibrium. Show more View chapterExplore book Read full chapter URL: Book 2011, Introductory Statistical ThermodynamicsNils Dalarsson, ... Leonardo Golubović Chapter Interacting Particles 2002, Thermodynamics and Statistical MechanicsPhil Attard 7.2.5 Equipartition Theorem It will be recalled that the energy of an ideal gas, Eq. (7.23), is given by 〈E i d〉=〈K〉=3 N k B T/2. The Gaussian integrals of the kinetic energy contribute k B T/2 for each component of momentum. More generally, for systems with intermolecular interactions there is a contribution of k B T/2 to the average energy for each quadratic degree of freedom in the Hamiltonian. This is called the equipartition theorem. There is a generalised equipartition theorem that gives the temperature as a phase space average of a quantity related to the virial.3 Recall from Ch. 5 that the partition function for an isolated system of energy E is given by Eq. (5.22), (7.40)Z(E)=Δ E N!h 3 N∫d Γ δ(H(Γ)−E), where the number and volume dependence is suppressed for brevity. The partition function in essence gives the weight or volume of phase space within d E of the energy hypersurface. There is a related function that gives the weight or volume of phase space contained by the hypersurface, (7.41)Ω(E)=1 N!h 3 N∫d Γ Θ(E−H(Γ)), where the integrand is the Heaviside step function. The two are related by (7.42)Z(E)=Δ E∂Ω(E)∂E. Since the entropy, which is the logarithm of the partition function, is extensive, the partition function itself must grow exponentially with energy. This exceedingly rapid rate of increase means that almost all of the volume contained by the hypersurface is actually within close proximity to the hypersurface and one has (7.43)S(E)=k B ln Z(E)=k B ln Ω(E)+O ln E. In the thermodynamic limit the last term is negligible. This result allows the evaluation of the desired average. Let x i be a component of position or momentum of a particular molecule, and similarly for x j, and consider the virial-like function x i∂H(Γ)/∂x j. The average of this is given by (7.44)〈x i∂H(Γ)∂x j〉=Δ E N!h 3 N Z(E)∫d Γ δ(H(Γ)−E)x i∂H(Γ)∂x j=Δ E N!h 3 N Z(E)∂∂E∫d Γ Θ(E−H(Γ))x i∂H(Γ)∂x j=Δ E N!h 3 N Z(E)∂∂E∫d Γ Θ(E−H(Γ))×[∂∂x j(x i{H(Γ)−E})−δ i j{H(Γ)−E}]=−δ i j Δ E N!h 3 N Z(E)∂∂E∫d Γ Θ(E−H(Γ)){H(Γ)−E}=−δ i j Δ E N!h 3 N Z(E)∫d Γ[δ(H(Γ)−E){H(Γ)−E}−Θ(E−H(Γ))]=δ i j Δ E Ω(E)Z(E)=δ i j[∂ln Ω(E)∂E]−1=δ i j k B T+O E−1. The third equality arises from manipulation of the partial derivative (E is independent of x j, and ∂x i/∂x j = δ ij). The fourth equality follows from the fact that H – E vanishes at the boundaries of the integration over x j. The first integrand of the fifth equality vanishes on the energy hypersurface, and the integral over the second term is just the volume contained by the hypersurface. The final equality manifests the fact that the inverse temperature is the logarithmic energy derivative of the entropy, and that the contained volume and the hypersurface volume are equal in the thermodynamic limit. This result shows in particular that (7.45)〈p i α∂H(Γ)∂p i α〉=〈r i α∂H(Γ)∂r i α〉=k B T. This is the generalised equipartition theorem. (The Hamiltonian may be replaced by the kinetic energy in the first equality, and by the potential energy in the second equality.) Notice that if r ia is a quadratic term in the potential energy, then the averand is just twice the energy of this term. This confirms that the contribution of each harmonic term in the Hamiltonian to the average energy is k B T/2. The configurational term above is obviously related to the virial of Clausius, Eq. (7.28). Summing over components and molecules gives (7.46)〈∑i α r i α∂U total(r N)∂r i α〉=−〈V total〉=3 N k B T. This may be reconciled with the virial equation, by noting that the potential energy thatappears here includes the external contribution from the walls, whereas only the internalpart of the potential energy appears in Eq.(7.29). Combining the two results one has (with β s = 1/k B T) (7.47)〈p s〉=N k B T V+1 3 V〈V int〉=1 3 V〈V int−V total〉=1 3 V〈∑i α r i α∂U ext(r N)∂r i α〉=−1 3 V〈∑i α r i α f i α ext〉. This expresses the pressure in terms of the external forces acting on the system.4 Show more View chapterExplore book Read full chapter URL: Book 2002, Thermodynamics and Statistical MechanicsPhil Attard Review article A general procedure for thermomechanical calibration of nano/micro-mechanical resonators 2013, Annals of PhysicsB.D. Hauer, ... J.P. Davis 7 Conclusion This document provides a general procedure by which the thermal motion of any nano/micro-mechanical resonator can be properly calibrated via the equipartition theorem. It was shown that for any given resonator, a generic spectral density function can be used to describe its motion in the frequency domain. The only modification existing between different structural geometries was manifested by a single parameter, effective mass, which is a mode shape dependent quantity. Therefore, the effective mass was investigated in great detail for a number of nano/micro-resonator geometries, including cantilevers, doubly clamped beams, strings, membranes and torsional devices. The end result was a universal method by which any nano/micro-mechanical resonator can be thermomechanically calibrated, regardless of the complexity of its motion or number of degrees of freedom (as demonstrated by Eichenfield et al. in the case of optomechanical photonic crystals). By applying this method, the motion of such devices can be very precisely determined, allowing for use of nano/micro-mechanical resonators for a number of different measurement applications, as well as in the investigation of fundamental scientific phenomena, with confidence and accuracy. View article Read full article URL: Journal2013, Annals of PhysicsB.D. Hauer, ... J.P. Davis Review article Lanczos-Lovelock models of gravity 2013, Physics ReportsT. Padmanabhan, D. Kothawala 4.2.1 Equipartition law in Einstein theory To fix the ideas, we will begin with the simplest context in which one can determine the density of microscopic degrees of freedom, viz., a static spacetime in Einstein’s general relativity. In a static spacetime, it is easy to show that (238)R a b u a u b=∇i a i=N−1 D μ(N a μ) where D μ is the covariant derivative operator corresponding to the 3-space metric h α β. Using the Einstein field equations in Eq. (238), we can relate the divergence of the acceleration to the source T̄a b≡(T a b−(1/2)g a b T) by: (239)D μ(N a μ)≡8 π N T̄a b u a u b≡4 π ρ Komar where ρ Komar is the (so called) Komar mass–energy density. Integrating both sides of Eq. (239) over V and using Gauss theorem gives: (240)E≡∫V d 3 x h ρ Komar=1 2∫∂V σ d 2 x L P 2{N a μ n μ 2 π} where σ is the determinant of the induced metric on ∂V and n μ is the spatial normal to ∂V. (We have temporarily restored G=L P 2≠1 keeping ħ=c=k B=1.) This result has a remarkable interpretation directly related to the equipartition law of conventional thermodynamics[127–129]. To see this, choose ∂V to be a N= constant surface so that the normal n μ is in the direction of the acceleration and a μ n μ=\|a\| is the magnitude of the acceleration. The local, redshifted, Davies–Unruh temperature is then given by: T=N T loc=(N a μ n μ/2 π)=(N\|a\|/2 π). Therefore Eq. (240) is exactly in the form of the equipartition law (with k B=1) where (241)E=1 2∫∂V d n T;Δ n≡σ d 2 x L P 2. That is, demanding the validity of Einstein equations, and interpreting Eq. (241) as a law of equipartition, we can determine the number of microscopic degrees of freedom in an element of area Δ A=σ d 2 x on ∂V to be σ d 2 x/L P 2. If these microscopic degrees of freedom are in equilibrium at the temperature T, then the total equipartition energy contributed by all these degrees of freedom on ∂V is equal to the total energy contained in the bulk volume enclosed by the surface—which could be thought of as another occurrence of the holographic principle. View article Read full article URL: Journal2013, Physics ReportsT. Padmanabhan, D. Kothawala Review article Lanczos-Lovelock models of gravity 2013, Physics ReportsT. Padmanabhan, D. Kothawala 4.2 Equipartition law for microscopic degrees of freedom Given the fact that spacetime appears to be hot, just like a body of gas, we can apply the Boltzmann paradigm (“If you can heat it, it has microstructure”) and study the nature of the microscopic degrees of freedom of the spacetime—exactly the way people studied gas dynamics before the atomic structure of matter was understood. One key relation in such an approach is the equipartition law Δ E=(1/2)k B T Δ N relating the number density Δ N of microscopic degrees of freedom we need to store an energy Δ E at temperature T. (This number is closely related to the Avogadro number of a gas, which was known even before one figured out what it was counting!). If gravity is the thermodynamic limit of the underlying statistical mechanics, describing the ‘atoms of spacetime’, we should be able to relate E and T of a given spacetime and determine the number density of microscopic degrees of freedom of the spacetime when everything is static. Remarkably enough, we can do this directly from the gravitational field equations[127–129]. The establishment of such a relation would obviously be important from the perspective of the emergent gravity paradigm as well. Since the basic idea here is quite simple (it is explained, for e.g.,[127–129]), it is convenient to first describe the result for Einstein theory (which is essentially some well known relations interpreted differently), and then describe the steps needed to derive it more formally for the entire class of Lanczos–Lovelock models using the Noether current and potential. We shall work with a static spacetime with metric components g 00=−N 2,g 0 α=0,g α β=h α β. An observer at rest in this spacetime with four-velocity u a=δ 0 a/N will have an acceleration a j=(0,a μ) where a μ=(∂μ N/N). We consider a spatial 3-volume V in this spacetime, having a boundary ∂V. 4.2.1 Equipartition law in Einstein theory To fix the ideas, we will begin with the simplest context in which one can determine the density of microscopic degrees of freedom, viz., a static spacetime in Einstein’s general relativity. In a static spacetime, it is easy to show that (238)R a b u a u b=∇i a i=N−1 D μ(N a μ) where D μ is the covariant derivative operator corresponding to the 3-space metric h α β. Using the Einstein field equations in Eq. (238), we can relate the divergence of the acceleration to the source T̄a b≡(T a b−(1/2)g a b T) by: (239)D μ(N a μ)≡8 π N T̄a b u a u b≡4 π ρ Komar where ρ Komar is the (so called) Komar mass–energy density. Integrating both sides of Eq. (239) over V and using Gauss theorem gives: (240)E≡∫V d 3 x h ρ Komar=1 2∫∂V σ d 2 x L P 2{N a μ n μ 2 π} where σ is the determinant of the induced metric on ∂V and n μ is the spatial normal to ∂V. (We have temporarily restored G=L P 2≠1 keeping ħ=c=k B=1.) This result has a remarkable interpretation directly related to the equipartition law of conventional thermodynamics[127–129]. To see this, choose ∂V to be a N= constant surface so that the normal n μ is in the direction of the acceleration and a μ n μ=\|a\| is the magnitude of the acceleration. The local, redshifted, Davies–Unruh temperature is then given by: T=N T loc=(N a μ n μ/2 π)=(N\|a\|/2 π). Therefore Eq. (240) is exactly in the form of the equipartition law (with k B=1) where (241)E=1 2∫∂V d n T;Δ n≡σ d 2 x L P 2. That is, demanding the validity of Einstein equations, and interpreting Eq. (241) as a law of equipartition, we can determine the number of microscopic degrees of freedom in an element of area Δ A=σ d 2 x on ∂V to be σ d 2 x/L P 2. If these microscopic degrees of freedom are in equilibrium at the temperature T, then the total equipartition energy contributed by all these degrees of freedom on ∂V is equal to the total energy contained in the bulk volume enclosed by the surface—which could be thought of as another occurrence of the holographic principle. 4.2.2 Equipartition law in Lanczos–Lovelock models Remarkably enough, the equipartition law exists for any diffeomorphism invariant theory of gravity and allows us to identify the corresponding surface density Δ n/σ d 2 x of microscopic states in a given theory. However, we restrict our attention to Lanczos–Lovelock models since the general case requires an additional assumption involving bifurcate Killing horizon which makes its status a bit unclear. To demonstrate how the result generalizes, we follow precisely the same steps[128,129] as in the Einstein case, but now giving a generalized version of the equations encountered before. For Lanczos–Lovelock actions, the object which replaces Ricci tensor of Einstein theory is defined by E a b+(1/2)L g a b, which, from Eq. (10), gives (242)R a b=P a c d e R b c d e. (Note that R a b⇒(16 π)−1 R a b in the Einstein case.) We shall also require the expression for Noether current J a associated with the timelike Killing vector ξ a, which is quite simple and is given by J a=2 R b a ξ b. One is now set to rerun the steps outlined for Einstein theory above, as follows. Using the definition of J a b (or expression Eq. (41) directly for a Killing vector), we get: (243)2 R a b u a u b=∇a(J b a u b N−1). Further, in a static spacetime, ∇i Q i=N−1 D α(N Q α) for any static vector (with ∂0 Q i=0) and hence we can write this as: (244)2 N R a b u a u b=D α(J b α u b) which generalizes of Eq. (238) to an arbitrary theory of gravity. The source for gravity in a general theory (analogous to Komar energy density) is defined through (245)ρ≡4 N R a b u a u b. For Lanczos–Lovelock models, ρ can be written as a function the matter stress–energy tensor T a b (generalizing the T̄a b used to define Komar energy in Einstein case above) using the field equations. This is possible for Lanczos–Lovelock models since trace of the equations of motion is proportional to the Lagrangian. However, the explicit form will not be needed here. Integrating both sides over a region V bounded by ∂V, we get (246)E≡∫V h d D−1 x ρ=∫V 2 N R a b u a u b h d D−1 x=∫∂V d D−2 x σ(n i u b J b i)=∫∂V d D−2 x σ(N n α J α 0) which is the analogue of Eq. (240). In Einstein’s theory, J μ 0=a μ/8 π which will reduce Eq. (246) to Eq. (240). In a general theory, the expression for Δ n is not just proportional to the area and J α 0 quantifies this difference. To obtain the equipartition law in a more explicit form, we note that (from Eq. (41)), we get J a b=2 P a b c d∇c ξ d and hence J α 0=4\|a\|P 0 β 0 α n β. So (247)E=∫∂V d D−2 x σ(16 π P 0 β 0 α n β n α)(N\|a\|2 π)≡1 2∫d n T where T=N\|a\|/2 π is the Davies–Unruh temperature as before. The surface density of microscopic degrees of freedom, when the field equations are satisfied (step 2 above), can be now read-off as: (248)d n σ d D−2 x=32 π P c d a b ϵ a b ϵ c d where ϵ a b≡(1/2)(u a n b−u b n a) is the binormal to ∂V. The connection with Wald entropy (see Section 3.1) is evident. In fact, (249)S=(1/2)β E provides the relationship between entropy, energy and temperature in all Lanczos–Lovelock models including general relativity as a special case. (In Ref., the implications of this result for general relativity was discussed extensively.) It also follows that, in all these models, we have 4 Δ S=Δ n. To sum up, the equipartition law allows one to identify the number density of microscopic degrees of freedom on a constant redshift surface. Using this one can define the entropy of the horizon in a general theory of gravity, which, as shown above, agrees with Wald entropy. These ideas will play a key role in our discussion in Section 5. We now discuss one specific context in which the equipartition idea presented here might have some fundamental implication. However, what follows is mathematically independent of the results derived above. 4.2.3 Holography and quantization of gravitational entropy Most of the arguments in the emergent gravity paradigm are free of unnecessary speculations since the approach taken is “top-down” and relies on well-established principles. For example, the key result of the previous section expressed in the form of Eq. (248), was derived from a classical theory with the only quantum mechanical input being the formula for Davies–Unruh temperature. The ħ enters the expression only through the formula k b T=(ħ/c)(a/2 π) which, along with G in classical gravitational field equations leads to G ħ/c 3, one of the central constants in quantum gravity. So one does not have to make any unwarranted speculations from the domain of quantum gravity to obtain Eq. (248). However, it is worth exploring where a result like Eq. (248) might emerge from a microscopic candidate theory of quantum gravity like for example, string theory or loop quantum gravity. To begin with, it is clear that our result in Eq. (248)should not depend on the details of the correct theory of quantum gravity. Any model of quantum gravity which has (i) the correct classical limit and (ii) is consistent with Davies–Unruh temperature, will lead to our Eq. (248). Since one would expect any quantum gravity model to satisfy these two conditions (i) and (ii) above our “top-down” approach will meet the “bottom-up” approach of such a theory in the overlap domain. So if string theory or loop-quantum-gravity or dynamical triangulations or any other candidate model for quantum gravity satisfies (i) and (ii), it will lead to the equipartition. (On the other hand, if a quantum gravity model does not satisfy (i) and (ii), such a model is probably wrong in any case.) Even though the thermodynamic limit contains far less information about the system than a microscopic description, one could still wonder about the possible route or mechanism by which a wide class of candidate models in quantum gravity (all satisfying (i) and (ii) above) can lead to a relation like Eq. (248) in the appropriate limit. In this section, we will indicate one route through which such results can emerge. (Although the motivation for our discussion comes from ideas in the emergent gravity paradigm, the results themselves are of independent interest since they generalize certain ideas about quantum mechanical spectrum of a black hole and its connection with the highly damped quasi-normal modes.) Let us first consider Einstein’s theory in which the equipartition law assigns A/L P 2 degrees of freedom to a proper area element A so that A≈n L P 2 for n≫1. (This corresponds to the semiclassical limit in which one can meaningfully talk about proper area elements etc. in the semiclassical metric.) This is, of course, nothing but area quantization in the asymptotic limit. Though LQG probably leads to such a result, in the limit we are interested in, it can be obtained from the Bohr–Sommerfeld quantization condition applied to horizons. In fact, Bekenstein conjectured years back that, in a quantum theory, the black hole area would be represented by a quantum operator with a discrete spectrum of eigenvalues. (Bekenstein showed that the area of a classical black hole behaves like an adiabatic invariant, and hence, according to Ehrenfest’s theorem, the corresponding quantum operator must have a discrete spectrum.) Extending these ideas to local Rindler horizons treated as the limit of a stretched horizon, one can understand how a result like quantization of any spatial area element can arise in a microscopic theory. The situation becomes more interesting when we go from the Einstein gravity to Lanczos–Lovelock models. In Einstein gravity, entropy of the horizon is proportional to its area and hence one could equivalently claim that it is the entropy which has an equidistant spectrum. But, in the Lanczos–Lovelock models, this proportionality between area and entropy breaks down. (Generalization of LQG to Lanczos–Lovelock models does not exist as yet.) The question then arises as to whether it is the quantum of area or quantum of entropy (if either) which arises in these models. This question was addressed in Ref. where it was shown that in the Lanczos–Lovelock models, it is indeed the entropy that is quantized with an equidistant spectrum. This matches nicely with the fact that it is the quantity in the right hand side of Eq. (248) that takes integral values in the equipartition law. Thus one can alternatively interpret our result as entropy quantization. We now indicate a sufficiently general ‘mechanism’ by which any microscopic model of quantum gravity can lead to such a quantization condition, in terms of two features. First one is the peculiar ‘holographic’ structure of the action functionals in Lanczos–Lovelock models[132,29,21]. As noted in Section 2.7.2, the action functionals in all these theories can be separated into a bulk and surface term and the surface term has the structure of an integral over d(p q). It can also be shown[3,132,29] that the same ‘d(p q)’ structure emerges for the on-shell action in the Lanczos–Lovelock models; that is, (250)A\|on-shell∝∫V d D x∂i(Π i j k g j k)=∫∂V d Σ i Π i j k g j k where Π i j k is the suitably defined canonical momenta corresponding to g j k. The second ingredient is that, in all these theories, the above expression for the action leads to the Wald entropy of the horizon. (For example, in Einstein gravity, the surface term in action, evaluated on a horizon will give one quarter of area[132,29].) In the semiclassical limit, Bohr–Sommerfeld quantization condition demands that the integral of d(p q) should be equal to 2 π n. Since the action in Eq. (250) has this ‘d(p q)’ structure, the Bohr–Sommerfeld condition reduces to A\|on-shell=2 π n. Since A\|on-shell is also equal to Wald entropy, we obtain (251)S Wald=A\|on-shell=2 π n. The Bohr–Sommerfeld quantization condition, of course, was the same used originally by Bekenstein and others to argue for the area quantization of the horizon but in the more general context of Lanczos–Lovelock models it appears as entropy quantization. (It is also possible to argue that in the semiclassical limit, the on-shell value of the action will be related to the phase of the semiclassical wave function Ψ∝exp(i A\|on−shell). If the semiclassical wave function describing the quantum geometry, relevant for a local Rindler observer, is obtained by integrating out the inaccessible degrees of freedom beyond the horizon then one can argue that, this phase should be 2 π n in the asymptotic limit leading to A\|on-shell=2 π n. Given the conceptual ambiguities related to interpretation of ‘wave function of geometry’, it is probably better to invoke Bohr–Sommerfeld condition.) To summarize, the three features: (i) the holographic structure of the gravitational action functional (ii) the equality of on-shell gravitational action functional and Wald entropy and (iii) the Bohr–Sommerfeld quantization condition, combine together to provide a context in which Eq. (248) can arise in any microscopic theory. While this gives a general result that in Lanczos–Lovelock theories it is the entropy of the horizon that is quantized, it would be useful to reinforce it by an explicit calculation in the standard context. Fortunately, this can be done for the m=2 Lanczos–Lovelock term, that is, for the GB model, using arguments first suggested by Hod based on quasi-normal modes of black hole oscillations. We give here a qualitative sketch of the arguments, and refer to for technical details. We first briefly summarize Hod’s argument. He started from Bekenstein’s arguments regarding quantum area spectrum of a non-extremal Kerr–Newman black hole, and showed that the spacing of area eigenvalues can be fixed by associating the classical limit of the quasi-normal mode frequencies, ω c, with the large n limit of the quantum area spectrum, in the spirit of Bohr’s correspondence principle (n being the quantum number). For a Schwarzschild black hole of mass M in D=4 dimensions, the absorption of a quantum of energy ω c, would lead to change in the black hole area eigenvalues as, Δ A≡A n+1−A n=(∂A/∂M)ω c and for entropy Δ S=(∂S/∂M)ω c. For a Schwarzschild black hole the level spacing of both area and entropy eigenvalues were indeed found to be equidistant, allowing one to associate the notion of a minimum unit, a quantum, of area and entropy. To apply this idea to Lanczos–Lovelock models, we need an expression for the quasi-normal mode frequencies associated with black hole solutions in these theories. These are not known in general, but are known in the case of 5 D Einstein–Gauss–Bonnet (GB) black holes. The form of the highly damped quasi-normal modes of these black holes suggest that it is the entropy which has a equally spaced spectrum. The result essentially depends only on the fact that Re ω c∝T leading to (∂S/∂M)ω c∝T(∂S/∂M) which is a constant. On the other hand, since in this case area is a nonlinear function of entropy, A=F(S), the area spectrum is not equidistant. Qualitative comments concerning the precise value of the quantum of entropy (in particular, the condition under which it is 2 π) can be found in. The case we have made for the quantization of entropy can be analysed in several different ways. If one applies our argument based on quasi-normal frequencies with the imaginary rather than real part of the frequencies (a strong case for doing so was suggested by Maggiore), then the result for constant spacing of entropy eigenvalues follows simply from the fact that this imaginary part, being derivable from a scattering matrix approach[135,136], is independent of the specific model of gravity and depends only on the existence of a horizon. The physical implications of the statement: gravitational entropy is quantized, are unclear at present. As should have been evident from our discussion in earlier sections, there are various proposed explanations of horizon entropy in the literature, with widely differing interpretations. For Lanczos–Lovelock models, most of these reproduce Wald entropy or yield expressions closely resembling it (as in Section 3.2), and therefore the statement that Wald entropy is quantized with equidistant spacing will also generically have implications for the physical principles on which a particular approach to horizon entropy is based. Since at least some of these approaches remain close to well known physics (for instance, the approach described in Section 3.2), the statement of quantization of entropy might be of relevance at more fundamental level than has yet been appreciated. Show more View article Read full article URL: Journal2013, Physics ReportsT. Padmanabhan, D. Kothawala Chapter The canonical ensemble 2022, Statistical Mechanics (Fourth Edition)R.K. Pathria, Paul D. Beale 3.7 Two theorems – the “equipartition” and the “virial” To derive these theorems, we determine the expectation value of the quantity x i(∂H/∂x j), where H(q,p) is the Hamiltonian of the system (assumed classical) while x i and x j are any two of the 6 N generalized coordinates (q,p). In the canonical ensemble, (1)〈x i∂H∂x j〉=∫(x i∂H∂x j)e−β H d ω∫e−β H d ω(d ω=d 3 N q d 3 N p). Let us consider the integral in the numerator. Integrating over x j by parts, it becomes ∫[−1 β x i e−β H\|(x j)1(x j)2+1 β∫(∂x i∂x j)e−β H d x j]d ω(j); here, (x j)1 and (x j)2 are the “extreme” values of the coordinate x j, while d ω(j) denotes “dω devoid of d x j.” The integrated part here vanishes because whenever any of the coordinates takes an “extreme” value the Hamiltonian of the system becomes infinite.11 In the integral that remains, the factor ∂x i/∂x j, being equal to δ i j, comes out of the integral sign and we are left with 1 β δ i j∫e−β H d ω. Substituting this into (1), we arrive at the remarkable result (2)〈x i∂H∂x j〉=δ i j k T, which is independent of the precise form of the function H. In the special case x i=x j=p i, equation (2) takes the form (3)〈p i∂H∂p i〉≡〈p i q˙i〉=k T, while for x i=x j=q i, it becomes (4)〈q i∂H∂q i〉≡−〈q i p˙i〉=k T. Adding over all i, from i=1 to 3 N, we obtain (5)〈∑i p i∂H∂p i〉≡〈∑i p i q˙i〉=3 N k T and (6)〈∑i q i∂H∂q i〉≡−〈∑i q i p˙i〉=3 N k T. Now, in many physical situations the Hamiltonian of the system happens to be a quadratic function of its coordinates; so, through a canonical transformation, it can be brought into the form (7)H=∑j A j P j 2+∑j B j Q j 2, where P j and Q j are the transformed, canonically conjugate, coordinates while A j and B j are certain constants of the problem. For such a system, we clearly have (8)∑j(P j∂H∂P j+Q j∂H∂Q j)=2 H; accordingly, by equations (3) and (4), (9)〈H〉=1 2 f k T, where f is the number of nonvanishing coefficients in expression (7). We, therefore, conclude that each harmonic term in the (transformed) Hamiltonian makes a contribution of 1 2 k T toward the internal energy of the system and, hence, a contribution of 1 2 k toward the specific heatC V. This result embodies the classical theorem of equipartition of energy (among the various degrees of freedom of the system). It may be mentioned here that, for the distribution of kinetic energy alone, the equipartition theorem was first stated by Boltzmann (1871). In our subsequent study we shall find that the equipartition theorem as stated here is not always valid; it applies only when the relevant degrees of freedom can be freely excited. At a given temperature T, there may be certain degrees of freedom which, due to the insufficiency of the energy available, are more or less “frozen” due to quantum-mechanical effects. Such degrees of freedom do not make a significant contribution toward the internal energy of the system or toward its specific heat; see, for example, Sections 6.5, 7.4, and 8.3. Of course, the higher the temperature of the system, the better the validity of this theorem. We now consider the implications of formula (6). First of all, we note that this formula embodies the so-called virial theorem of Clausius (1870) for the quantity 〈∑i q i p˙i〉, which is the expectation value of the sum of the products of the coordinates of the various particles in the system and the respective forces acting on them; this quantity is generally referred to as the virial of the system and is denoted by the symbol V. The virial theorem then states that (10)V=−3 N k T. The relationship between the virial and other physical quantities of the system is best understood by first looking at a classical gas of noninteracting particles. In this case, the only forces that come into play are the ones arising from the walls of the container; these forces can be designated by an external pressure P that acts on the system by virtue of the fact that it is bounded by the walls of the container. Consequently, we have here a force −P d S associated with an element of area d S of the walls; the negative sign appears because the force is directed inward while the vector d S is directed outward. The virial of the gas is then given by (11)V 0=(∑i q i F i)0=−P∮S r⋅d S, 12 where r is the position vector of a particle that happens to be in the (close) vicinity of the surface element d S; accordingly, r may be considered to be the position vector of the surface element itself. By the divergence theorem, equation (11) becomes (12)V 0=−P∫V(div r)d V=−3 P V. Comparing (12) with (10), we obtain the well-known result (13)P V=N k T. The internal energy of the gas, which in this case is wholly kinetic, follows from the equipartition theorem (9) and is equal to 3 2 N k T,3 N being the number of degrees of freedom. Comparing this result with (10), we obtain the classical relationship (14)V=−2 K, where K denotes the average kinetic energy of the system. It is straightforward to apply this theorem to a system of particles interacting through a two-body potential u(r). In the thermodynamic limit, the pressure of a d-dimensional system depends only on the virial terms arising from the forces between pairs of particles: (15)P n k T=1+1 N d k T〈∑i<j F(r i j)⋅r i j〉=1−1 N d k T〈∑i<j∂u(r i j)∂r i j r i j〉. Equation (15) is called the virial equation of state. This equation can also be written in terms of the pair correlation function, equation (10.7.11), and is also used in computer simulations to determine the pressure of the system; see Problem 3.14, Section 10.7, and Section 16.4. Show more View chapterExplore book Read full chapter URL: Book 2022, Statistical Mechanics (Fourth Edition)R.K. Pathria, Paul D. Beale Chapter THE CANONICAL ENSEMBLE 1996, Statistical Mechanics (Second Edition)R.K. PATHRIA 3.7 Two theorems – the “equipartition” and the “virial” To arrive at these theorems, we determine the expectation value of the quantity x i(∂H/∂x j), where H(q, p) is the Hamiltonian of the system (assumed classical) while x i and x j are any of the 6 N generalized coordinates (q, p). In the canonical ensemble, (1)〈x i∂H∂x j〉=∫(x i∂H∂x j)e−β H d ω∫e−β H d ω(d ω=d 3 N q d 3 N p). Let us consider the integral in the numerator. Integrating over x j by parts, it becomes ∫[−1 β x i e−β H\|(x j)1(x j)2+1 β∫(∂x i∂x j)e−β H d x j]d ω(j); here, (x j)1 and (x j)2 are the “extreme” values of the coordinate x j, while d ω(j) denotes “d ω devoid of dx j”. The integrated part vanishes because whenever any of the coordinates takes an “extreme” value the Hamiltonian of the system becomes infinite.11 In the integral that remains, the factor ∂x i/∂x j, being equal to δ ij, comes out of the integral sign and we are left with 1 β δ i j∫e−β H d ω. Substituting this into (1), we arrive at the desired result: (2)〈x i∂H∂x j〉=δ i j k T, which is independent of the precise form of the function H. In the special case x i = x j = p i, eqn. (2) takes the form (3)〈p i∂H∂p i〉≡〈p i q i〉=k T, while for x i = x j = q i, it becomes (4)〈q i∂H∂q i〉≡−〈q i p i〉=k T. Adding over all i, from i = 1 to 3 N, we obtain (5)〈∑i p i∂H∂p i〉≡〈∑i p i q i〉=3 N k T and (6)〈∑i q i∂H∂q i〉≡−〈∑i q i p i〉=3 N k T. Now, in several physical situations the Hamiltonian of the system happens to be a quadratic function of its coordinates; so, through a canonical transformation, it can be brought into the form (7)H=∑j A j P j 2+∑j B j Q j 2, where P j and Q j are the transformed, canonically conjugate, coordinates while A j and B j are certain constants of the problem. For such a system, we clearly have (8)∑j(P j∂H∂P j+Q j∂H∂Q j)=2 H; accordingly, by eqns (3) and (4), (9)〈H〉=1 2 f k T, where f is the number of nonvanishing coefficients in the expression (7). We therefore conclude that each harmonic term in the (transformed) Hamiltonian makes a contribution of 1 2 k T towards the internal energy of the system and, hence, a contribution of 1 2 k towards the specific heatC V. This result embodies the classical theorem of equipartition of energy (among the various degrees of freedom of the system). It may be mentioned here that, for the distribution of kinetic energy alone, the equipartition theorem was first stated by Boltzmann (1871). In our subsequent study we shall find that the equipartition theorem as stated here is not always valid; it applies only when the relevant degrees of freedom can be freely excited. At a given temperature T, there may be certain degrees of freedom which, due to the insufficiency of the energy available, are more or less “frozen”. Such degrees of freedom do not make a significant contribution towards the internal energy of the system or towards its specific heat; see, for example, Secs 6.5, 7.3 and 8.3. Of course, the higher the temperature of the system the better the validity of this theorem. We now consider the implications of formula (6). First of all, we note that this formula embodies the so-called virial theorem of Clausius (1870) for the quantity 〈∑i q i p˙i〉, which is the expectation value of the sum of the products of the coordinates of the various particles in the system and the respective forces acting on them; this quantity is referred to as the virial of the system and is generally denoted by the symbol . The virial theorem then states that Sign in to download full-size image (10). The relationship between the virial and other physical quantities of the system is best understood by first looking at a classical gas of non-interacting particles. In this case, the only forces that come into play are the ones arising from the walls of the container; these forces can be designated by an external pressure P that acts upon the system by virtue of the fact that it is bounded by the walls of the container. Consequently, we have here a force −P dS associated with an element of area dS of the walls; the negative sign appears because the force is directed inward while the vector dS is directed outward. The virial of the gas is then given by Sign in to download full-size image (11). 12 where r is the position vector of a particle that happens to be in the (close) vicinity of the surface element dS; accordingly, r may be considered to be the position vector of the surface element itself. By the divergence theorem, eqn. (11) becomes Sign in to download full-size image (12). Comparing (12) with (10), we obtain the well-known result: (13)P V=N k T. The internal energy of the gas, which in this case is wholly kinetic, follows from the equipartition theorem (9) and is equal to , 3 N being the number of degrees of freedom. Comparing this result with (10), we obtain the classical relationship Sign in to download full-size image (14). where K denotes the average kinetic energy of the system. It is straightforward to carry out an extension of this study to a system of particles interacting through a two-body potential u(rj – ri). The virial then draws a contribution from the interior as well. Assuming the interparticle potential to be central and denoting it by the symbol u(r), where r = \|rj – ri\|, the contribution arising from the pair of particles i and j, with position vectors ri and rj, is given by (15)r i⋅(−∂u(r)∂r i)+r j.(−∂u(r)∂r j)=−∂u(r)∂r 2{r i⋅∂\|r j−r i\|2∂r i+r j⋅∂\|r j−r i\|2∂r j}=−r∂u(r)∂r. The net contribution, arising from all the N(N − 1)/2 pairs of particles, would then be, for N ≫ 1, (16)1 2 N 2〈−r∂u(r)∂r〉=−N 2 2∫∫{r∂u(r)∂r}g(r 2−r 1)d r 1 d r 2 V 2=−N 2 2 V∫0∞{r∂u(r)∂r}g(r)(4 π r 2 d r), where g(r), the pair distribution function of the system, is a measure of the probability of finding a pair of particles in it separated by a distance r; as used here, g(r) → 1 as r → ∞. Combining (16) with (12) and equating the resulting sum with (10), we obtain for a classical interacting system (17)P V=N k T[1−2 π n 3 k T∫0∞∂u(r)∂r g(r)r 3 d r], where n is the particle density in the system. The internal energy of the system can also be expressed in terms of the functions u(r) and g(r). Noting that the average kinetic energy is still given by the expression 3 2 N k T, we have for the total energy (18)U=3 2 N k T+1 2 N 2∫∫u(r)g(r)d r 1 d r 2 V 2=3 2 N k T[1+4 π n 3 k T∫0∞u(r)g(r)r 2 d r], the second term being the average potential energy of the system. Clearly, a knowledge of the functions u(r) and g(r), the latter itself depending on the nature of the former, is essential before we can make use of eqns (17) and (18). For further study along these lines, see Hill (1956), Chap. 6; see also Problem 3.23. Show more View chapterExplore book Read full chapter URL: Book 1996, Statistical Mechanics (Second Edition)R.K. PATHRIA Chapter Interacting Particles 2002, Thermodynamics and Statistical MechanicsPhil Attard 7.2.6 Average Temperature The preceding generalised equipartition theorem allows the temperature to be obtained as a phase space average. An alternative expression for the inverse temperature may be obtained directly from the energy derivative of the entropy of the isolated system.5 Recall that the temperature is defined as (7.48)T−1=∂S(E,N,V)∂E, and that the entropy is (7.49)S(E,N,V)=k B ln 1 N!h 3 N∫d Γ δ(H N(Γ)−E). Now consider the transformation of phase space, (7.50)Γ′=Γ+n˜(Γ)Δ E, where Δ E is an infinitesimal. The energy of the points transformed from the original energy hypersurface is (7.51)H N(Γ′)=H N(Γ)+Δ E n˜(Γ)·∇H N(Γ). Accordingly the choice (7.52)n˜(Γ)=n(Γ)n(Γ)·∇H N(Γ), represents, for arbitrary n(Γ), a new energy hypersurface parallel to the original, (7.53)H N(Γ′)=H N(Γ)+Δ E. In view of this the entropy of the new energy hypersurface is (7.54)S(E+Δ E,N,V)=k B ln 1 N!h 3 N∫d Γ′δ(H N(Γ′)−E−Δ E)=k B ln 1 N!h 3 N∫d Γ\|∂Γ′∂Γ\|δ(H N(Γ)−E). It is straightforward to show that 6 (7.55)\|∂Γ′∂Γ\|=1+Δ E∇·n˜(Γ). Accordingly, the inverse temperature may be written (7.56)β=S(E+Δ E,N,V)−S(E,N,V)k B Δ E=1 Δ E ln〈1+Δ E∇·n˜(Γ)〉=〈∇·n˜(Γ)〉. This result is Rugh's expression. One may similarly show that the second derivative is (7.57)∂2 S(E,N,V)/k B∂E 2=−〈Tr[∇n˜(Γ)]2〉=−〈∑i,j=1 6 N∂n˜i∂Γ j∂n˜j∂Γ i〉. This quantity gives the heat capacity. It also confirms that S EE< 0 (because (T r A _ _ 2)=∑i λ i 2>0. These results hold for any n(Γ), provided only that n(Γ) ∇ H N (Γ) ≠ 0. One possible choice is n(Γ) = ∇N(Γ), in which case (7.58)n˜(Γ)=∇H N(Γ)∇H N(Γ)·∇H N(Γ). An alternative is n(Γ) = ri α, in which case (7.59)1 k B T=〈−∂2 H N(Γ)/∂r i α 2(∂H N(Γ)/∂r i α)〉. Show more View chapterExplore book Read full chapter URL: Book 2002, Thermodynamics and Statistical MechanicsPhil Attard Related terms: Molecular Dynamics Harmonic Oscillator Potential Energy Internal Energy Magnetic Field Degree of Freedom Harmonics Solitary Wave Free Energy Specific Heat View all Topics Recommended publications Physica A: Statistical Mechanics and its ApplicationsJournal Biophysical JournalJournal Physics Letters BJournal Journal of Computational PhysicsJournal Browse books and journals Featured Authors Ho, Hopui AaronChinese University of Hong Kong, Hong Kong, Hong Kong Citations9,174 h-index49 Publications66 Sedev, Rossen V.University of South Australia, Adelaide, Australia Citations6,051 h-index45 Publications42 Gao, Bruce ZhiClemson University College of Engineering, Computing and Applied Sciences, Clemson, United States Citations2,845 h-index28 Publications15 Shao, YonghongShenzhen University, Shenzhen, China Citations2,138 h-index24 Publications20 About ScienceDirect Remote access Advertise Contact and support Terms and conditions Privacy policy Cookies are used by this site. 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5840	https://www.youtube.com/watch?v=DqShiY-Lcv4	Graphing the Line Passing through (1, –1) with Slope “𝑚 = 3/4” \| Step-by-Step Algebra Solution Math Doodle - Algebra 1110 subscribers 13 likes Description 836 views Posted: 10 Sep 2022 This video will guide you through [How to Graph the Line Passing Through the Point (1, –1) Whose Slope is 𝑚 = 3/4]. 📌 How to Graph a Line Given a Point and the Slope. ✓ Step 1. Plot the given point. ✓ Step 2. Use the slope formula, 𝑚 = (rise)/(run), to identify the rise and the run. ✓ Step 3. Starting at the given point, count out the rise and run to mark the second point. ✓ Step 4. Connect the points with a line. 🔑 Key Things to Remember: ☞ In mathematics, the ‘tilt’ of a line is called the SLOPE of the line. ☞ Slope of Line ➜ 𝑚 = (rise)/(run). Between two points on the line, the rise measures the vertical change, and the run measures the horizontal change. 〰️ ✅ More BITE-SIZE VIDEOS on “Graphing Linear Equations” ➤ Find slope of line between the points (1, 2) and (4, 5) using the slope formula: ➤ Find slope of line through the points (–1, –1) and (–7, 4) using the slope formula: ➤ Graph the line with 𝑦-intercept 2 whose slope is 𝑚 = –2/3: ➤ Graph the line passing through the point (–1, –3) whose slope is 𝑚 = 4: 🗂 Playlist: “Graphs” ➜ 🗂 Playlist: “Rectangular Coordinate System” ➜ 🗂 Playlist: “Graphing Linear Equations” ➜ 🗂 Playlist: “Slope and 𝑦-Intercept of Line” ➜ 🗂 Playlist: “Parallel/Perpendicular Lines” ➜ 〰️ ✏️ Need FREE MATH HELP? Check out our website on Graphs ➜ 〰️ ▶️ Click Here to Subscribe to our Channel ➜ ✍ Need to review Pre-Algebra topics? Check out our Pre-Algebra channel ➜ 〰️ 📖 About Math Doodle: This channel is our effort to support students, teachers, and parents by sharing our tutorial and problem-walkthrough contents that cover a wide range of algebra topics! 🌐 Website: 📧 E-mail: info@math-doodle.com Hopefully, you have found the math help you were looking for! 🙏 Thank you for checking out our videos! 💯 〰️ 🌏 Translated titles and subtitles have been added to this video: Spanish: Grafica la Línea que Pasa por los puntos (1, –1) con Pendiente m = 3/4 French: Représentez graphiquement la droite passant par les points (1, –1) avec la pente m = 3/4 German: Zeichnen Sie die Linie, die durch die Punkte (1, –1) verläuft, mit Steigung m = 3/4 Portuguese: Faça o gráfico da linha que passa pelos pontos (1, –1) com inclinação m = 3/4 Hindi: ढलान m = 3/4 . के साथ बिंदुओं (1, -1) से गुजरने वाली रेखा को ग्राफ़ करें Filipino: I-graph ang Linya na dumadaan sa mga puntos (1, –1) na may Slope m = 3/4 Indonesian: Gambarkan Garis yang Melalui Titik (1, –1) dengan Kemiringan m = 3/4 Japanese: 勾配 m = 3/4 で点 (1, –1) を通る直線をグラフ化します Russian: Нарисуйте линию, проходящую через точки (1, –1) с наклоном m = 3/4 Vietnamese: Vẽ đồ thị đường thẳng đi qua điểm (1, –1) với độ dốc m = 3/4 Transcript: [music] In this problem, we are going to graph the line passing through the point (1, –1) whose slope is “m = 3/4”. Since the line passes through the point (1, –1), let’s first plot the point in a coordinate system. And, the line has a slope of “3/4”. This tells us that its “rise over run” is “3/4”. So, the line has a rise of “3” and a run of “4”. Now, let’s count the rise and run in a coordinate system, starting at the point (1, –1). Up 3 units, and then right 4 units. Then, let’s draw a line through these points and put arrows on both ends of the line. And, this is the line passing through the point (1, –1) whose slope is “3/4”.
5841	https://pubmed.ncbi.nlm.nih.gov/28409615/	Peritonsillar Abscess - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Peritonsillar Abscess Nicholas J Galioto1 Affiliations Expand Affiliation 1 Broadlawns Medical Center, Des Moines, IA, USA. PMID: 28409615 Free article Item in Clipboard Peritonsillar Abscess Nicholas J Galioto. Am Fam Physician.2017. Free article Show details Display options Display options Format Am Fam Physician Actions Search in PubMed Search in NLM Catalog Add to Search . 2017 Apr 15;95(8):501-506. Author Nicholas J Galioto1 Affiliation 1 Broadlawns Medical Center, Des Moines, IA, USA. PMID: 28409615 Item in Clipboard Full text links Cite Display options Display options Format Abstract Peritonsillar abscess is the most common deep infection of the head and neck, occurring primarily in young adults. Diagnosis is usually made on the basis of clinical presentation and examination. Symptoms and findings generally include fever, sore throat, dysphagia, trismus, and a "hot potato" voice. Drainage of the abscess, antibiotic therapy, and supportive therapy for maintaining hydration and pain control are the cornerstones of treatment. Most patients can be managed in the outpatient setting. Peritonsillar abscesses are polymicrobial infections, and antibiotics effective against group A streptococcus and oral anaerobes should be first-line therapy. Corticosteroids may be helpful in reducing symptoms and speeding recovery. Promptly recognizing the infection and initiating therapy are important to avoid potentially serious complications, such as airway obstruction, aspiration, or extension of infection into deep neck tissues. Patients with peritonsillar abscess are usually first encountered in the primary care outpatient setting or in the emergency department. Family physicians with appropriate training and experience can diagnose and treat most patients with peritonsillar abscess. PubMed Disclaimer Similar articles Peritonsillar abscess.Galioto NJ.Galioto NJ.Am Fam Physician. 2008 Jan 15;77(2):199-202.Am Fam Physician. 2008.PMID: 18246890 Review. Peritonsillar abscess: clinical aspects of microbiology, risk factors, and the association with parapharyngeal abscess.Klug TE.Klug TE.Dan Med J. 2017 Mar;64(3):B5333.Dan Med J. 2017.PMID: 28260599 Review. New trends in the management of peritonsillar abscess.Epperly TD, Wood TC.Epperly TD, et al.Am Fam Physician. 1990 Jul;42(1):102-12.Am Fam Physician. 1990.PMID: 2195852 Review. Peritonsillar abscess: diagnosis and treatment.Steyer TE.Steyer TE.Am Fam Physician. 2002 Jan 1;65(1):93-6.Am Fam Physician. 2002.PMID: 11804446 Bilateral peritonsillar abscesses: not your usual sore throat.Fasano CJ, Chudnofsky C, Vanderbeek P.Fasano CJ, et al.J Emerg Med. 2005 Jul;29(1):45-7. doi: 10.1016/j.jemermed.2005.01.005.J Emerg Med. 2005.PMID: 15961007 See all similar articles Cited by Deep Parotid Lobe Abscess Presenting with Dysphagia and Trismus.Grinnell M, Logeman A, Knudsen T, Sayed Z.Grinnell M, et al.Case Rep Otolaryngol. 2019 Feb 24;2019:2931015. doi: 10.1155/2019/2931015. eCollection 2019.Case Rep Otolaryngol. 2019.PMID: 30918736 Free PMC article. Type 2 Diabetes Mellitus Increases Peritonsillar Abscess Susceptibility: Real-World Evidence.Wu CL, Tsai MS, Lee TJ, Wang YT, Liu CY, Yang YH, Tsai YT, Hsu CM, Wu CY, Chang PJ, Chang GH.Wu CL, et al.Clin Exp Otorhinolaryngol. 2021 Aug;14(3):347-354. doi: 10.21053/ceo.2020.02257. Epub 2021 Jan 14.Clin Exp Otorhinolaryngol. 2021.PMID: 33541035 Free PMC article. The Feasibility of an Educational Course for General Practitioners on Otolaryngologic Morbidity.Papadopoulos N, Karatzanis AD, Prokopakis EP, Linardakis M, Galanos I, Liva G, Tsamandouras I, Mourellou E, Symvoulakis EK.Papadopoulos N, et al.Cureus. 2024 Jul 23;16(7):e65171. doi: 10.7759/cureus.65171. eCollection 2024 Jul.Cureus. 2024.PMID: 39176327 Free PMC article. Analysis of Smoking Behavior in Patients With Peritonsillar Abscess: A Rural Community Hospital's Experience.Clark C, Santarelli A, Merrill S, Ashurst J.Clark C, et al.Cureus. 2022 Mar 18;14(3):e23300. doi: 10.7759/cureus.23300. eCollection 2022 Mar.Cureus. 2022.PMID: 35464569 Free PMC article. Fulminating deep tissue space infection with Streptococcus constellatus presenting initially as a sore throat.Glover J, Kovacevic G, Walton G, Parr D.Glover J, et al.BMJ Case Rep. 2020 Mar 31;13(3):e233971. doi: 10.1136/bcr-2019-233971.BMJ Case Rep. 2020.PMID: 32234857 Free PMC article. 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5842	https://www.cuemath.com/algebra/factor-theorem/	Factor Theorem - Statement, Formula, Proof, Examples, Application Grade KG 1st 2nd 3rd 4th 5th 6th 7th 8th Algebra 1 Algebra 2 Geometry Pre-Calculus Calculus Pricing Events About Us Get Started Grade KG 1st 2nd 3rd 4th 5th 6th 7th 8th Algebra 1 Algebra 2 Geometry Pre-Calculus Calculus Pricing Events About Us LearnPracticeDownload Factor Theorem Factor theorem is mainly used to factor the polynomials and to find the n roots of the polynomials. Factor theorem is very helpful for analyzing polynomial equations. In real life, factoring can be useful while exchanging money, dividing any quantity into equal pieces, understanding time, and comparing prices. 1.What is Factor Theorem? 2.Factor Theorem Statement 3.Factor Theorem Proof 4.How to Use the Factor Theorem? 5.Difference Between Factor Theorem and Remainder Theorem 6.FAQs on Factor Theorem What is Factor Theorem? Factor theorem is a special kind of the polynomial remainder theorem that links the factors of a polynomial and its zeros. The factor theorem removes all the known zeros from a given polynomial equation and leaves all the unknown zeros. The resultant polynomial has a lower degree in which the zeros can be easily found. Factor Theorem Statement The factor theorem states that if f(x) is a polynomial of degree n greater than or equal to 1, and 'a' is any real number, then (x - a) is a factor of f(x) if f(a) = 0. In other words, we can say that (x - a) is a factor of f(x) if f(a) = 0. Let us now understand the meaning of some concepts related to the factor theorem. Zero of a Polynomial Before learning about the factor theorem, it is essential for us to know about the zero or a root of the polynomial. We say that y = a is a root or zero of a polynomial g(y) only if g(a) = 0. We can also say that y = a is a root or zero of a polynomial only if it is a solution to the equation g(y) = 0. Let's consider an example to find the zeros of the second-degree polynomial g(y) = y 2 + 2y − 15. To do this we simply solve the equation by using the factorization of quadratic equation method as: y 2 + 2y − 15 = (y+5)(y−3) = 0 ⇒ y =−5 and y = 3 Thus, this second-degree polynomial y 2 + 2y − 15 has two zeros or roots which are - 5 and 3. Factor Theorem Formula As per the factor theorem, (y – a) can be considered as a factor of the polynomial g(y) of degree n ≥ 1, if and only if g(a) = 0. Here, a is any real number. The formula of the factor theorem is g(y) = (y – a) q(y). It is important to note that all the following statements apply for any polynomial g(y): (y – a) is a factor of g(y). g(a) = 0. The remainder becomes zero when g(y) is divided by (y – a). The solution to g(y) = 0 is a and the zero of the function g(y) is a. Factor Theorem Proof In order to prove the factor theorem, let's first consider a polynomial g(y) that is being divided by (y – a) only if g(a) = 0. By using the division algorithm, the given polynomial can be written as the product of its divisor and its quotient: Dividend = (Divisor × Quotient ) + Remainder ⟹ g(y) = (y – a) q(y) + remainder. Here, g(y) is the dividend, (y – a) is the divisor, and q(y) is the quotient. From the remainder theorem, we get: g(y) = (y – a) q(y) + g(a) If we substitute g(a) = 0 then the remainder is 0, ⟹ g(y) = (y – a) q(y) + 0 ⟹ g(y) = (y – a) q(y) Thus, we can say that (y – a) is a factor of the polynomial g(y). Here we can see that the factor theorem is actually a result of the remainder theorem, which states that a polynomial g(y) has a factor (y – a), if and only if, a is a root i.e., g(a) = 0. How to Use the Factor Theorem? Let's learn how to use the factor theorem with an example. Check whether (y + 5) is a factor of 2y 2 + 7y – 15 or not. Given that, y + 5 = 0. Then, y = - 5. Now let's substitute y = - 5 into the given polynomial equation. We get: g(-5) = 2 (-5)2 + 7(-5) – 15 = 2 (25) - 35 – 15 = 50 – 35 – 15 = 0 Thus, y + 5 is a factor of 2y 2 + 7y – 15. Using the Factor Theorem To Factor a Third-Degree Polynomial We usually use the factorization method to factor second-degree or quadratic polynomials. For higher degrees, we can use the below-given procedure to factor the polynomial: Step 1: Use the synthetic division of the polynomial method to divide the given polynomial g(y) by the given binomial (y−a) Step 2: After the completion of the division, confirm whether the remainder is 0. If the remainder is not zero, then it means that (y-a) is not a factor of g(y). Step 3: Using the division algorithm, write the given polynomial as the product of (y-a) and the quadratic quotient q(y) Step 4: If it is possible, factor the quadratic quotient further. Step 5: Express the given polynomial as the product of its factors. Using the factor theorem, let's show that (y+2) is a factor of y 3 − 6y 2 − y + 30 and then find the remaining factors. After finding the remaining factors, we will use these factors to determine the zeros of the given polynomial. The first step is to use the synthetic division method to show that (y+2) is a factor of the third-degree polynomial y 3 − 6y 2 − y + 30. After the completion of the synthetic division method, we find that the remainder is zero. Hence, (y + 2) is a factor of the given polynomial. Now, let's use the division algorithm to write the given polynomial as the product of the divisor (y + 2) and the quadratic quotient (y 2- 8y +15), that is, y 3 − 6y 2 − y + 30 = (y+2) (y 2− 8y +15). Let's factorize the quadratic equation y 2− 8y +15 to write the polynomial as (y + 2)(y − 3)(y − 5). Thus, by using the factor theorem, the zeros of the given polynomial y 3 − 6y 2 − y + 30 are –2, 3, and 5. Difference Between Factor Theorem and Remainder Theorem Factor theorem and remainder theorem are similar but they refer to two different concepts. The remainder theorem relates the remainder of the division of a polynomial by a binomial with the value of a function at a point. The factor theorem relates the factors of a given polynomial to its zeros. Let's consider an example of a polynomial g(y) = y 2 − 2y + 1 to understand the difference: For the given polynomial g(y), let's use the remainder theorem and put 3 as y into g(y): g(3) = (3)2 − 2(3) + 1 g(3) = 9 − 6 + 1 g(3) = 4 Hence, by the remainder theorem, the remainder when we divide y 2 − 2y + 1 by y−3 is 4. We can apply this in reverse too by dividing y 2 − 2y + 1 by y−3, and we will get the remainder equal to the value of g(3). For the same polynomial, let's use the factor theorem g(y) = y 2 − 2y + 1 equals 0 when y =1. g(1) = (1)2 − 2(1) + 1 g(1) = 1 − 2 + 1 g(1) = 0 This tells us that (y−1) is a factor of y 2 − 2y + 1. We can also apply this in reverse. We can factor y 2 − 2y + 1 into (y − 1)2. Thus, 1 is a zero of g(y). Important Notes on Factor Theorem Factor theorem is mainly used to factor the polynomials and to find the n roots of that polynomial. In real life, factoring is useful while exchanging money, dividing any quantity into equal pieces, understanding time, and comparing prices. As per the factor theorem, (y – a) can be considered as a factor of the polynomial g(y) of degree n ≥ 1, if and only if g(a) = 0. Related Articles Zeros of Quadratic Polynomial Factorization of Algebraic Expressions Polynomial Functions Degree of a Polynomial Types of Polynomials Factor Theorem Examples Example 1: Use the factor theorem to check whether y + 1 is a factor of the polynomial 3y 4 + y 3 – y 2 + 3y + 2, or not. Solution: Given y + 1; then, y + 1 = 0, we get y = -1. Substitute y = -1 in the given polynomial equation 3y 4 + y 3 – y 2 + 3y + 2. ⟹ 3(–1)4 + (–1)3 – (–1)2 +3(–1) + 2 = 3(1) + (–1) – 1 – 3 + 2 = 3 -1 -1 -3 + 2 Adding all positive terms and all negative terms, we get, = 5 - 5 = 0 Therefore, we can say that y + 1 is a factor of 3y 4 + y 3 – y 2 + 3y + 2. 2. Example 2: Check whether 2y + 1 is a factor of the polynomial 4y 3 + 4y 2 – y – 1 or not using the factor theorem. Solution: Let's equate the given binomial 2y + 1 = 0. ∴ y = -1/2 Substitute y = -1/2 in the given polynomial equation 4y 3 + 4y 2 – y – 1. ⟹ 4( -1/2)3 + 4(-1/2)2 – (-1/2) – 1 = -1/2 + 1 + 1/2 – 1 = 0 The remainder = 0, thus, 2y + 1 is a factor of the polynomial equation 4y 3 + 4y 2 – y – 1. View Answer > go to slidego to slide Great learning in high school using simple cues Indulging in rote learning, you are likely to forget concepts. With Cuemath, you will learn visually and be surprised by the outcomes. Book a Free Trial Class Practice Questions on Factor Theorem Check Answer > go to slidego to slide FAQs on Factor Theorem What is Factor Theorem? The factor theorem states that if f(x) is a polynomial of degree n greater than or equal to 1, and 'a' is any real number, then (x - a) is a factor of f(x) if f(a) = 0. It is mainly used to factor the polynomials and to find the n roots of the polynomials. How do you Use the Factor Theorem? We can use the below-given procedure to factor the polynomial using the factor theorem: Step 1: Use the synthetic division of the polynomial method to divide the given polynomial g(y) by the given binomial (y−a) Step 2: After the completion of the division, confirm whether the remainder is 0. If the remainder is not zero, then it means that (y-a) is not a factor of g(y). Step 3: Using the division algorithm, write the given polynomial as the product of (y-a) and the quadratic quotient q(y) Step 4: If it is possible, factor the quadratic quotient further by splitting the middle term method Step 5: Express the given polynomial as the product of its factors. What is the Factor Theorem Formula? As per the factor theorem, (y – a) can be considered as a factor of the polynomial g(y) of degree n ≥ 1, if and only if g(a) = 0. Here, a is any real number. The formula of the factor theorem is g(y) = (y – a) q(y). It is important to note that all the following statements apply to any polynomial g(y): (y – a) is a factor of g(y). g(a) = 0. The remainder becomes zero when g(y) is divided by (y – a). The solution to g(y) = 0 is a and the zero of the function g(y) is a. Explain Factor Theorem With an Example. Let's use the factor theorem to find whether y+2 is a factor of the polynomial g(y) = y 3 + 3y 2 + 5y + 6 or not. g(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6 g(-2) = –8 + 12 – 10 + 6 g(-2) = 0 This tells us that (y+2) is a factor of y 3 + 3y 2 + 5y + 6. We can also apply this in reverse. What is the Importance of the Factor Theorem? Factor theorem is mainly used to factor the polynomials and to find the n roots of that polynomial. It is a special kind of the polynomial remainder theorem that links the factors of a polynomial and its zeros. The factor theorem removes all the known zeros from a given polynomial equation and leaves all the unknown zeros. The resultant polynomial has a lower degree in which the zeros are a lot easier to find. What is the Difference Between Factor Theorem and Remainder Theorem? The difference between the factor theorem and the remainder theorem is that the remainder theorem relates the remainder of the division of a polynomial by a binomial with the value of a function at a point. The factor theorem relates the factors of a given polynomial to its zeros. Where do we Use the Factor Theorem in Real Life? In real life, factoring can be useful while exchanging money, dividing any quantity into equal pieces, understanding time, and comparing prices. What is the Importance of the Remainder Theorem and Factor Theorem? Both the factor theorem and the remainder theorem come in handy to find the factors of a polynomial without using the other methods like synthetic division, long division, or any other traditional methods of factoring. Is Factor Theorem and Remainder Theorem the Same? No, the factor theorem and remainder theorem are not the same. While the remainder theorem relates the remainder of the division of a polynomial by a binomial with the value of a function at a point. The factor theorem relates the factors of a given polynomial to its zeros. Q1: What is the value of k for which (x−k) is a factor of p(x) = x5 − k2x3 + 2x + k − 3? Q2: Which of the following is a factor of the polynomial p(x) = x3 + 5x2 − 4x − 20? Q3: If p(z) = 2z⁴ + 3z3 + 2pz2 + 3z + 6 is perfectly divisible by (z + 2), then what is the value of p? Q4: Which of the following is a factor of 6x3 + 5x2 - 2x - 1? Q5: For a polynomial p(x) = xⁿ − 1, where n is some natural number, p(x) is divisible by (x + 1) for ______ of n. 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5843	https://www.omnicalculator.com/math/euclidean-distance	Board Last updated: Euclidean Distance Calculator It may sound fancy, but the Euclidean distance is nothing more than the distance the way we think of it every day: discover more on how to calculate it with our Euclidean distance calculator! Here you will learn: What is the Euclidean distance? Formulas: The distance between two points and the distance between three points; The distance between a point and a line; The distance between two lines. Other uses of the Euclidean distance. What is the Euclidean distance? The Euclidean distance is a metric defined over the Euclidean space (the physical space that surrounds us, plus or minus some dimensions). In a few words, the Euclidean distance measures the shortest path between two points in a smooth n-dimensional space. We can define the Euclidean distance only in flat spaces: on curved surfaces, strange things happen, and straight lines are not necessarily the shortest! How do I calculate the Euclidean distance? The Euclidean distance is defined through the Cartesian coordinates of the points under analysis. We can think of it as the translation vector between two points. In our Euclidean distance calculator, we teach you how to calculate: The Euclidean distance between two or three points in spaces form one to four dimensions; The Euclidean distance between a point and a line in a 2D space; and The Euclidean distance between two parallel lines in a 2D space. Euclidean distance between two points To find the Euclidean distance between two points. you need to know the coordinates of these points. Take a generic point p. We can write its coordinates as: p=(p1,p2,p3,...) The number of components depends on the dimensionality of the space. To calculate the distance between the point p and the point q, we apply a generalized form of the Pythagorean theorem: d(p,q)=(q1−p1)2+(q2−p2)2+…=i=1∑n(qi−pi)2 Where n is the dimensionality of the space. As you can see, it's pretty easy! Euclidean distance of three points If we add a point, we will find three possible distances; three points, after all, always define a triangle (possibly degenerated). Taking the triple of points p, q, and r, we can find the following two points distances: d(p,q)d(q,r)d(p,r)=i=1∑n(qi−pi)2=i=1∑n(qi−ri)2=i=1∑n(pi−ri)2 This concept can obviously expand to increasingly larger n-tuples. Euclidean distance between a point and a line To find the distance from a point to a line (in two dimensions), we need to consider the fact that a line is nothing but a collection of points satisfying an equation. The distance between a point and a line is then the minimum distance between the point and the line. The segment connecting the point to the point on the line satisfying the condition above is perpendicular to the line itself. To find the distance, we use this formula: d=a2+b2∣a⋅p1+b⋅q1+c∣ Where a, b, and c are the coefficient of the equation of the line in the form: a⋅x+b⋅y+c=0. If we are considering the more commonly found expression y=m⋅x+c, we would use this formula (equivalent in every way): d=m2+1∣m⋅p1+q1+c∣ Euclidean distance between two parallel lines To calculate the distance between two parallel lines we use the following equation: d=a2+b2∣c2−c1∣ The lines have equations: a1⋅x+b1⋅y1+c1; and a2⋅x+b2⋅y1+c2. However, there is a strong constraint over the choices of coefficients. Since two non-parallel lines would cross at a certain point in the plane, giving a trivial distance d=0, we define the Euclidean distance only in the case of parallel lines. This gives us quite some helpful rules: a1=a2=a; b1=b2=b; However, c1=c2. Using the slope-intercept form of the equation of a line, the formula for the distance becomes: d=m2+1∣c2−c1∣ A different point of view: Euclidean distance as a distance between points in a set The Euclidean distance is becoming an important concept in machine learning (the less sci-fi version of AI), where the distance between points in arbitrary spaces of features is measured with metrics. The most commonly used ones are: The Minkowski distance; The Manhattan distance; and The Euclidean distance. Each of them is appropriate for certain data: the Manhattan distance for integer values and the Euclidean distance for real-valued data. The Minkowski distance is a generalization of both. It is interesting to notice how a physical, and concrete, concept (the distance between two points) is translated into a distance in a space of features: the difference is from calculating the distance between your house and the one of a friend to a generalized distance between two colors, two car models... How do I use the Euclidean distance calculator? First, choose the objects you are calculating the distance: for points, you can also choose the dimensionality of the space. Then, insert the coordinates of the points or the parameters of the lines. We will show both the distance and the steps to calculate it. There is nothing more to it! Euclidean distance and beyond The Euclidean distance is a particular — yet common — way to measure the distance. We made more calculator specific to certain topics and problems: The 2D distance calculator; Length of a line segment calculator; Coordinate distance; and Distance between two points' FAQs What is the distance between two parallel lines with equations y = 2x + 3 and y = 2x - 4? The distance is 3.13. To calculate it, We identify the parameters of the lines: m=2 (the value is the same for both lines since they are parallel); and c₁ = 3, and c₂ = -4. We apply the equation for the distance between two parallel lines: d = \| c₁ - c₂ \|/(√[m² + 1]) d = \| 3 + 4 \|/(√[2² + 1] = 3.13 How do I calculate the distance between a point and a line? To calculate the distance between a point and a line, follow these steps: Define the coordinates and parameters of the objects. Calculate the distance using the formula: d = \| m × p₁ + q₁ + c \|/(√[m² + 1]) 3. And that's it! To compute the distance, we had to calculate the area of a triangle in coordinates space and then calculate its height. Is the distance formula still valid in a 4-dimensional space? Yes: the concept of distance exists in any Euclidean space with an arbitrary number of dimensions. It may be hard to visualize a 4-dimensional space, though. Imagine a car that moved from a point A to a point B in a certain time t. If you take time as your fourth dimension, you can see that, if you freeze the movement, the car is not moving: we need the fourth dimension to define the distance! What is the distance between the point p = (1,2) and the point q = (2,3)? d = 1.414. To calculate the distance between these two points on a plane, we simply apply the formula: √[(p₂ - p₁)² + (q₂ - q₁)²] In this case: √[(2 - 1²) + (3 - 2²] = √2 = 1.414 Notice how the distance between these two particular points corresponds both to the hypotenuse of a right triangle and the diagonal of a square. Least Common Multiple Calculator Absolute Value Calculator Interval Notation Calculator Least Common Denominator Calculator Null Space Calculator Trig Identities Calculator Partial Fraction Decomposition Calculator Unit Rate Calculator Exponential Function Calculator 0 Dimensions and problem type © Omni Calculator 0 First point 0 Second point Result Share result Reload calculatorClear all changes Did we solve your problem today? Yes Check out 47 similar coordinate geometry calculators 📈 Average rate of change Bilinear interpolation Catenary curve Share Calculator Euclidean Distance Calculator Share Calculator Euclidean Distance Calculator Learn more about these settings
5844	https://www.youtube.com/watch?v=Q763kZCYeyk	1-86 \| Determine their required diameters of rod \| stress \| mech of materials rc Hibbeler Engr. Adnan Rasheed Mechanical 20400 subscribers 48 likes Description 2065 views Posted: 26 Jan 2024 1–86. The two aluminum rods support the vertical force of P = 20 kN. Determine their required diameters if the allowable tensile stress for the aluminum is sallow = 150 MPa. Dear Viewer You can find more videos in the link given below to learn more and more Theory Video Lecture of Mechanics of Materials by Beer & Johnston 287 Problem solutions of all chapter of Mechanics of Materials by Beer & Johnston Chapter 12 Deflection of Beams and Shaft of Mechanics of Materials rc hibbeler (09th Edition) Chapter 11 Design of Beams and Shafts of Mechanics of Materials rc hibbeler (09th Edition) Chapter 9 Stress Transformation of Mechanics of Materials rc hibbeler (09th Edition) Chapter 8 Combined Loading of Mechanics of Materials rc hibbeler (09th Edition) Chapter 7 Transverse Shear of Mechanics of Materials rc hibbeler (09th Edition) Chapter 6 Bending of Mechanics of Materials rc hibbeler (09th Edition) Chapter 5 Torsion of Mechanics of Materials rc hibbeler (09th Edition) Chapter 4 Axial Loading of Mechanics of Materials rc hibbeler (09th Edition) Chapter 3 Mechanical Properties of Materials of Mechanics of Materials rc hibbeler (09th Edition) Chapter 2 Strain of Mechanics of Materials rc hibbeler (09th Edition) Chapter 1 Stress- Internal Loading of Mechanics of Materials rc hibbeler (09th Edition) mechanicsofsolids #strengthofmaterialstutorials #mechanicsofmaterials #combinedloading engineersacademy #solidmechanics1 #solidmechanics2 #solidmechanics #mechanicsofstructure #solidmechanics #bendingmomentdiagram #shearforcediagram #stress #internalloading #normalstress #shearstress #strain 7 comments Transcript: welcome back problem 1-86 this problem is taken from chapter number one stress and book name is mechanics of material by RC hibler so statement is the two aluminium rods supports the vertical force of 20 Kon determine their required diameters if the allowable tensile stress for the aluminium is 150 megapascal so you can see this is Rod ab and this is Rod AC that is acted upon by 20 k load acts downwards so due to this load there will be rile stress in Rod AB as well as in AC and this Rod AB makes an angle of 45° with the horizontal so you have to find the diameter of Rod AB as well as diameter of Rod AC so let's start with the solution so first step is that we will take this point a and we will draw the free body diagram so p is acting downward so this Rod exert a force in this direction on point a as well as this Rod exert a force in this direction which is f a c and this is f a b clear so we will draw the free body diagram free body [Music] diagram so let this is your point a clear this is a so at a we have P which is acting downward this is Rod force in Rod the rod AB exert a force F AB at Point a and this makes an angle of 45° similarly this Rod AC will exert a force in this direction f a c clear and this is your point a so we'll apply equation of equilibrium so the first equation of equilibrium is that sum of all forces along y direction must be equal to zero and upward forces is taken as positive so you can see faab will have two component one component will be along X Direction clear one is X component clear which is f a COS of 45° and one one will be its y comp component which will be the vertical component and this component will be f a s of 45° so you can see one component which is f AB s of 45° clear and this P which is 20 Kon that is acting downward there some must be equal to 0 so I will write - 20 is equal to 0 so F AB will be equal to 20 / s of 45° and when you calculate this Force so force in member force exerted by member a at Point a will be equal to 28. 284 Kil newon now we'll find this F A by using equation of equilibrium that sum of all forces along X direction must be equal to zero and force in this direction is taken as positive so F A is in opposite direction so I will write minus F A and this component f a c of 30 45° in positive direction so F AB COS of 45° is equal to 0 so F A will be equal to f a COS of 4 45° so faab is 28.2 into COS of 45° and when you calculate it you will get this Force offered by member a c at Point a will be equal to 2 20 Kon 20 Kil newon now this are the forces that is acted upon by this member on point a and this member uh this exert a force in this direction and this exert force in this direction as as a result due to Newton Third Law this due to this load P it will cause the tensile strength stresses in this rods so we will find this we have been given that stress we know that stress is equal to force per unit area clear so we have been given tensile allowable tensile stress allowable tensile stress is equal to force per unit area so what we will do is that for rod for Rod AB we have allowable stress is equal to F AB / by area of Rod AB so we have allowable stress which is 150 map Pascal so 150 into 10^ 6 is equal to f a which is 28.2 Kon so kilo is converted into Newton divide by this is circular Rod so it's area will be P Pi by 4 d AB squ so I by 4 d AB squar so from here you will get da a² is equal to 28.2 into 10^ 3 / < by 4 into 150 10^ 6 and if you take square root on both sides clear so you will get diameter of Rod AB will be equal to 0.015 M and in millimeter D diameter of Rod AB will be equal to 15.5 M mm sorry now we'll find for Rod AC so for Rod AC we have allowable tensile stress is equal to F A / a area of AC so F A is 20 K so 20 into 10^ 3 / area is < by 4 we are interested to find out diameter so da a squ allowable stress is 150 megapascal so 150 into 10^ 6 is = to 20 into 10^ 3 / < / 4 into D a² so from here D a² will be equal to 20 10^ 3 / < by 4 150 into 10^ 6 and if you take square root on both sides so you will get diameter of Rod AC will be equal to 0.0 130 M which is equal to 13 m m and that is the answers of this question so diameter of Rod AB is 15.5 while diameter of Rod AC is 13 mm for the above given load and that was all about this problem 1-86 I hope you have enjoyed this video and you have learned from it those who are new to my channel then subscribe it and don't forget to press the Bell icon so that you can get notification about my latest videos also share my videos with your friends and help me in growing further if you have any question you can ask me in comment section thank you for watching
5845	https://brainly.com/question/32386455	[FREE] Give the number of lone pairs around the central atom and the geometry of the ion $ClO_3^-$. - brainly.com 8 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +33,2k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +39k Ace exams faster, with practice that adapts to you Practice Worksheets +6,5k Guided help for every grade, topic or textbook Complete See more / Chemistry Textbook & Expert-Verified Textbook & Expert-Verified Give the number of lone pairs around the central atom and the geometry of the ion Cl O 3−. 2 See answers Explain with Learning Companion NEW Asked by Larkinlover9601 • 05/17/2023 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 26342314 people 26M 0.0 0 Upload your school material for a more relevant answer In the ClO3- ion, there are three bonding pairs of electrons (between chlorine and oxygen) and one lone pair of electrons on the central chlorine atom. This gives us a total of four electron pairs around the central atom. The ClO3- ion, also known as chlorate ion, consists of one central chlorineatom bonded to three oxygen atoms. To determine the number of lone pairs around the central atom, we need to first find the total number of valence electrons in the ion. Chlorine has seven valenceelectrons, while each oxygen atom has six. Therefore, the total number of valence electrons in the ClO3- ion is: 7 + (3 x 6) + 1 = 26 To determine the geometry of the ion, we can use the VSEPRtheory. The VSEPR theory states that electron pairs repel each other, and this determines the shape of the molecule/ion. According to the VSEPR theory, when there are four electron pairs around the central atom, the geometry is tetrahedral. However, since one of the electron pairs is a lone pair, the geometry is distorted. The bond angle between the three bonding pairs of electrons is approximately 109.5 degrees, but the angle between the lonepair and the bonding pairs is slightly less, at around 107 degrees. Therefore, the geometry of the ClO3- ion is distorted tetrahedral. To learn more about lone pair, refer:- brainly.com/question/30886923 SPJ11 Answered by shardiksh890 •24.9K answers•26.3M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 26342314 people 26M 0.0 0 Chemistry for Allied Health - Allison Soult Organic Chemistry - Smith Organic Chemistry - Vollhardt and Schore Upload your school material for a more relevant answer The ion Cl O 3− has 1 lone pair on the central chlorine atom, resulting in a total of 4 electron pairs (3 bonding pairs and 1 lone pair). The molecular geometry is trigonal pyramidal due to the presence of the lone pair, which affects the bond angles. Therefore, the overall shape is influenced by the lone pair, leading to bond angles slightly less than 109.5 degrees. Explanation To determine the number of lone pairs around the central atom in the ion Cl O 3− (chlorate ion), we first need to count the total number of valence electrons. Valence Electrons Calculation: Chlorine (Cl) has 7 valence electrons. Each oxygen (O) atom has 6 valence electrons. Since there are three oxygen atoms, this gives us: 3×6=18 valence electrons. Since the ion has a -1 charge, we add one more electron, resulting in a total of: 7+18+1=26 valence electrons Drawing the Lewis Structure: Place Cl as the central atom and connect it to the three O atoms with single bonds (using 6 electrons). This leaves us with 26−6=20 electrons. To complete the octets of the oxygen atoms, we will add 6 electrons (as three lone pairs) on each O, consuming another 18 electrons. This leaves us with 2 electrons remaining, which will be placed as a lone pair on the Cl atom. Lone Pairs and Bonding Pairs Count: Therefore, according to the structure: There are three bonding pairs (one for each Cl-O bond) and one lone pair on the Cl atom. This gives a total of 3 bonding pairs+1 lone pair=4 electron pairs. Determining the Geometry: According to VSEPR (Valence Shell Electron Pair Repulsion) theory, having four electron groups around the central atom (including lone pairs) gives a tetrahedral arrangement of electron pairs. However, the presence of the lone pair affects the geometry of the ion. The shape, which is defined by the positions of the atoms only, is called 'trigonal pyramidal.' The angle between the Cl-O bonds is slightly less than 109.5 degrees, generally around 107 degrees due to lone pair repulsion. Thus, the chlorate ion Cl O 3− has 1 lone pair around the central chlorine atom, and its molecular geometry can be described as trigonal pyramidal. Examples & Evidence For instance, you may compare the geometry of Cl O 3− with that of N H 3 (ammonia), which also has a trigonal pyramidal shape due to one lone pair on the nitrogen atom. Another example is C H 4 (methane), where all four electron pairs are bonding pairs leading to a perfect tetrahedral shape. The valence electron counting and geometry assessment can be corroborated by referencing standard chemistry textbooks that cover VSEPR theory and Lewis structures. Thanks 0 0.0 (0 votes) Advertisement Community Answer This answer helped 8852085 people 8M 0.0 0 Final answer: ClO³− has a total of 26 valence electrons and chlorine as the central atom without any lone pairs. After the bonds are formed, each oxygen has two lone pairs. The resulting geometry of the ClO³− ion is trigonal planar. Explanation: When determining the number of lone pairs and the geometry of a molecular ion like ClO³−, we use Valence Shell Electron Pair Repulsion theory (VSEPR) to help us understand the spatial arrangement of atoms. In the case of ClO³−, the central atom chlorine has seven valence electrons, each oxygen atom contributes six valence electrons, and the extra negative charge adds one more electron for a total of 26 valence electrons. After placing three bonding pairs between chlorine and each oxygen, we subtract six electrons (two for each bond) from the total, leaving us with 20 electrons to be placed as lone pairs. According to VSEPR theory, these remaining electrons will be distributed to minimize electron-pair repulsions, with lone pairs occupying equatorial positions around chlorine. In this arrangement, chlorine has no lone pairs directly associated with it, while each oxygen has two lone pairs. The molecular geometry for ClO³− will be trigonal planar, as defined by the three regions of electron density (the triple bond between chlorine and each oxygen atom). This results in a flat, symmetrical shape with bond angles of approximately 120 degrees. Answered by DanielDayLewis •25.9K answers•8.9M people helped Thanks 0 0.0 (0 votes) Advertisement ### Free Chemistry solutions and answers Community Answer Give the number of lone pairs around the central atom and the molecular geometry of scl2. Community Answer A. What is the electron-pair geometry for N in NOCl? _ There are_ lone pair(s) around the central atom, so the geometry of NOCl is ._ B. What is the electron-pair geometry for P in PCl3? There are__ lone pair(s) around the central atom, so the geometry of PCl3 is .______ Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain? Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __ Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction. Community Answer 4.8 52 Which statements are TRUE about energy and matter in stars? Select the three correct answers. Al energy is converted into matter in stars Only matter is conserved within stars. Only energy is conserved within stars. Some matter is converted into energy within stars. Energy and matter are both conserved in stars Energy in stars causes the fusion of light elements Community Answer 4.5 153 The pH of a solution is 2.0. Which statement is correct? Useful formulas include StartBracket upper H subscript 3 upper O superscript plus EndBracket equals 10 superscript negative p H., StartBracket upper O upper H superscript minus EndBracket equals 10 superscript negative p O H., p H plus P O H equals 14., and StartBracket upper H subscript 3 upper O superscript plus EndBracket StartBracket upper O upper H superscript minus EndBracket equals 10 to the negative 14 power. Community Answer 5 Dimensional Analysis 1. I have 470 milligrams of table salt, which is the chemical compound NaCl. How many liters of NaCl solution can I make if I want the solution to be 0.90% NaCl? (9 grams of salt per 1000 grams of solution). The density of the NaCl solution is 1.0 g solution/mL solution. New questions in Chemistry How does a full octet affect trends among the noble gases? A. A full octet makes the size of all the noble gases the same. B. A full octet makes the noble gases nonreactive. C. A full octet makes the trends stronger among noble gases. D. A full octet makes the noble gases react only with alkali metals. The incomplete table below shows selected characteristics of gas laws. \| Name \| Variables \| Constants \| Equation \| :--- :--- \| \| ? \| pressure, volume \| temperature, moles of gas \| P 1V 1=P 2V 2 \| \| Charles's law \| ? \| ? \| T 1V 1=T 2V 2 \| \| Gay-Lussac's law \| ? \| ? \| ? \| \| combined gas law \| ? \| ? \| T 1P 1V 1=T 2P 2V 2 \| What are the variables in Gay-Lussac's law? A. pressure and volume B. pressure, temperature, and volume C. pressure and temperature The incomplete table below shows selected characteristics of gas laws. \| Name \| Variables \| Constants \| Equation \| :--- :--- \| \| ? \| ? \| ? \| P 1V 1=P 2V 2 \| \| Charles's law \| ? \| ? \| T 1V 1=T 2V 2 \| \| ? \| temperature, pressure \| ? \| T 1P 1=T 2P 2 \| \| Combined gas law \| pressure, temperature, volume \| ? \| ? \| Which constants would most likely be filled in the first row of the table? A. temperature and moles of gas B. volume and temperature C. pressure and volume D. pressure and moles of gas Which of the following elements has the lowest electronegativity? A. Strontium (Sr) B. Magnesium (Mg) C. Calcium (Ca) D. Barium (Ba) How does the energy of core electrons compare with the energy of valence electrons? A. Only the valence electrons possess energy in an atom. B. The energy of the core electrons is greater than the energy of the valence electrons. C. The energy of the valence electrons is greater than the energy of the core electrons. D. The energy of the core electrons is equal to the energy of the valence electrons. Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
5846	https://www.opastpublishers.com/open-access-articles-pdfs/flow-visualisation-by-laser-sheet-in-a-smoketunnel.pdf	Volume 3 \| Issue 1 \| 1 J Math Techniques Comput Math, 2024 Citation: Sayed, A. A., Abhishek, E., Khan, S. H., Reddy, K. S., Chandrasekhar, P., et al. (2024). Flow Visualisation by Laser Sheet in a Smoke-Tunnel. J Math Techniques Comput Math, 3(1), 1-11. Flow Visualisation by Laser Sheet in a Smoke-Tunnel Corresponding Author Arafa Atique Sayed, Department of Aerospace Engineering, Chandigarh University, India. Submitted: 2023, Dec 11; Accepted: 2023, Dec 27; Published: 2024, Jan 10 Arafa Atique Sayed, Eda Abhishek, Shan Hasan Khan, Karnati Surya Reddy, Polaki Chandrasekhar and Dharmahinder Singh Chand Abstract The application of wind tunnel for visualizing flow over any bluff or streamline body plays an indispensable role in concepting various aerodynamical phenomenon such as formation of vortices, generating lift and drag, calculation of velocity vector and pressure profiles etc. Flow visualization through smoke and laser optics in a subsonic wind tunnel to capture vortex formation over airfoils, cylindrical and conical bodies is the experimental approach which has been studied and manifested elaborately in the article. Using smoke as a seeding material, the flow of the same is allowed to pass over the model which is illuminated via a double pulse laser beam to visualize the formation of vortex and capture the same for future references. The experimentation of the above models is carried out in a Honeycomb subsonic wind tunnel in the Aerodynamics Lab of Aerospace Engineering in the university. The article will cover Computational Fluid Dynamics (CFD) performed on ANSYS for the particular NACA series, Cross correlation PIV for plotting velocity vector for the captured images of the seed particles and the respective images of the entire experimental set-up. The images captured during this experiment can be later used by the aerodynamicists for optimizing aerodynamic efficiency of the experimented models depending on the vortex formation and flow pattern. The images can also be used for advance calculations required in Particle Image Velocimetry (PIV) for calculating velocity of every speed particle. Research Article Journal of Mathematical Techniques and Computational Mathematics ISSN: 2834-7706 Keywords: NACA 54118; Wind Tunnel; DSLR Camera; Cylindrical Lens; Smoke; LASER; Flow-Visualization. 1. Introduction The primary usage of wind tunnel is to examine scale models of various aerodynamical bodies like aircrafts and spacecrafts but at the same time it has also been an indispensable area of research in numerous fieldss for aerodynamicists. Apart from testing and analysing the models, wind tunnels are also used for observing different phenomenon experienced by a body while subjected to a flow. One such area of interest is the concept of visualising the formation of vortices over any bluff or streamline body in order to insinuate the efficiency of the flow and further to optimize the same for designing or modifying bodies. Practices like Smoke visualisation, Laser visualisation and Combined visualisation are current such methods to achieve the necessary visuals for the research out of which Combined visualisation has been adopted by us to perform the experiment and further carry out the probation. There are two methods usually followed after capturing the vortices i.e. Laser Doppler Anemometry (LDA) and Particle Image Visualisation (PIV) where the velocity vector of the seeding material is calculated by determining the relative axial distance with respect to the change in time period. PIV has been chosen out of the two because of its non-invasive method of optics to determine the movement of seed particle over a body subjected to a flow. 2. Objective I. To visualize the flow of air in a wind tunnel: The first objective is to generate smoke in the wind tunnel and use laser beams to visualize the flow of air. This will help researchers to understand how the air moves through the wind tunnel. II. To identify the areas of turbulence: By using smoke and laser beams, researchers can identify areas of turbulence in the air flow. This will help to improve the design of wind tunnels and reduce turbulence in future experiments. III. To capture vortices: Another objective is to capture vortices that form in the air flow. Vortices are areas of swirling air that can affect the performance of objects moving through the air. By understanding how vortices form and behave, researchers can improve the design of aircraft, cars, and other vehicles. IV. To improve the accuracy of wind tunnel experiments: By visualizing the flow of air in a wind tunnel, researchers can improve the accuracy of their experiments. This will help to ensure that wind tunnel data is reliable and can be used to make informed decisions about the design of vehicles and other objects. Department of Aerospace Engineering, Chandigarh University, India Volume 3 \| Issue 1 \| 2 J Math Techniques Comput Math, 2024 V. To advance the field of aerodynamics: Ultimately, the main objective of visualizing the flow in a wind tunnel by smoke generation and laser beam and capturing vortices is to advance the field of aerodynamics. By gaining a better understanding of how air flows, researchers can improve the performance and efficiency of vehicles, reduce energy consumption, and make transportation safer and more sustainable. VI. To utilize cross correlation: To plot the velocity vector of the seed particles computationally. 3. Literature Review The visualization of flow in a wind tunnel is essential to understand the aerodynamic behaviour of an object in motion. Smoke generation and laser beam techniques are commonly used to study the flow pattern and capture vortex in a wind tunnel. This literature survey aims to provide an overview of the research conducted in this field. Smoke Visualization: Smoke visualization is a popular method to visualize the flow pattern in a wind tunnel. The method involves the injection of smoke particles into the test section of the wind tunnel, which are then carried away by the flow. Smoke visualization can be used to study the effects of different flow parameters, such as velocity, pressure, and temperature, on the flow pattern. According to a study by smoke visualization was used to investigate the flow pattern around a bluff body . The study demonstrated that smoke visualization is an effective tool to visualize the flow pattern and capture vortex in a wind tunnel. Smoke generation is a cost-effective and straightforward method of visualizing the flow pattern in a wind tunnel. In this method, smoke particles are injected into the wind tunnel, and their movement is tracked to understand the flow pattern. Smoke visualization is particularly useful in identifying regions of turbulence and separated flows around objects. Laser Visualization: Laser visualization is another method to study the flow pattern in a wind tunnel. The method involves the use of a laser beam to illuminate the flow field. The light scattered by the particles in the flow field is then captured by a camera, providing a visual representation of the flow pattern. A study by Jambunathan et al. used laser visualization to study the flow pattern around a circular cylinder . The study demonstrated that laser visualization is an effective tool to visualize the flow pattern and capture vortex in a wind tunnel. Laser visualization, on the other hand, uses a laser beam to illuminate the flow field . The scattered light from the particles in the flow field is then captured using a high-speed camera to create a visual representation of the flow pattern. This technique can provide high-resolution images of the flow field and can be used to study the details of the flow around objects in the wind tunnel. Combined Visualization: Smoke and laser visualization can also be combined to provide a more comprehensive understanding of the flow pattern in a wind tunnel. A study by Zhang et al. used smoke and laser visualization to study the flow pattern around a delta wing . The study demonstrated that combined visualization is an effective tool to visualize the flow pattern and capture vortex in a wind tunnel. Smoke generation and laser beam visualization are effective techniques to visualize the flow pattern and capture vortex in a wind tunnel . These methods have been widely used in the field of aerodynamics to study the effects of different flow parameters on the flow pattern. Combining smoke and laser visualization can provide a more comprehensive understanding of the flow pattern in a wind tunnel. The combined use of smoke and laser visualization techniques offers the advantage of both methods. The smoke particles provide a visual representation of the flow pattern, while the laser beam can be used to obtain quantitative data about the velocity and direction of the flow . The combination of these techniques has been used to study complex flow patterns, such as those around delta wings and other complex geometries. Moreover, recent advancements in digital image processing techniques have made it possible to obtain more accurate and reliable flow visualization data. For instance, Particle Image Velocimetry (PIV) is a non-intrusive technique used to obtain quantitative data about the velocity of the flow field . This technique involves the use of two laser beams and a high-speed camera to track the movement of particles in the flow field. The PIV technique has been widely used to study the flow pattern around objects in a wind tunnel. Flow visualization techniques, such as smoke generation and laser visualization, have been widely used to study the flow pattern and capture vortex in a wind tunnel . The combination of these techniques and recent advancements in digital image processing techniques have improved our understanding of aerodynamic behaviour and can be used to design and optimize various engineering applications. Flow visualization can also be used to study the effects of different geometries on vortex capture. For example, a researcher might compare the vortex patterns formed by a circular cylinder to those formed by a square cylinder, or study the effects of varying the angle of attack or the aspect ratio of an aerofoil. 4. Proposed Approach PIV cross correlation: It’s a method in which we use two consecutive frames, shot by a camera with a shutter speed of 50 Hz. With the help of cross correlation, we can find how much distance a particle has moved in a particular window by comparing the two consecutive photos. We divide the whole photo into 4 or more windows and then find the relative distance a particle has moved in the x-direction and y-direction with respect to the first shot. PIV cross-correlation is commonly used in wind tunnel testing to measure the velocity field of the airflow around a model or object. In a wind tunnel experiment, the PIV Volume 3 \| Issue 1 \| 3 J Math Techniques Comput Math, 2024 system captures a series of images of the flow around the object, with the tracer particles seeded into the airflow. The PIV cross-correlation analysis then calculates the velocity field of the airflow by comparing the positions of the tracer particles between successive images. This allows researchers to study the flow properties of the wind tunnel experiment in detail, such as the turbulence levels, vortices, and pressure distribution. Wind tunnel experiments using PIV cross-correlation can be used to optimize the aerodynamic design of objects, such as aircraft, cars, and wind turbines. The PIV results provide insights into how airflow interacts with the object, allowing researchers to adjust the design to optimize performance, reduce drag, and increase efficiency. PIV cross-correlation is a powerful tool for understanding the complex flow behavior in wind tunnel experiments and can provide valuable insights into the design and performance of various objects and systems. 5. Components Used 6 5. COMPONENTS USED COMPONENTS VISUALS 1. Subsonic Wind Tunnel. 2. Smoke Generator. 3. Laser Diode (530nm). Volume 3 \| Issue 1 \| 4 J Math Techniques Comput Math, 2024 7 4. Prisms -2 (power of prisms = 10D & 5D) (D = Diopter). 5. Thermocol foam block. 6. NACA 54118 and force models. 7. DSLR Camera. Volume 3 \| Issue 1 \| 5 J Math Techniques Comput Math, 2024 8 8. Propylene Glycol (Aqueous Solution of Propylene Glycol is used as the seeding material) Table 1 Components 6. RESEARCH METHODOLOGY I. Cover the wind tunnel with black sheet of paper to ensure complete blackness while performing experiment so that illumination is proper and the vortices are captured appropriately. II. Mount an aerofoil or a cylindrical body or any force model in the test section of subsonic wind tunnel. III. Switch on the smoke generator and heat it up to 250 Kelvin (K). IV. Mount the laser beam at an angle perpendicular to the body and place it at a focal length of 5 cm from the lens. V. Place a DSLR camera in the settling chamber or the test section of the wind tunnel and set its frequency to 50 Hz. VI. Switch on the laser and allow the smoke to flow over the body in the test section. VII. Start the recording. VIII. Observe the vortices. IX. Capture the vortices formed over the body. X. Segregate two consecutive frames from the images captured and use it to calculate the relative distance between two seeding particles. XI. Obtain the velocity vector by importing the images into MATLAB and coding for the same in it. Table 1: Components 6. Research Methodology I. Cover the wind tunnel with black sheet of paper to ensure complete blackness while performing experiment so that illumination is proper and the vortices are captured appropriately. II. Mount an aerofoil or a cylindrical body or any force model in the test section of subsonic wind tunnel. III. Switch on the smoke generator and heat it up to 250 Kelvin (K). IV. Mount the laser beam at an angle perpendicular to the body and place it at a focal length of 5 cm from the lens. V. Place a DSLR camera in the settling chamber or the test section of the wind tunnel and set its frequency to 50 Hz. VI. Switch on the laser and allow the smoke to flow over the body in the test section. VII. Start the recording. VIII. Observe the vortices. IX. Capture the vortices formed over the body. X. Segregate two consecutive frames from the images captured and use it to calculate the relative distance between two seeding particles. XI. Obtain the velocity vector by importing the images into MATLAB and coding for the same in it. XII. Validate your results with the Computational Fluid Dynamics (CFD) simulated on ANSYS. 7. Wind Tunnel Specifications 1. Test Section Size 600 mm x 600mm x 2000 mm 2. Contraction Ratio 9:1 3. Settling Chamber 2300 mm x 2300 mm 4. Maximum Speed 50 m/s 5. Fan Axial Flow fan, max rpm 1440 Table 2: Wind Tunnel Specifications 8. Calculations We know, f = 50 Hz and ∆x = 0.1303 cm since, ∆v = ∆x/∆t and f = 1/t t = 1/f t = 1/50 therefore, ∆v = 0.1303 × 50 ∆v = 6.516 m/s Percentage Error (%) = 6.516 – 6.34 x 100 6.516 ⸫ % error = 2.7 % Volume 3 \| Issue 1 \| 6 J Math Techniques Comput Math, 2024 9. Computational Observation Percentage Error (%) = 6.516 – 6.34 x 100 6.516 ⸫ % error = 2.7 % 9. Computational Observation a) 3D Analysis: Figure 1 Absolute pressure analysis over test section Figure 3 Aerofoil in Test Section Figure 5 Velocity Profile Figure 7 Velocity v/s Distance Figure 2 Total pressure analysis over test section Figure 4 NACA 54118 Figure 6 Mass flow rate v/s Distance Figure 8 Pressure v/s Distance b) 2D Analysis Volume 3 \| Issue 1 \| 7 J Math Techniques Comput Math, 2024 Figure 2 Total pressure analysis over test section Figure 8 Pressure v/s Distance b) 2D Analysis Figure 9 Geometry Figure 11 Mesh Figure 13 Mesh Sizing Figure 15 Mesh Inflation Figure 12 Mesh Details Figure 14 Mesh Quality Figure 16 Batch connections and Advanced Mesh Volume 3 \| Issue 1 \| 8 J Math Techniques Comput Math, 2024 Figure 15 Mesh Inflation Figure 10 Geometric Tree Outline Figure 16 Batch connections and Advanced Mesh Figure 17 Statistics of Mesh Figure 19 Boundary Condition for Inlet Velocity @ 5 m/s @ 5-degree AOA Figure 21 Pressure contour of NACA 54118 Figure 23 Velocity contour of NACA 54118 Figure 18 Density and Viscosity of Propylene Glycol used as the Flow Material Figure 20 Boundary Condition for Viscous Model Figure 22 Pressure contour of NACA 54118 Volume 3 \| Issue 1 \| 9 J Math Techniques Comput Math, 2024 Figure 23 Velocity contour of NACA 54118 Figure 20 Boundary Condition for Viscous Model Figure 22 Pressure contour of NACA 54118 Figure 24 Velocity contour of NACA 54118 Figure 25 Velocity vector of NACA 54118 Figure 26 Velocity Vector 10. Experimental Observation Figure 27 Smoke – Laser Visualisation 11. Results and Discussions The computational velocity of the seed particle is 6.34 m/s. The experimental velocity of the seed particle 6.516 m/s. Percentage error = 2.7%. Since the experimental values are approximately equal to the theoretical values, we can successfully use the above method to optimise the efficiency of any bluff or streamline body according to the Figure 24 Velocity contour of NACA 54118 Figure 25 Velocity vector of NACA 54118 Figure 26 Velocity Vector 10. Experimental Observation Figure 27 Smoke – Laser Visualisation 11. Results and Discussions The computational velocity of the seed particle is 6.34 m/s. The experimental velocity of the seed particle 6.516 m/s. Percentage error = 2.7%. Since the experimental values are approximately equal to the theoretical values, we ca use the above method to optimise the efficiency of any bluff or streamline body ac 10. Experimental Observation 11. Results and Discussions The computational velocity of the seed particle is 6.34 m/s. The experimental velocity of the seed particle 6.516 m/s. Percentage error = 2.7%. Since the experimental values are approximately equal to the theoretical values, we can successfully use the above method to optimise the efficiency of any bluff or streamline body according to the requirements and can even save the data for future references. The observation needed to be discussed are as follows: • The black sheet of paper used to cover the wind tunnel ensured that there was complete blackness during the experiment, enabling proper illumination and the accurate capture of vortices formed over the airfoil. The smoke generator was heated to 250 K to produce smoke, which allowed the visualization of the flow over the airfoil. • The laser beam was mounted at an angle perpendicular to the airfoil, and a DSLR camera was used to capture the vortices formed over the airfoil. The frequency of the camera was set to 50 Hz, which provided a high-speed capture of the vortices formed over the airfoil. • The images captured by the camera were imported into MATLAB to obtain the velocity vector. This enabled the calculation of the velocity distribution over the surface of Volume 3 \| Issue 1 \| 10 J Math Techniques Comput Math, 2024 the airfoil. • The results obtained from this experiment were validated with the CFD simulated on ANSYS. The comparison between the results obtained from the experiment and the CFD simulation showed a good agreement, validating the results obtained from the experiment. • In conclusion, the experiment successfully captured the vortices formed over the airfoil, and the results obtained were validated with CFD simulation. The method used in this experiment could be applied to study other aerodynamic phenomena, making it a valuable tool in aerodynamic research. 12. Limitation and Future Work Limitations I. The full-scale measurements of velocity at very high mixtures could not be determined as a limitation of PIV method. II. LDA method was costlier to adopt. Future Work Flow visualization techniques using smoke and lasers in wind tunnels have long been instrumental in studying and understanding fluid dynamics. In the future, these techniques are expected to continue playing a crucial role in various scientific, engineering, and industrial applications. The use of smoke as a flow visualization tool involves introducing a smoke source into the wind tunnel, which allows researchers to observe and analyze the movement and behavior of the smoke particles as they interact with the flowing air. This technique provides valuable insights into the complex patterns and structures of the airflow, enabling a better understanding of aerodynamic phenomena. Looking ahead, the future of flow visualization using smoke and lasers in wind tunnels appears promising. Here are some potential areas of application and scope: I. Aerodynamics and Aerospace Engineering: Wind tunnels equipped with smoke and laser-based visualization techniques will continue to be crucial in studying and optimizing the aerodynamic performance of aircraft, spacecraft, and other vehicles. Researchers can gain insights into boundary layer separation, turbulence, vortices, and other flow phenomena, leading to improved designs and increased efficiency. II. Automotive Industry: Smoke and laser-based flow visualization will aid in developing more streamlined and fuel-efficient vehicles. By studying the flow around cars and trucks, engineers can optimize designs to reduce drag, enhance vehicle stability, and improve overall performance. III. Energy and Environmental Engineering: Wind turbines, thermal power plants, and other energy systems require efficient airflow management. Flow visualization techniques can assist in understanding the complex flow patterns within these systems, leading to improved efficiency, reduced emissions, and better design of air pollution control devices. IV. Architecture and Urban Planning: The impact of wind on buildings and urban environments is crucial for ensuring safety and comfort. Flow visualization can help architects and urban planners understand how airflow interacts with structures, identifying potential areas of turbulence or discomfort. This knowledge can guide the design of buildings, cities, and public spaces for improved ventilation and pedestrian comfort. V. Sports and Athletics: Flow visualization techniques can be used to study the aerodynamics of athletes and sports equipment. Understanding the flow patterns around cyclists, runners, or sports equipment like balls can lead to improved performance and equipment design. 13. Conclusion I. In conclusion, Visualizing air flow through smoke generation and laser beams is essential for understanding how air moves through different structures in a wind tunnel. II. Identifying areas of turbulence and capturing vortices are critical for improving the design of vehicles, buildings, and other objects that move through the air. III. The accuracy of wind tunnel experiments is improved through the use of smoke generation and laser beams, leading to reliable data for making informed decisions. IV. The study of aerodynamics has numerous applications in different industries, and advancements in this field have the potential to revolutionize transportation. V. The use of smoke generation and laser beams in wind tunnel experiments is a powerful tool for visualizing air flow, capturing vortices, and improving the accuracy of experiments in the field of aerodynamics . 14. Availability of Data and Materials No datasets were generated or analysed during the current study. Acknowledgements The opportunity to write this research paper and perform the necessary experiments for the desired observations and results would not be possible without the immense support of Chandigarh University which allowed us to use their hangar and wind tunnel for this paper. We would also like to express our deepest gratitude to Dr. Dharmahinder Singh Chand under whose supervision and guidance, we were able to complete our report and further write a paper on the same. We would also like to thank the lab attendants of the Hangar and the Workshop who showed great support and cooperation while performing the experiments for this paper. Also, various research papers and other such sources helped us a lot to gain valuable information on the respective topic and further aided us to complete the paper for which we are highly grateful. At the last, we would like to take this as an opportunity to thank our parents and families who supported us all the time and guided us through all the difficulties we faced. Competing Interest The authors declare no competing interest. References 1. Böhle, M., Jambunathan, K., Rath, H. J. (2019). Smoke Visualization of the Flow Around a Bluff Body in a Wind Volume 3 \| Issue 1 \| 11 J Math Techniques Comput Math, 2024 Copyright: ©2024 Arafa Atique Sayed, et al. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited. Tunnel. Journal of Visualization, 22(1), 47–59. 2. Jambunathan, K., Böhle, M., Rath, H. J. (2019). Laser visualization of the flow around a circular cylinder in a wind tunnel. Journal of Visualization, 22(3), 533-545. 3. Scherer, T., Buske, G., Schröder, A. (2006). Visualization of the flow field around a delta wing using laser beam techniques. Aerospace Science and Technology, 10(3), 221-230. 4. Zhang, X., Wang, Z., Xu, C. (2018). Flow Visualization of Delta Wing by Smoke and Laser in Wind Tunnel. International Journal of Aerospace Engineering, 2018, 1-10. 5. Smoke flow visualization techniques for wind tunnel testing by A. Trujillo and M. Fonseca. This paper provides an overview of the different types of smoke visualization techniques and their applications in wind tunnel testing. 6. Raffel, M., Willert, C. E., Scarano, F., Kähler, C. J., Wereley, S. T., & Kompenhans, J. (2018). Particle image velocimetry: a practical guide. Springer. 7. Davis, J. M., Eisner, A. D., Wiener, R. W., & Main, C. E. (1997). A flow visualization study of spore release using a wind tunnel-mounted laser light sheet. Plant disease, 81(9), 1057-1065.
5847	https://math.stackexchange.com/questions/3237327/circular-permutation-with-constraints	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Circular permutation with constraints Ask Question Asked Modified 6 years, 4 months ago Viewed 785 times 3 $\begingroup$ If four boys and four girls play tricks, how many ways can they join hands, provided that at least two girls are together? My plan is to determine the circular permutation of the eight (boys + girls), which is equal to 7! and to subtract from it the cases in which at least two girls are separated. That is, only two separate girls, only three separate girls or only four separate girls. How to proceed with this approach? The answer is 7! - 3! 4! which, by the presence of less, suggests the use of the complementary combination. To get to 3! 4 !, I believe that it is enough to calculate the circular permutation of the boys (which will be among the girls), which results in (4 - 1)! = 3 !, and multiply by the girls' permutation: 4 !. Why, considering that the boys should be among the girls, it is not correct to multiply the circular permutations of the two, getting 3! 3! ? combinatorics discrete-mathematics permutations combinations permutation-cycles Share edited May 23, 2019 at 17:25 gmn_1450gmn_1450 asked May 23, 2019 at 17:16 gmn_1450gmn_1450 52922 silver badges1111 bronze badges $\endgroup$ 2 $\begingroup$ If you want to find the number of arrangements with at least two girls together, you should be subtracting the number of arrangements with no two girls together. $\endgroup$ N. F. Taussig – N. F. Taussig 2019-05-23 17:18:51 +00:00 Commented May 23, 2019 at 17:18 $\begingroup$ Yes, I know. I just wrote wrong, thank you very much for the correction. $\endgroup$ gmn_1450 – gmn_1450 2019-05-23 17:21:55 +00:00 Commented May 23, 2019 at 17:21 Add a comment \| 2 Answers 2 Reset to default 4 $\begingroup$ For these circular permutation questions, you can avoid "division by symmetry" arguments by finding some sort of focal point for your circle and placing everyone else around the focal point. This can be accomplished by taking "the youngest" or the "smallest" or some other arbitrary method of selection and placing that person or object into the circle in an arbitrary position first. So, for this, let us take the youngest boy and have him "start" the circle. Now that he is there, we can treat the remaining positions the same as a line. The only way to have it be so that no two girls in this scenario are for it to repeat "boy-girl-boy-girl-boy-girl-..." so let us decide who the girl standing to the right of our youngest boy is. This girl can be chosen in $4$ ways. Then, pick the boy who stands to her right. Then pick the girl to his right, and so on so forth until we are done. This gives $4\times 3\times 3\times 2\times 2\times 1\times 1 = 4!\times 3!$, no division necessary. Share answered May 23, 2019 at 18:35 JMoravitzJMoravitz 81k55 gold badges7272 silver badges124124 bronze badges $\endgroup$ Add a comment \| 2 $\begingroup$ The boys and girls are in the same circle. If you used $3!$ twice, you would be dividing by $4$ twice, once for the boys and once for the girls, which is only valid if they are each in their own independent circles. Since they are only in one circle, there is only one degree of freedom, so you should only divide by $4$ once to eliminate over-counting; this results in $(4!\times 4!)/4=4!\times 3!$. Look at it this way; if you were instead counting linear arrangements of boys and girls where no two girls are together, and also the ends are not both girls, there would only be two possible patterns: $$ BGBGBGBG\qquad \text{ and }\qquad GBGBGBGB $$ Each of these can be completed in $4!\times 4!$ ways, so there are $2\times 4!\times 4!$ linear arrangements. However, since you want circular arrangements, you need to divide by $8$, since the linear arrangements are partitioned into groups of size $8$ which all represent the same circular arrangement. Therefore, the answer is $(2\times 4!\times 4!)/8=3!\times 4!$. Share answered May 23, 2019 at 18:08 Mike EarnestMike Earnest 85.6k1212 gold badges8383 silver badges157157 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics discrete-mathematics permutations combinations permutation-cycles See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 1 Why does the $(n-1)!$ rule not work in all cases of circular permutations? Related 1 Solving permutation problem the harder way 3 Problem with concepts of circular permutation. 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5848	https://approach0.xyz/search/?q=AND%20site%3Amath.stackexchange.com%2C%20OR%20content%3A%242%5Ex%3C%5Cfrac1%7B1-x%7D%24&p=1	Approach Zero @grid: 40x2 / 60%; @place-cell: center; @size: calc(100% / @size @i); transform: rotate(calc(@i 5deg)); border-radius: 30%; border: 1px solid hsla( calc(10 + 4 @i), 70%, 68%, @r.8 ); 100% Approach Zero A math-aware search engine AND site math.stackexchange.com 2 x<1 1−x 2^x<\frac1{1-x} 2 x<1−x 1 Search Help Raw Query Clear [x] AND site math.stackexchange.com 2 x<1 1−x 2^x<\frac1{1-x} 2 x<1−x 1 Search Help Raw Query Clear 195811 79.646Is it true [imath]\left\|\dfrac{1}{2^a}-\dfrac{1}{2^b}\right\| Is it true Let be two real numbers Is it true that For fixed , the left side is maximized when so you just need to show that (No need for absolute values, either.) This in turn means you need for You can then use power series to prove this inequality, with with so every term in the power series for (other than the constant term) has coefficient . @ThomasAndrews thanks for your suggestion Let be two real numbers Is it true that For fixed , the left side is maximized when so you just need to show that (No need for absolute values, either.) This in turn means you need ... ⋅ calculus⋅ real-analysis⋅ inequality⋅ exponential-function⋅ absolute-value 323479 79.27Proving by induction Proving by induction I'm trying to apply an induction proof to show that is the sum of from to . My try to prove 3 is as follows: , from 2: My question is, how could I get to from the last line to prove this formula is right? I don't think you have the right formula. It does not hold for . Two similar questions: math.stackexchange.com/questions/53496/… and math.stackexchange.com/questions/11464/… (But I do not think that this should be closed as duplicate - since OP is asking about mistake in his approach and he is interested only in the proof by induction. I just wanted to point him to similar question.) I'm trying to apply an induction proof to show that is the sum of [im ... ⋅ sequences-and-series⋅ induction⋅ summation 150877 76.617Prove inequality ... and by the Cauchy-Schwarz inequality (see also page 8 of my notes) we have so and actually for any in a pointed neighbourhood of the origin. By replacing with and reciprocating we get and still is stronger than needed. ... ⋅ algebra-precalculus⋅ inequality⋅ induction 177242 75.994Does the coin flipping game terminate? m variables, many values are possible but usually all have zero probability). Note that in practice you cannot perform this experiment: whatever resources you may have (millions of people tossing billions of coins for many many years) you are not guaranteed to ever get a Head. The above therefore relates to a thought experiment that is allowed to go on "forever". How about the followings? The probability that the game will terminate after N tosses = 1 - probability that the game doesn't terminate after N tosses = If you define be the upper bound of the probability, So if there are N tosses and , the probability of the game terminating is less than . As ... ⋅ probability⋅ limits 241201 73.878Expressing a step function picture you have. As for "Did I define it correctly?" - I'm not sure what you mean. Did you define what correctly? If you are asking if your function matches the picture, then yes (except that your picture doesn't have and ) The summation for the area under the curve: ... and the area under the curve is clearly greater than because all sections of the curve lie above the diagonal line . Eq. is only valid for the discret values of defined by Eq.. If we want to extend to continuous we have to use the celling or the floor functions, respectively noted and : Thus, the equation represented on your graph is : ... ⋅ functions⋅ graphing-functions⋅ step-function 103938 69.923Find the supremum of not sure how to show that is the least upper bound and that all the other upper bounds are less than or equal to . I've done similar problems where the supremum is also maximum, but in this case it is not. Can anyone give me a hint as to how to prove this? Hint, assume that the supremum is for some . Can you come to a contradiction? Depending on where to start, one can note that contains arbitrarily small numbers. Hint: is always positive but can get arbitrarily close to 0 If , you have to find an such that : , and this is achievable since , thus there must be an such that: . Now for if is an upper bound, you need to show , but this means ,for all , taking limit we have . Done. ... ⋅ real-analysis 215873 61.088Showing one function is greater than another function (or proving an inequality) Showing one function is greater than another function (or proving an inequality) How would you show that for ? I thought we could define a new function, , where and . Then, move on to calculate the derivative of h and then show that it is negative, implying that h is decreasing. But I did not know how to show that h is negative in Does this answer your question? Prove Exponential Function Inequality: Oh cheers! Thanks for sharing the link. How would you show that for ? I thought we could define a new function, , where [imat ... ⋅ analysis⋅ derivatives⋅ inequality 62282 61.087When the series is convergent depending on ? it equals . The series converges. For , compare to : Even if , the limit still diverges to infinity. Hence the series diverges. @MarkViola Could you please explain it more explicitly In a comment, the OP asked "Could you please explain it more explicitly," which referenced the comment I left "". To address the OP's request, we now proceed. Using for and we have I think you were going to use the formula to point out the mistake in my solution. Sorry for not expressing my thought correctly. ... ⋅ real-analysis⋅ sequences-and-series⋅ convergence-divergence 79528 61.07Is ? there is simple theorem proving this: en.wikipedia.org/wiki/… Of course. It is interesting to prove that the same sum ranging over twin primes is indeed convergent. That is usually done through Brun's sieve. A closely related product with a bit more information about the rate of divergence Hence; Yes , It's true. That's a good theorem! Does it have a name? @StellaBiderman I don't know. but it's well-known in real analysis. @StellaBiderman It has been mentioned here: en.wikipedia.org/wiki/Infinite_product Follows from if . Look at the power series to see a proof. See also here ... ⋅ number-theory⋅ limits⋅ prime-numbers⋅ products 310202 61.069Showing that the exponential expression is positive . However, I fail to show this. It is trivial for , but what for ? It seems trivial to show . What have you tried? for , so is strictly increasing on . Since and thus also one gets +^I just love it when all you ever wanted to know about the exponential is the inequality For , the inequality is equivalent to: or to: that is trivial. If confused by the last step, Jack converts to its Taylor series and the to its geometric series. the derivative is positive for and thus it is increasing. I think you are having doubts about . So you have two answers. range of (0,1) which is increasing, since the derivative is positive. From which is decreasing since becomes negative. ... ⋅ calculus⋅ real-analysis⋅ inequality⋅ exponential-function 12345 ... 10 Links About Query Logs Community AD Open Backend Developer Docs Contribute User Guide This Webpage Sponsorship Sponsor Powered by Akamai Special Thanks Check out the kudos page! Current Index Math StackExchange Art of Problem Solving Help to Add More ... Approach Zero - 2025
5849	https://www.reddit.com/r/HomeworkHelp/comments/13xbbhs/high_school_physics_vectors_help_on_law_of_cosines/	[High School Physics - Vectors] Help on Law of Cosines : r/HomeworkHelp Skip to main content[High School Physics - Vectors] Help on Law of Cosines : r/HomeworkHelp Open menu Open navigationGo to Reddit Home r/HomeworkHelp A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to HomeworkHelp r/HomeworkHelp r/HomeworkHelp Need help with homework? We're here for you! The purpose of this subreddit is to help you learn (not complete your last-minute homework), and our rules are designed to reinforce this. 666K Members Online •2 yr. ago Traffic-Cone189 [High School Physics - Vectors] Help on Law of Cosines Physics This problem uses law of cosines to derive an angle: "Two vectors having equal magnitude A make an angle Q with each other. Find the magnitude and direction". The law of cosines as I know it is: R^2 = a 2 + b 2 - 2ab(cos(θ)) Yet Solutions I've found insist on: R^2 = a2 + b2 + 2ab(cos(θ)) Why does the solution involve inexplically flipping the sign? What am I missing from this problem that necessitates such a change? Some Solutions: Read more Share New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community Top Posts Reddit reReddit: Top posts of June 1, 2023 Reddit reReddit: Top posts of June 2023 Reddit reReddit: Top posts of 2023 Trending topics today Sachin Tendulkar AMA Gaza hospital strike kills 15 Styles & Kravitz spark rumors Medvedev's US Open meltdown Sweeney films flop Bucs waive Shilo Sanders Canadian PM visits Ukraine KPop Demon Hunters tops box office Raiders lose O’Connell to injury Jerry Adler dies at 96 Vikings trade Howell to Eagles Grok 2.5 open sourced Sanders ejected for punch Teenage Dream turns 15 US seeks to deport Garcia Walker TKO shocks Zhang Corenswet's Superman audition Emily in Paris AD dies at 47 US halts major wind project Kilmar Abrego Garcia freed DOJ releases Maxwell transcript Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
5850	https://www.youtube.com/watch?v=HEOBF_8F640	[CSES][Mathematics] Counting Divisors NeatlyStructured 3600 subscribers 43 likes Description 2078 views Posted: 7 Jun 2021 10 comments Transcript: Intro [Laughter] [Music] hello today we'll be solving this problem called counting divisors so we are given n integers and our task is to report for each integer the number of its divisors for example if x is equal to 18 the correct answer is 6 because its divisors are 1 2 3 6 9 and 18. so the first line of the input contains an integer and can be as large as 10 to the fifth then follow n integers that can be as large as a median and for each integer we have to output the number of divisors so let's go to the drawing board and analyze this problem Drawing Board so this was our example let's go ahead and try to find the number of divisors for each number and in order to do that let us introduce a famous theorem in arithmetic and that is the fundamental theorem of arithmetic so let's see what this theorem is about Fundamental Theorem so the fundamental theorem of arithmetic states that every integer greater than one either is a prime number itself or can be represented as a product of prime numbers and that moreover this representation is unique and that's very important so given any number like here uh 1200 we can represent this as 2 raised to the 4th power times 3 times 5 squared and if we order the prime factors in increasing order this representation is unique so let's go ahead and factor our numbers in this manner so this will be equal to 2 raised to the fourth power 17 is a prime number so it would be just equal to 17 and 18 is 2 times 9 so 2 times 3 raised to the second power great now we want to find the number of divisors of 16 okay so let me ask you a question can 3 be a divisor of 16 and the answer is no because 3 does not show up in the factorization of 16. in other words any divisor of 16 can has to belong to this list and since 2 is the only prime factor of 16 then any divisor i come up with can only contain two and the second question i would ask you to raise to the seventh power for example be a divisor of 16 and the answer is again no because 16 is equal to two to the fourth and thus the largest power a divisor of 16 can have is 4. so basically all divisors of 16 will be of the form 2 raised to some alpha and alpha is between 0 and 4 included it can be equal to 0 and that's the case where the divisor is equal to 1 1 is indeed a divisor of 16 and it can be represented as 2 raised to the 0 power and the maximum value alpha can take is 4. so basically 16 will have 5 divisors that are 2 raised to the 0 2 to the 1 to the second power two to the third and two to the fourth and you won't have any other divisor the same goes for 17 so all the divisors of 17 can only contain and the alpha here is between zero and one because here it's topped by one and therefore the number of divisors for 17 will be just two namely one and 17 what about 18 now our divisors can either contain a2 or a3 we have a choice here and we have two powers here alpha and beta and alpha here it can be anything between zero and one whereas beta can be anything between zero and two so here we have two choices for alpha and three choices for beta and since our choice of alpha is independent of our choice of beta then the total number of possibilities will be just the product of these two and consequently 18 would have 2 times 3 which is equal to 16 and consequently 18 would have 2 times 3 which is equal to 6 divisors and these are all the combination of two raised to some alpha times three raised to some beta where these are the constraints on alpha and beta so in general any number x that is equal to p 1 raised to the alpha 1 times p 2 raised to the alpha 2 times p 3 raised to the alpha 3 times p n raised to the alpha 1 has a number of divisors that is equal to the product so this is equal to the product we represent the product with pi here as we represent the sum with sigma so this is equal to the product from i equals one to n of alpha i plus one and we we have a plus one here to account for the zero so remember like here we had four and we had five divisors because five is equal to four plus one so basically this alpha i plus one is represents our choices for alpha one so for alpha one for example we can have zero one two three up to alpha one so the total number of choices is alpha i plus one and that's pretty much it our number of devices will be just equal to this product and now all what is left is to find this factorization for the numbers that we will be given and as you may know there is an algorithm that runs in all of square root of n that finds all the factors of any given number so basically all what we have to do is go from i equals 2 up to square root of n i less than or equal to the square root of n and actually we don't use this uh comparison because taking the square root of n requires the use of some function that takes the square root of n and introducing this may lead to a precision issues so instead we check that i squared is less than or equal to n so i times i is less than or equal to n i plus plus and for each value of i we check if n modulo i is equal to 0 that is i divides n and if that's the case we will just keep dividing so while and modulo i is equal to zero we'll keep dividing n by i and each time we'll increment alpha so alpha plus plus alpha here representing our power and at the same time we will divide n by i and after we do this we just record this value of alpha and that's all what we're going to require in our program problem here but we can also record the despair that is the prime number along with its exponent if we want to retrieve the whole factorization but in our problem here all what we are interested in is are these values of advance so we will require all of square root of n for finding this factorization and for this computation here for finding the product of the adverbs in order to find our answer this won't require that much computation because the worst case is that our number here is composite that is it contains many prime numbers so the worst case is that our number looks something like 2 times 3 times 5 times 7 times 11 and so on and since our number can be only as large as a million this can have at most a few prime factors so here if we add a few more we can see that having 10 or up to 20 is gonna go beyond 10 to the six pretty quickly so here we gonna perform at most 20 to 30 multiplications so this won't account much for the for computing the complex for for our complexity so our total complexity will be square root of n plus whatever this is so let's say 30 plus uh and all this times the number of queries we have here which can be up to 10 to the fifth so our total complexity here will be 10 to the third times 10 to the fifth which is about 10 38 and this is within our threshold so we are fine now let's check out our code so this is Coding our program we'll start by reading the number of queries we'll have to answer and then we'll process them and for each query we'll read our number and for each number we will declare a vector of ends that will store the alphas that we mentioned so it's going to store the powers of prime factors then we're gonna factorize our number so as we said we're gonna have a for loop that goes from two up to the square root of our number and we'll check for this by checking that i square is less than or equal to our number and each time we're gonna check if the number is divisible by i that is the number modulo i is equal to zero and if that's the case we're gonna declare our power that we're gonna keep incrementing and while the number is divisible by i we're gonna increment our power and divide the number by i and at the end we'll just push this power into our vector and when we are done we might have that the number is different from one so for example say that the number here is 17 17 is not divisible by any number that is uh smaller than it or that is different than it's because it is a prime number so at the end here we're gonna have that number is different from one and that's why we have to push one into our vector because it accounts for the power of 17 in the prime factorization of 17 whereas if we had some composite number here like let's say four we're gonna start with two and two times two is less than or equal to four and we're gonna find that four modulo two is indeed equal to zero so we're gonna initialize our power with zero and then while four is divisible by two then we're gonna increment our power once and divide four by two to get two and then again we're gonna have that two is divisible by two so we're gonna increment our power a second time and divide number by two and this time uh the number is equal to one so when we get to this point we're gonna find that the number is indeed equal to one so we want to do anything so basically this is only applicable for numbers that are prime numbers or for four numbers that have a prime factor that is larger than its square root and at this point we would have gotten all the powers of prime factors of our number and all what we have to do now is to compute the answer and i'm gonna call it the number of divisors and i'm gonna initialize it with one then i'm gonna look through all the numbers in my powers of prime factors vector and i'm gonna multiply my answer by z plus one as we saw and at the end i'll just print the answer so that's pretty much it let's go ahead and submit so that worked thank you very much for watching see you in the next video bye bye
5851	https://math.stackexchange.com/questions/1415786/right-english-wording-for-counterexamples-to-a-theorem	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Right English wording for "counterexamples to a theorem" Ask Question Asked Modified 10 years ago Viewed 2k times $\begingroup$ This question is about the right English wording. I give here what I call "counterexamples to Banach fixed-point theorem". What I do, is that I look to what happen if some hypothesis of the theorem are not fulfilled. I call this "counterexamples to Banach fixed-point theorem". However, I was told that this is not a proper English wording as a theorem cannot have a counterexample! I'm a French native speaker and it seems that a straight translation of "contre-exemples au thÃ©orÃ¨me du point fixe de Banach" is not good. What would be the appropriate wording to this situation? terminology examples-counterexamples Share edited Sep 1, 2015 at 11:25 Alex M. 36.1k1717 gold badges5050 silver badges101101 bronze badges asked Aug 31, 2015 at 11:50 mathcounterexamples.netmathcounterexamples.net 71.9k55 gold badges3939 silver badges107107 bronze badges $\endgroup$ 16 14 $\begingroup$ I don't think this an English vs. French issue, I think it's more to do with how rigorous you are in your wording and whether the audience understands what you mean. Perhaps it is clearer to say that you're giving counterexamples to a "strengthened version of HB". $\endgroup$ Ben Grossmann – Ben Grossmann 2015-08-31 12:01:02 +00:00 Commented Aug 31, 2015 at 12:01 2 $\begingroup$ While it may not be strictly correct, "Counterexamples to the Banach fixed-point theorem" is certainly catchy! Specially if you're giving a talk. $\endgroup$ lhf – lhf 2015-08-31 12:01:22 +00:00 Commented Aug 31, 2015 at 12:01 10 $\begingroup$ It is not an English vs French issue. A counterexample to a statement (logical sentence) is a an example that shows the statement is not true. A theorem is known to be true. A better wording would be "examples that the hypothesis of Banach fixed-point theorem can not be dropped". $\endgroup$ Ramiro – Ramiro 2015-08-31 12:21:08 +00:00 Commented Aug 31, 2015 at 12:21 4 $\begingroup$ I do not believe that the French expression says what you claim, but the English one needs an article or a genitive. $\endgroup$ Carsten S – Carsten S 2015-08-31 17:12:52 +00:00 Commented Aug 31, 2015 at 17:12 5 $\begingroup$ @Ramiro: Mathematics, like other languages, often has idioms which are widely understood as having a meaning different from their literal meaning, but which might not appear in a dictionary. $\endgroup$ Nate Eldredge – Nate Eldredge 2015-08-31 19:11:45 +00:00 Commented Aug 31, 2015 at 19:11 \| Show 11 more comments 7 Answers 7 Reset to default 18 $\begingroup$ How about simply "counterexamples related to the Banach fixed point theorem"? Or if you want to be more precise, "counterexamples to possible strengthenings of the Banach fixed point theorem". Share answered Aug 31, 2015 at 12:53 Nate EldredgeNate Eldredge 102k1414 gold badges152152 silver badges284284 bronze badges $\endgroup$ 4 2 $\begingroup$ Wouldn't "regarding" be more concise than "related to"? $\endgroup$ guest – guest 2015-08-31 18:18:23 +00:00 Commented Aug 31, 2015 at 18:18 2 $\begingroup$ @guest: Same number of letters :-) And I actually think "related" fits better here, though it's hard to explain why. $\endgroup$ Nate Eldredge – Nate Eldredge 2015-08-31 19:09:45 +00:00 Commented Aug 31, 2015 at 19:09 $\begingroup$ I thank you all for your constructive comments and proposals. It helped me to: (1) better understand the meaning of counterexample in English, (2) have proposals with various points of view and (3) highlight what can be catchy (I discovered the word) which is one of the objective of a website post. $\endgroup$ mathcounterexamples.net – mathcounterexamples.net 2015-08-31 20:13:57 +00:00 Commented Aug 31, 2015 at 20:13 1 $\begingroup$ +1 The "precise" statement is good but verbose. Seems to me that an agreed formal term for this would be useful. Seems similar to "edge cases" and "boundary cases". $\endgroup$ Keith – Keith 2015-09-01 04:20:28 +00:00 Commented Sep 1, 2015 at 4:20 Add a comment \| 13 $\begingroup$ I am also not a native English speaker, so not sure it this is a good suggestion, but I would use something like Necessity proofs or Necessity demonstrations or Necessity examples (meaning: examples that prove that every assumption of the theorem is necessary). Share answered Aug 31, 2015 at 13:30 Erel Segal-HaleviErel Segal-Halevi 11.5k44 gold badges4747 silver badges115115 bronze badges $\endgroup$ 1 2 $\begingroup$ Nice. A slightly different perspective, but much less misleading than "counter-examples". $\endgroup$ Colm Bhandal – Colm Bhandal 2015-08-31 19:41:17 +00:00 Commented Aug 31, 2015 at 19:41 Add a comment \| 9 $\begingroup$ A possible English wording is "what happens if we drop the hypothesis of the Banach fixed-point theorem?" Share edited Aug 31, 2015 at 12:34 user940 answered Aug 31, 2015 at 11:55 uniquesolutionuniquesolution 19.2k11 gold badge1717 silver badges3636 bronze badges $\endgroup$ 1 $\begingroup$ Or "hypotheses", if there's more than one of them :) $\endgroup$ psmears – psmears 2015-08-31 17:49:58 +00:00 Commented Aug 31, 2015 at 17:49 Add a comment \| 7 $\begingroup$ I have looked at your website before, and I remember being a bit dissatisfied with the terminology "counterexamples"- from a technical point of view. For me, a counter-example is some constructible object that demonstrates the falsehood of some statement. Equivalently, a counter example demonstrates the truth of the negation of the original statement. So for (counter-)example, take the statement "All functions are invertible". That is, for every $f$ there is some $f^{-1}$ such that for all $x$ in the domain of $f$, it is true that $f^{-1}(f(x)) = x$. A counter example to this is the function $f(x) = x^2$. Because, assuming an inverse function $f^{-1}$ we have $$1 = f^{-1}(f(1)) = f^{-1}(1) = f^{-1}(f(-1)) = -1$$ which is clearly a contradiction. So this is a counter example to the claim. On the other hand, as one commenter points out, from a "marketing" point of view, "counter-example" sure sounds catchy. But ultimately I find it misleading. Edit: I realise I didn't really answer your question! In your case, I would say "Counter-examples to variations on the theorem statement" or something along those lines. Or go with the answer above. There are really a lot of ways to say it, but I would certainly avoid "counter examples to the ____ theorem". Share edited Aug 31, 2015 at 12:27 answered Aug 31, 2015 at 12:21 Colm BhandalColm Bhandal 4,8471616 silver badges4040 bronze badges $\endgroup$ 2 1 $\begingroup$ I like "Counter-examples to variations on the theorem statement"; it suggests very nicely that the theorem itself is correct, but making slight changes produces something incorrect. $\endgroup$ gnasher729 – gnasher729 2015-08-31 19:17:05 +00:00 Commented Aug 31, 2015 at 19:17 1 $\begingroup$ Thanks, you like it more than me in that case! Technically it's OK but I think verbally we could do with something a bit more succinct. $\endgroup$ Colm Bhandal – Colm Bhandal 2015-08-31 19:24:30 +00:00 Commented Aug 31, 2015 at 19:24 Add a comment \| 1 $\begingroup$ This is much more a mathematical question than an English language question. (Note that I didn't even look at the maths) On your website you say "We look here at counterexamples to the Banach fixed-point theorem when some hypothesis are not fulfilled." Basically this theorem, like most theorems, takes the form "if conditions A, B and C are fulfilled, then we get the result X". And you are giving results that demonstrate that if A is not fulfilled, or B is not fulfilled, then we don't get that result X. You call A, B, and C "hypothesis". That's wrong. They are not hypotheses. They are preconditions of the theorem. You then demonstrate that you don't get the result X. That is not a counterexample to the theorem, since the theorem never claimed you would get the result X without all the preconditions. I'd suggest "Examples for the necessity of the preconditions of the Banach fixed-point theorem". There are probably many possible ways to put this. All that said, when you were told "a theorem cannot have counterexamples", that is not quite right. "Theorem" is not used for a mathematical statement that is true, but usually for one that is widely believed to be proven. Sometimes not even that is necessary; Fermat's Last Theorem was called Fermat's Last Theorem for many, many years when no proof was known, and when it was not even know if a proof existed. So it is entirely possible that something called a "Theorem" is actually wrong and that there are counterexamples. If you managed to give an actual counterexample and that counterexample has been widely accepted, the name would probably get changed. Share answered Aug 31, 2015 at 19:14 gnasher729gnasher729 10.7k2020 silver badges3838 bronze badges $\endgroup$ 1 $\begingroup$ I have a few issues with the above, all just notation though, nothing serious. 1) I have seen A, B, C called hypothesis before. I worked with the Coq theorem prover and in that tool, your "preconditions" or "arguments" or "parameters" to a theorem are exactly called "hypothesis" 2) I'm not sure I entirely agree on your classification of "Theorem" as "believed to be true". But it's an interesting standpoint, and I can't fault you on it. Personally I think it should have been "Fermat's last conjecture" though. Usually something like that is called a conjecture e.g. Goldbach. $\endgroup$ Colm Bhandal – Colm Bhandal 2015-08-31 19:31:36 +00:00 Commented Aug 31, 2015 at 19:31 Add a comment \| 0 $\begingroup$ I would go for non-inclusive examples or exclusive examples, as in essence any violation to the hypotheses/preconditions of the theorem are supposed to produce results that are excluded from when every condition is fulfilled. For instance, on the level of definitions, when defining the term 'rectangle', you can state that a square and a triangle are inclusive and exclusive examples of a rectangle respectively. Share answered Aug 31, 2015 at 21:51 Kevin JohnsonKevin Johnson 10922 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ Great question, great discussion. Perhaps "special case" or "classic example" is appropriate. For instance when discussing the implications between continuity and differentiability the special case or classic example is the Weierstrass function continuous everywhere, differentiable nowhere. This demonstrates that continuity does not imply differentiability. Share answered Sep 1, 2015 at 20:17 dantopadantopa 10.8k1010 gold badges4646 silver badges8585 bronze badges $\endgroup$ 4 $\begingroup$ Please avoid making trivial edits to very-low-quality posts since this removes them from the very-low-quality queue, so reduces the chance that they will be properly handled (that "answer" should either be converted to a comment or deleted). If you can re-flag it as VLQ that would be appreciated. $\endgroup$ Bill Dubuque – Bill Dubuque 2019-03-07 22:58:17 +00:00 Commented Mar 7, 2019 at 22:58 $\begingroup$ Same here and here. $\endgroup$ Bill Dubuque – Bill Dubuque 2019-03-07 23:12:55 +00:00 Commented Mar 7, 2019 at 23:12 $\begingroup$ @Bill Dubuque: You make a good point on VLQ. The other two posts came up in the First Posts queue. As a courtesy, the text was converted to MathJax after posting the "Welcome, Please use MathJax" comment. There is some queue malady on MSE VLQ and FP seem to blend. And thanks for being so specific. $\endgroup$ dantopa – dantopa 2019-03-07 23:32:36 +00:00 Commented Mar 7, 2019 at 23:32 1 $\begingroup$ Ah, I see. I rarely use them so I don't know their quirks. I though it would be worth pointing out the VLQ issue in case you didn't know that the edit terminates the review. $\endgroup$ Bill Dubuque – Bill Dubuque 2019-03-07 23:34:52 +00:00 Commented Mar 7, 2019 at 23:34 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions terminology examples-counterexamples See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 0 Evaluate $\iint_{S} z^2\, dS$ 0 General Annuity practice question Related 2 A question about weakening the conditions of Schauder's fixed point theorem 10 Derivable doesn't exist in english? 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5852	https://www.nagwa.com/en/plans/123189585901/	Lesson Plan: Distance and Displacement \| Nagwa Lesson Plan: Distance and Displacement \| Nagwa Sign Up Sign In English English العربية English English العربية My Wallet Sign Up Sign In My Classes My Messages My Reports My Wallet My Classes My Messages My Reports Lesson Plan: Distance and Displacement Physics • First Year of Secondary School This lesson plan includes the objectives, prerequisites, and exclusions of the lesson teaching students how to define distance as the length of a path between two positions and define displacement as the straight line distance between two positions. Objectives Students will be able to recognize that distance is not directional, recognize that distance can only be positive valued, recognize that displacement has a specific direction, recognize that displacement can be positive or negative valued, recognize that the magnitude of the displacement of a point from another is the straight-line distance between the points. Prerequisites Students should already be familiar with magnitude. Exclusions Students will not cover any rate of motion. Lesson Menu Lesson Lesson Plan Lesson Presentation Lesson Video Lesson Explainer Lesson Playlist Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! Interactive Sessions Chat & Messaging Realistic Exam Questions Nagwa is an educational technology startup aiming to help teachers teach and students learn. Company About Us Contact Us Privacy Policy Terms and Conditions Careers Tutors Content Lessons Lesson Plans Presentations Videos Explainers Playlists Copyright © 2025 Nagwa All Rights Reserved Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy Accept
5853	https://www.webqc.org/molecular-weight-of-ni.html	Ni (Nickel) molar mass Printed from Molar Mass, Molecular Weight and Elemental Composition Calculator Enter a chemical formula to calculate its molar mass and elemental composition: Molar mass of Ni (Nickel) is 58.69340 ± 0.00040 g/mol Discover more Hydrochloric acid HCl Convert between Ni weight and moles \| Compound \| Moles \| Weight, g \| --- \| Ni \| \| \| \| \| Elemental composition of Ni \| Element \| Symbol \| Atomic weight \| Atoms \| Mass percent \| --- --- \| Nickel \| Ni \| 58.6934 \| 1 \| 100.0000 \| \| Computing molar mass step by step \| \| First, compute the number of each atom in Ni: Ni: 1 Then, lookup atomic weights for each element in periodic table: Ni: 58.6934 Now, compute the sum of products of number of atoms to the atomic weight: Molar mass (Ni) = ∑ Count i Weight i = Count(Ni) Weight(Ni) = 1 58.6934 = 58.69340 g/mol \| \| Mass Percent Composition \| Atomic Percent Composition \| --- \| \| Ni Nickel (100.00%) \| Ni Nickel (100.00%) \| \| Mass Percent Composition \| \| Ni Nickel (100.00%) \| \| Atomic Percent Composition \| \| Ni Nickel (100.00%) \| \| Appearance \| \| Nickel is a lustrous, silvery-white metal with a slight golden tinge. \| Sample reactions for Ni \| Equation \| Reaction type \| --- \| \| AgNO 3 + Ni = Ni(NO 3)2 + Ag \| single replacement \| \| Ni + HCl = NiCl 2 + H 2 \| single replacement \| \| HNO 3 + Ni = Ni(NO 3)2 + H 2 \| single replacement \| \| Ni + CO = Ni(CO)4 \| synthesis \| \| Ni + CuSO 4 = NiSO 4 + Cu \| single replacement \| \| Related \| \| Oxidation state calculator \| \| Compound properties \| Computing molar mass (molar weight) To calculate molar mass of a chemical compound enter its formula and click 'Compute'. In chemical formula you may use: Any chemical element. Capitalize the first letter in chemical symbol and use lower case for the remaining letters: Ca, Fe, Mg, Mn, S, O, H, C, N, Na, K, Cl, Al. Functional groups: D, T, Ph, Me, Et, Bu, AcAc, For, Tos, Bz, TMS, tBu, Bzl, Bn, Dmg parenthesis () or brackets []. Common compound names. Examples of molar mass computations: NaCl, Ca(OH)2, K4[Fe(CN)6], CuSO45H2O, nitric acid, potassium permanganate, ethanol, fructose, caffeine, water. Molar mass calculator also displays common compound name, Hill formula, elemental composition, mass percent composition, atomic percent compositions and allows to convert from weight to number of moles and vice versa. Computing molecular weight (molecular mass) To calculate molecular weight of a chemical compound enter it's formula, specify its isotope mass number after each element in square brackets. Examples of molecular weight computations: CO2, SO2. Definitions Molecular mass (molecular weight) is the mass of one molecule of a substance and is expressed in the unified atomic mass units (u). (1 u is equal to 1/12 the mass of one atom of carbon-12) Molar mass (molar weight) is the mass of one mole of a substance and is expressed in g/mol. Mole is a standard scientific unit for measuring large quantities of very small entities such as atoms and molecules. One mole contains exactly 6.022 ×10 23 particles (Avogadro's number) Steps to calculate molar mass Identify the compound: write down the chemical formula of the compound. For example, water is H 2 O, meaning it contains two hydrogen atoms and one oxygen atom. Find atomic masses: look up the atomic masses of each element present in the compound. The atomic mass is usually found on the periodic table and is given in atomic mass units (amu). Calculate molar mass of each element: multiply the atomic mass of each element by the number of atoms of that element in the compound. Add them together: add the results from step 3 to get the total molar mass of the compound. Example: calculating molar mass Let's calculate the molar mass of carbon dioxide (CO 2): Carbon (C) has an atomic mass of about 12.01 amu. Oxygen (O) has an atomic mass of about 16.00 amu. CO 2 has one carbon atom and two oxygen atoms. The molar mass of carbon dioxide is 12.01 + (2 × 16.00) = 44.01 g/mol. Lesson on computing molar mass Practice what you learned: Practice calculating molar mass Weights of atoms and isotopes are from NIST article. Related: Molecular weights of amino acids molecular weights calculated today Please let us know how we can improve this web app. Chemistry tools Gas laws Unit converters Periodic table Chemical forum Constants Symmetry Contribute Contact us Choose languageDeutschEnglishEspañolFrançaisItalianoNederlandsPolskiPortuguêsРусский中文日本語한국어 How to cite? 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5854	https://www.studocu.com/latam/document/universidad-de-costa-rica/fisica-iv/chapter-05-nice/5894455	Chapter 05 - Nice - University Physics, 13e (Young/Freedman) Chapter 5 Applying Newton's Laws 5 - Studocu Saltar al documento Profesores Universidad Instituto Descubre Iniciar sesión Te damos la bienvenida a Studocu Inicia sesión para acceder a los recursos de estudio Iniciar sesión Regístrate Usuario invitado Añade tu universidad o instituto 0 seguidores 0 Archivos subidos 0 upvotes Nuevo Página de inicio My Biblioteca AI Notes Ask AI Quiz IA Chats Reciente Todavía no tienes ningún elemento reciente. My Biblioteca Materias Todavía no tienes ninguna asignatura. Añade Cursos Libros Todavía no tienes ningún libro. Studylists Todavía no tienes ninguna Studylists. Crear un Studylist Página de inicio My Biblioteca Descubre Descubre Universidades Institutos Recursos de enseñanza Generador de planes de lecciones Generador de exámenes Generador de quizzes en directo Ask AI Chapter 05 - Nice Nice Materia Física Iv (FS0424) Universidad Universidad de Costa Rica Año académico:2019/2020 Subido por: Eduardo Arias Universidad Estatal a Distancia Costa Rica 0 seguidores 1 Archivos subidos0 upvotes Seguir Comentarios Inicia sesión (Iniciar sesión) o regístrate (Registrarse) para publicar comentarios. Reportar documento Otros estudiantes también vieron Conferencias - Nota: 10 Extracci n reactiva - Nota: 80 Ed sexual - Proyecto de investigación sobre las conformaciones familiares en Costa Rica, Emergencias-1996 8 1 46-49-49 Lovelock Marketing de Servicios: Casos 1-3 Análisis y Estrategias 4.1.1.7 Lab - Ruta de red y análisis con Traceroute y Ping Otros documentos relacionados Arqueología, volcanismo y teledetección en la región del Arenal, Costa Rica Proyecto arenal notas Práctica 1 II-22 - Examen I: Parámetros de Solubilidad y Polímeros Guía sobre Obesidad y Salud - Módulo Adultez Sana (2023) Ensayo sobre la Función del Arte y Diseño en la Expresión Humana Tarea 4 - Diseño de Muro de Retención: Análisis y Cálculos Vista previa del texto University Physics, 13e (Young/Freedman) Chapter 5 Applying Newton's Laws 5 Conceptual Questions A 615 N student standing on a scale in an elevator notices that the scale reads 645 N. From this information, the student knows that the elevator must be moving A) downward. B) upward. C) You cannot tell if it is moving upward or downward. Answer: C Var: 1 Which one of the following free-body diagrams best represents the free-body diagram, with correct relative force magnitudes, of a person in an elevator that is traveling upward with an unchanging velocity? f is the force of the floor on the person and g is the force of gravity on the person. A) B) C) D) E) Answer: B Var: 1 1 Which one of the following free-body diagrams best represents the free-body diagram, with correct relative force magnitudes, of a person in an elevator that is traveling upward but is gradually slowing down at a rate of 9 m/s 2? f is the force of the floor on the person and g is the force of gravity on the person. A) B) C) D) E) Answer: B Var: 1 2 Two objects having masses m 1 and m 2 are connected to each other as shown in the figure and are released from rest. There is no friction on the table surface or in the pulley. The masses of the pulley and the string connecting the objects are completely negligible. What must be true about the tension T in the string just after the objects are released? A) T = m 2 g B) T > m 2 g C) T < m 2 g D) T = m 1 g E) T > m 1 g Answer: C Var: 1 Two bodies P and Q on a smooth horizontal surface are connected by a light cord. The mass of P is greater than that of Q. A horizontal force (of magnitude F) is applied to Q as shown in the figure, accelerating the bodies to the right. The magnitude of the force exerted by the connecting cord on body P will be A) zero. B) less than F but not zero. C) equal to F. D) greater than F. Answer: B Var: 1 4 Two unequal masses M and m (M > m) are connected by a light cord passing over a pulley of negligible mass, as shown in the figure. When released, the system accelerates. Friction is negligible. Which figure below gives the correct free-body force diagrams for the two masses in the moving system? A) B) 5 A box slides down a frictionless plane inclined at an angle θ above the horizontal. The gravitational force on the box is directed A) parallel to the plane in the same direction as the movement of the box. B) parallel to the plane in the opposite direction as the movement of the box. C) perpendicular to the plane. D) vertically. E) at an angle θ below the inclined plane. Answer: D Var: 1 A crate is sliding down an inclined ramp at a constant speed of 0 m/s. The vector sum of all the forces acting on this crate must point down the ramp. A) True B) False Answer: B Var: 1 A brick is resting on a rough incline as shown in the figure. The friction force acting on the brick, along the incline, is A) zero. B) equal to the weight of the brick. C) greater than the weight of the brick. D) less than the weight of the brick. Answer: D Var: 1 A woman is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follows: P is the upward force the woman exerts on the crate, C is the vertical contact force exerted on the crate by the floor, and W is the weight of the crate. How are the magnitudes of these forces related while the woman is trying unsuccessfully to lift the crate? A) P + C = W B) P + C < W C) P + C > W D) P = C Answer: A Var: 1 7 If you swing a bucket of water fast enough in a vertical circle, at the highest point the water does not spill out because an outward force balances the pull of gravity on the water. A) True B) False Answer: B Var: 1 A string is attached to the rear-view mirror of a car. A ball is hanging at the other end of the string. The car is driving around in a circle, at a constant speed. Which of the following lists gives all of the forces directly acting on the ball? A) tension and gravity B) tension C) tension, gravity, and the centripetal force D) tension, gravity, the centripetal force, and friction Answer: A Var: 1 Suppose a highway curve is properly banked to eliminate friction for a speed of 45 mph. If your tires were bald and you wanted to avoid sliding on the road, you would have to drive A) somewhat below 45 mph. B) somewhat above 45 mph. C) at exactly 45 mph. Answer: C Var: 1 When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reached terminal speed A) his acceleration is equal to g. B) the force of air drag on him is equal to zero. C) the force of air drag on him is equal to g. D) his speed is equal to g. E) the force of air drag on him is equal to his weight. Answer: E Var: 1 Suppose the force of the air drag on an object is proportional to the speed of the object and in the direction opposite the object's velocity. If you throw an object upward, the magnitude of its acceleration is greatest A) right after the object is released. B) at the top of its trajectory. C) The acceleration of the object is the same throughout the entire trajectory. Answer: A Var: 1 8 Three boxes in contact rest side-by-side on a smooth, horizontal floor. Their masses are 5- kg, 3-kg, and 2-kg, with the 3-kg box in the center. A force of 50 N pushes on the 5-kg box, which pushes against the other two boxes. (a) Draw the free-body diagrams for each of the boxes. (b) What magnitude force does the 3-kg box exert on the 5-kg box? (c) What magnitude force does the 3-kg box exert on the 2-kg box? Answer: (a) The following forces act on the 5-kg box: the force due to gravity, normal force, contact force between 5-kg mass and 3-kg mass, the force of 50 N pushing on the box. The following forces act on the 3-kg box: the force due to gravity, normal force, contact force between the 5-kg box and the 3-kg box, the contact force between the 3-kg box and the 2- kg box. The following forces act on the 2-kg box: the force due to gravity, normal force, contact force between the 3-kg box and the 2-kg box. (b) 25 N (c) 10 N Var: 1 The figure shows a 100-kg block being released from rest from a height of 1 m. It then takes it 0 s to reach the floor. What is the mass m of the other block? The pulley has no appreciable mass or friction. A) 60 kg B) 54 kg C) 48 kg D) 42 kg Answer: A Var: 1 10 A wooden block A of mass 4 kg slides on a frictionless table when pulled using a massless string and pulley array by a hanging box B of mass 5 kg, as shown in the figure. What is the acceleration of block A as it slides on the frictionless table? Hint: Think carefully about the acceleration constraint. A) 4 m/s 2 B) 3 m/s 2 C) 3 m/s 2 D) 2 m/s 2 Answer: A Var: 1 The figure shows two 1 kg-blocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at 2. m/s 2 by force. What is the tension at the top end of rope 1? A) 18 N B) 15 N C) 2 N D) 3 N Answer: A Var: 1 11 Three objects are connected by massless wires over a massless frictionless pulley as shown in the figure. The tension in the wire connecting the 10-kg and 15-kg objects is measured to be 133 N. What is the tension in wire A? A) 87 N B) 245 N C) 280 N D) 333 N E) 517 N Answer: D Var: 1 A 20-N box rests on a 50-N box on a perfectly smooth horizontal floor. When a horizontal 15-N pull to the right is exerted on the lower box (see figure), both boxes move together. Find the magnitude and direction of the net external force on the upper box. Answer: 4 N to the right Var: 1 A block is given a very brief push up a 20° frictionless incline to give it an initial speed of 12 m/s. (a) How far along the surface of the plane does the block slide before coming to rest? (b) How much time does it take to return to its starting position? Answer: (a) 21 m (b) 7 s Var: 1 A child on a sled starts from rest at the top of a 15° slope. If the trip to the bottom takes 13 15 s, how long is the slope? Assume that frictional forces may be neglected. A) 293 m B) 586 m C) 1130 m D) 147 m Answer: A Var: 50+ A system comprising blocks, a light frictionless pulley, a frictionless incline, and connecting ropes is shown in the figure. The 9-kg block accelerates downward when the system is released from rest. The tension in the rope connecting the 6-kg block and the 4-kg block is closest to A) 30 N. B) 33 N. C) 36 N. D) 39 N. E) 42 N. Answer: E Var: 1 A 50-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0 and the coefficient of kinetic friction is 0. What is the friction force on the box if (a) a horizontal 140-N push is applied to it? (b) a horizontal 175-N push is applied to it? Answer: (a) 140 N (b) 98 N Var: 1 14 A driver in a 1000 kg car traveling at 20 m/s slams on the brakes and skids to a stop. If the coefficient of friction between the tires and the horizontal road is 0, how long will the skid marks be? A) 26 m B) 21 m C) 33 m D) 24 m Answer: A Var: 27 Jason takes off from rest across level water on his jet-powered skis. The combined mass of Jason and his skis is 75 kg (the mass of the fuel is negligible). The skis have a thrust of 200 N and a coefficient of kinetic friction on water of 0. Unfortunately, the skis run out of fuel after only 48 s. What is Jason's top speed? A) 81 m/s B) 130 m/s C) 13 m/s D) 48 m/s Answer: A Var: 50+ Jason takes off from rest across level water on his jet-powered skis. The combined mass of Jason and his skis is 75 kg (the mass of the fuel is negligible). The skis have a thrust of 200 N and a coefficient of kinetic friction on water of 0. Unfortunately, the skis run out of fuel after only 67 s. How far has Jason traveled when he finally coasts to a stop? A) 10,000 m B) 7600 m C) 5400 m D) 3800 m Answer: A Var: 50+ Kieran takes off from rest down a 50 m high, 10° slope on his jet-powered skis. The skis have a thrust of 280 N parallel to the surface of the slope. The combined mass of skis and Kieran is 50 kg (the fuel mass is negligible). Kieran's speed at the bottom is 40 m/s. What is the coefficient of kinetic friction of his skis on snow? A) 0. B) 0. C) 0. D) 0. Answer: A Var: 34 16 A factory robot drops a 10 kg computer onto a conveyor belt running at 3 m/s. The materials are such that μs = 0 and μk = 0 between the belt and the computer. How far is the computer dragged before it is riding smoothly on the belt? A) 1 m B) 0 m C) 2 m D) 3 m Answer: A Var: 50+ You push downward on a box at an angle 25° below the horizontal with a force of 750 N. If the box is on a flat horizontal surface for which the coefficient of static friction with the box is 0, what is the mass of the heaviest box you will be able to move? A) 59 kg B) 68 kg C) 54 kg D) 82 kg Answer: A Var: 31 A person is dragging a packing crate of mass 100 kg across a rough horizontal floor where the coefficient of kinetic friction is 0. He exerts a force F sufficient to accelerate the crate forward. At what angle above horizontal should his pulling force be directed in order to achieve the maximum acceleration? A) 21° B) 27° C) 30° D) 34° E) 45° Answer: A Var: 1 A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right, and friction. The weight of the crate is 400 N. If the 600-N force is removed, the resultant force acting on the crate is A) zero. B) 200 N to the right. C) 200 N to the left. D) 400 N to the left. E) impossible to determine from the information given. Answer: A Var: 1 17 A rope pulls on the lower block in the figure with a tension force of 20 N. The coefficient of kinetic friction between the lower block and the surface is 0. The coefficient of kinetic friction between the lower block and the upper block is also 0. The pulley has no appreciable mass or friction. What is the acceleration of the 2 kg block? A) 4 m/s 2 B) 5 m/s 2 C) 8 m/s 2 D) 9 m/s 2 Answer: A Var: 1 Three objects are connected as shown in the figure. The strings and frictionless pulleys have negligible masses, and the coefficient of kinetic friction between the 2-kg block and the table is What is the acceleration of the 2-kg block? A) 2 m/s 2 B) 1 m/s 2 C) 3 m/s 2 D) 4 m/s 2 Answer: A Var: 50+ 19 A system comprised blocks, a light frictionless pulley, and connecting ropes is shown in the figure. The 9-kg block is on a perfectly smooth horizontal table. The surfaces of the 12-kg block are rough, with μk = 0 between the block and the table. If the 5-kg block accelerates downward when it is released, find its acceleration. A) 1 m/s 2 B) 1 m/s 2 C) 1 m/s 2 D) 1 m/s 2 E) 1 m/s 2 Answer: A Var: 1 20 Chapter 05 - Nice Descarga Descarga Herramientas de IA Ask AI Opción múltiple Tarjetas didácticas Video quiz Lección de audio 0 0 Guardar Chapter 05 - Nice Materia: Física Iv (FS0424) Universidad: Universidad de Costa Rica Información Más información Descarga Descarga Herramientas de IA Ask AI Opción múltiple Tarjetas didácticas Video quiz Lección de audio 0 0 Guardar University Physics, 13e (Young/Freedman) Chapter 5 Applying Newton's Laws 5.1 Conceptual Questions 1) A 615 N student standing on a scale in an elevat or notices that the scale reads 645 N. From this information, the student knows that the elevator must be moving A) downward. B) upward. C) You cannot tell if it is moving upward or downward. Answer: C Var: 1 2) Which one of the following free-body diagrams best represents the free-body diagra m, with correct relative force magnitudes, of a person in an elevat or that is traveling upward with an unchanging velocity? f is the force of the floor on the person and g is the force of gravity on the person. A) B) C) D) E) Answer: B Var: 1 1 Copyright © 2012 Pearson Education, Inc. 3) Which one of the following free-body diagrams best represents the free-body diagra m, with correct relative force magnitudes, of a person in an elevat or that is traveling upward but is gradually slowing down at a rate of 9 m/s 2? f is the force of the floor on the person and g is the force of gravity on the person. A) B) C) D) E) Answer: B Var: 1 2 Copyright © 2012 Pearson Education, Inc. 4) Two weights are connected by a massless wire and pull ed upward with a constant speed of 1.50 m/s by a vertical pull P. The tension in the wire is T (see figure). Which one of the following relationshi ps between T and P must be true? A) T>P B) T = P C) P + T = 125 N D) P = T + 25 N E) P = T + 100 N Answer: D Var: 1 3 Copyright © 2012 Pearson Education, Inc. ¿Demasiado largo para leerlo en el teléfono? Guárdalo para leerlo más tarde en el ordenador Guardar en una Studylist 5) Two objects having masses m 1 and m 2 are connected to each other as shown in the figure and are released from rest. There is no friction on the table surfac e or in the pulley. The masses of the pulley and the string connecting the objects are comple tely negligible. What must be true about the tension T in the string just after the objects are relea sed? A) T = m 2 g B) T>m 2 g C) T<m 2 g D) T = m 1 g E) T>m 1 g Answer: C Var: 1 6) Two bodies P and Q on a smooth horizontal surface are connected by a light cord. The mass of P is greater than that of Q. A horizonta l force (of magnitude F) is applied to Q as shown in the figure, accelerating the bodies to the right. The ma gnitude of the force exerted by the connecting cord on body P will be A) zero. B) less than F but not zero. C) equal to F. D) greater than F. Answer: B Var: 1 4 Copyright © 2012 Pearson Education, Inc. 7) Two unequal masses M and m (M > m) are connected by a light cord passing over a pulley of negligible mass, as shown in the figure. When released, the system acce lerates. Friction is negligible. Which figure below gives the correct free-body force diagrams for the two masse s in the moving system? A) B) 5 Copyright © 2012 Pearson Education, Inc. C) D) Answer: C Var: 1 8) Two objects have masses m and 5 m, respectively. They both are placed side by side on a frictionless inclined plane and allowed to slide down from rest. A) It takes the lighter object 5 time s longer to reach the bottom of the incline than the heavier object. B) It takes the lighter object 10 times longer to reach the bottom of the inc line than the heavier object. C) It takes the heavier object 5 time s longer to reach the bottom of the incline than the lighter object. D) It takes the heavier object 10 times longer to reach the bottom of the incl ine than the lighter object. E) The two objects reach the bottom of the incl ine at the same time. Answer: E Var: 1 6 Copyright © 2012 Pearson Education, Inc. 9) A box slides down a frictionless plane inclined at an angle θ above the horizontal. The gravitational force on the box is directe d A) parallel to the plane in the same direct ion as the movement of the box. B) parallel to the plane in the opposite directi on as the movement of the box. C) perpendicular to the plane. D) vertically. E) at an angle θ below the inclined plane. Answer: D Var: 1 10) A crate is sliding down an inclined ramp at a constant speed of 0.55 m/s. The vect or sum of all the forces acting on this crate must point down the ramp. A) True B) False Answer: B Var: 1 11) A brick is resting on a rough incline as shown in the figure. The fricti on force acting on the brick, along the incline, is A) zero. B) equal to the weight of the brick. C) greater than the weight of the brick. D) less than the weight of the brick. Answer: D Var: 1 12) A woman is straining to lift a large crate, without suc cess because it is too heavy. We denote the forces on the crate as follows: P is the upward force the woman exerts on the crate, C is the vertical contact force exerted on the crate by the floor, and W is the weight of the crate. How are the magnitudes of these forces related while the woman is trying unsucce ssfully to lift the crate? A) P + C = W B) P + C < W C) P + C > W D) P = C Answer: A Var: 1 7 Copyright © 2012 Pearson Education, Inc. 1 de 41 Compartir Descarga Descarga Otros estudiantes también vieron Conferencias - Nota: 10 Extracci n reactiva - Nota: 80 Ed sexual - Proyecto de investigación sobre las conformaciones familiares en Costa Rica, Emergencias-1996 8 1 46-49-49 Lovelock Marketing de Servicios: Casos 1-3 Análisis y Estrategias 4.1.1.7 Lab - Ruta de red y análisis con Traceroute y Ping Otros documentos relacionados Arqueología, volcanismo y teledetección en la región del Arenal, Costa Rica Proyecto arenal notas Práctica 1 II-22 - Examen I: Parámetros de Solubilidad y Polímeros Guía sobre Obesidad y Salud - Módulo Adultez Sana (2023) Ensayo sobre la Función del Arte y Diseño en la Expresión Humana Tarea 4 - Diseño de Muro de Retención: Análisis y Cálculos Obtén ayuda de IA para tus tareas con la aplicación de Studocu Abre la aplicación Español Latinoamérica Empresa Sobre nosotros Studocu Premium Integridad académica Empleos Blog Dutch Website Herramientas de estudio Todas las herramientas Ask AI AI Notes Generador de quizzes IA Notes to Quiz Videos Notes to Audio Generador de infografías Contacto y Ayuda Preguntas frecuentes Contacto Noticias Legal Términos Política de Privacidad Configuración de cookies Declaración de cookies Copyright & DSA Español Latinoamérica Studocu no está afiliado ni respaldado por ninguna escuela, colegio o universidad. 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5855	https://medium.com/@utsavmadaan823/concurrency-vs-parallelism-lets-finally-clear-up-the-confusion-d927a66a0e42	Sitemap Open in app Sign in Sign in Concurrency vs Parallelism: Let’s Finally Clear Up the Confusion 🤔 Utsav Madaan 8 min readApr 18, 2025 Okay, so I’ve been coding for years now, and I STILL sometimes mix up concurrency and parallelism. Anyone else? Just me? These two concepts have tripped me up more times than I’d like to admit. I remember sitting in a tech interview once, confidently explaining parallelism when the interviewer was clearly asking about concurrency. Talk about awkward! 😬 Look, these terms sound similar (and kinda boring, let’s be honest), but understanding the difference has seriously leveled up my programming game. It’s like suddenly realizing you’ve been using a screwdriver to hammer in nails all these years. So many devs use these words interchangeably — and I get it! But here’s the thing: concurrency is about juggling lots of tasks, while parallelism is about having multiple hands to work with. Not the same thing at all! Getting this wrong isn’t just about sounding smart at meetups (tho that’s a nice bonus). It actually changes how you design your code from the ground up. I’ve seen projects crash and burn because someone thought throwing more threads at a problem would magically make it faster. Spoiler alert: sometimes it makes things waaaaay worse. So grab your coffee ☕ (or tea, no judgment), and let’s break this down in plain English. I promise by the end of this article, you’ll actually understand the difference AND know when to use each approach. What’s the Deal with Concurrency? ☕ The Concept: Concurrency is basically about structure — how you organize your program to handle multiple tasks that seem to happen at once. Think about it like this: have you ever been to one of those tiny coffee shops with just a single barista? I used to work at one in college, and lemme tell you, it was CHAOS during rush hour. 🧑‍🍳 As that lone barista, I couldn’t physically make three different drinks simultaneously (unless I suddenly grew extra arms). But I could: Start taking your order While you’re digging for your wallet, I’d quickly grind beans for the previous customer Jump back to you to take your payment While the card machine does its thing, I’d steam milk for someone else’s latte This constant switching between tasks? That’s basically concurrency in action. I’m not doing multiple things at the EXACT same time — I’m just switching between them so quickly it SEEMS like I am. This is what your computer does with a single CPU core. It’s rapidly switching contexts, giving each task a tiny slice of attention before moving to the next one. The switching happens so fast that it creates the illusion of simultaneity. In programming, we achieve this with things like: Threading — different execution paths within the same program Async programming — using stuff like async/await to pause tasks that are waiting for something (like a slow network response) and do other useful work instead Use Cases: 👍 Concurrency really shines when your program spends a lot of time WAITING for stuff. These are called I/O-bound tasks: Web Servers: OMG, this is the classic example. Your server might be handling thousands of users, but most of them are just sitting there while the database fetches their data. Why not handle other requests during that waiting time? User Interfaces: Ever used an app that freezes when it’s doing something in the background? Super annoying, right? Concurrency keeps the UI responsive while other stuff happens. Network Operations: If you’re downloading 10 files, you don’t have to wait for #1 to finish before starting #2. Start ’em all and juggle! Data Scraping: When I built my first web scraper, it was painfully slow until I made it concurrent. Game changer! Code Example (Python Asyncio): Let me show you a simple example using Python’s asyncio. Don't worry if you're not a Python dev - the concept is what matters: ``` import asyncioimport time# Define an asynchronous function (coroutine)async def brew_coffee(task_name, wait_time): print(f"{time.strftime('%X')} - {task_name}: Starting to brew... ☕") # Simulate waiting (e.g., network I/O) without blocking await asyncio.sleep(wait_time) # Use asyncio.sleep, not time.sleep! print(f"{time.strftime('%X')} - {task_name}: Coffee ready! ✅")async def main(): start_time = time.perf_counter() print(f"{time.strftime('%X')} - Opening the coffee shop!") # Schedule multiple coffee brewing tasks to run concurrently await asyncio.gather( brew_coffee("Espresso", 3), brew_coffee("Latte", 2), brew_coffee("Cappuccino", 4) ) end_time = time.perf_counter() print(f"{time.strftime('%X')} - Coffee shop closed! Total time: {end_time - start_time:.2f} seconds.")# Run the main asynchronous functionif name == "main": asyncio.run(main()) # Manages the event loop import import Define an asynchronous function (coroutine) async def brew_coffeetask_name, wait_time printf"{time.strftime('%X')} - {task_name}: Starting to brew... ☕"{time.strftime('%X')} '%X'{task_name} Simulate waiting (e.g., network I/O) without blocking await Use asyncio.sleep, not time.sleep! printf"{time.strftime('%X')} - {task_name}: Coffee ready! ✅"{time.strftime('%X')} '%X'{task_name} async def main printf"{time.strftime('%X')} - Opening the coffee shop!"{time.strftime('%X')} '%X' Schedule multiple coffee brewing tasks to run concurrently await "Espresso" 3 "Latte" 2 "Cappuccino" 4 printf"{time.strftime('%X')} - Coffee shop closed! Total time: {end_time - start_time:.2f} seconds."{time.strftime('%X')} '%X'{end_time - start_time:.2f}.2 Run the main asynchronous function if "main" Manages the event loop ``` When you run this code, you’ll notice something cool — all three coffee tasks start almost immediately, and the total time is only about 4 seconds (the longest task), not 9 seconds (the sum of all tasks). That’s because while one coffee is “brewing” (waiting), we can work on the others! This is EXACTLY why concurrency is so powerful for I/O-bound tasks. While waiting for one thing, do something else useful! Okay, Then What is Parallelism? 👯‍♀️ The Concept: So parallelism is totally different — it’s about ACTUALLY doing multiple things at the exact same time. Not switching between tasks — literally handling multiple tasks simultaneously. Back to our coffee shop example. Let’s say business is booming, and the owner finally caves and hires a second barista. They also buy a second espresso machine. 🧑‍🍳🧑‍🍳 Now you’ve got two people making drinks at the SAME TIME. That’s parallelism! The key difference? Parallelism requires actual additional resources — more baristas, more machines. In computer terms, this means multiple CPU cores or processors. You can’t do true parallelism on a single-core machine, no matter how clever your code is. This is typically achieved using multiprocessing in programming, where your operating system runs multiple independent processes, potentially on different CPU cores. Use Cases: 👍 Parallelism absolutely shines when you have tasks that require heavy computation — what we call CPU-bound tasks: Heavy Data Processing: I once had to analyze a 10GB dataset that took FOREVER until I split it across all 8 cores of my machine. Went from hours to minutes! Video Encoding: Ever wonder how video editing software renders your masterpiece? It typically chops the video into chunks and processes multiple segments in parallel. Scientific Simulations: Climate models, physics simulations, etc. — these number-crunching monsters eat CPU for breakfast. Machine Learning: Training complex models on large datasets — perfect candidate for parallelization. Code Example (Python Multiprocessing): Here’s how you might use Python’s multiprocessing module to do CPU-intensive work in parallel: ``` import multiprocessingimport timeimport math# A function that simulates CPU-intensive workdef heavy_calculation(num): print(f"Process {multiprocessing.current_process().pid}: Calculating for {num}...") # Example: Calculate factorial (can be CPU intensive for large numbers) # Note: For truly large numbers, other approaches might be needed, # this is just illustrative. A simple loop can also show CPU work. result = math.factorial(num % 20 + 10) # Keep numbers reasonable for example print(f"Process {multiprocessing.current_process().pid}: Done calculating for {num}. Result starts with {str(result)[:5]}") return resultdef main(): start_time = time.perf_counter() print(f"Main Process {multiprocessing.current_process().pid}: Starting calculations...") # Numbers to process numbers_to_process = list(range(1, 9)) # Let's process 8 numbers # Create a pool of worker processes (usually matches CPU cores) # Let's use 4 processes for this example # Using 'with' ensures the pool is properly closed with multiprocessing.Pool(processes=4) as pool: # Map the function to the data across the process pool results = pool.map(heavy_calculation, numbers_to_process) end_time = time.perf_counter() # print(f"Results: {results}") # Results contains the return values print(f"Main Process {multiprocessing.current_process().pid}: All calculations finished! Total time: {end_time - start_time:.2f} seconds.")# Standard practice for multiprocessingif name == "main": main() import import import A function that simulates CPU-intensive work def heavy_calculation num printf"Process {multiprocessing.current_process().pid}: Calculating for {num}..."{multiprocessing.current_process().pid}{num} Example: Calculate factorial (can be CPU intensive for large numbers) Note: For truly large numbers, other approaches might be needed, this is just illustrative. A simple loop can also show CPU work. 20 10 Keep numbers reasonable for example printf"Process {multiprocessing.current_process().pid}: Done calculating for {num}. Result starts with {str(result)[:5]}"{multiprocessing.current_process().pid}{num}{str(result)[:5]} str 5 return def main printf"Main Process {multiprocessing.current_process().pid}: Starting calculations..."{multiprocessing.current_process().pid} Numbers to process list range 1 9 Let's process 8 numbers Create a pool of worker processes (usually matches CPU cores) Let's use 4 processes for this example Using 'with' ensures the pool is properly closed with 4 as Map the function to the data across the process pool map print(f"Results: {results}") # Results contains the return values printf"Main Process {multiprocessing.current_process().pid}: All calculations finished! Total time: {end_time - start_time:.2f} seconds."{multiprocessing.current_process().pid}{end_time - start_time:.2f}.2 Standard practice for multiprocessing if "main" ``` Run this on a multi-core machine, and you’ll see different Process IDs (PIDs) working AT THE SAME TIME. The calculations are distributed across multiple worker processes, potentially running on different CPU cores. If these calculations were truly intensive, this would be WAY faster than doing them one after another. I remember the first time I implemented this pattern on a data processing job — it was like magic watching my CPU usage jump from 25% to 100% and the job finishing in a quarter of the time! The Big Difference: Concurrency vs Parallelism Summarized 📊 Let me break this down super clearly: The KEY takeaway? Concurrency is about dealing with many things at once (managing tasks), while Parallelism is about doing many things at once (executing tasks simultaneously). And here’s a cool insight: concurrency can actually enable parallelism. If you structure your program well for concurrency, it’s often easier to make it run in parallel across multiple cores when needed. When to Choose Which? 🤔 Alright, so here’s the million-dollar question: when should you use concurrency, and when should you reach for parallelism? Choose Concurrency (Threads/Asyncio) when: Your program spends most of its time waiting for stuff — network requests, database queries, file operations, etc. These are I/O-bound tasks. You need to keep an app responsive (like a UI) while doing background work. Your tasks need to share memory easily (threads share memory, processes don’t by default). NOTE: Watch out for the GIL! If you’re using Python, there’s this thing called the Global Interpreter Lock (GIL) that means only one thread can execute Python bytecode at a time, even on multi-core machines. This makes threading in Python great for I/O-bound tasks but pretty useless for CPU-bound tasks. That’s why many Python devs prefer asyncio for I/O-bound work these days. Choose Parallelism (Multiprocessing) when: Your program is doing heavy calculations or data processing that can be broken into chunks. These are CPU-bound tasks. You want to max out those CPU cores you paid good money for! Your tasks don’t need to share memory often (because inter-process communication is slower). Pro Tip: Sometimes the BEST solution uses BOTH! I worked on a web service once where we used concurrency (asyncio) to handle thousands of incoming connections efficiently, but when a request needed heavy computation, we'd hand that specific task off to a separate pool of worker processes for parallel execution. Best of both worlds! Wrapping Up ✨ Phew! We made it! Concurrency and parallelism — two powerful but different approaches to making your code faster and more efficient. Concurrency is like being that skilled barista juggling multiple orders — perfect for when your program spends a lot of time waiting. Parallelism is like having multiple baristas with multiple machines — ideal for heavy computational work. Understanding this difference has saved me countless hours of debugging and performance tuning. It helps you pick the right tools for the job (threading, asyncio, multiprocessing), design better systems, and avoid common pitfalls (like trying to use threads for CPU-bound work in Python... learned that one the hard way!). Remember: Concurrency manages many tasks, Parallelism executes many tasks. I hope this clears things up! It took me months if not years of making mistakes to really get this, so don’t feel bad if it doesn’t click immediately. Programming is a journey, and we’re all learning as we go. What about you? Have you ever confused these concepts or have a funny story about using the wrong approach? Maybe you have a different analogy that helped it click for you? 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5856	http://www.timberlane.net/hs/files/Lecture_Ch_17.pdf	© 2005 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. Lecture PowerPoints Chapter 17 Physics: Principles with Applications, 6th edition Giancoli Chapter 17 Electric Potential Units of Chapter 17 • Electric Potential Energy and Potential Difference •Relation between Electric Potential and Electric Field •Equipotential Lines •The Electron Volt, a Unit of Energy •Electric Potential Due to Point Charges •Potential Due to Electric Dipole; Dipole Moment Units of Chapter 17 • Capacitance • Dielectrics • Storage of Electric Energy • Cathode Ray Tube: TV and Computer Monitors, Oscilloscope • The Electrocardiogram (ECG or EKG) 17.1 Electrostatic Potential Energy and Potential Difference The electrostatic force is conservative – potential energy can be defined Change in electric potential energy is negative of work done by electric force: (17-1) 17.1 Electrostatic Potential Energy and Potential Difference Electric potential is defined as potential energy per unit charge: (17-2a) Unit of electric potential: the volt (V). 1 V = I J/C. 17.1 Electrostatic Potential Energy and Potential Difference Only changes in potential can be measured, allowing free assignment of V = 0. (17-2b) 17.1 Electrostatic Potential Energy and Potential Difference Analogy between gravitational and electrical potential energy: 17.2 Relation between Electric Potential and Electric Field Work is charge multiplied by potential: Work is also force multiplied by distance: 17.2 Relation between Electric Potential and Electric Field Solving for the field, (17-4b) If the field is not uniform, it can be calculated at multiple points: 17.3 Equipotential Lines An equipotential is a line or surface over which the potential is constant. Electric field lines are perpendicular to equipotentials. The surface of a conductor is an equipotential. 17.3 Equipotential Lines 17.4 The Electron Volt, a Unit of Energy One electron volt (eV) is the energy gained by an electron moving through a potential difference of one volt. 17.5 Electric Potential Due to Point Charges The electric potential due to a point charge can be derived using calculus. (17-5) 17.5 Electric Potential Due to Point Charges These plots show the potential due to (a) positive and (b) negative charge. 17.5 Electric Potential Due to Point Charges Using potentials instead of fields can make solving problems much easier – potential is a scalar quantity, whereas the field is a vector. 17.6 Potential Due to Electric Dipole; Dipole Moment The potential due to an electric dipole is just the sum of the potentials due to each charge, and can be calculated exactly. 17.6 Potential Due to Electric Dipole; Dipole Moment Approximation for potential far from dipole: (17-6a) Or, defining the dipole moment p = Ql, (17-6b) 17.6 Potential Due to Electric Dipole; Dipole Moment 17.7 Capacitance A capacitor consists of two conductors that are close but not touching. A capacitor has the ability to store electric charge. 17.7 Capacitance Parallel-plate capacitor connected to battery. (b) is a circuit diagram. 17.7 Capacitance When a capacitor is connected to a battery, the charge on its plates is proportional to the voltage: (17-7) The quantity C is called the capacitance. Unit of capacitance: the farad (F) 1 F = 1 C/V 17.7 Capacitance The capacitance does not depend on the voltage; it is a function of the geometry and materials of the capacitor. For a parallel-plate capacitor: (17-8) 17.8 Dielectrics A dielectric is an insulator, and is characterized by a dielectric constant K. Capacitance of a parallel-plate capacitor filled with dielectric: (17-9) 17.8 Dielectrics Dielectric strength is the maximum field a dielectric can experience without breaking down. 17.8 Dielectrics The molecules in a dielectric tend to become oriented in a way that reduces the external field. 17.8 Dielectrics This means that the electric field within the dielectric is less than it would be in air, allowing more charge to be stored for the same potential. 17.9 Storage of Electric Energy A charged capacitor stores electric energy; the energy stored is equal to the work done to charge the capacitor. (17-10) 17.9 Storage of Electric Energy The energy density, defined as the energy per unit volume, is the same no matter the origin of the electric field: (17-11) The sudden discharge of electric energy can be harmful or fatal. Capacitors can retain their charge indefinitely even when disconnected from a voltage source – be careful! 17.9 Storage of Electric Energy Heart defibrillators use electric discharge to “jump-start” the heart, and can save lives. 17.10 Cathode Ray Tube: TV and Computer Monitors, Oscilloscope A cathode ray tube contains a wire cathode that, when heated, emits electrons. A voltage source causes the electrons to travel to the anode. 17.10 Cathode Ray Tube: TV and Computer Monitors, Oscilloscope The electrons can be steered using electric or magnetic fields. 17.10 Cathode Ray Tube: TV and Computer Monitors, Oscilloscope Televisions and computer monitors (except for LCD and plasma models) have a large cathode ray tube as their display. Variations in the field steer the electrons on their way to the screen. 17.10 Cathode Ray Tube: TV and Computer Monitors, Oscilloscope An oscilloscope displays en electrical signal on a screen, using it to deflect the beam vertically while it sweeps horizontally. 17.11 The Electrocardiogram (ECG or EKG) The electrocardiogram detects heart defects by measuring changes in potential on the surface of the heart. Summary of Chapter 17 • Electric potential energy: • Electric potential difference: work done to move charge from one point to another • Relationship between potential difference and field: Summary of Chapter 17 • Equipotential: line or surface along which potential is the same • Electric potential of a point charge: • Electric dipole potential: Summary of Chapter 17 • Capacitor: nontouching conductors carrying equal and opposite charge •Capacitance: • Capacitance of a parallel-plate capacitor: Summary of Chapter 17 • A dielectric is an insulator • Dielectric constant gives ratio of total field to external field • Energy density in electric field:
5857	https://statisticseasily.com/glossario/what-is-fair-bet-understanding-fair-gambling/	What is: Fair Bet - Understanding Fair Gambling Skip to content LEARN STATISTICS EASILY Learn Data Analysis Now! Home Articles Statistical Analysis ANCOVA ANOVA Chi-Square Generalized Linear Models Kendall Correlation Kruskal-Wallis Test Linear Regression Logistic Regression Mann-Whitney U Test MANOVA Normality Test Pearson Correlation Spearman Correlation Student’s t-test Descriptive Statistics Absolute Mean Deviation Kurtosis Mean Median Mode Skewness Standard Deviation Data Science Artificial Intelligence Deep Learning Machine Learning Bayesian Statistics Books General Generators Graphs Important Concepts Multivariate statistics Results Reporting Sample Size Softwares eBooks Books About Us Contact Us . Click Here to Master Data Analysis! LEARN STATISTICS EASILY Learn Data Analysis Now! What is: Fair Bet What is a Fair Bet? A fair bet is a concept in gambling and probability that refers to a wager where the expected value is zero. This means that the odds offered by the bookmaker or casino accurately reflect the true probability of the outcome occurring. In a fair bet, neither the player nor the house has an inherent advantage, leading to a balanced and equitable betting environment. Understanding fair bets is crucial for gamblers who wish to make informed decisions and maximize their potential returns. Advertisement Simplify the Analysis of Your Data! Gain mastery of statistics and analyze your data with confidence. Start your journey right now! Analysis consulting services Data science course Learn More The Importance of Fair Bets in Gambling Fair bets play a significant role in maintaining the integrity of gambling activities. When bets are fair, players can trust that they are not being exploited by the house. This trust is essential for the sustainability of the gambling industry, as it encourages more players to participate. Additionally, fair bets ensure that the outcomes are based purely on chance, rather than manipulation or unfair practices, which can lead to a more enjoyable experience for all involved. Calculating Fair Odds To determine whether a bet is fair, one must calculate the fair odds based on the probabilities of the possible outcomes. For instance, if a coin toss has two possible outcomes (heads or tails), the fair odds would be 1:1, as each outcome has a 50% chance of occurring. If a bookmaker offers odds of 1.8:1 for either outcome, the bet is not fair, as the expected value is negative for the player. Understanding how to calculate fair odds is essential for anyone looking to engage in betting activities. Statistics books online Expected Value and Fair Bets The concept of expected value (EV) is integral to understanding fair bets. EV is a mathematical calculation that helps bettors determine the potential profitability of a wager. A fair bet has an EV of zero, meaning that over time, players can expect to neither win nor lose money. Conversely, bets with negative EV indicate that players are likely to lose money in the long run. By focusing on fair bets, gamblers can improve their overall betting strategy and minimize losses. Examples of Fair Bets Common examples of fair bets can be found in various gambling scenarios, such as sports betting, poker, and casino games. For instance, in a fair game of poker, the odds of winning a hand should accurately reflect the players’ skill levels and the cards dealt. Similarly, in sports betting, a fair bet would involve odds that correspond to the actual likelihood of a team winning a match. Recognizing these examples helps bettors identify fair opportunities in their gambling endeavors. Best video game consoles Advertisement Simplify the Analysis of Your Data! Gain mastery of statistics and analyze your data with confidence. Start your journey right now! Learn More Unfair Bets and Their Consequences Unfair bets occur when the odds do not accurately reflect the true probabilities of an outcome. This can happen due to various factors, including bookmaker bias, misinformation, or market manipulation. Engaging in unfair bets can lead to significant financial losses for players, as the house edge increases. It is essential for gamblers to be aware of the signs of unfair betting practices and to avoid them to protect their bankrolls. Analysis consulting services Data science course The Role of Regulation in Ensuring Fair Bets Regulatory bodies play a crucial role in ensuring that fair bets are available to players. These organizations establish guidelines and standards for gambling operators, requiring them to offer fair odds and transparent practices. By enforcing regulations, authorities help maintain a level playing field for all participants, fostering trust and confidence in the gambling industry. Players should always seek out licensed and regulated operators to ensure they are engaging in fair betting practices. Strategies for Identifying Fair Bets Identifying fair bets requires a combination of research, analysis, and intuition. Bettors should familiarize themselves with the odds offered by various bookmakers and compare them to their own calculations of the true probabilities. Additionally, keeping track of market trends and player performance can provide valuable insights into potential fair betting opportunities. By employing these strategies, gamblers can enhance their chances of finding and capitalizing on fair bets. Statistics books online The Future of Fair Betting As the gambling industry continues to evolve, the concept of fair betting will remain a cornerstone of responsible gambling practices. With the rise of online betting platforms and advanced data analytics, players have more tools than ever to assess the fairness of their bets. The future of fair betting will likely see increased transparency and accountability from operators, further empowering players to make informed decisions and engage in fair gambling activities. Advertisement Transform Your Research with Data Analysis! Enhance your research outcomes with data analysis. Learn the necessary techniques now! 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Available budget for your marketing activities: Full Name Company Name Business Email Country Whatsapp Comment Pitch Deck or business plan Business Email submissions will be answered within 1 or 2 business days. Personal Email submissions will take longer Submit 1234 The topic applications of arithmetic progression in real life has 98 sections. Narrow your search by using keyword search and selecting one of the keywords below: common difference (14) nth term (13) arithmetic progression arithmetic progression (10) absolute function (9) real-life scenarios (8) fundamental concept (8) powerful tool (8) consecutive terms (8) informed decisions (7) crucial role (7) valuable insights (7) practical application (7) 1.Applications of Arithmetic Progression in Real Life[Original Blog] Applications of Arithmetic Progression in Real Life Arithmetic Progression (AP) is a fundamental concept in mathematics that finds its applications in various real-life scenarios. From calculating simple interest to predicting future trends, AP plays a crucial role in many fields. In this section, we will explore some of the practical applications of AP and how it helps us in different aspects of our lives. Financial Planning: AP is extensively used in financial planning and investment management. For instance, when determining the monthly installments for a loan, AP helps us calculate the fixed amount that needs to be paid over a specific period. By understanding the concept of AP, individuals can effectively plan their budgets and make informed financial decisions. Time and Distance: AP is also useful in solving problems related to time and distance. Consider a scenario where a person travels a fixed distance every day. By analyzing the distance covered each day, we can determine if the person is maintaining a consistent speed or if there are variations. This analysis can be beneficial in monitoring the progress of an athlete or evaluating the efficiency of a delivery service. Population Growth: AP can be applied to study population growth patterns. By analyzing the increase or decrease in population over a specific period, we can identify if the growth rate is consistent. This knowledge is crucial for urban planners, demographers, and policymakers to make informed decisions about resource allocation, infrastructure development, and social programs. Stock Market Analysis: AP is widely used in analyzing stock market trends. By studying the historical data of a stock's price, analysts can identify if the stock follows an arithmetic progression. This insight helps investors predict future trends and make informed decisions regarding buying or selling stocks. Understanding AP in stock market analysis can potentially lead to profitable investments. Sports Performance Analysis: AP is employed in analyzing sports performance to assess the progress of athletes over time. For example, in athletics, the improvement in timing for a sprinter or the increase in the number of goals scored by a soccer player can be analyzed using AP. Coaches and trainers can use this analysis to identify areas of improvement and develop tailored training programs. Construction and Engineering: AP plays a vital role in various construction and engineering projects. For instance, when constructing a road or railway track, engineers need to ensure a consistent slope or gradient. AP helps them calculate the change in elevation over a specific distance, ensuring a smooth and safe construction process. Project Management: AP finds applications in project management, especially in scheduling tasks and allocating resources. By understanding the concept of AP, project managers can effectively plan and monitor the progress of tasks, ensuring timely completion of projects. This knowledge also helps in optimizing resource allocation and minimizing project delays. The applications of Arithmetic Progression in real life are diverse and far-reaching. From financial planning to sports performance analysis, AP aids in making informed decisions, predicting trends, and optimizing various aspects of our lives. By understanding and utilizing the principles of AP, individuals can enhance their problem-solving skills and make better-informed choices in different fields of study and work. Applications of Arithmetic Progression in Real Life - Arithmetic Progression: Advancing with Arithmetic Progression 2.Real-World Applications of Arithmetic Mean Return[Original Blog] One of the most common ways to measure the performance of an investment is to calculate its arithmetic mean return, also known as the simple average return. This is the sum of all the periodic returns divided by the number of periods. For example, if an investment returns 10%, 15%, and 20% in three consecutive years, its arithmetic mean return is (10% + 15% + 20%) / 3 = 15%. However, this does not account for the effects of compounding, which means that the actual return may be higher or lower than the arithmetic mean return. In this section, we will explore some real-world applications of arithmetic mean return and how it can be used to compare different investments, evaluate portfolio performance, and estimate future returns. We will also discuss some of the limitations and drawbacks of using arithmetic mean return as a metric. Some of the real-world applications of arithmetic mean return are: Comparing different investments: Arithmetic mean return can be used to compare the performance of different investments over the same time period. For example, if you want to compare the performance of two mutual funds over the past five years, you can calculate their arithmetic mean returns and see which one has a higher value. However, this method assumes that the returns are independent and identically distributed, which may not be the case in reality. Moreover, arithmetic mean return does not account for the risk or volatility of the investments, which may affect your preference. Therefore, you should also consider other metrics such as standard deviation, Sharpe ratio, or alpha when comparing different investments. evaluating portfolio performance: Arithmetic mean return can be used to evaluate the performance of your portfolio over a given time period. For example, if you have a portfolio of stocks, bonds, and cash, you can calculate the arithmetic mean return of each asset class and then weight them according to their proportion in your portfolio. This will give you the arithmetic mean return of your portfolio, which you can compare with your target return or a benchmark index. However, this method does not account for the correlation or diversification effects among the asset classes, which may affect your portfolio's risk and return. Therefore, you should also consider other metrics such as variance, beta, or R-squared when evaluating portfolio performance. Estimating future returns: Arithmetic mean return can be used to estimate the expected return of an investment or a portfolio in the future. For example, if you have historical data on the returns of a stock or a fund, you can calculate the arithmetic mean return and use it as a proxy for the future return. However, this method assumes that the future returns will follow the same distribution as the past returns, which may not be the case in reality. Moreover, arithmetic mean return does not account for the uncertainty or variability of the future returns, which may affect your confidence level. Therefore, you should also consider other metrics such as confidence intervals, standard error, or margin of error when estimating future returns. As you can see, arithmetic mean return is a useful metric to measure the simple average return of your investments, but it also has some limitations and drawbacks. It does not account for the effects of compounding, risk, volatility, correlation, or diversification, which may affect your investment decisions. Therefore, you should always use arithmetic mean return in conjunction with other metrics and tools to get a more comprehensive and realistic picture of your investments. 3.Real-Life Applications of Arithmetic Mean[Original Blog] The arithmetic mean, also known as the average, is a concept that we encounter in our daily lives more often than we realize. From calculating our grades in school to determining the average temperature of a city, the arithmetic mean is a powerful tool that helps us make sense of the world around us. In this section, we will explore some real-life applications of the arithmetic mean and delve into how it can be used to draw valuable insights. Grades: One of the most common uses of the arithmetic mean is in calculating grades. Teachers use the arithmetic mean to determine the average score of a student over a set of exams or assignments. This provides a fair representation of a student's overall performance and helps in evaluating their progress. For example, if a student scores 80, 90, and 70 in three exams, the arithmetic mean of their grades would be 80+90+70/3 = 80. This average score gives a clear indication of the student's overall performance. Sports Statistics: The arithmetic mean is extensively used in analyzing sports statistics. Whether it's calculating a player's batting average in baseball or a team's shooting percentage in basketball, the arithmetic mean provides a concise summary of performance. For instance, in basketball, a player's shooting percentage is calculated by dividing the total number of successful shots by the total number of attempted shots. This average gives us an idea of the player's accuracy and effectiveness on the court. Market Research: Companies often use the arithmetic mean to analyze market research data. For example, a company may conduct a survey to determine the average age of their target audience or the average income of potential customers. This information helps businesses tailor their products or services to meet the needs of their target market. By understanding the average characteristics of their customers, companies can make informed decisions and develop effective marketing strategies. Weather Analysis: Meteorologists utilize the arithmetic mean to analyze weather data and make predictions. By calculating the average temperature, rainfall, or wind speed over a specific period, meteorologists can identify weather patterns and trends. For instance, if the average temperature in a region is consistently rising over the years, it may indicate the presence of climate change. These insights derived from the arithmetic mean help in understanding and predicting weather patterns, which is crucial for planning various activities and mitigating potential risks. Financial Analysis: The arithmetic mean is widely used in financial analysis to evaluate investment returns or stock market performance. By calculating the average return of an investment portfolio or a specific stock over a period, investors can assess its profitability. For example, if an investor has invested in multiple stocks and wants to determine the average return, they can calculate the arithmetic mean of the individual stock returns. This average return provides a measure of the overall performance of the investment. Population Studies: In the field of demography, the arithmetic mean is employed to analyze population data. By calculating the average age or income of a population, researchers can gain insights into the socioeconomic characteristics and trends. This information is crucial for policy-making, resource allocation, and understanding the needs of different demographic groups within a society. The arithmetic mean is a powerful tool that helps us summarize data, draw conclusions, and make informed decisions. From education to sports, market research to weather analysis, and finance to population studies, the applications of the arithmetic mean are vast and diverse. By understanding and utilizing this concept, we can better comprehend the world around us and harness its power for practical purposes. Real Life Applications of Arithmetic Mean - Arithmetic Mean: The Power of Geometric Mean in Comparison 4.Introduction to Arithmetic Progression[Original Blog] Arithmetic Progression: Introduction to Arithmetic Progression Arithmetic Progression (AP) is a fundamental concept in mathematics that plays a crucial role in various fields such as algebra, calculus, and number theory. It is a sequence of numbers in which the difference between any two consecutive terms is constant. In simpler terms, it is a sequence where each term is obtained by adding a fixed number to the previous term. From a mathematical standpoint, APs offer a systematic way of understanding the relationship between numbers and their patterns. They provide a framework for solving problems involving sequences and series, making them an essential tool for mathematicians, engineers, and scientists. When discussing APs, it is important to understand the key components that define them: Common Difference: The common difference (d) is the fixed value added to each term to obtain the next term in the sequence. It determines the rate of change between consecutive terms. For example, in the sequence 2, 5, 8, 11, 14, the common difference is 3. First Term: The first term (a) is the initial value of the sequence. It serves as the starting point from which subsequent terms are derived. In the sequence mentioned above, the first term is 2. General Term: The general term (a) represents the nth term of the AP. It is given by the formula a = a + (n-1)d, where n represents the position of the term in the sequence. Now, let's delve deeper into the various aspects of arithmetic progression: Finding the nth Term: One of the most common tasks in APs is determining the nth term of a given sequence. To find the nth term, we can use the general term formula mentioned earlier. By substituting the appropriate values for a, d, and n, we can easily calculate any term in the sequence. For instance, in the sequence 3, 7, 11, 15, 19, the 10th term can be found using a = 3, d = 4, and n = 10. Finding the Sum of an AP: Another important aspect of APs is calculating the sum of a given number of terms. This concept is often used to solve real-world problems involving the accumulation of values over time. The sum of an AP can be determined using the formula S = n/2[2a + (n-1)d]. This equation allows us to find the sum of any given number of terms efficiently. Applications of Arithmetic Progression: APs have numerous applications in various fields. For instance, in finance, they are used to calculate compound interest and annuities. In physics, APs help in understanding the motion of objects with constant acceleration. Furthermore, APs find applications in computer science algorithms, data analysis, and even music composition. Alternative Progressions: While APs are widely used and versatile, it is worth mentioning other types of progressions that possess unique characteristics. Geometric Progressions (GPs) are sequences in which each term is obtained by multiplying the previous term by a fixed ratio. Fibonacci sequences, on the other hand, have each term derived by adding the two preceding terms. These alternative progressions offer different patterns and properties that can be explored and compared to APs. Understanding arithmetic progression is essential for building a strong foundation in mathematics. It provides a systematic approach to solving problems involving sequences and series. By grasping the concepts and formulas associated with APs, one can confidently tackle a wide range of mathematical and real-world challenges. Introduction to Arithmetic Progression - Arithmetic Progression: Advancing with Arithmetic Progression 5.Understanding the Basics of Arithmetic Progression[Original Blog] Understanding the Basics of Arithmetic Progression Arithmetic Progression (AP) is a fundamental concept in mathematics that plays a crucial role in various fields, from finance to physics. It is a sequence of numbers in which the difference between any two consecutive terms is constant. By understanding the basics of arithmetic progression, we can unlock its potential to solve problems and analyze patterns. In this section, we will explore the key elements of AP and delve into its significance. Definition and Notation: Arithmetic Progression is denoted by the term "AP" and is represented as {a, a + d, a + 2d, a + 3d, ...}. Here, 'a' is the first term, 'd' is the common difference, and the terms following the first term are obtained by adding 'd' to the previous term. Finding the nth Term: The nth term of an arithmetic progression can be calculated using the formula: an = a + (n - 1)d. For example, consider an AP with a = 3 and d = 2. To find the 5th term, we substitute n = 5 into the formula: a5 = 3 + (5 - 1)2 = 11. Sum of Terms in an AP: The sum of the first 'n' terms in an arithmetic progression can be determined using the formula: Sn = (n/2)(2a + (n - 1)d). For instance, let's consider an AP with a = 2, d = 3, and n = 4. Using the formula, we find: S4 = (4/2)(2(2) + (4 - 1)3) = 26. Applications of AP: Arithmetic progression s are widely used in financial calculations, such as calculating compound interest, annuities, and amortization schedules. In physics, APs are utilized to describe uniformly accelerated motion, where the displacement or velocity changes by a constant amount over equal intervals of time. Comparing Arithmetic Progression with Other Sequences: Arithmetic progression differs from geometric progression (GP) and harmonic progression (HP) in terms of the relationship between consecutive terms. While AP has a constant difference between terms, GP has a constant ratio, and HP has a constant difference between the reciprocals of terms. Choosing the Best Option: When dealing with a sequence that involves a constant change, such as time or distance, arithmetic progression is often the most appropriate choice. If the sequence involves exponential growth or decay, geometric progression might be more suitable. Harmonic progression is typically used in situations where the reciprocal of the terms follows a constant difference pattern. Understanding the basics of arithmetic progression equips us with a powerful tool for solving problems and analyzing patterns. By recognizing the defining elements and formulas associated with AP, we can confidently tackle a wide range of mathematical and real-world scenarios. Whether it's calculating the nth term or sum of terms, AP provides a structured approach to understanding and interpreting sequential data. Understanding the Basics of Arithmetic Progression - Arithmetic Progression: Advancing with Arithmetic Progression 6.Finding the Common Difference in Arithmetic Progression[Original Blog] Arithmetic Progression (AP) is a fundamental concept in mathematics that finds applications in various fields, including finance, physics, and computer science. understanding the key components of an AP, such as the first term, common difference, and nth term, is essential for solving problems related to this sequence. In this section, we will focus on finding the common difference in an arithmetic progression, exploring different perspectives and providing detailed information on various methods. Formula-based approach: One of the most straightforward ways to find the common difference in an arithmetic progression is by using the formula: D = (a - a) / (n - 1) Here, d represents the common difference, a is the nth term of the sequence, a is the first term, and n is the number of terms. This formula allows us to calculate the common difference directly, without having to find all the terms in the sequence. Example: Consider an AP with the first term a = 3 and the 10th term a = 23. Using the formula, we can find the common difference as follows: D = (23 - 3) / (10 - 1) = 2 Hence, the common difference in this AP is 2. Observational approach: In some cases, it may be possible to observe the pattern in an arithmetic progression and determine the common difference without using any formulas. This approach relies on identifying the regular increment or decrement between consecutive terms. Example: Suppose we have an AP with the first term a = 5, and the terms are 5, 9, 13, 17, 21, and so on. By observing the sequence, we can see that each term is obtained by adding 4 to the previous term. Hence, the common difference in this AP is 4. Difference between consecutive terms: Another method to find the common difference is by calculating the difference between consecutive terms. By subtracting any term from its previous term, we can determine the common difference. Example: Let's consider an AP with the terms 3, 7, 11, 15, 19. By subtracting each term from its previous term, we obtain the differences: 7 - 3 = 4, 11 - 7 = 4, 15 - 11 = 4, and 19 - 15 = 4. It is evident that the common difference in this AP is 4. Utilizing the nth term formula: The nth term formula for an arithmetic progression is given by: A = a + (n - 1) d By rearranging this formula, we can isolate the common difference: D = (a - a) / (n - 1) This method is similar to the formula-based approach mentioned earlier but involves rearranging the formula to find the common difference. Comparing the options: Each method discussed above offers a different approach to finding the common difference in an arithmetic progression. The best option depends on the given information and the problem at hand. Here's a comparison of the methods: Formula-based approach: This method is useful when the first and nth terms of the AP are known, allowing for direct calculation of the common difference. Observational approach: When a pattern is easily recognizable in the sequence, this approach provides a quick and intuitive way to determine the common difference. Difference between consecutive terms: This method is handy when the terms are given, but the first term is not explicitly mentioned. It helps identify the regular increment or decrement between consecutive terms. Utilizing the nth term formula: This approach is beneficial when the nth term is known, but the first term is not explicitly given. It allows for rearranging the formula to find the common difference. By considering the given information and the nature of the problem, one can choose the most suitable method to find the common difference in an arithmetic progression. Finding the Common Difference in Arithmetic Progression - Arithmetic Progression: Advancing with Arithmetic Progression 7.Calculating the nth Term in Arithmetic Progression[Original Blog] understanding the Importance of calculating the nth Term in Arithmetic Progression In arithmetic progression, calculating the nth term is a crucial skill that allows us to determine any term in the sequence, regardless of its position. This calculation helps us understand the pattern and predict future terms, making it an essential tool in various mathematical and real-world scenarios. From a mathematical perspective, finding the nth term enables us to express the relationship between consecutive terms in a concise and systematic manner. It allows us to explore the structure of the sequence and identify any underlying patterns or trends. Moreover, this skill is particularly useful in solving problems involving arithmetic progressions, such as finding missing terms or determining the sum of a given number of terms. Looking at it from a practical standpoint, calculating the nth term in arithmetic progression can be applied to real-life situations. For instance, it can help in analyzing financial investments, where the interest or growth rate follows a constant increment. By understanding the progression, we can estimate future values and make informed decisions. Methods for Calculating the nth Term There are several methods available for calculating the nth term in arithmetic progression. Each method has its advantages and may be more suitable depending on the given problem or context. Let's explore some of the common approaches: A) Formula: The most straightforward method involves using the formula for the nth term of an arithmetic progression. The formula is given as: Nth term (an) = a + (n - 1)d Where 'a' represents the first term, 'n' is the position of the term, and 'd' denotes the common difference between consecutive terms. Example: Consider an arithmetic progression with a first term of 2 and a common difference of 3. To find the 6th term, we substitute the values into the formula: 6th term (a6) = 2 + (6 - 1)3 = 2 + 15 = 17 The 6th term in this progression is 17. B) Recursive Approach: Another method involves using a recursive approach to calculate the nth term. This method is useful when we know the first term and the common difference but don't have a direct formula for the nth term. We can use the relationship between consecutive terms to iteratively find the desired term. Example: Let's consider the same arithmetic progression as before, with a first term of 2 and a common difference of 3. To find the 6th term using the recursive approach, we start with the first term and repeatedly add the common difference: A1 = 2 A2 = a1 + d = 2 + 3 = 5 A3 = a2 + d = 5 + 3 = 8 A4 = a3 + d = 8 + 3 = 11 A5 = a4 + d = 11 + 3 = 14 A6 = a5 + d = 14 + 3 = 17 Again, we find that the 6th term is 17. Comparing the Options Both the formulaic and recursive approaches have their merits, but the formulaic method is generally more efficient and convenient. It provides a direct and immediate solution without the need for iterative calculations. However, the recursive approach can be useful when the formula is not known or when finding multiple terms in a sequence. Calculating the nth term in arithmetic progression is a vital skill that allows us to understand and predict the behavior of a sequence. While both the formulaic and recursive approaches can be used, the formulaic method offers a more straightforward and efficient solution in most cases. By mastering this skill, we can confidently navigate arithmetic progression s and apply them to various mathematical and real-world scenarios. Calculating the nth Term in Arithmetic Progression - Arithmetic Progression: Advancing with Arithmetic Progression 8.Summing up the Terms in an Arithmetic Progression[Original Blog] Summing up the terms in an arithmetic progression is an essential skill that allows us to find the total value of a series of numbers that follow a specific pattern. Whether you're a student learning about arithmetic progressions or someone using them in real-life scenarios, understanding how to calculate the sum of terms is crucial. In this section, we will explore different approaches to summing up the terms in an arithmetic progression, comparing various methods and determining the best option. The Formula Approach: One of the most common and efficient ways to find the sum of terms in an arithmetic progression is by using the formula for the sum of an arithmetic series. The formula is given by: 's = (n/2)(2a + (n-1)d), Where's represents the sum of the terms, n is the number of terms, a is the first term, and d is the common difference between the terms. This formula allows us to calculate the sum directly without having to manually add up each term. Example: Consider the arithmetic progression with a first term of 3, a common difference of 2, and 10 terms. Using the formula, we can find the sum as follows: 's = (10/2)(2 3 + (10-1) 2) = 5(6 + 18) = 120. The formula approach is straightforward and efficient, especially when dealing with a large number of terms. Pairing the Terms: Another method to calculate the sum of terms in an arithmetic progression is by pairing the terms in reverse order. This technique involves adding the first and last terms, the second and second-to-last terms, and so on, until all the terms are paired. Example: Let's consider the arithmetic progression with a first term of 1, a common difference of 3, and 5 terms. Pairing the terms, we get: (1 + 13) + (4 + 10) + (7 + 7) = 14 + 14 + 14 = Summing up the Terms in an Arithmetic Progression - Arithmetic Progression: Advancing with Arithmetic Progression 9.Solving Problems using Arithmetic Progression[Original Blog] Solving These Problems Solving Problems using Arithmetic Progression Arithmetic Progression (AP) is not just a mathematical concept that we learn in school; it is a powerful tool that can be used to solve a variety of real-life problems. From calculating the time it takes for a train to reach its destination to determining the number of terms in a sequence, AP can provide us with valuable insights and solutions. In this section, we will explore how AP can be applied to solve problems, providing a deep understanding of its practical applications. Finding the nth term of an AP: One common problem is to find the nth term of an AP when only a few terms are given. For example, let's say we have an AP with the first term (a) as 2 and the common difference (d) as 3. To find the 10th term, we can use the formula: nth term = a + (n-1)d. Plugging in the values, we get: 10th term = 2 + (10-1)3 = 29. Therefore, the 10th term of this AP is 29. Calculating the sum of an AP: Another useful application of AP is determining the sum of a given number of terms. This can be particularly helpful when dealing with financial calculations or series of numbers. The sum of an AP can be found using the formula: sum = (n/2)(2a + (n-1)d), where n represents the number of terms. For instance, if we have an AP with a first term of 5, a common difference of 2, and we want to find the sum of the first 15 terms, we can substitute these values into the formula: sum = (15/2)(2 5 + (15-1) 2) = 15(10 + 28) = 15 38 = 570. Therefore, the sum of the first 15 terms of this AP is 570. Solving problems involving time, distance, and speed: AP can also be applied to solve problems involving time, distance, and speed. For example, let's consider a scenario where a car travels a certain distance at a constant speed. If we know the time it took for the car to travel a certain distance and the average speed, we can use AP to find the total distance covered. By representing the time as the nth term of an AP and using the formula for calculating the sum of an AP, we can determine the total distance traveled. optimizing business strategies: AP can be a valuable tool in optimizing business strategies, particularly when it comes to pricing and profit calculations. By understanding the concept of AP, businesses can set prices in a way that maximizes profit while considering factors such as Solving Problems using Arithmetic Progression - Arithmetic Progression: Advancing with Arithmetic Progression 10.Exploring the Connection between Arithmetic Progression and Algebra[Original Blog] Exploring the Connection between Arithmetic Progression and Algebra Arithmetic progression (AP) is a fundamental concept in mathematics that has been studied for centuries. It provides a framework for understanding patterns and sequences of numbers, making it a valuable tool in various mathematical and real-world applications. One fascinating aspect of arithmetic progression is its connection to algebra. By exploring this connection, we can gain deeper insights into both subjects and unlock new possibilities for problem-solving. From an algebraic perspective, arithmetic progression can be represented using variables and equations. Consider the general form of an arithmetic progression: a, a + d, a + 2d, a + 3d, ..., where 'a' represents the first term and 'd' denotes the common difference between consecutive terms. By introducing algebraic notation, we can express the nth term of an arithmetic progression as a + (n-1)d. This algebraic expression allows us to analyze and manipulate arithmetic progression s in a more abstract and general manner. To further explore the connection between arithmetic progression and algebra, let's delve into some key insights: Algebraic Formulas: Algebra provides us with powerful tools to derive formulas for arithmetic progression s. These formulas enable us to find the nth term, the sum of the first 'n' terms, and other important properties of an arithmetic progression. For example, the formula for the sum of the first 'n' terms of an arithmetic progression is given by Sn = (n/2)(2a + (n-1)d). This formula allows us to quickly calculate the sum without having to manually add up each term. Solving Equations: Algebraic techniques can be employed to solve problems involving arithmetic progression s. By setting up equations based on the given information, we can find missing terms, determine the common difference, or solve for other variables of interest. For instance, consider the problem: "The sum of the first 10 terms of an arithmetic progression is 200, and the common difference is 5. Find the first term." By using algebraic equations, we can set up the equation 10/2(2a + 9d) = 200 and solve for 'a', which gives us the value of the first term. Graphical Representation: Algebraic concepts can also be visualized through graphical representations. When we plot the terms of an arithmetic progression on a graph, we observe a straight line. The slope of this line corresponds to the common difference, while the y-intercept represents the first term. This graphical representation helps us understand the relationship between arithmetic progression and linear functions, reinforcing the connection between algebra and AP. Bridging Concepts: Exploring the connection between arithmetic progression and algebra can help bridge concepts from different areas of mathematics. For example, the concept of slope in algebra is closely related to the common difference in arithmetic progression. By understanding this connection, students can deepen their understanding of both subjects and see how different mathematical ideas intersect and reinforce each other. The connection between arithmetic progression and algebra is a rich and fruitful one. By leveraging algebraic techniques, we can derive formulas, solve problems, graphically represent AP, and bridge concepts between these two areas of mathematics. This exploration allows us to gain a deeper understanding of arithmetic progression and opens up new avenues for problem-solving and mathematical exploration. Exploring the Connection between Arithmetic Progression and Algebra - Arithmetic Progression: Advancing with Arithmetic Progression 11.Conclusion and Further Exploration of Arithmetic Progression[Original Blog] The study of arithmetic progression has taken us on a journey through the fascinating world of numbers and patterns. We have explored the basics of arithmetic progression, delved into its properties and formulas, and even learned how to solve complex problems using this powerful tool. Now, as we reach the end of our exploration, it is time to draw some conclusions and consider the possibilities for further study. The Importance of Arithmetic Progression: Arithmetic progression plays a crucial role in various fields, including mathematics, physics, computer science, and economics. Its simplicity and predictability make it an invaluable tool for analyzing and understanding numerical sequences. Whether it is calculating the future value of an investment, modeling the motion of objects, or designing algorithms, arithmetic progression provides a solid foundation for problem-solving. Applications in Real Life: Arithmetic progression finds numerous applications in our everyday lives. One such example is the concept of compound interest. Suppose you deposit a certain amount of money in a bank account that offers an annual interest rate. The amount of interest earned each year forms an arithmetic progression. By understanding this pattern, you can calculate the future value of your investment accurately. 3. Limitations and Extensions: While arithmetic progression is a powerful tool, it does have its limitations. One limitation is that it assumes a constant difference between consecutive terms. However, in some scenarios, the difference may vary. In such cases, we can explore the concept of a non-constant difference, known as a variable arithmetic progression. This extension allows us to analyze more complex sequences and opens up new possibilities for exploration. Geometric Progression vs. Arithmetic Progression: Throughout our study, we have encountered another type of sequence called geometric progression. Geometric progression follows a different pattern, where each term is obtained by multiplying the previous term by a constant ratio. While both arithmetic and geometric progressions have their merits, arithmetic progression is often preferred for its simplicity and ease of use. However, it is essential to recognize that geometric progression offers its unique insights and applications, particularly in exponential growth scenarios. Further Topics of Exploration: Having gained a solid understanding of arithmetic progression, there are several avenues for further exploration. One area of interest is the study of infinite arithmetic progressions, where the sequence extends indefinitely. Investigating the sum of an infinite arithmetic progression leads us to the concept of series, a fascinating topic with its own set of properties and formulas. 6. Practical Problem-Solving: To truly master arithmetic progression, it is crucial to apply the concepts to practical problem-solving. By practicing with a variety of problems, such as finding missing terms, determining the sum of a given sequence, or solving real-life scenarios, we Conclusion and Further Exploration of Arithmetic Progression - Arithmetic Progression: Advancing with Arithmetic Progression 12.Implementing the Basics in Real Life[Original Blog] Basics in Real In this section, we will explore practical applications of the basics in real-life scenarios. Understanding how to apply the knowledge gained from the basics is crucial for solving problems and achieving tangible results in various fields. 7.1 Practical Application 1: Explanation and Example Provide a clear explanation of the first practical application and provide an example that demonstrates its implementation in real life. Compare different approaches or strategies if relevant. For example, let's consider the practical application of effective communication in a workplace setting: Effective Communication: Effective communication involves the exchange of information, ideas, and emotions between individuals or groups in a clear, concise, and respectful manner. It is a key skill in the workplace and plays a vital role in building relationships, resolving conflicts, and achieving organizational goals. Example: In a team meeting, effective communication can be observed when team members actively listen to each other, ask clarifying questions, and provide constructive feedback. By fostering an environment of open communication, team members can share ideas, collaborate effectively, and ultimately achieve better outcomes. 7.2 Practical Application 2: Explanation and Example Provide a clear explanation of the second practical application. Again, use examples to illustrate its implementation and provide context. Compare different approaches or strategies if relevant. Continue this pattern for each practical application you choose to include in this section. Ensure that each application is explained in a practical and relatable manner, using examples that the reader can easily understand and apply in their own lives. Implementing the Basics in Real Life - A comprehensive guide to understanding the basics 13.What are the main takeaways and lessons learned from the blog and how to apply them in real life situations?[Original Blog] Life situations Real Life Situations Conclusion: What are the main takeaways and lessons learned from the blog and how to apply them in real life situations? In this comprehensive exploration of abandonment and breach, we have delved into the intricacies of these legal concepts and deciphered their complexities. Throughout the blog, we have examined various perspectives and gained valuable insights into the implications of abandonment and breach in different scenarios. Now, let us consolidate our understanding and extract the key takeaways and lessons that can be applied to real-life situations. Understanding the distinction: One crucial takeaway is the importance of distinguishing between abandonment and breach. Abandonment refers to the intentional relinquishment of a right or claim, while breach pertains to the failure to fulfill contractual obligations. Recognizing this difference is essential as it impacts the legal consequences and remedies available to the parties involved. For example, imagine a scenario where a tenant vacates a rental property without notice, leaving behind personal belongings. In this case, the landlord may consider it an abandonment, allowing them to reclaim possession of the property. However, if the tenant fails to pay rent for several months, it would be considered a breach of the lease agreement, entitling the landlord to pursue legal action for unpaid rent. Intent and communication: A significant lesson learned is the significance of intent and communication in determining abandonment or breach. Intent plays a pivotal role in establishing whether a party has truly abandoned their rights or breached a contract. Clear and unambiguous communication between the parties is crucial to avoid misunderstandings and potential legal disputes. For instance, suppose a company enters into a partnership agreement with another firm. If one party ceases all communication abruptly without any indication of their intentions, it may be perceived as abandonment. On the other hand, if the same party communicates their inability to fulfill their obligations due to unforeseen circumstances, it might be seen as a breach, prompting the other party to explore legal remedies. Documentation and evidence: Documentation and evidence are vital tools in determining abandonment or breach and proving one's case in a legal setting. Maintaining thorough records, including contracts, correspondence, invoices, and any other relevant documentation, can significantly strengthen a party's position when faced with a dispute. Consider a situation where an artist commissions a painter to create a custom artwork. If the artist fails to provide clear instructions or specifications for the painting and later claims abandonment, the painter can present documented evidence of their attempts to seek clarification. This evidence would demonstrate that the artist's claim of abandonment is unfounded, potentially protecting the painter's rights. Seek legal advice: When confronted with complex legal issues such as abandonment and breach, it is advisable to seek professional legal advice. Consulting an attorney who specializes in contract law can provide invaluable guidance and ensure that your rights and interests are protected. For instance, if you find yourself in a situation where you suspect abandonment or breach, an attorney can help you assess the circumstances, evaluate the strength of your case, and advise you on the most appropriate course of action. Their expertise can prove instrumental in navigating the intricacies of the legal system and achieving a favorable outcome. Mitigation and resolution: Lastly, a key lesson learned from our exploration is the importance of mitigation and resolution in abandonment and breach cases. Instead of immediately resorting to litigation, parties should consider alternative dispute resolution methods, such as negotiation, mediation, or arbitration. These approaches can often lead to more efficient and cost-effective resolutions while preserving relationships between the parties involved. For example, imagine a situation where a vendor fails to deliver goods within the agreed-upon timeframe, constituting a breach of contract. Rather than immediately filing a lawsuit, the buyer may choose to engage in negotiations with the vendor to explore possible remedies, such as extending the delivery deadline or seeking compensation for the delay. This approach allows both parties to find a mutually beneficial resolution without the need for protracted legal proceedings. Our exploration of abandonment and breach has provided valuable insights into these legal concepts. By understanding the distinctions, recognizing the importance of intent and communication, documenting evidence, seeking legal advice, and considering alternative dispute resolution methods, individuals and businesses can navigate abandonment and breach situations more effectively in real-life scenarios. Armed with this knowledge, we can better protect our rights, mitigate risks, and strive for fair resolutions when faced with such legal challenges. What are the main takeaways and lessons learned from the blog and how to apply them in real life situations - Abandonment and breach: Breach or Abandonment: Decoding the Legalities 14.Applications of Absolute Convergence in Real Life[Original Blog] Absolute convergence is a fundamental concept in mathematics that deals with infinite series. It is a powerful tool that enables mathematicians to determine whether a series converges or diverges. Absolute convergence has many practical applications in real life, including in finance, physics, and engineering. In this section, we will explore some of the most common applications of absolute convergence in real-life scenarios. Financial Analysis: In finance, absolute convergence is used to calculate the value of investments and financial instruments. For example, the present value of a bond is calculated using the absolute convergence of the bond's cash flows. Similarly, the value of a stock is calculated using the absolute convergence of its future cash flows. Absolute convergence is also used to calculate the net present value (NPV) of a project, which helps businesses determine whether a project is worth investing in. NPV is calculated by summing the absolute convergence of all future cash flows associated with the project. Signal Processing: In signal processing, absolute convergence is used to analyze and manipulate digital signals. Digital signals are represented as sequences of numbers, and absolute convergence is used to determine whether the signal is stable or not. Stable signals are those that do not grow to infinity as time goes on. Absolute convergence is also used to filter out unwanted noise from digital signals, which is essential in applications such as audio and video processing. Engineering: In engineering, absolute convergence is used to design and analyze systems that involve infinite series. For example, in electrical engineering, absolute convergence is used to analyze alternating current (AC) circuits. AC circuits involve infinite series of sinusoidal functions, and absolute convergence is used to determine the behavior of the circuit. Absolute convergence is also used in mechanical engineering to analyze systems that involve infinite series, such as heat transfer and fluid flow. Quantum Mechanics: In quantum mechanics, absolute convergence is used to calculate the probability of an event occurring. Quantum mechanics involves the study of particles at the atomic and subatomic level, and absolute convergence is used to calculate the probability of a particle being in a certain state. Absolute convergence is also used to calculate the wave function of a particle, which is essential in understanding the behavior of particles at the quantum level. Optimization: In optimization, absolute convergence is used to determine the optimal solution to a problem. Optimization involves finding the best solution to a problem given a set of constraints. Absolute convergence is used to determine whether a solution is optimal or not. For example, in linear programming, absolute convergence is used to determine the optimal solution to a problem involving linear equations. Absolute convergence has many practical applications in real life, including in finance, signal processing, engineering, quantum mechanics, and optimization. It is a powerful tool that enables mathematicians and scientists to analyze and manipulate infinite series. Understanding absolute convergence is essential in many fields, and its applications are far-reaching and diverse. Applications of Absolute Convergence in Real Life - Absolute convergence: Absolute Value and Absolute Convergence: A Deep Dive 15.Applications of the Absolute Value Function in Real Life[Original Blog] The absolute value function, denoted as \|x\|, is a mathematical concept that has numerous applications in real life. It is a fundamental tool used to measure the distance between two points on a number line, regardless of their direction. While it may seem like a simple concept, the absolute value function plays a crucial role in various fields, from physics and engineering to economics and computer science. In this section, we will explore some of the practical applications of the absolute value function and how it helps us understand and solve real-life problems. 1. Distance and Magnitude: One of the most straightforward applications of the absolute value function is measuring distances. Whether it's calculating the distance between two cities on a map or determining how far an object has traveled, the absolute value function provides a reliable method to obtain accurate results. For instance, if you want to find out how far you are from your destination while driving, you can subtract your current position from the target location and take the absolute value of the difference. This ensures that you get a positive value representing the actual distance traveled. Temperature Changes: In meteorology and everyday weather forecasting, understanding temperature changes is essential. The absolute value function comes into play when dealing with temperature differences. For example, if you want to determine how much colder or hotter it will be tomorrow compared to today, you would subtract today's temperature from tomorrow's temperature and take the absolute value of that difference. This allows meteorologists to accurately predict temperature fluctuations and provide valuable information for planning outdoor activities or preparing for extreme weather conditions. Financial Analysis: The absolute value function finds its application in financial analysis as well. When analyzing stock market data or evaluating investment portfolios, it is common to calculate percentage changes in asset values over time. By taking the absolute value of these changes, investors can focus on the magnitude rather than the direction of fluctuations. This helps them assess risk levels accurately and make informed decisions regarding their investments. 4. Optimization Problems: Optimization problems arise in various fields, including engineering, manufacturing, and logistics. These problems involve finding the maximum or minimum value of a function within a given set of constraints. The absolute value function often appears in such scenarios when dealing with absolute differences or deviations from an ideal state. For instance, in production planning, manufacturers aim to minimize the difference between actual and target output levels. By using the absolute value function, they can formulate mathematical models that optimize production processes and reduce inefficiencies. Signal Processing: In signal processing, Applications of the Absolute Value Function in Real Life - Absolute difference: A Closer Look at the Absolute Value Function update 16.Applications of the Absolute Value Function in Real Life[Original Blog] The absolute value function is one of the most important functions in mathematics, and it has many real-life applications. In this section, we will explore some of the most common applications of the absolute value function. Distance and Displacement One of the most common applications of the absolute value function is in measuring distance and displacement. In physics, distance is the total length traveled by an object, while displacement is the difference between the final and initial positions of an object. Both distance and displacement are always positive, so the absolute value function is used to ensure that the values are positive. Example: If you drive 10 miles north and then 5 miles south, your displacement is 5 miles north. However, your distance traveled is 15 miles (10 miles + 5 miles). The absolute value function can be used to find the distance traveled: \|10\| + \|(-5)\| = 15. Temperature Another common application of the absolute value function is in measuring temperature. Temperature can be measured in Celsius or Fahrenheit, but both scales have negative values. To avoid negative values, the absolute value function is used to ensure that the temperature is always positive. Example: If the temperature is -5 degrees Celsius, the absolute value function can be used to find the actual temperature: \|(-5)\| = 5 degrees Celsius. Optimization Problems Optimization problems are common in many fields, including economics, engineering, and physics. These problems involve finding the maximum or minimum value of a function. The absolute value function is often used in these problems to ensure that the function is always positive. Example: A company wants to minimize the cost of producing a product. The cost function is C(x) = 2x + \|x-10\|, where x is the number of units produced. The absolute value function is used to ensure that the function is always positive. To minimize the cost, we need to find the value of x that minimizes the cost function. Inequality Problems Inequality problems are common in mathematics and physics. These problems involve finding the range of values that satisfy an inequality. The absolute value function is often used in these problems to ensure that the values are positive. Example: Solve the inequality \|x-3\| < 5. The absolute value function is used to ensure that the values are positive. To solve the inequality, we need to consider two cases: x-3 < 5 and x-3 > -5. The solution is -2 < x < 8. Signal Processing Signal processing is a field of study that deals with analyzing and processing signals. Signals can be represented as functions, and the absolute value function is often used to extract the amplitude of a signal. Example: A microphone records a sound wave. The sound wave can be represented as a function. The absolute value function can be used to extract the amplitude of the sound wave. The absolute value function is a powerful tool with many applications in real life. It is used in measuring distance and displacement, temperature, optimization problems, inequality problems, and signal processing. Understanding the absolute value function is essential for solving many real-life problems. Applications of the Absolute Value Function in Real Life - Absolute function: Unraveling the Absolute Value Mystery 17.Applications of Absolute Value in Real Life[Original Blog] Absolute value is an essential mathematical concept that has several applications in real-life situations. It is a measure of the magnitude of a number or quantity, independent of its sign. This property makes the absolute value an essential tool in solving problems related to distance, temperature, and other physical quantities. In this section, we will discuss the different applications of absolute value in real-life situations. Distance and Displacement The absolute value is used to measure the distance and displacement of an object. The distance between two points can be calculated by finding the absolute value of the difference between their coordinates. For example, if a person moves from point A (-5, 3) to point B (2, 7), the distance between them can be calculated by finding the absolute value of (2-(-5)) + (7-3) = 9 + 4 = 13. The displacement of the person can also be calculated by finding the absolute value of the difference between the final and initial positions. In this case, the displacement is 7. Temperature Another application of absolute value is in measuring temperature. The absolute value of temperature is used to calculate the difference between two temperatures. The temperature scale used in everyday life is the Celsius scale, which is based on the freezing and boiling points of water. However, in scientific calculations, the Kelvin scale is used, which is based on absolute zero, the lowest possible temperature. The absolute value of temperature is used to convert between these scales. For example, to convert 25C to Kelvin, we add 273.15 to it and get 298.15K. Modelling Real-Life Situations Absolute value is also used to model real-life situations. For example, in finance, the absolute value function is used to represent the magnitude of changes in stock prices. The absolute value function is also used in physics to model the motion of a particle in a magnetic field. In statistics, the absolute value is used to calculate the absolute deviation, which is a measure of the spread of a data set. Solving Equations Absolute value is used to solve equations that involve the absolute value function. The absolute value equation has two solutions, one positive and one negative. For example, the equation \|x\| = 4 has two solutions, x = 4 and x = -4. In some cases, the absolute value equation can be rewritten as a piecewise function, which has different expressions for different intervals of x. Finding Limits The absolute value is also used to find limits in calculus. The limit of a function is the value that the function approaches as the input approaches a certain value. The absolute value function is used to find the limit of a function when the input approaches a certain value from both sides. For example, the limit of f(x) = \|x\| as x approaches 0 is 0. Absolute value has several applications in real-life situations, including distance and displacement, temperature, modelling real-life situations, solving equations, and finding limits. Understanding the concept of absolute value and its applications can help us solve problems in different fields, including finance, physics, and statistics. Applications of Absolute Value in Real Life - Absolute precision: Navigating the World of Absolute Value 18.Applications of Absolute Value in Real Life[Original Blog] Absolute value is a mathematical concept that has many real-life applications. It is a powerful tool that helps us understand and solve problems that involve both positive and negative numbers. Absolute value is the distance between a number and zero on the number line. It is always positive or zero. In this section, we will explore some of the most common applications of absolute value in real life. 1. Distance and Magnitude: One of the most fundamental applications of absolute value is in measuring distance and magnitude. For example, if you want to measure the distance between two points, you can use the absolute value of the difference between their coordinates. Similarly, if you want to measure the magnitude of a vector, you can use the absolute value of its components. Absolute value is also used to measure the magnitude of physical quantities such as force, velocity, and acceleration. Temperature and Weather: Another common application of absolute value is in measuring temperature and weather. In the Celsius scale, zero degrees represents the freezing point of water, and negative values represent temperatures below freezing. However, in the Kelvin scale, zero represents absolute zero, the coldest possible temperature. Absolute value is used to convert temperatures between these scales and to measure the temperature difference between two points. Finance and Economics: Absolute value is also used in finance and economics. For example, if you want to calculate the absolute return on an investment, you can use the absolute value of the difference between the final and initial values. Absolute value is also used to measure the deviation of a stock or portfolio from its benchmark index. In economics, absolute value is used to measure the elasticity of demand and supply, which is the responsiveness of quantity demanded or supplied to changes in price. Optimization and Constraints: Absolute value is a useful tool in optimization and constraints. For example, if you want to minimize the absolute value of a function, you can use the absolute value function as a constraint. Similarly, if you want to maximize the absolute value of a function, you can use the absolute value function as an objective function. Absolute value is also used to model constraints in linear programming, which is a method for solving optimization problems. Statistics and Probability: Absolute value is used in statistics and probability to measure the deviation of data from the mean or expected value. For example, the mean absolute deviation is a measure of the average distance between each data point and the mean. Absolute value is also used to calculate the absolute difference between two random variables, which is a measure of their correlation. Absolute value is also used to calculate the absolute value of the z-score, which is a measure of the distance between a data point and the mean in terms of the standard deviation. Absolute value is a versatile and powerful tool that has many real-life applications. It is used to measure distance, magnitude, temperature, finance, economics, optimization, constraints, statistics, and probability. Understanding the concept and applications of absolute value is essential for solving problems in various fields and industries. Applications of Absolute Value in Real Life - Nonnegative: Embracing the Nonnegative: An Absolute Value Journey 19.Applications of Absolute Value in Real Life[Original Blog] Absolute value is a mathematical concept that represents the distance between a number and zero on the number line. It is a crucial tool in analyzing real numbers because it helps to determine the magnitude of numbers irrespective of their sign. The concept of absolute value is not only useful in mathematics, but it also has many real-life applications. In this section, we will explore some of the applications of absolute value in real life. Distance Absolute value is an essential tool in measuring distance. It is used to determine the distance between two points on a number line or in a coordinate plane. For instance, if you want to calculate the distance between two cities, you can use their coordinates and apply the distance formula, which involves absolute values. Absolute value ensures that the distance is always positive, regardless of the direction of movement. Temperature Temperature is a physical quantity that can be negative or positive. Absolute value is used to convert negative temperatures to positive ones. For instance, if the temperature is -10 degrees Celsius, taking the absolute value of -10 will give you 10 degrees Celsius, which is a positive value. Absolute value is also useful in calculating temperature differences, such as the difference between the highest and lowest temperatures in a day. Stock Prices Absolute value is used in finance to determine the percentage change in stock prices. The percentage change is calculated by dividing the absolute value of the difference between the current price and the previous price by the previous price. This calculation ensures that the percentage change is always positive, regardless of whether the stock price increased or decreased. Improper Fractions Improper fractions are fractions where the numerator is greater than the denominator. Absolute value is used to convert improper fractions to mixed numbers. For instance, if you have the fraction 7/4, you can convert it to a mixed number by dividing the numerator by the denominator and taking the absolute value of the remainder. The quotient becomes the whole number, and the absolute value of the remainder becomes the numerator of the fraction. Electrical Engineering Absolute value is used in electrical engineering to calculate the magnitude of alternating currents. Alternating currents are currents that change direction periodically. The magnitude of an alternating current is the absolute value of its maximum amplitude. Absolute value is also used in calculating the voltage drop across a resistor, which is proportional to the absolute value of the current flowing through it. Absolute value is a fundamental concept in mathematics, but it also has many real-life applications. It is used in measuring distance, converting negative temperatures to positive ones, calculating percentage changes in stock prices, converting improper fractions to mixed numbers, and calculating the magnitude of alternating currents in electrical engineering. Absolute value ensures that the results are always positive, regardless of the direction of movement or the sign of the number. Applications of Absolute Value in Real Life - Real numbers: Absolute Value: An Essential Tool for Analyzing Real Numbers 20.Applications of Absolute Value in Real Life[Original Blog] The concept of absolute value is a fundamental tool in mathematics that has numerous applications in real life. It allows us to measure the distance between two points on a number line, regardless of their direction. Absolute value is not only useful for solving equations or inequalities, but it also plays a crucial role in various fields such as physics, economics, and computer science. By understanding the applications of absolute value in real life, we can gain valuable insights into how this mathematical concept impacts our daily lives. From a practical standpoint, absolute value finds its application in determining distances. For instance, consider a scenario where you are driving from one city to another. The distance between these two cities can be represented by the absolute value of the difference between their respective mile markers on the highway. Regardless of whether you are traveling east or west, the distance covered remains the same. This illustrates how absolute value helps us measure distances without considering direction. In physics, absolute value is used to calculate magnitudes. Magnitude refers to the size or quantity of a physical property, such as velocity or force. By taking the absolute value of these properties, we can determine their magnitude without considering their direction. For example, when calculating speed, we disregard whether an object is moving forward or backward and focus solely on its numerical value. In economics and finance, absolute value plays a significant role in analyzing data related to profits and losses. When evaluating financial performance, it is essential to consider both positive and negative values. Absolute value allows us to ignore the sign and focus solely on the magnitude of gains or losses. This enables economists and investors to make informed decisions based on objective data. Temperature: Absolute value is commonly used in weather forecasting to represent temperature differences accurately. By taking the absolute value of temperature changes, meteorologists can analyze patterns and predict weather conditions more effectively. Stock Market: In the world of finance, absolute value is crucial for analyzing stock market trends. By considering the absolute value of stock price changes, investors can assess the magnitude of fluctuations and make informed decisions about buying or selling stocks. Error Analysis: Absolute value is extensively used in error analysis to measure the difference between an observed value and its true value. This allows scientists and researchers to quantify uncertainties and evaluate the accuracy of their measurements. Optimization Problems: Absolute value plays a vital role in solving optimization problems, where we aim to find the maximum or minimum value of a function. Applications of Absolute Value in Real Life - Real numbers: Absolute Value: An Essential Tool for Analyzing Real Numbers update 21.Applications of Absolute Value in Real Life[Original Blog] Absolute value is one of the fundamental concepts in mathematics. It is a measure of the distance between a number and zero on a number line. Absolute value has several applications in real life, from measuring distances to calculating the difference between two quantities. In this blog, we will explore some of the most common applications of absolute value in real life. Distance and Displacement Absolute value is commonly used to measure distance and displacement. For example, if you want to calculate the distance between two points on a map, you can use the absolute value of the difference between their coordinates. Similarly, if you want to calculate the displacement of an object, you can use the absolute value of the difference between its initial and final positions. This is useful in a wide range of applications, from navigation to physics. Temperature Absolute value is also used to measure temperature differences. In this case, the absolute value of the difference between two temperatures represents the magnitude of the temperature difference, regardless of whether the temperatures are positive or negative. This is particularly useful in scientific applications where precise temperature measurements are required. Financial Calculations Absolute value is also used in financial calculations, particularly in the calculation of profit and loss. For example, if you buy a stock at $50 and it drops to $40, the absolute value of the difference between the two prices is $10. This represents the loss you have incurred. Similarly, if the stock rises to $60, the absolute value of the difference between the two prices is $10, representing the profit you have made. Error Analysis Absolute value is also used in error analysis, particularly in the calculation of the absolute error and relative error. The absolute error is the difference between the measured value and the true value, while the relative error is the absolute error divided by the true value. Both of these values are useful in determining the accuracy of a measurement or calculation. Optimization Absolute value is also used in optimization problems, particularly in linear programming. In this case, the absolute value of a variable is used to ensure that it remains positive, even if its coefficient in the objective function is negative. This is particularly useful in applications such as resource allocation or production planning. Absolute value is a fundamental concept in mathematics that has several applications in real life. It is used to measure distances, temperature differences, financial calculations, error analysis, and optimization problems. Understanding the applications of absolute value can help us solve a wide range of problems in different fields. Applications of Absolute Value in Real Life - Real numbers: Unlocking the Power of Absolute Value with Real Numbers 22.Applications of Absolute Value in Real Life Scenarios[Original Blog] Real Life Scenarios Calculating Distance and Magnitude One of the most common applications of absolute value in real life scenarios is in calculating distance and magnitude. Absolute value allows us to determine the distance between two points on a number line or in a coordinate plane, regardless of whether the points are positive or negative. For example, when measuring the distance between two cities, we use absolute value to ignore the direction and focus solely on the magnitude of the distance. Similarly, in physics, absolute value is crucial in determining the magnitude of vectors, such as velocity or acceleration. Evaluating Errors and Deviations Absolute value is also invaluable in evaluating errors and deviations. In fields such as statistics and data analysis, it is common to encounter situations where we need to determine the difference between an observed value and the expected value. Absolute value allows us to ignore the sign and focus solely on the magnitude of the difference. For instance, when measuring the accuracy of a prediction model, we can calculate the absolute value of the difference between the predicted and actual values to determine the degree of error. Solving Equations and Inequalities Absolute value plays a significant role in solving equations and inequalities, particularly when dealing with absolute value equations or inequalities. By isolating the absolute value expression and considering both the positive and negative cases, we can find all possible solutions. For example, when solving the equation \|x - 5\| = 3, we consider both x - 5 = 3 and x - 5 = -3, which leads to two possible solutions: x = 8 and x = 2. This concept is widely used in various fields, including engineering, economics, and finance. analyzing Stock Market trends In the world of finance, absolute value is frequently used to analyze stock market trends. By calculating the absolute value of the daily changes in stock prices, investors can focus on the magnitude of the change rather than the direction. This allows them to identify significant market movements and make informed decisions. For instance, if a stock's price rises by $5 one day and falls by $5 the next, the absolute value of the change is $5 both times, indicating a consistent level of volatility. Determining Electrical Current Absolute value is also employed in determining electrical current in alternating current (AC) circuits. AC circuits involve the flow of current that continuously changes direction. The instantaneous value of the current can be represented as a sinusoidal wave, and the absolute value of the wave indicates the magnitude of the current at any given point. This information is crucial for understanding and analyzing the behavior of electrical circuits, as well as for designing and troubleshooting electrical systems. Overall, absolute value plays a crucial role in various real-life scenarios, from calculating distances and magnitudes to analyzing stock market trends and solving equations. By understanding and applying the concept of absolute value, we can gain valuable insights and make informed decisions in a wide range of fields. Applications of Absolute Value in Real Life Scenarios - Absolute Value: The Importance of Absolute Value in G24 23.Real Life Examples of Accelerated Death Benefit[Original Blog] Real life examples Accelerated Death The accelerated death benefit is an option that can provide financial assistance to individuals who have been diagnosed with a terminal illness. This benefit can be used to cover medical expenses, hospice care, and other end-of-life costs. To better understand how this benefit works, it's helpful to explore some real-life case studies. By examining different scenarios, we can gain insights into how the accelerated death benefit can be used to support individuals and their families during a difficult time. Here are some examples: Case Study 1: Sarah's Story Sarah was diagnosed with stage 4 cancer and given a prognosis of 6 months to live. She had a life insurance policy with an accelerated death benefit rider, which allowed her to receive a portion of her death benefit while she was still alive. Sarah used this money to pay for her medical treatments and to make modifications to her home so that she could receive hospice care in a comfortable environment. This benefit provided her with peace of mind and allowed her to focus on spending time with her loved ones. Case Study 2: John's Story John was diagnosed with a terminal illness and had a life insurance policy with an accelerated death benefit rider. He decided to use this benefit to take a once-in-a-lifetime trip with his family. This trip provided him with a sense of closure and allowed him to create lasting memories with his loved ones. Case Study 3: Maria's Story Maria was diagnosed with a terminal illness and had a life insurance policy with an accelerated death benefit rider. She used this benefit to pay for her medical expenses and to provide financial support to her family. This benefit allowed her to maintain her quality of life and to provide for her family even as her health declined. In each of these case studies, the accelerated death benefit provided individuals with much-needed financial support during a difficult time. By exploring these scenarios, we can see how this benefit can be used to help individuals and their families manage the financial costs associated with a terminal illness. Real Life Examples of Accelerated Death Benefit - Accelerated death benefit: Financial Support in Times of Terminal Illness 24.Applying the Modified Dietz Method in Real Life[Original Blog] Applying Modified Applying the Modified Dietz Modified Dietz Method Applying the Modified Dietz Method In this section, we will delve into a case study that demonstrates the practical application of the Modified Dietz method in real-life scenarios. By examining a specific example, we can gain a deeper understanding of how this performance measurement technique can be utilized effectively. Background of the Case Study: To provide context, let's consider a hypothetical investment portfolio managed by a financial advisor named Sarah. Sarah's client, Emily, has invested in a diverse range of assets, including stocks, bonds, and mutual funds. Emily wants to evaluate the performance of her portfolio over a specific period to assess its profitability and make informed investment decisions moving forward. Implementing the Modified Dietz Method: Sarah decides to employ the Modified Dietz Method to measure the performance of Emily's portfolio accurately. She starts by gathering the necessary data, including the market values of each asset at the beginning and end of the evaluation period, as well as any cash flows or contributions made during that time. calculating the Daily returns: Sarah calculates the daily returns for each asset by dividing the difference in market values by the previous day's market value. This step allows her to capture the fluctuation in the portfolio's value accurately. Example: Let's say Emily's portfolio had a market value of $100,000 at the beginning of the period and $110,000 at the end. If the market value increased by $1,000 on the first day, the daily return would be 0.01 (1,000/100,000). Sarah repeats this calculation for all assets in the portfolio. Weighting the Daily Returns: Sarah then weights the daily return s by the proportion of each asset's market value in the portfolio. This step is crucial as it accounts for the varying contributions of different assets to the overall performance. Example: If stocks constitute 60% of Emily's portfolio, bonds 30%, and mutual funds 10%, Sarah multiplies the daily return s of each asset by their respective weightings. Aggregating the Daily Returns: Next, Sarah aggregates the weighted daily returns to calculate the portfolio's overall return for the evaluation period. She sums up the weighted returns of all assets to derive a single measure of performance. Adjusting for Cash Flows: If there were any cash flows or contributions during the evaluation period, Sarah adjusts the portfolio's return accordingly. She factors in the timing and magnitude of these cash flows to ensure an accurate representation of the portfolio's performance. Example: If Emily made a $5,000 contribution to her portfolio halfway through the evaluation period, Sarah would account for this cash flow by adjusting the return calculation. Interpreting the Results: Finally, Sarah presents the calculated return to Emily, providing her with a clear understanding of her portfolio's performance during the evaluation period. Sarah can compare this return to relevant benchmarks or industry standards to assess the portfolio's relative performance. This case study exemplifies how the Modified Dietz Method can be applied to measure investment performance accurately. By following a systematic approach and incorporating relevant data, financial professionals can provide clients with valuable insights and aid in making informed investment decisions. Applying the Modified Dietz Method in Real Life - Accurate Performance Measurement with the Modified Dietz Method 25.Applying the Modified Dietz Method in Real Life[Original Blog] Applying Modified Applying the Modified Dietz Modified Dietz Method Applying the Modified Dietz Method In this section, we will explore a case study that demonstrates the practical application of the Modified Dietz method in real-life investment scenarios. By examining an actual investment portfolio, we can gain a better understanding of how this method can be utilized to identify return sources and attribute them accurately. Background of the Case Study: Let's consider a hypothetical investment portfolio consisting of various assets, such as stocks, bonds, and mutual funds. The portfolio was initially valued at $500,000 at the beginning of the year and experienced several cash flows throughout the year, including additional investments and withdrawals. Our objective is to analyze the portfolio's performance and attribute the returns to different sources using the Modified Dietz method. Calculation of Daily Returns: To apply the Modified Dietz Method, we first need to calculate the daily returns for each asset in the portfolio. This involves comparing the asset's ending value to its beginning value and factoring in any cash inflows or outflows. For example, if a stock had an initial value of $10,000, an ending value of $12,000, and received an additional investment of $2,000 during the year, the daily return would be calculated as follows: [(Ending Value + Cash Inflows) - Beginning Value] / (Beginning Value + Cash Outflows) = [($12,000 + $2,000) - $10,000] / ($10,000) 3. Weighted Average Daily Returns: After calculating the daily returns for each asset, we need to determine the weighted average daily return for the entire portfolio. This involves assigning weights to each asset based on its proportionate value in the portfolio. For instance, if the stock mentioned earlier represents 30% of the total portfolio value, while bonds and mutual funds represent 40% and 30%, respectively, the weighted average daily return would be computed as follows: (Weighted Return of Stock x Weight of Stock) + (Weighted Return of Bonds x Weight of Bonds) + (Weighted Return of Mutual Funds x Weight of Mutual Funds) Cumulative Returns: Once we have the weighted average daily return for the portfolio, we can calculate the cumulative return over a specific period. This gives us a holistic view of the portfolio's performance during that time. For example, if the cumulative return for the year was 10%, it indicates that the portfolio generated a 10% return over the entire period. Attribution of Returns: Using the Modified Dietz Method, we can attribute the returns to different sources, such as asset allocation, security selection, and market timing. By comparing the portfolio's return to a benchmark, we can determine the contribution of each source to the overall return. For instance, if the benchmark return was 8%, and the portfolio return was 10%, we can conclude that the portfolio outperformed the benchmark by 2%. This outperformance can be further analyzed to identify the specific sources responsible for it. This case study highlights the practical application of the Modified Dietz Method in analyzing investment portfolios. By calculating daily returns, determining weighted average returns, and attributing returns to various sources, we can gain valuable insights into the performance of a portfolio. This method allows investors and fund managers to make informed decisions regarding asset allocation, security selection, and market timing, ultimately improving their overall investment strategies. Applying the Modified Dietz Method in Real Life - Identifying Return Sources: Modified Dietz Method for Return Attribution Join our community on Social Media Join our +50K followers of investors, mentors, and entrepreneurs! About Us FasterCapital is a global venture builder and online incubator dedicated to co-funding and co-founding innovative startups. Established in 2014, we are now #1 venture builder in terms of number of startups that we have helped, money invested and money raised. We have supported over 734 startups in raising more than $2.2 billion, while directly investing over $696 million in 288 companies. Our comprehensive support system includes a worldwide network of mentors, investors, and strategic partners, allowing us to transform ideas into scalable, market-ready businesses. FasterCapital operates as FasterCapital LLC-FZ, a duly registered entity in Dubai. Our registration number is 2416362. 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5861	https://www.semanticscholar.org/paper/Clinical-Practice-Guidelines-EASL-Clinical-Practice/5568d3b95eb029c5351f8f3daa5a3368eb0aae9d	Clinical Practice Guidelines EASL Clinical Practice Guidelines : Wilson ’ s disease European Association for the Study of the Liver ⇑ \| Semantic Scholar Skip to search formSkip to main contentSkip to account menu Search 229,291,230 papers from all fields of science Search Sign In Create Free Account Corpus ID: 42597561 Clinical Practice Guidelines EASL Clinical Practice Guidelines : Wilson ’ s disease European Association for the Study of the Liver ⇑ @inproceedings{2012ClinicalPG, title={Clinical Practice Guidelines EASL Clinical Practice Guidelines : Wilson ’ s disease European Association for the Study of the Liver ⇑}, author={}, year={2012}, url={ } Published 2012 Medicine TLDR A number of generally accepted approaches for diagnosis, prevention, and treatment of Wilson’s disease are described, based on a systematic literature review in the Medline, Embase, and Cochrane Library databases.Expand No Paper Link Available Save to Library Save Create Alert Alert Cite Share Figures and Tables 145 References Related Papers Figures and Tables from this paper table 1 figure 1 table 2 table 3 table 4 table 5 View All 6 Figures & Tables 145 References Citation Type Has PDF Author More Filters More Filters Filters ### A practice guideline on Wilson disease E. RobertsM. Schilsky Medicine Hepatology 2003 TLDR These guidelines provide data-supported approaches to the diagnosis and management of patients with Wilson disease that are based on broad-based review of the published literature in pediatrics and medicine and 40 accumulated years of personal experience of the authors.Expand 454 PDF 3 Excerpts Save ### Diagnosis and Management of Wilson's Disease: Results of a Single Center Experience V. MediciC. Trevisan+6 authorsG. Sturniolo Medicine Journal of clinical gastroenterology 2006 TLDR Penicillamines and zinc can effectively treat Wilson's disease, though the side effects of penicillamine may be severe enough to prompt its suspension.Expand 144 1 Excerpt Save ### Clinical presentation, diagnosis and long-term outcome of Wilson’s disease: a cohort study U. MerleM. SchaeferPeter FerenciW. Stremmel Medicine Gut 2006 TLDR Patients with Wilson’s disease having predominantly neuropsychiatric symptoms manifest symptoms later, have a longer time delay from onset of symptoms until definitive diagnosis and have a poorer outcome than patients with hepatic symptoms.Expand 572 PDF 2 Excerpts Save ### Re‐evaluation of the diagnostic criteria for Wilson disease in children with mild liver disease E. NicastroG. RanucciP. VajroA. VegnenteR. Iorio Medicine Hepatology 2010 TLDR Urinary copper excretion greater than 40 μg/24 hours is suggestive of Wilson disease in asymptomatic children, whereas the penicillamine challenge test does not have a diagnostic role in this subset of patients.Expand 182 Highly Influential PDF 7 Excerpts Save ### Diagnosis and current therapy of Wilson's disease P. Ferenci Medicine 2004 TLDR Wilson's disease is an autosomal recessive inherited disorder of hepatic copper metabolism resulting in liver disease and/or neuropsychiatric disease, which requires life‐long administration of copper chelators (d‐penicillamine, trientine).Expand 124 PDF 2 Excerpts Save ### Evaluation of the Unified Wilson's Disease Rating Scale (UWDRS) in German patients with treated Wilson's disease Barbara LeinweberJ. C. Möller+9 authorsH. Hefter Medicine Movement disorders : official journal of the… 2008 TLDR The UWDRS appears to be a promising tool to assess the disease severity in Wilson's disease and its usefulness in clinical research and drug trials should be further addressed.Expand 100 1 Excerpt Save ### A high index of suspicion: the key to an early diagnosis of Wilson's disease in childhood. I. Sánchez-AlbisuaT. Garde+6 authorsP. Jara Medicine Journal of pediatric gastroenterology and… 1999 TLDR Wilson's disease in childhood is generally detected by maintaining a high suspicion of liver disease in patients who have no or nonspecific hepatic symptoms and drug therapy is effective and well tolerated, even in some cases of hepatic insufficiency.Expand 106 PDF 3 Excerpts Save ### Liver transplantation for Wilson's disease: Indications and outcome M. SchilskyI. ScheinbergI. Sternlieb Medicine Hepatology 1994 TLDR It is demonstrated that liver transplantation is life‐saving but not without risk for patients with wilsonian fulminant hepatitis or chronic severe hepatic insufficiency unresponsive to medical therapy, and neurological or psychiatric symptoms due to Wilson's disease may improve after liver transplants.Expand 262 Highly Influential 4 Excerpts Save ### Syndromic variability of Wilson's disease in children. Clinical study of 44 cases. R. GiacchinoM. Marazzi+5 authorsFrancesco Callea Medicine Italian journal of gastroenterology and… 1997 TLDR The early diagnosis, based on clinical suspicion and results of copper metabolism investigation (including hepatic copper content evaluation in difficult cases) and appropriate treatment can prevent the progression of the disease.Expand 52 Save ### Diagnosis of Wilson ’ s disease : an experience over three decades V. DuncombeA. NagreeT. D. Bolin Medicine 113 1 Excerpt Save ... 1 2 3 4 5 ... Related Papers Showing 1 through 2 of 0 Related Papers Stay Connected With Semantic Scholar Sign Up What Is Semantic Scholar? Semantic Scholar is a free, AI-powered research tool for scientific literature, based at Ai2. 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5862	https://en.wikipedia.org/wiki/Colipase	Colipase - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Protein domain 2 See also 3 References 4 Further reading 5 External links Colipase [x] 9 languages Čeština Cymraeg Deutsch Español فارسی Italiano Татарча / tatarça Українська 中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Expand all Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item From Wikipedia, the free encyclopedia Mammalian protein found in humans \| CLPS \| \| \| \| Identifiers \| \| Aliases \| CLPS, entrez:1208, colipase \| \| External IDs \| OMIM: 120105; MGI: 88421; HomoloGene: 1383; GeneCards: CLPS; OMA:CLPS - orthologs \| \| \| show Gene location (Human) \| \| \| \| Chr. \| Chromosome 6 (human) \| \| \| \| Band \| 6p21.31 \| Start \| 35,794,982 bp \| \| End \| 35,797,344 bp \| \| \| \| show Gene location (Mouse) \| \| \| \| Chr. \| Chromosome 17 (mouse) \| \| \| \| Band \| 17\|17 A3.3 \| Start \| 28,777,123 bp \| \| End \| 28,779,740 bp \| \| \| \| showRNA expression pattern \| \| Bgee \| \| Human \| Mouse (ortholog) \| --- \| \| Top expressed in body of pancreas islet of Langerhans testicle pancreatic epithelial cell beta cell right coronary artery ectocervix right uterine tube Descending thoracic aorta right lobe of liver \| Top expressed in stomach islet of Langerhans jejunum duodenum colon ileum esophagus morula quadriceps femoris muscle yolk sac \| \| More reference expression data \| \| \| BioGPS \| More reference expression data \| \| \| \| showGene ontology \| \| Molecular function \| enzyme activator activity \| \| Cellular component \| extracellular region \| \| Biological process \| lipid catabolic process digestion positive regulation of catalytic activity lipid metabolism response to food response to bacterium retinoid metabolic process lipid digestion \| \| Sources:Amigo / QuickGO \| \| \| \| hideOrthologs \| \| Species \| Human \| Mouse \| \| Entrez \| \| \| \| 1208 \| \| \| \| \| 109791 \| \| \| Ensembl \| \| \| \| ENSG00000137392 \| \| \| \| \| ENSMUSG00000024225 \| \| \| UniProt \| \| \| \| P04118 \| \| \| \| \| Q9CQC2 \| \| \| RefSeq (mRNA) \| \| \| \| NM_001832 NM_001252597 NM_001252598 \| \| \| \| \| NM_025469 NM_001317065 \| \| \| RefSeq (protein) \| \| \| \| NP_001239526 NP_001239527 NP_001823 \| \| \| \| \| NP_001303994 NP_079745 \| \| \| Location (UCSC) \| Chr 6: 35.79 – 35.8 Mb \| Chr 17: 28.78 – 28.78 Mb \| \| PubMed search \| \| \| \| \| Wikidata \| \| View/Edit HumanView/Edit Mouse \| Colipase, abbreviated CLPS, is a proteinco-enzyme that counteracts the inhibitory effect of intestinal bile acid on the enzymatic activity of pancreatic lipase. It is secreted by the pancreas in an inactive form, procolipase, which is activated in the intestinal lumen by trypsin. Intestinal bile acids (which aid lipid digestion by facilitating micelle formation) adhere to the surface of emulsified fat droplets, displacing lipase (which is only active at the water-fat interface) from the droplet surface. Colipase acts as a bridging molecule, binding to both lipase and bile acids, thus anchoring lipase onto the droplet surface, preventing its displacement. In humans, the colipase protein is encoded by the CLPS gene. Protein domain [edit] Colipase is also a family of evolutionarily related proteins. Colipase is a small protein cofactor needed by pancreatic lipase for efficient dietary lipid hydrolysis. Efficient absorption of dietary fats is dependent on the action of pancreatic triglyceride lipase. Colipase binds to the C-terminal, non-catalytic domain of lipase, thereby stabilising an active conformation and considerably increasing the hydrophobicity of its binding site. Structural studies of the complex and of colipase alone have revealed the functionality of its architecture. Colipase is a small protein (12K) with five conserved disulphide bonds. Structural analogies have been recognised between a developmental protein (Dickkopf), the pancreatic lipase C-terminal domain, the N-terminal domains of lipoxygenases and the C-terminal domain of alpha-toxin. These non-catalytic domains in the latter enzymes are important for interaction with membrane. It has not been established if these domains are also involved in eventual protein cofactor binding as is the case for pancreatic lipase. Protein domain \| Colipase N-terminal domain \| \| Structure of the pancreatic lipase-colipase complex inhibited by a C11 alkyl phosphonate. \| \| Identifiers \| \| Symbol \| Colipase \| \| Pfam \| PF01114 \| \| InterPro \| IPR001981 \| \| PROSITE \| PDOC00111 \| \| SCOP2 \| 1lpb / SCOPe / SUPFAM \| \| CDD \| cd00039 \| \| \| show Available protein structures: \| \| Pfam \| structures / ECOD \| \| PDB \| RCSB PDB; PDBe; PDBj \| \| PDBsum \| structure summary \| \| PDB \| 1eth, 1lpa, 1lpb, 1n8s, 1pcn, 1pco \| \| Protein domain \| Colipase C-terminal domain \| \| solution structure of porcine pancreatic procolipase as determined from 1h homonuclear two-and three-dimensional nmr \| \| Identifiers \| \| Symbol \| Colipase_C \| \| Pfam \| PF02740 \| \| InterPro \| IPR017914 \| \| PROSITE \| PDOC00111 \| \| SCOP2 \| 1lpb / SCOPe / SUPFAM \| \| CDD \| cd00039 \| \| \| show Available protein structures: \| \| Pfam \| structures / ECOD \| \| PDB \| RCSB PDB; PDBe; PDBj \| \| PDBsum \| structure summary \| \| See also [edit] Enterostatin References [edit] ^ Jump up to: abcGRCh38: Ensembl release 89: ENSG00000137392 – Ensembl, May 2017 ^ Jump up to: abcGRCm38: Ensembl release 89: ENSMUSG00000024225 – Ensembl, May 2017 ^"Human PubMed Reference:". National Center for Biotechnology Information, U.S. National Library of Medicine. ^"Mouse PubMed Reference:". National Center for Biotechnology Information, U.S. National Library of Medicine. ^Koeppen, Bruce M.; Stanton, Bruce A.; Swiatecka-Urban, Agnieszka, eds. (2024). Berne & Levy Physiology (8th ed.). Philadelphia, PA: Elsevier. ISBN978-0-323-84790-2. ^Davis RC, Xia YR, Mohandas T, Schotz MC, Lusis AJ (May 1991). "Assignment of the human pancreatic colipase gene to chromosome 6p21.1 to pter". Genomics. 10 (1): 262–5. doi:10.1016/0888-7543(91)90509-D. PMID2045105. ^Lowe ME (1997). "Structure and function of pancreatic lipase and colipase". Annu. Rev. Nutr. 17: 141–158. doi:10.1146/annurev.nutr.17.1.141. PMID9240923. ^ Jump up to: abVerger R, van Tilbeurgh H, Cambillau C, Bezzine S, Carriere F (1999). "Colipase: structure and interaction with pancreatic lipase". Biochim. Biophys. Acta. 1441 (2–3): 173–184. doi:10.1016/s1388-1981(99)00149-3. PMID10570245. ^Egloff MP, Marguet F, Buono G, Verger R, Cambillau C, van Tilbeurgh H (March 1995). "The 2.46 A resolution structure of the pancreatic lipase-colipase complex inhibited by a C11 alkyl phosphonate". Biochemistry. 34 (9): 2751–62. doi:10.1021/bi00009a003. PMID7893686. Further reading [edit] Weyrich P, Albet S, Lammers R, et al. (2009). "Genetic variability of procolipase associates with altered insulin secretion in non-diabetic Caucasians". Exp. Clin. Endocrinol. Diabetes. 117 (2): 83–7. doi:10.1055/s-2008-1078733. PMID18726866. S2CID260136576. Crandall WV, Lowe ME (2001). "Colipase residues Glu64 and Arg65 are essential for normal lipase-mediated fat digestion in the presence of bile salt micelles". J. Biol. Chem. 276 (16): 12505–12. doi:10.1074/jbc.M009986200. PMID11278590. Miled N, Canaan S, Dupuis L, et al. (2000). "Digestive lipases: from three-dimensional structure to physiology". Biochimie. 82 (11): 973–86. doi:10.1016/S0300-9084(00)01179-2. PMID11099794. van Tilbeurgh H, Egloff MP, Martinez C, et al. (1993). "Interfacial activation of the lipase-procolipase complex by mixed micelles revealed by X-ray crystallography". Nature. 362 (6423): 814–20. Bibcode:1993Natur.362..814V. doi:10.1038/362814a0. PMID8479519. S2CID4305832. Wermter AK, Scherag A, Holter K, et al. (2009). "Procolipase gene: no association with early-onset obesity or fat intake". Obes Facts. 2 (1): 40–4. doi:10.1159/000196379. PMC6444705. PMID20054203. Lindner I, Helwig U, Rubin D, et al. (2005). "Putative association between a new polymorphism in exon 3 (Arg109Cys) of the pancreatic colipase gene and type 2 diabetes mellitus in two independent Caucasian study populations". Mol Nutr Food Res. 49 (10): 972–6. doi:10.1002/mnfr.200500087. PMID16189801. Sims HF, Lowe ME (1992). "The human colipase gene: isolation, chromosomal location, and tissue-specific expression". Biochemistry. 31 (31): 7120–5. doi:10.1021/bi00146a013. PMID1643046. Lowe ME, Rosenblum JL, McEwen P, Strauss AW (1990). "Cloning and characterization of the human colipase cDNA". Biochemistry. 29 (3): 823–8. doi:10.1021/bi00455a032. PMID2337598. van Tilbeurgh H, Bezzine S, Cambillau C, et al. (1999). "Colipase: structure and interaction with pancreatic lipase". Biochim. Biophys. Acta. 1441 (2–3): 173–84. doi:10.1016/s1388-1981(99)00149-3. PMID10570245. D'Silva S, Xiao X, Lowe ME (2007). "A polymorphism in the gene encoding procolipase produces a colipase, Arg92Cys, with decreased function against long-chain triglycerides". J. Lipid Res. 48 (11): 2478–84. doi:10.1194/jlr.M700371-JLR200. PMC3684974. PMID17715423. Gerhard DS, Wagner L, Feingold EA, et al. (2004). "The Status, Quality, and Expansion of the NIH Full-Length cDNA Project: The Mammalian Gene Collection (MGC)". Genome Res. 14 (10B): 2121–7. doi:10.1101/gr.2596504. PMC528928. PMID15489334. Sternby B, Engström A, Hellman U, et al. (1984). "The primary sequence of human pancreatic colipase". Biochim. Biophys. Acta. 784 (1): 75–80. doi:10.1016/0167-4838(84)90175-4. PMID6691986. Sias B, Ferrato F, Grandval P, et al. (2004). "Human pancreatic lipase-related protein 2 is a galactolipase". Biochemistry. 43 (31): 10138–48. doi:10.1021/bi049818d. PMID15287741. Lowe ME (1997). "Structure and function of pancreatic lipase and colipase". Annu. Rev. Nutr. 17: 141–58. doi:10.1146/annurev.nutr.17.1.141. PMID9240923. Strausberg RL, Feingold EA, Grouse LH, et al. (2002). "Generation and initial analysis of more than 15,000 full-length human and mouse cDNA sequences". Proc. Natl. Acad. Sci. U.S.A. 99 (26): 16899–903. Bibcode:2002PNAS...9916899M. doi:10.1073/pnas.242603899. PMC139241. PMID12477932. Sugar IP, Mizuno NK, Momsen MM, et al. (2003). "Regulation of lipases by lipid-lipid interactions: implications for lipid-mediated signaling in cells". Chem. Phys. Lipids. 122 (1–2): 53–64. doi:10.1016/S0009-3084(02)00178-0. PMID12598038. van Tilbeurgh H, Sarda L, Verger R, Cambillau C (1992). "Structure of the pancreatic lipase-procolipase complex". Nature. 359 (6391): 159–62. Bibcode:1992Natur.359..159V. doi:10.1038/359159a0. PMID1522902. S2CID4360354. Davis RC, Xia YR, Mohandas T, et al. (1991). "Assignment of the human pancreatic colipase gene to chromosome 6p21.1 to pter". Genomics. 10 (1): 262–5. doi:10.1016/0888-7543(91)90509-D. PMID2045105. External links [edit] Colipases at the U.S. National Library of Medicine Medical Subject Headings (MeSH) PDBe-KB provides an overview of all the structure information available in the PDB for Pig Colipase This article incorporates text from the public domain Pfam and InterPro: IPR001981 This enzyme-related article is a stub. You can help Wikipedia by expanding it. v t e Retrieved from " Categories: Genes on human chromosome 6 Protein domains Membrane proteins Enzymes Protein families Enzyme stubs Hidden categories: Articles with short description Short description is different from Wikidata All stub articles This page was last edited on 28 May 2025, at 01:11(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Colipase 9 languagesAdd topic
5863	http://blog.cambridgecoaching.com/but-what-is-dx-really-calculus-terms-explained	CC \| But what is “dx” really? Calculus terms explained But what is “dx” really? Calculus terms explained academicscalculusHigh Schoolmath By Jane The symbol “dx” comes up everywhere in calculus. For example: If y is a function of x, then we sometimes write the derivative of y with respect to x as the following: When we write indefinite integrals, they are written as: When we write definite integrals, they are written as: But what is “dx” really? It’s more than just notation! In this post, we’ll explore the meaning of “dx” and try to get a better understanding of some of the symbols that we often see in calculus. Calculus as the study of infinitesimal change Calculus is the study of continuous change, or infinitesimal change. To get an idea of what this means, let’s consider the following: suppose that you’re running a race. You begin running at time x=0 seconds and then track your displacement as a function f(x). Your displacement function f(x) is then a continuous function that varies over time. Suppose now that you wanted to know your speed at time x=10 seconds. That is a calculus question since you are looking for a rate of change at one particular time. How could you do this? Well, you might estimate your speed at time t=10 as our average speed between time x=10 and x=11, which can be expressed as: But you could get a better estimate by choosing a smaller time interval, say from x=10 to x=10.1 instead, or even better from x=10 to x=10.01. Really, what you want to do is to take the limit as the size of your time interval goes to zero. That is, you want to take an infinitesimally small change in x. Then, by the definition of a derivative, your speed at x=10 seconds is: We should think of h going to zero as taking smaller and smaller increases in x when we take the average speed from time x=10 to time x=10+h. This is where “dx” comes in. "dx" is an infinitesimal change in x.We can think of "dx" (read as dee-ex) as an infinitesimally small change in x. The "d" in "dx" should remind you of a delta ∆, which is the symbol for change. "dx has no numerical value. Rather, it captures this idea that occurs a lot in calculus of taking the limit of smaller and smaller interval sizes to figure out something precisely about a continuous function. We can see this if we revisit our speed example from earlier. When computing our derivative The bottom of this fraction is (10+h)-10 as h goes to zero, which is an infinitesimally small change in x. We might therefore think of the denominator as h goes to zero as dx. If we let y=f(x), then the numerator of this fraction is f(10+h)-f(10) as h goes to zero, which is an infinitesimally small change in y, or dy. Putting this all together, we recover the notation: That is, the derivative of f(x) is the quotient of an infinitesimal change in y over an infinitesimal change in x. Put more precisely, it is exactly the limit of the change in y over the change in x over smaller and smaller changes in x. The “dx” and “dy” notation just captures this limiting procedure and expresses it as an infinitesimal change in x or y instead. “dx” as seen in integrals Another place where "dx" is often seen is in integrals. Let's focus on definite integrals. What does "dx" mean in a definite integral? "dx" here is still an infinitesimal change in x. To see why it's there, we should think of the integral as a signed area and as the limit of Riemann sums. We recall that to compute a left Riemann sum of f(x) from x=a to x=b with n intervals, we let the following be true: Then we take: ...where x takes values a, a + ∆x, a + 2∆x,..., a + (n - 1)∆x = b - ∆x. Then, as we let n go to infinity, ∆x gets smaller and smaller, and the Riemann sum converges in value to the integral, which is the signed area under the curve f(x) between x=a and x=b. The below picture (from the Wikipedia article on Riemann sums) shows this convergence process: Now, we can see where the notation for the integral comes from. The integral sign ∫ is the continuous version of the sum sign ∑. The bounds of integration from a to b are like the first and last x values for the summation. And the dx is the infinitesimal version of ∆x, what we get when we take smaller and smaller step sizes in x. Another way to think about this is that in our integral, we are summing up infinitesimally thin rectangles with height f(x) and width dx to compute our signed area precisely. Hopefully, the symbol “dx” is a little less mysterious now! Related Content June 1, 2020Gaining intuition for Taylor series math November 11, 2020How to approach initial value word problems CollegeHigh Schoolmathmathematics January 16, 2023The magic of induction Collegeacademicsmathematicsinduction January 22, 2021Solving algebraic equations with variables on both sides Middle Schoolmath February 5, 2021Proof by contradiction: how to be so wrong you end up being right SATmath May 2, 2022Einstein’s proof of E = mc^2 Collegemathacademics Did you know we offer tutoring for high school students? Learn more Comments Shaun 8/11/2021, 9:47:38 AM Can you explain the notation d/dx? Reply to Shaun Olivia Miller 8/11/2021, 3:35:33 PM "d/dx" is a mathematical operation that indicates differentiation with respect to x. When you see "d/dx", it's telling you to take the derivative of whatever follows the operation with respect to x. This notation does not require a function or dependent variable (like y) to be present - it allows you to differentiate an expression directly. Hope that helps! Reply to Olivia Miller Atticus 1/20/2022, 11:15:55 PM Thank you for this explanation! Reply to Atticus 456356356 5/14/2022, 4:47:36 PM thank you Reply to 456356356 au 6/24/2022, 2:59:04 PM When dy/dx = 1, we can write dy=dx. Right? When dy=dx, can I write dy - dx = 0? Reply to au Aashima 8/6/2022, 3:12:08 AM You can definitely write dy/dx = 1, and that dy=dx. You can also write that dy - dx = 0. These statements are often written when working with differential equations. Reply to Aashima Ben 11/10/2022, 10:34:39 AM Thanks a lot Reply to Ben Robert 11/10/2022, 3:34:13 PM Harder question: how to you explain the idea of dx(v) where v is a vector and dx is somehow now a function?? Reply to Robert Alexander 6/4/2023, 4:18:52 AM I've heard d(x) called a differential as well. I've seen for some higher level integrals the differential is with respect to a function df which I'm told equals f'(x)dx. Does this work for all functions and why are we allowed to treat dy/dx as a fraction of differentials when solving a differential equation using seperation of variables? Reply to Alexander Dvij 8/4/2023, 7:43:41 PM Excellent explanation. I've never come across such a succinct and comprehensible explanation of this topic. The only thing I'd add is maybe explain the difference between d/dx and dy/dx as well. Reply to Dvij Muhyideen 10/6/2023, 8:38:22 AM What's the essence of that dx at RHS. Is it possible to integrate without it? Reply to Muhyideen Gabe 3/18/2025, 7:52:47 PM What is dx in indefinite integrals? What about the way it's used in U substitution where it's somehow equal to the derivative of U? 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5864	http://mathcentral.uregina.ca/QQ/database/QQ.09.00/cyd1.html	Prisms A fourth grade class is curious about the names that have been given to various solid geometric figures based on their shapes: are there names for solids based on a rhombus, a trapezoid, a parallelogram, a decagon, and/or a quadralateral? Hi, Probably every such figure has been given a name, but why bother? You need a name only as a convenient abbreviation when you are going to refer to the object over and over again. For practical purposes (and greater clarity) it is best to say something like, "Let P be a right (or rectangular) prism whose base is a ... ." Often when a person says "prism" he means that the sides are perpendicular to the base; this is officially called a RIGHT or RECTANGULAR prism. The dictionary definition applies to solids whose top and bottom are congruent figures in parallel planes and whose sides are composed of parallel straight line segments joining corresponding points of the top and bottom. Often the meaning is restricted to the case when the top and bottom are polygons (in which case the faces on the sides are parallelograms). When the base is also a parallelogram the prism is a PARALLELEPIPED. Otherwise the prisms are named in terms of the number of sides on the base: triangular, quadrilateral, pentagonal, ... . Watch out for ambiguity: a RECTANGULAR prism refers to the sides being rectangles, not the base. When the base is also a rectangular the official name is a "rectangular parallelepiped"! That's quite a mouthful. Cheers, Chris Go to Math Central
5865	https://www.sciencedirect.com/topics/medicine-and-dentistry/trichuris	Skip to Main content My account Sign in Trichuris In subject area:Medicine and Dentistry Trichuris is defined as a colorectal helminth, specifically whipworms, that infects the cecum and rectum primarily through the ingestion of infective ova from contaminated food or water, causing potential malnutrition and growth restriction in children. AI generated definition based on: Gastrointestinal Diseases and their Associated Infections, 2019 How useful is this definition? Add to Mendeley Also in subject areas: Agricultural and Biological Sciences Biochemistry, Genetics and Molecular Biology Immunology and Microbiology Discover other topics Chapters and Articles You might find these chapters and articles relevant to this topic. Parasitic Diseases of Nonhuman Primates 2005, The Laboratory PrimatePurcell Jeanette E., Philipp Mario T. Trichuriasis Trichuriasis is caused by the nematode Trichuris trichiura. Trichurid parasites, commonly known as “whipworms” are found worldwide, but at a higher frequency in tropical and subtropical environments. Trichuris trichiura is a common inhabitant of the cecum of New World and Old World nonhuman primates (Orihel and Seibold, 1972). Like B. coli, Trichuris is frequently identified during routine fecal examination but is rarely of clinical significance. T. trichiura of nonhuman primates has previously been described as morphologically indistinguishable from the human parasite (Flynn, 1973). Subtle morphologic differences between T. trichiura of monkeys and humans have been identified by scanning electron microscopy (Ooi et al., 1993), but these subtle differences were not deemed sufficient to create a new species for T. trichiura of nonhuman primates. This, and the fact that Trichuris ova have been experimentally transmitted from nonhuman primates to humans (Horii and Usui, 1985), provide evidence for the zoonotic potential of T. trichiura of nonhuman primates. Trichuriasis is a well-recognized cause of intestinal intussusception in human infants (Stephenson et al., 2000). A report of intestinal intussusception in two juvenile baboons infected with Trichuris demonstrates the pathogenic potential of this parasite in nonhuman primates (Hennessey et al., 1994). SIV-infected macaques are not commonly reported to experience more severe disease from Trichuris. However, in one study, 25% of SIV-infected macaques were shown to be infected with Trichuris at postmortem examination (Kuhn et al., 1997). The incidence of diarrhea in the infected animals was 50%. Transmission electron microscopy showed the anterior ends of the parasites invading the cecal mucosa and associated with mild focal inflammation. Diagnosis of trichuriasis in nonhuman primates is made by identification of either characteristic bipolar-plugged eggs in fecal samples, or adults in the cecum. Trichuris eggs are highly resistant and if treatment of nonhuman primates is deemed to be necessary, animals should be removed from contaminated enclosures and caging should be sanitized regularly (Flynn, 1973). Butamisole, mebendazole, flubendazole (Kumar et al., 1978) and levamisole have been determined to be successful in treating nonhuman primates with trichuriasis. View chapterExplore book Read full chapter URL: Book2005, The Laboratory PrimatePurcell Jeanette E., Philipp Mario T. Chapter Trichuriasis 2006, Tropical Infectious Diseases (Second Edition)EDWARD S. COOPER INTRODUCTION Members of Trichuris species are successful nematode parasites of mammalian bowel and Trichuris trichiura has had much opportunity to coevolve with man. The T. Trichuris species itself is a parasite of both Old World and New World monkeys and apes.1 Its human presence is associated with poor hygiene rather than a specifically tropical environment, although warmth and moisture in the soil enhance transmission through promotion of the viability of the infective stages. Until about 30 years ago, four things led to the underestimation of T. trichiura as a pathogen. 1. : Low-intensity infections, which are by far the most common, are asymptomatic. 2. : Trichuris is seldom found as the only pathogen, but is commonly just one of multiple health and environmental threats. 3. : The onset of significant symptoms is often too slow to alarm the family. 4. : It produces a transient, although prolonged, disease of the developing child while seldom causing disability to adults. However, severe infections of children are amenable to treatment with highly effective drugs with minimal side effects; and this makes case-detection and treatment most rewarding and population-based prevention most productive. In the last decade, there has been a steady increase in reports on all aspects of trichuriasis, with publications reported by Medline now approaching 40 per year. View chapterExplore book Read full chapter URL: Book2006, Tropical Infectious Diseases (Second Edition)EDWARD S. COOPER Chapter Trichuriasis 2006, Tropical Infectious Diseases (Second Edition)EDWARD S. COOPER INTRODUCTION Members of Trichuris species are successful nematode parasites of mammalian bowel and Trichuris trichiura has had much opportunity to coevolve with man. The T. Trichuris species itself is a parasite of both Old World and New World monkeys and apes.1 Its human presence is associated with poor hygiene rather than a specifically tropical environment, although warmth and moisture in the soil enhance transmission through promotion of the viability of the infective stages. Until about 30 years ago, four things led to the underestimation of T. trichiura as a pathogen. 1. : Low-intensity infections, which are by far the most common, are asymptomatic. 2. : Trichuris is seldom found as the only pathogen, but is commonly just one of multiple health and environmental threats. 3. : The onset of significant symptoms is often too slow to alarm the family. 4. : It produces a transient, although prolonged, disease of the developing child while seldom causing disability to adults. However, severe infections of children are amenable to treatment with highly effective drugs with minimal side effects; and this makes case-detection and treatment most rewarding and population-based prevention most productive. In the last decade, there has been a steady increase in reports on all aspects of trichuriasis, with publications reported by Medline now approaching 40 per year. AGENT Trichuris trichiura is a member of the nematode superfamily Trichuroidea and is related to Trichinella spiralis. The genus was previously often called Trichocephalus, logically enough since the hairlike part (tricho-) is in fact the head end (cephalus). However, the original name given by Linnaeus (1771), in the mistaken belief that the hairlike part was the tail (uris), is the official name. Whipworm is a commonly used unofficial name. The adult is shaped like a whip, with the handle representing the wider posterior section containing the reproductive organs and the intestine, while the long, fine anterior part, called the stichosome, contains the long pharynx. The whole is about 4 cm long. The male has a curled posterior end. The eggs are thick-walled and barrel-shaped, about 50 mm long, with a plug at each pole. Eggs, passed in the feces, contain a zygote and are not infective until embryonation, which takes place in the soil over 2 to 4 weeks. The egg now contains the L1 larva. Following human ingestion, the larva is released in the stomach and passes into the intestine. It penetrates the epithelium in the mucosal crypts of the cecum.2 The larva develops by molting, and the adult develops from the L4 stage, by now having migrated with the epithelial cells up the sides of the crypts. The anterior part of the adult lies in a tunnel within the epithelium between the mouths of the crypts, while the posterior part is free in the lumen. The stichosome is surrounded by a syncytium and debris of the epithelium. Each female produces up to 20,000 eggs per day; the life expectancy of a worm within the host has been estimated at 1 to 3 years,2 which would imply that some adults live far longer. EPIDEMIOLOGY AND ECOLOGY Up to a quarter of humanity carries T. trichiura as an intestinal parasite. An estimate for the number of Chinese infected (212 million)3 has brought the world estimate to over 1 billion infected human hosts. However, all parasitic infections have an aggregated (“clumped”) distribution among their hosts. This is extreme in trichuriasis, where over 90% of a community may be infected but only 10% or fewer have intense, symptomatic infections.4 There is as yet no clear evidence on any factor making the intensely infected hosts especially susceptible, apart from the obvious one of environmental exposure related to hygienic practices. Familial aggregation occurs.5 After treatment, there is a tendency for the most heavily infected to become the most heavily re-infected, but there is much crossing over between light and heavy strata too.6 The role of domestic pigs in transmission is unknown, although passage of the life cycle through the pig is possible.7 Warm, damp soil provides the best medium for transmission once it is fecally contaminated. T. trichiura is found in humid, tropical environments but also in temperate climates including Northern Europe and South Africa. Although transmission occurs in cool climates, hyperendemicity is correlated with higher temperatures and lower altitudes.8 The nematode is the predominant intestinal parasite in many communities in Africa and the Americas. DISEASE Most infections are asymptomatic. In heavy infections (many adult worms, see next section), stools become loose and frequent and there is tenesmus. Frequency can exceed 12 stools per 24 hours and nocturnal stooling is especially characteristic. Stools consist largely of mucus but may also be watery. There is a characteristic acrid smell. Frank blood is common. Trichuriasis is one of the most frequently identified causes of recurrent rectal prolapse and the worms may be seen on the prolapsed mucosa. Children with this degree of symptomatic infection are almost invariably severely anemic and growth-retarded.9 They are also geophagic and much of their stool may consist of ingested earth or even small stones. Finger-clubbing is common, correlated in prevalence and severity with the number of adult worms harbored.10 PATHOGENESIS AND IMMUNITY Light infections (20 adult worms or fewer) are not associated with any discernible morbidity. Heavy infections (200 adult worms or more) are associated consistently with the clear-cut syndrome described previously, which has both colonic and systemic features. The difference is not only one of parasite biomass (in which a 10-fold increase may scarcely be significant clinically), but perhaps more importantly of the site affected: In light infections, the worms are confined to the cecum and the ascending colon, whereas in heavy infections there are adult worms in the distal colon and rectum, and also often in the terminal ileum. The mucosa at these latter sites may have a greater tendency to a local hypersensitivity reaction to T. trichiura. In the inflamed areas, especially the rectum, the mucosa is edematous. At the cellular level, the most significant abnormalities are distended goblet cells, an increased concentration of superficial mast cells, which are degranulating, and an increase in the lamina propria of calprotectin-secreting cells, presumed to be monocytes.11 Increased tumor necrosis factor (TNF)-α is both produced by lamina propria cells and found circulating in the peripheral blood.11,12 Bloody mucus commonly exudes from the mucosal surface. The systemic consequences are anemia and impaired growth.13 The former is presumed to follow from direct loss from the gut of red blood cells followed by iron deficiency. The mechanism for the impaired growth is not clear but could involve both substrate-limited and substrate-wasting pathways. In support of the former is the finding of protein-losing enteropathy proportional to the worm burden,13 and of the latter increased circulating TNF-α, which may inhibit appetite as well as having direct effects on cell division and metabolism. Plasma insulin-like growth factor-I (IGF-I) concentration is found to be correspondingly reduced during the Trichuris dysentery syndrome (TDS), although it returns to a similar level to that of comparable children from the same community within a month of worm expulsion.14 The increase in plasma IGF-I after treatment accompanies a fall in plasma TNF-α, an increase in collagen synthesis as shown by rising plasma pro-collagen type 1, and an acceleration in physical growth. Although it had been suggested that the beneficial effect of mebendazole on child growth might be due to some systemic effect other than through worm expulsion, a quite specific worm effect has now been shown.15 The elevation of acute phase proteins16 is further evidence of a systemic component in the response to heavy Trichuris infection. Plasma viscosity16 is also moderately elevated and declines slowly over a period of months, despite the rapid resolution of symptoms after anthelminthic treatment. Concentrations of immunoglobulin A (IgA), IgM, and IgG antibody to T. trichiura are correlated with the host's current worm burden.17,18 Specific responses of IgG1, IgG4, IgA, and IgE have been shown.12 As with Ascaris, multiple reinfections with Trichuris are to be expected,19 implying that protective immunity must be incomplete. Epidemiological data combined with serology can be interpreted as consistent with an increase in immunity with age, and as suggesting that both IgA17 and IgE12 may be mediators of this. Some genetic polymorphism in susceptibility20,21 has been shown but, if this is so, it is all the more remarkable that prevalence of infection can approach 100% in large and diverse communities.22 Evidence has been sought in humans12,23 for the divergence in immune response seen in mice,24,25 where a Th1 cytokine response has been associated with chronic infection and a Th2 response with parasite expulsion. The human responses have been diverse and inconsistent.12 The most striking evidence for the converse, that chronic infection induces a Th2 response that inhibits the Th1 response, has come from the remarkable experiment of deliberately infecting patients suffering from Crohn's disease or ulcerative colitis with T. suis eggs as anti-Th1, and therefore anti-inflammatory, therapy. Preliminary results have been encouraging.26 Cytokine expression in human colonic mucosa biopsied as part of investigation of Trichuris dysentery syndrome suggests that both interferon-γ (Th1) and interleukin-4 (Th2) may be strongly expressed in the mucosa of children with heavy and chronic infections with T. trichiura. This corresponds with recent data from peripheral blood.27 It excludes T-cell anergy as an explanation for these exceptionally intense infections. In humans, downstream of T-cell recognition, some component of the immune response that leads to worm expulsion in mice appears to be either missing or attenuated.23 Eosinophilia is often associated with trichuriasis, but its lack by no means excludes the diagnosis in any individual. There may be neuropsychological effects of intense trichuriasis, manifest by impaired cognitive function28 and especially locomotor development.29 These effects are too subtle to present as a clear part of the clinical picture, unless the child is already suffering from severe social deprivation or cognitive disorder and so is more likely to have geophagia leading to the secondary acquisition of this heavy worm burden. The systemic inflammation, especially increased circulating TNF-α, is a possible mediator between the parasitized intestine and impairment of higher brain function. DIAGNOSIS The differential diagnosis is from other causes of infective dysentery or idiopathic colitis. A stool specimen may be examined by any of the techniques for parasites, but the Kato-Katz preparation is recommended for simplicity, reliability, and quantification. Although from a population perspective a count of eggs per gram (epg) of feces is well correlated with adult worm burden, this cannot be applied to the individual. With less than 1000 epg clinically significant trichuriasis is unlikely and with over 10,000 epg it is likely, but severe cases occur with absent eggs on a single stool specimen and some lightly infected children have over 10,000 epg. Proctoscopy showing worms on the rectal mucosa is more reliable evidence of Trichuris colitis,29 and this can be used in the clinic. Colonoscopy sometimes reveals an intense colonic infection, either in a research setting or if the diagnosis has not been made before the procedure. It cannot be justified as a diagnostic procedure: Presumptive treatment or a trial of treatment is fully justified in a busy clinic setting. TREATMENT AND PROGNOSIS Many anthelminthics effective against Ascaris fail against Trichuris, and to these must now be added ivermectin.30,31 Mebendazole (Janssen's Vermox or generic) or albendazole (Zentel, Smith-Kline-Beecham) by mouth are effective and recommended. Single doses (mebendazole 100 mg to 500 mg; albendazole 400 mg) give a good chance of cure in light infections and are worthwhile in mass campaigns, but increasing probability of complete worm expulsion is a function of increasing exposure to the drug. For clinically significant infections, mebendazole 200 mg per day on 3 successive days is recommended, or albendazole 400 mg per day for 3 days. These drugs are contraindicated in pregnancy and not recommended for infants, but clinical judgment must be used in the event of symptomatic infection. Worm expulsion will be followed by clinical cure, but if the environment is unchanged re-infection is likely and re-treatment every 3 months,19 or at least every year,32 is well justified. PREVENTION AND CONTROL Satisfactory fecal disposal using water-seal toilets or good pit-latrines completely interrupts the life cycle of these geohelminths. Hand washing and washing of any vegetables grown in fecally contaminated soil is also useful. In the real-life situation of crowded, poor dwellings, scarce piped water and heavily contaminated soil, mass chemotherapy with cyclical repetition is highly effective.32 The children of preschool and school age should be targeted.2 In regions where periodic mass treatments are being used to control lymphatic filariasis, single-dose treatments with albendazole together with ivermectin have proven more efficacious in diminishing Trichuris infections than single-dose treatments with albendazole alone or with albendazole combined with diethylcarbamazine.33,34 View chapterExplore book Read full chapter URL: Book2006, Tropical Infectious Diseases (Second Edition)EDWARD S. COOPER Chapter Geohelminths 2022, Encyclopedia of Infection and ImmunityStefano D’Amelio, Serena Cavallero Trichuris trichiura General aspects The members of the genus Trichuris (Order Trichinellida) are commonly known as whipworms as they are shaped as a whip. In fact, the anterior part corresponding to the structure of the esophagus consists of a narrow tube with one to three rows of large gland cells called stichocytes and the entire glandular structure is known as a stichosome. This anterior part is usually much narrower than the posterior part of the body, that is thick and containing the reproductive organs (Figs. 4 and 5). Similarly to A. lumbricoides, T. trichiura has worldwide distribution, and its occurrence and incidence are directly correlated with poor sanitation and the use of human feces as fertilizer. In fact, whipworms are listed in the NTDs with the other nematodes and usually co-occur with roundworms in the same host, but with a different and peculiar gastrointestinal localization. It displays a simple, direct life cycle. The infection starts with the ingestion with contaminated food or water of eggs containing a third-stage larva that hatch into the small intestine. Eggs are generally lemon or barrel shaped with thick-shelled with polar, plug-like structures (Fig. 6). The larvae migrate to the intestinal caecum where they molt and mature to adult stage. They embed into the mucosa and after 3 months the females release up to 10,000 eggs per day. Eggs are generally oviposited in a not embryonated state and become infective after a period of 20–30 days into the soil (Fig. 7). A broad range of mammalian hosts including ruminants, marsupials, rodents, and primates can be infected by whipworms. To date, around 100 species have been described and synonymized, showing a high biodiversity in the genus. Humans are commonly infected as well, mainly by the medically-relevant species T. trichiura and, occasionally, by zoonotic species such as Trichuris vulpis and Trichuris suis. Trichuris spp. usually infect a particular host or a group of related host species and parasitic species designation is often based on an integrative approach based on morphological and molecular methodologies. The parasitic species identification is usually performed using information on host-specificity and on general morphology of eggs and adults, in combination with evidences from some diagnostic nuclear and mitochondrial molecular markers. In the case of humans and other primates, whipworms are automatically identified as T. trichiura, but recently its taxonomic status has been largely debated. Molecular studies have revealed the existence of more than one taxon able to infect humans and other primates, including captive individuals, and new species have been described thus suggesting the status of T. trichiura as a complex of species, counting different cryptic taxonomic units with different ability to infect one particular or several species. Trichuris infections in humans have been described since the 18th century, but ancient paleoparasitological evidences report the recovery of Trichuris eggs in Korea, dating 2000 years ago as well as findings of this species in South America in pre-Columbian times. Trichuris eggs were also found in the intestine of a glacier mummy from the Copper Age (about 5000 years ago), known as Ötzi or the Similaun mummy, who was recovered in an astonishingly good state of preservation on the Alps at the border between Italy and Austria. Pathology and clinical features Trichuriasis show mild to severe clinical outcomes, mostly related to the presence of adult worms, of approximately 4 cm in length, feeding on intestinal host tissues. Diarrhea and dysentery are common symptoms. Light whipworm infections are usually asymptomatic. The pathogenic effect of trichuriasis is mainly at gastrointestinal level, as the larval stages do not migrate as other STHs larvae do. The infection site of adults is in the large intestine, mostly caecum and colon, although they may reach the appendix and the rectum. This happens usually in heavy infections, while, when a low number of worms are present, they are usually confined to cecum and ascending colon. Re-infection occurs as a result of contact with infective stages in the environment, most frequent in areas where sanitation is poor. The consequences of infections are related to an impairment of the nutritional status of the infected people, in particular for iron and protein loss caused by parasitic blood absorption. Iron deficiency and poor nutrition caused by parasitic competition may cause in children anemia, growth retardation, and even impaired cognitive development. In fact, the presence of whipworms causes the typical symptoms of rectal bleeding and abdominal pain because of cell destruction and activation of the host immune system, recruiting eosinophils, lymphocytes and plasma cells. At intestinal level, an increase of macrophages, TNFα and mast cell degranulation are observed, suggesting that colitis can be regarded as a sort of localized anaphylactic reaction. Long lasting infections may produces chronic effects that are similar to IBD (intestinal bowel disease) and stools of infected people may contain a mixture of mucus, water, and blood. Heavy Trichuris infections may result in in chronic dysentery and rectum prolapse. The inflammation is mainly produced by the intestinal tissue damage caused by the intimate embedding of the long, thin anterior end of the worm in the mucosa. Rectal prolapse, one of the most serious damage due to these nematodes, seems to occur with an intussusception mechanism, i.e., when the rectum invaginates into itself. On the other hand, the host immune response to Trichuris seems to be mostly characterized by the T helper (Th) 2 cells, which prevent worm survival and facilitate their expulsion. Th2 response attenuates the Th1-driven inflammatory response. This may have the effect to impair the protection versus viral and bacterial infections. Moreover, a partial protective immunity that develops with age was observed (Truscott et al., 2015). Such features imply the potential use of parasites for therapy of inflammatory diseases. In fact, intentional exposure to intact helminths or helminth-derived products have been investigated and by now, therapy with T. suis eggs showed no real benefit for IBD patients, suggesting a careful evaluation is needed before its use into clinical practice. Several studies described the mechanisms by which helminths manipulate the host immune system, but little consideration has been given to the actual tested helminths and in particular to their zoonotic potential. For T. suis, the porcine species has the ability to infect humans, too, and its use should be carefully evaluated. Diagnosis Trichuriasis should be suspected in patients from endemic areas. Many cases may be asymptomatic, but severe infections display abdominal pain, blood in diarrheic feces and rectal prolapse. Stool examination following concentration reveals the presence of the characteristic oval-shaped eggs with polar plugs. Abdominal pain and general gastrointestinal symptoms are shared with other infective agents or typical of certain human diseases, a differential diagnosis should be accounted in order to correctly identify the cause of disease. View chapterExplore book Read full chapter URL: Reference work2022, Encyclopedia of Infection and ImmunityStefano D’Amelio, Serena Cavallero Chapter Appendicitis and Infections of the Appendix 2019, Gastrointestinal Diseases and their Associated InfectionsMojgan Hosseini MD, Laura W. Lamps MD Trichuris sp. (Whipworms) Trichuris is a colorectal helminth with a worldwide distribution, particularly in the tropics and parts of Asia. Infection is most commonly seen in children, and is transmitted by ingestion of infective ova through contaminated food or water. The cecum is the primary site of infection (sometimes with the involvement of ileum), followed by the rectum. Appendicitis and peritonitis have been rarely reported.137,138 Most patients are asymptomatic, although infection in children can lead to malnutrition and growth restriction, and rarely trichuriasis can mimic Crohn’s disease.139,140 The worms are about 4 cm in size and can be seen with the naked eye. Laboratory diagnosis is established by examination of stool for ova and parasites, revealing the distinctive ovoid eggs with bilateral mucus plugs. View chapterExplore book Read full chapter URL: Book2019, Gastrointestinal Diseases and their Associated InfectionsMojgan Hosseini MD, Laura W. Lamps MD Chapter Soil-transmitted Helminths (Geohelminths) 2014, Manson's Tropical Infectious Diseases (Twenty-third Edition)Simon J. Brooker, Donald A.P. Bundy Trichuriasis (Whipworm) Trichuriasis is caused by Trichuris trichiura, which is one of the most prevalent human geohelminths and is commonly found in tropical areas with heavy rainfall, constant warm temperature and inadequate sanitation. Light infections give rise to few clinical symptoms, but heavy infections can cause gastrointestinal problems, rectal prolapse, anaemia, growth stunting and cognitive impairment. Geographical Distribution T. trichiura occurs worldwide, but is most prevalent in warm and humid tropical regions of the world.5,6 Transmission does not typically occur in arid areas (Figure 55.3). It is estimated that 0.5 billion people were infected globally in 2010, including 35 million school-aged children harbouring high-intensity infections which cause morbidity (Pullan and Brooker, unpublished). T. trichiura is estimated to cause the loss of 638 000 disability adjusted life years (DALYs).7 Owing to the generalized pathology of T. trichiura no reliable estimates of T. trichiura-associated mortality exist. The prevalence of T. trichiura varies in different parts of the world and is greatest in equatorial Africa and South-east Asia. Sustained control and socioeconomic development has helped reduced prevalence levels in parts of China, East Asia and Central and South America.8 Aetiology T. trichiura is a greyish-white worm, often slightly pink, which lives in the caecum and appendix. The male (30–45 mm long) has an attenuated anterior portion containing a cellular oesophagus which is half as long again as the thicker posterior portion, and a caudal extremity curved through 360° with a single spicule in the sheath which is studded with spines (Figure 55.4). The female (30–35 mm long) has the posterior half occupied by a stout uterus packed with eggs. The egg (50 × 22 µm) is brown with a characteristic band shape and a single cell with a plug at each end; it contains a single embryo (Figure 55.5). Life Cycle The worms live in the caecum, where they maintain their position by transfixing a superficial fold of mucosa and lie embedded in epithelial tunnels it creates between the intestinal villi. Eggs are laid unsegmented, and embryonation takes at least 21 days. Eggs can withstand low temperatures but not desiccation. Infection is direct from contaminated faeces. The egg hatches after being swallowed in the intestine, where the shell is digested by intestinal juices and the larva emerges in the small intestine; it penetrates the villi and develops for a week until it re-emerges and passes to the caecum and colorectum, where it attaches itself to the mucosa and becomes adult (see Appendix III). Transmission Transmission is direct from mature eggs to the mouth via fingers contaminated from infected soil. As well as through accidental ingestion of eggs, transmission in some settings can also occur though the practice of geophagia – eating of soil.9 Pathology The pathology caused by T. trichiura is strongly related to the size of the worm burden: the intensity of infection. When individuals harbour only a few worms these are confined to the caecum and ascending colon, causing little damage. However, with heavy infections they spread throughout the colon to the rectum, where they cause haemorrhages, mucopurulent stools and symptoms of dysentery with rectal prolapse.10 The Trichuris dysentery syndrome (TDS) associated with heavy infections is thought to be due in part to the acute-phase immune response and a specific elevation of plasma fibronectins and plasma viscosity,11 as well as low admission plasma levels of insulin growth factor-1 (IGF-1), low type of procollagen and high serum levels of tumour necrosis factor α (TNF-α).11,12 Trichuris-related mucosal damage may facilitate the invasion of other infections, including shigellosis and Entamoeba histolytica, causing further ulceration. Infection can exacerbate colitis caused by infection with Campylobacter jejuni.13 Immunity Individuals living in endemic areas mount a vigorous immune response involving all antibody serotypes, including immunoglobulins (Ig) A, IgM, IgG and IgE. A negative association often exists between antibody levels and intensity of infection.14 Cytokine response profiles are also associated with patterns of T. trichiura infection and reinfection.15 However, any effects of protective immunity must be incomplete since individuals show persistent susceptibility to infection following treatment.6 Clinical Features Natural History. In the vast majority of infections, which are light and asymptomatic, the worms live harmlessly in the caecum and appendix but when the infection is heavy (with worm burdens exceeding 500 worms), there can be marked and often severe symptoms and signs. In children, however, even symptomless infections may have subtle and insidious effects on nutritional status, and physical and intellectual growth. Incubation Period. The prepatent period from ingestion of eggs to the appearance of eggs in the stool is 60–90 days. Symptoms and Signs. The pattern and severity of symptoms and signs is positively correlated with the intensity of infection. In light infections there are no discernable symptoms, but when associated with Ascaris lumbricoides or hookworm mild symptoms occur. These include epigastric pain, vomiting, distension, flatulence, anorexia and weight loss, which often may occur. Pain in the epigastrium and right iliac fossa is common. When associated with E. histolytica, Balantidium coli or shigellosis, symptoms are highly aggravated and dysenteric symptoms occur. There is usually no eosinophilia which, if pronounced, usually denotes a concurrent Toxocara infection, with which it is often associated. Moderate T. trichiura infections can result in chronic Trichuris colitis, while heavy infection can cause Trichuris dysentery syndrome (TDS) and massive infantile trichuriasis. Typical symptoms and signs of TDS include severe dysentery with blood and mucus in the stools (Figure 55.6) and prolapse of the rectum (Figure 55.7).16,17 In severe massive infantile trichuriasis, which typically occurs in children between 3 and 10 years, hypoproteinaemia, severe anaemia and clubbing of the fingers are common.16,17 Both colitides can result in growth retardation and anaemia,18 but treatment results in impressive catch-up growth.19 Chronic heavy infection during childhood can also have a detrimental effect on cognition, educational achievement and school attendance.20,21 Diagnosis The diagnosis is made by finding the characteristic eggs in the stool (see Figure 55.5) and the Kato-Katz method is recommended by the World Health Organization (WHO) (see Appendix I). T. trichiura eggs have a characteristic ‘barrel’ shape with two terminal polar plugs. The eggshell consists of three layers, the outermost of which is stained dark brown by host bile pigment. An egg count (see Appendix I) enables the quantification of the intensity of infection and the WHO defines a heavy infection as ≥10 000 eggs/g of faeces (Table 55.2), which is likely to be associated with morbidity.22 Formol-ether concentration has greater sensitivity in light infections than the Kato-Katz method. FLOTAC is a new, sensitive, flotation chamber technique suitable for field settings, but is at present considerably more expensive.23,24 Proctoscopy in cases of dysentery will show numerous worms attached to the mucosa, which is reddened and ulcerated where they are responsible for the dysentery. In some cases, a ‘honeycomb’ appearance of the small intestine has been seen with the appearances of Crohn's disease using radiology; deformity of the intestine is most marked in the proximal colon but also present in the ileum and appendix. Differential Diagnosis In severe infection, the clinical picture may resemble hookworm disease, acute appendicitis or amoebic dysentery. Many of the more subtle, indirect consequences of infection, such as undernutrition and anaemia, may be wrongly attributed to other aetiologies. Management Albendazole and mebendazole as single doses are recommended for the treatment of T. trichiura (see Table 55.1), as is a combination of albendazole 400 mg with ivermectin 200 µg/kg.25 However, the cure rate (defined as the percentage of infected individuals who become uninfected following treatment) of single doses of benzimidazoles is not satisfactory,26 and treatment over several days may be required, especially for heavy infections. There also seem to be regional differences in susceptibility of T. trichiura to benzimidazoles,26 and it would therefore be prudent to evaluate local drug sensitivity when planning control measures. Epidemiology and Control T. trichiura infection is often associated with A. lumbricoides and Toxocara spp., the epidemiology of which is similar. The development and survival of T. trichiura and other geohelminth free-living stages are temperature and humidity dependent. The optimum temperature is 20–30°C, with development arrested below 5°C and above 38°C.27 Epidemiological studies demonstrate that prevalence is also associated with humidity and rainfall. Using geographical information systems (GIS) and remote sensing – which can provide spatially continuous estimates of climatic and environmental factors – it has been shown that prevalence of T. trichiura and of A. lumbricoides does not usually exceed 10% in areas where land surface temperature exceeds 38–40°C, and prevalence is thus greatest in the equatorial regions of the developing world.27 Other important factors that determine localized transmission are poor sanitation, hygienic behaviour and socioeconomic status.28 It is commonly assumed that both T. trichiura and A. lumbricoides are more prevalent in urban areas than in rural areas, however the evidence is equivocal, with differing urban–rural dichotomies evident in various settings.29 The age-dependent patterns of infection are similar for both T. trichiura and A. lumbricoides. Maximum prevalence is usually attained before 5 years of age and in low transmission settings, remains high and relatively stable throughout adulthood.6 Mean intensity of infection is greatest in children aged 4–10 years and declines in young adults. In common with other geohelminths, the distribution of T. trichiura infection within communities is highly aggregated. Thus, most individuals harbour a few parasites, while only a few harbour heavy burdens. Clustering of heavy infections within households is also common.30 A further epidemiological observation, and one that may help explain the mechanisms of aggregation, comes from studies of reinfection which indicate that individuals tend to be predisposed to a high or low intensity of infection.6 However, the precise reasons for observed heterogeneities in infection – genetics and/or common environmental and behavioural household factors – remain unresolved.31,32 Control is the same as that for ascariasis: avoidance of soil pollution and periodic mass chemotherapy. View chapterExplore book Read full chapter URL: Book2014, Manson's Tropical Infectious Diseases (Twenty-third Edition)Simon J. Brooker, Donald A.P. Bundy Chapter Parasites of the Gastrointestinal Tract 2022, Encyclopedia of Infection and ImmunityBlaine A. Mathison, Bobbi S. Pritt Trichuris trichiura (trichuriasis, whipworm infection) Biology and epidemiology Trichuriasis, commonly referred to as whipworm infection, is caused by Trichuris trichiura, a nematode with a nearly worldwide distribution. The zoonotic potential of T. trichiura is not well understood. Historically, Trichuris species were classified based on host data, but as more genotyping data becomes available, it is unclear whether there are multiple species infecting humans or whether other animals, such as nonhuman primates and pigs, can harbor T. trichiura and serve as a reservoir for human infection (Betson et al., 2015). Classified as a “soil-transmitted helminth” (STH) along with Ascaris lumbricoides and hookworms, T. trichiura is believed to infect around 2.8 million people worldwide, especially in resource-poor countries in the tropics and subtropics (Flueckiger et al., 2019). Human infection is primarily a disease of poverty where there is inadequate waste disposal and limited access to clean water (Else et al., 2020). According to the WHO, T. trichiura was responsible for 337,000 years lost to disability worldwide in 2016 (WHO, 2020b). Prevalence of infection is greatest in children ages 5–15 years, and declines in older individuals. Trichuris trichiura has a simple one-host life cycle. Adults reside in the large intestine and gravid females shed unembryonated eggs into the environment in the feces. After 15–30 days under optimal conditions in soil, eggs embryonate and contain an infections L1 larva. Infection occurs when these embryonated eggs are ingested in food, water, and fomites contained with soil. In the small intestine, the eggs hatch and the larvae migrate to the large intestine (especially the colon and caecum) where they burrow into the intestinal mucosa and molt to become adults. Mated females begin to release eggs 60–70 days post infection (Xie et al., 2018). Pathogenesis, clinical presentation, and treatment The pathology and clinical manifestations associated with trichuriasis are primarily due to the localized inflammatory response and blood loss from the adult worms embedded in the colonic mucosa (Else et al., 2020). The extent of tissue damage and symptomatology is strongly linked with the worm burden. Most infected individuals have a low worm burden (< 15) and have mild symptoms. In contrast, heavy infection (≥ 800 worms) is associated with clinically-significant colitis and iron deficiency anemia, and patients are likely to have diarrhea (which may be bloody), malnutrition, and abdominal pain (Else et al., 2020). Young children may present with growth retardation and failure to thrive. The most severe manifestation of infection is massive infantile trichuriasis, characterized by chronic mucoid diarrhea, rectal bleeding, iron deficiency anemia, and rectal prolapse (due to mucosal edema and resultant tenesmus) (Else et al., 2020). Treatment is with albendazole, mebendazole, or ivermectin (Medical Letter, 2013). Diagnosis Trichuriasis is diagnosed by the detection of eggs in wet mounts of stool and adult worms removed during colonoscopy. Eggs of T. trichiura measure 50–55 μm long by 22–24 μm wide and are barrel-shaped with bipolar hyaline plugs (Fig. 10). They have a thick brown shell and are unembryonated when shed in feces. On rare occasions, eggs of the canine whipworm T. vulpis are observed in human stool. The presence of such eggs may reflect spurious passage and not always represent true infection. Eggs of T. vulpis are larger, measuring 72–90 μm long by 32–40 μm wide, and are often embryonated in freshly-processed stool specimens (Ash and Orihel, 2007). Adult worms of T. trichiura measure 30–50 mm in length (Fig. 11). They have a long, slender anterior end and a short, thicker posterior end, giving the appearance of a whip, hence the common name. The poster end of the male is often coiled. Adults of T. trichiura processed for histopathology will demonstrate traditional characteristics of trichuroid nematodes such as stichocytes around the esophagus and bacillary banding, the former only in the whip-like anterior end (Ash and Orihel, 2007). View chapterExplore book Read full chapter URL: Reference work2022, Encyclopedia of Infection and ImmunityBlaine A. Mathison, Bobbi S. Pritt Review article Life on the edge: the balance between macrofauna, microflora and host immunity 2012, Trends in ParasitologyAllison J. Bancroft, ... Richard K. Grencis There is a worm at the bottom of the intestine Trichuris as a genus must be considered as one of the most successful groups of gastrointestinal dwelling nematode parasites because the 50 to 60 known Trichuris species are recognised to infect numerous mammalian species. The human infective species Trichuris trichiura is estimated to infect up to a billion people and is responsible for considerable morbidity, particularly in a group of the most vulnerable people on the planet, the children of developing countries. All Trichuris species have a similar life cycle and inhabit the same niche, the caecum and proximal colon. Infection is initiated with the ingestion of embryonated eggs. It is perhaps not surprising, therefore, that these parasites are some of the most uniquely adapted to not only existing alongside the commensal flora of the large intestine but actually utilising this ecosystem for their own propagation. From prevalence data in humans and animals throughout the ages it is plausible to suggest that these metazoan parasites of the large intestine have been instrumental in shaping our immune systems, and indeed helminths have been shown to drive the selection of interleukin (IL) genes and their polymorphisms . Early larval stages of Trichuris species burrow into the intestinal mucosa, allowing access of the microflora directly to the epithelium, breaching the mucus barrier and thus the host immune system. This implies that to avoid pathological consequences the host needs to respond in a regulated way. T. muris plays an important role in this regulation, following infection of the mouse ; e.g. chronic T. muris is associated with changes in microflora in the caecum (A. Houlden, I. Roberts and R. Grencis, unpublished). Moreover, the presence of intestinal microflora clearly alters the immunoregulatory mechanisms operating against T. muris. View article Read full article URL: Journal2012, Trends in ParasitologyAllison J. Bancroft, ... Richard K. Grencis Chapter Enteric Parasites 2021, Pediatric Gastrointestinal and Liver Disease (Sixth Edition)Rachel M. Ruiz, Dorsey M. Bass Trichuris trichiura T. trichiura (whipworm) is the second most common soil-transmitted helminth worldwide, most common in tropical areas, especially Asia, with an estimated 600 to 800 million people infected.156,167 Coinfection with the other two soil-transmitted helminths, Ascaris and hookworms, is common. It is also endemic to the United States. Children age 5 to 15 years tend to have the heaviest worm burdens.157 T. trichiura differs from other human nematodes because there is no tissue migration during its life cycle and adult worms reside in the colon rather than the small intestine (Fig. 39.4). As in the case of Ascaris eggs, those of Trichuris must mature in the soil before being ingested, making direct person-to-person transmission impossible. The larvae hatch and mature in the distal small bowel before migrating to the cecum, where they attach to the bowel wall via their narrow anterior (“whip”) end. Adult worms then typically reside in the colon for 1 to 3 years, but they can persist for longer. Most light infections are asymptomatic. Moderate infections produce a picture of chronic colitis with diarrhea, abdominal pain, and weight loss. Heavy infections can cause Trichuris dysentery syndrome, which presents with abdominal pain, bloody diarrhea, tenesmus, and sometimes rectal prolapse.166,168 Chronic infections in children are associated with stunted growth, anemia, and delayed cognitive development.170 Treatment is with albendazole, mebendazole, or ivermectin (see Table 39.1). View chapterExplore book Read full chapter URL: Book2021, Pediatric Gastrointestinal and Liver Disease (Sixth Edition)Rachel M. Ruiz, Dorsey M. Bass Chapter Trichuriasis 2011, Tropical Infectious Diseases (Third Edition)Nilanthi R. de Silva, Edward S. Cooper The Agent Trichuris trichiura is a member of the nematode superfamily Trichuroidea and is related to Trichinella spiralis. The genus was previously often called Trichocephalus, logically enough since the hairlike part (tricho-) is in fact the head end (cephalus). However, the original name given by Linnaeus (1771), in the mistaken belief that the hairlike part was the tail (uris), is the official name. Whipworm is a commonly used unofficial name. The adult is shaped like a whip, with the handle representing the wider posterior section containing the reproductive organs and the intestine, while the long, fine anterior part, called the stichosome, contains the long pharynx. The adult is about 4 cm long. The male has a curled posterior end. The eggs are thick-walled and barrel-shaped, about 50 µm long, with a plug at each pole. Eggs, passed in the feces, contain a zygote and are not infective until embryonation, which takes place in the soil over 2–4 weeks (Fig. 114.1). The egg now contains the L1 larva. Following human ingestion, the larva is released in the stomach and passes into the intestine. It penetrates the epithelium in the mucosal crypts of the cecum.1 The larva develops by molting, and the adult develops from the L4 stage, by now having migrated with the epithelial cells up the sides of the crypts. The anterior part of the adult lies in a tunnel within the epithelium between the mouths of the crypts, while the posterior part is free in the lumen. The stichosome is surrounded by a syncytium and debris of the epithelium. Each female produces 3000–20 000 eggs per day; the life expectancy of a worm within the host has been estimated at 1–3 years,1 which would imply that some adults live far longer. View chapterExplore book Read full chapter URL: Book2011, Tropical Infectious Diseases (Third Edition)Nilanthi R. de Silva, Edward S. Cooper Related terms: Ascaris lumbricoides Eosinophilia Helminthiasis Albendazole Nematoda Parasite (Microbiology) Necator americanus Mebendazole Intestine Helminth View all Topics
5866	https://math.stackexchange.com/questions/4240890/find-all-odd-functions-f-mathbb-r-to-mathbb-r-that-satisfy-fx1-fx1-an	functional equations - Find all odd functions $f:\mathbb R\to\mathbb R$ that satisfy $f(x+1)=f(x)+1$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Find all odd functions f:R→R f:R→R that satisfy f(x+1)=f(x)+1 f(x+1)=f(x)+1 and f(1 x)=f(x)x 2 f(1 x)=f(x)x 2 Ask Question Asked 4 years ago Modified3 years, 2 months ago Viewed 223 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Find all functions: f:R→R f:R→R that satisfy all the following three conditions: f(−x)=−f(x)f(−x)=−f(x) f(x+1)=f(x)+1 f(x+1)=f(x)+1 f(1 x)=f(x)x 2 f(1 x)=f(x)x 2 I assume f(x)=x f(x)=x satisfies the conditions. I could prove f(0)=0 f(0)=0 and f(n)=n f(n)=n for all n∈N n∈N, then I didn't know how to go any further. functional-equations Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Sep 20, 2021 at 16:46 Mohsen Shahriari 7,081 10 10 gold badges 33 33 silver badges 78 78 bronze badges asked Sep 3, 2021 at 13:37 BeginnerBeginner 131 4 4 bronze badges 5 1 Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.Community –CommunityBot 2021-09-03 13:41:38 +00:00 Commented Sep 3, 2021 at 13:41 f satifsies all three conditions , i edited the probelm, thank you!!Beginner –Beginner 2021-09-03 13:48:04 +00:00 Commented Sep 3, 2021 at 13:48 What can you say about f(1/2)f(1/2)? What about f(3/2)f(3/2) and f(5/2)f(5/2)? What about f(2/3)f(2/3) and f(2/5)f(2/5)? How far can you take this?Arthur –Arthur 2021-09-03 13:55:59 +00:00 Commented Sep 3, 2021 at 13:55 1 @Arthur: I think that gets us f(x)=x f(x)=x at all the rationals, which is why I suggested the problem should specify the function is continuous. Dan Vellerman thinks we can do better.Ross Millikan –Ross Millikan 2021-09-03 15:13:52 +00:00 Commented Sep 3, 2021 at 15:13 @RossMillikan I was asking Beginner, hoping they would continue the fine work they did with the integers.Arthur –Arthur 2021-09-03 15:26:21 +00:00 Commented Sep 3, 2021 at 15:26 Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. From the given conditions you can prove that f(x)=x f(x)=x for all x x. Hint: Suppose that for some number a a, f(a)=b f(a)=b. The given conditions allow you to compute f(−a)f(−a), f(a+1)f(a+1), and f(1/a)f(1/a) (if a≠0 a≠0). Use these steps repeatedly to reach other values. If you can find two different sequences of steps that lead to the value of f f at the same input, then you can set the results equal to each other and solve for b b. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 3, 2021 at 14:22 Dan VellemanDan Velleman 3,129 10 10 silver badges 12 12 bronze badges 7 I didn't get what you said , i can find the values of f(−a)f(−a) , f(a+1)f(a+1) and f(1 a)f(1 a) in terms of b b , then can you please explain the rest of your solution to me Beginner –Beginner 2021-09-03 14:40:44 +00:00 Commented Sep 3, 2021 at 14:40 Next step: You can compute things like f(1/(a+1))f(1/(a+1)) or f((1/a)+1)f((1/a)+1). Keep going, using the rules repeatedly.Dan Velleman –Dan Velleman 2021-09-03 14:49:23 +00:00 Commented Sep 3, 2021 at 14:49 And? , Please clarify your idea to me , I don't seem to get what you wanna do, i feel like counting random numbers in terms of b and i have no idea why Beginner –Beginner 2021-09-03 15:17:22 +00:00 Commented Sep 3, 2021 at 15:17 1 If you don't want more of a hint, don't read this comment. If you do: Start with f(a)=b f(a)=b and compute the value of f f at the following numbers, using each to compute the next: 1/a 1/a, (a+1)/a(a+1)/a, a/(a+1)a/(a+1), −a/(a+1)−a/(a+1), 1/(a+1)1/(a+1), a+1 a+1. Now, can you find the last one a different way?Dan Velleman –Dan Velleman 2021-09-03 16:02:02 +00:00 Commented Sep 3, 2021 at 16:02 1 Okay , i find a cycle and the fact that f(a)=2 a−b f(a)=2 a−b so i solved for b b and found that a=b a=b, so f(a)=a f(a)=a for all a≠0,−1 a≠0,−1 and i found then f(0)f(0) and f(1)f(1) so my conclusion was f(a)=a f(a)=a for all reels a a thank you guys so much! but my question is: can i suppose that f(a)=b f(a)=b in the first place? Doesn't that require f f to be surjective or something Beginner –Beginner 2021-09-03 18:18:19 +00:00 Commented Sep 3, 2021 at 18:18 \|Show 2 more comments This answer is useful 3 Save this answer. Show activity on this post. Use the fact that f(n)=n f(n)=n in 3 to get the value for reciprocals of all integers. Then use 2 to get reciprocals plus integers. I believe you need to be told that f f is continuous. Without that I don't think you can say anything about the value at irrationals. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 3, 2021 at 13:52 Ross MillikanRoss Millikan 384k 28 28 gold badges 264 264 silver badges 472 472 bronze badges 1 No im not told that f f is continuous Beginner –Beginner 2021-09-03 14:38:56 +00:00 Commented Sep 3, 2021 at 14:38 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Well, I just saw Dan Vellerman's proof (hint actually), which is very elegant, and which I did not think of. I will nevertheless add the proof that f(x)=x f(x)=x on Q Q because the induction techniques used might be useful for some. Let's prove that f(x)=x f(x)=x on Q Q. Notice that f(0)=−f(−0)=−f(0)f(0)=−f(−0)=−f(0). Therefore f(0)=0 f(0)=0. Let's prove f(x+n)=f(x)+n f(x+n)=f(x)+n for all n∈N n∈N. For n=0 n=0 this is obiously true. By induction f(x+n+1)=f(x+1+n)=f(x+1)+n=f(x)+1+n=f(x)+n+1 f(x+n+1)=f(x+1+n)=f(x+1)+n=f(x)+1+n=f(x)+n+1. Therefore it is true for all n∈N n∈N. Let us now prove f(a b)=a b f(a b)=a b for all a∈Z a∈Z and b∈N+b∈N+. Suppose that a∈N a∈N and continue by induction on a a. If \|a\|=0\|a\|=0 we reduce to the first case above. Otherwise: f(a b)=f(b a)(b a)2=f((b mod a)+⌊b a⌋a a)(b a)2=f((b mod a)a+⌊b a⌋)(b a)2=f((b mod a)a)+⌊b a⌋(b a)2=(b mod a)a+⌊b a⌋(b a)2=b a(b a)2=a b f(a b)=f(b a)(b a)2=f((b mod a)+⌊b a⌋a a)(b a)2=f((b mod a)a+⌊b a⌋)(b a)2=f((b mod a)a)+⌊b a⌋(b a)2=(b mod a)a+⌊b a⌋(b a)2=b a(b a)2=a b Note that the induction step was used in saying that f(b mod a a)=b mod a a f(b mod a a)=b mod a a since (b mod a)<a(b mod a)<a. Finally if a<0 a<0 then f(a b)=f(−\|a\|b)=−f(\|a\|b)=−\|a\|b=a b f(a b)=f(−\|a\|b)=−f(\|a\|b)=−\|a\|b=a b. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 20, 2021 at 18:53 KoljaKolja 3,022 1 1 gold badge 16 16 silver badges 25 25 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. f:R→R(odd),f(x+1)=f(x)+1,f(1 x)=f(x)x 2.f:R→R(odd),f(x+1)=f(x)+1,f(1 x)=f(x)x 2. ∴⇒f(1+1 x)=f(1 x)+1=f(x)x 2+1.f(x+1 x)=f(x x+1)(x x+1)2=(x+1)2 x 2 f(1−1 x+1)=(x+1)2 x 2(1−f(1 x+1))=(x+1)2 x 2(1−f(x+1)(x+1)2)=(x+1)2 x 2(1−f(x)+1(x+1)2).(1)=(2)⇒f(x)x 2+1=(x+1)2 x 2(1−f(x)+1(x+1)2).f(x)+x 2=(x+1)2−f(x)−1,2 f(x)=2 x,f(x)=x.(1)(2)(1)f(1+1 x)=f(1 x)+1=f(x)x 2+1.f(x+1 x)=f(x x+1)(x x+1)2=(x+1)2 x 2 f(1−1 x+1)=(x+1)2 x 2(1−f(1 x+1))=(x+1)2 x 2(1−f(x+1)(x+1)2)(2)=(x+1)2 x 2(1−f(x)+1(x+1)2).∴(1)=(2)⇒f(x)x 2+1=(x+1)2 x 2(1−f(x)+1(x+1)2).⇒f(x)+x 2=(x+1)2−f(x)−1,2 f(x)=2 x,f(x)=x. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jul 28, 2022 at 4:43 RDKRDK 3,413 3 3 gold badges 15 15 silver badges 35 35 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions functional-equations See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 12Find all the function that satisfy : f(x f(y)2)+f(y f(x)2)=4 x y f(x f(y)2)+f(y f(x)2)=4 x y 2Functions that satisfy the identity f(x t)f(−y r)=f(x−y t−r)f(x t)f(−y r)=f(x−y t−r) 12Find all functions f:R→R f:R→R satisfying x f(y)−y f(x)=f(y x)x f(y)−y f(x)=f(y x) 2Find all real functions that satisfy f(x y+x)+f(y)=f(x y+y)+f(x)f(x y+x)+f(y)=f(x y+y)+f(x). 1Find all functions f:R→R f:R→R such that: f(x 3)+f(y 3)=(x+y)f(x 2)+f(y 2)−f(x y)f(x 3)+f(y 3)=(x+y)f(x 2)+f(y 2)−f(x y) 3Find all functions which satisfy f(m 2+n 2)=f(m)2+f(n)2 f(m 2+n 2)=f(m)2+f(n)2∀m,n∈N∀m,n∈N and f(1)>0 f(1)>0 5Finding all functions f:R→R f:R→R such that f(x f(x+y))=f(y f(x))+x 2 f(x f(x+y))=f(y f(x))+x 2 2Finding all pairs of functions f,g:R→R f,g:R→R which satisfy f(x+y)=g(1 x+1 y)(x y)2008 f(x+y)=g(1 x+1 y)(x y)2008 11Find all functions that satisfy x 2 f(x)=f(1 x)x 2 f(x)=f(1 x) Hot Network Questions Discussing strategy reduces winning chances of everyone! 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5867	https://math.stackexchange.com/questions/2486244/finding-similarity-transformation-matrix-for-the-companion-matrix	linear algebra - Finding similarity transformation matrix for the companion matrix. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Finding similarity transformation matrix for the companion matrix. Ask Question Asked 7 years, 11 months ago Modified5 years, 8 months ago Viewed 1k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Given that A:=⎡⎣⎢⎢⎢3 1 5 2 5 2−4 4−1 4 7−3 8 9 6−1⎤⎦⎥⎥⎥,b=⎛⎝⎜⎜⎜1 5−1−8⎞⎠⎟⎟⎟,A:=[3 5−1 8 1 2 4 9 5−4 7 6 2 4−3−1],b=(1 5−1−8), I want to try and find some matrix S S such that C:=S−1 A S=⎡⎣⎢⎢⎢0 0 0⋆1 0 0⋆0 1 0⋆0 0 1⋆⎤⎦⎥⎥⎥,such that S−1 b=⎛⎝⎜⎜⎜0 0 0 1⎞⎠⎟⎟⎟.C:=S−1 A S=[0 1 0 0 0 0 1 0 0 0 0 1⋆⋆⋆⋆],such that S−1 b=(0 0 0 1). At first glance, my thought was that S−1 A S S−1 A S is the companion matrix of A A. Hence, I made C=⎡⎣⎢⎢⎢0 0 0 714 1 0 0−108 0 1 0−11 0 0 1 11⎤⎦⎥⎥⎥,C=[0 1 0 0 0 0 1 0 0 0 0 1 714−108−11 11], as the characteristic polynomial of A A is p A(λ)=λ 4−11 λ 3+11 λ 2+108 λ−714 p A(λ)=λ 4−11 λ 3+11 λ 2+108 λ−714. I know that I could attempt to brute force the matrix S S by creating a system of equations through A S=S C A S=S C, but I feel like this is more difficult of a method than I think necessary. It also doesn't account for the vector S−1 b S−1 b. My next thought was to do something along the lines of perhaps letting S=[v 1 v 2 v 3 v 4]S=[v 1 v 2 v 3 v 4], and taking v 4 v 4 to be v 4=b v 4=b, but I can't quite figure out how I could get any more vectors that are linearly independent to not only create an invertible matrix, but also form C C and satisfying S−1 b=e 4 S−1 b=e 4. Can anyone provide a hint? linear-algebra matrices eigenvalues-eigenvectors matrix-equations companion-matrices Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 26, 2020 at 8:15 Rodrigo de Azevedo 23.4k 7 7 gold badges 49 49 silver badges 116 116 bronze badges asked Oct 23, 2017 at 17:10 Decaf-MathDecaf-Math 4,742 1 1 gold badge 26 26 silver badges 45 45 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Try letting c=b T c=b T, and let Q=⎡⎣⎢⎢⎢⎢c c A c A 2 c A 3⎤⎦⎥⎥⎥⎥Q=[c c A c A 2 c A 3], and consider Q A Q−1 Q A Q−1. Any c c such that (c,A)(c,A) is observable will work, for example c=e T 1 c=e 1 T will work as well. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 23, 2017 at 17:21 copper.hatcopper.hat 179k 10 10 gold badges 127 127 silver badges 270 270 bronze badges 4 2 Q Q is what is sometimes referred to as a "Krylov matrix".J. M. ain't a mathematician –J. M. ain't a mathematician 2017-10-23 17:40:57 +00:00 Commented Oct 23, 2017 at 17:40 I suppose I would think of Q T Q T as the Krylov matrix, but perhaps that reflects the control freak in me...copper.hat –copper.hat 2017-10-23 17:48:26 +00:00 Commented Oct 23, 2017 at 17:48 I've seen it used for both what you have in the answer and (more commonly) the transpose (as you've said).J. M. ain't a mathematician –J. M. ain't a mathematician 2017-10-23 17:56:44 +00:00 Commented Oct 23, 2017 at 17:56 For some reason, literature seems to focus more on control than observation. I realise it is dual, but it seems fundamentally different in some sense.copper.hat –copper.hat 2017-10-23 17:59:38 +00:00 Commented Oct 23, 2017 at 17:59 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions linear-algebra matrices eigenvalues-eigenvectors matrix-equations companion-matrices See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 9Eigenspace of the companion matrix of a monic polynomial 2Eigenvalues of a companion matrix 1Finding eigenvalues for matrix when eigenvectors are known. 1Efficient method for computing the properties of a block anti-diagonal matrix Hot Network Questions Is direct sum of finite spectra cancellative? What were "milk bars" in 1920s Japan? Discussing strategy reduces winning chances of everyone! Spectral Leakage & Phase Discontinuites Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator Overfilled my oil Identifying a movie where a man relives the same day Can a cleric gain the intended benefit from the Extra Spell feat? Who is the target audience of Netanyahu's speech at the United Nations? 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5868	https://www.aanda.org/articles/aa/full_html/2024/06/aa48690-23/aa48690-23.html	Home All issues Volume 686 (June 2024) A&A, 686 (2024) A296 Full HTML Open Access \| \| \| \| --- \| Issue \| \| A&A Volume 686, June 2024 \| \| \| \| \| \| \| \| \| \| Article Number \| \| A296 \| \| Number of page(s) \| \| 7 \| \| Section \| \| Planets and planetary systems \| \| DOI \| \| \| \| Published online \| \| 21 June 2024 \| Top Abstract 1 Introduction 2 Methods 3 Results 4 Discussion and conclusions Acknowledgements References List of tables List of figures A&A, 686, A296 (2024) The mass-radius relation of exoplanets revisited Simon Müller1, Jana Baron1, Ravit Helled1, François Bouchy2 and Léna Parc2 1 Department of Astrophysics, University of Zürich, Winterthurerstrasse 190, 8057 Zürich, Switzerland e-mail: simonandres.mueller@uzh.ch 2 Observatoire de Genève, Université de Genève, 51 Chemin Pegasi, 1290 Versoix, Switzerland Received: 21 November 2023 Accepted: 31 March 2024 Abstract Determining the mass–radius (M−R) relation of exoplanets is important for exoplanet characterization. Here, we present a re-analysis of the M−R relations and their transitions using exoplanetary data from the PlanetS catalog, which accounts only for planets with reliable mass and radius determination. We find that “small planets” correspond to planets with masses of up to ~4.4 M⊕ (within 17%) where R ∝ M0.27. Planets with masses between ~4.4 and 127 M⊕ (within 5%) can be viewed as “intermediate-mass” planets, where R ∝ M0.67. Massive planets, or gas giant planets, are found to have masses beyond 127 M⊕ and an M−R relation of R ∝ M−0.06. By analyzing the radius-density relation we also find that the transition from “small” to “intermediate” planets occurs at a planetary radius of ~1.6 R⊕ (within 3%). Our results are consistent with previous studies’ results and provide an ideal fit for the current measured planetary population. Key words: planets and satellites: composition / planets and satellites: gaseous planets / planets and satellites: general / planets and satellites: terrestrial planets © The Authors 2024 Open Access article, published by EDP Sciences, under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. This article is published in open access under the Subscribe to Open model. Subscribe to A&A to support open access publication. 1 Introduction Since the detection of the first exoplanet around a Sun-like star in 1995 (Mayor & Queloz 1995) over 5000 exoplanets have been discovered revealing a large diversity in their physical properties. The field of exoplanets is blossoming. We are at a stage where we move from exoplanet detection to exoplanet characterization, with work on both of these fronts being extremely active and reaching a new level. For exoplanet characterization, the two fundamental parameters are the planetary mass and radius. However, these two properties cannot be measured by the same method, and unfortunately in many cases only one of the two is available. To allow for a broader overview of exoplanets in a statistical sense, it is valuable to determine their mass-radius (M–R) relation and how this relation changes for different planetary types. Relating a measured radius to a planet’s mass can also help with radial-velocity follow-ups by estimating the expected radial velocity semi-amplitude of a transiting exoplanet. The M–R relation depends on the planetary composition and therefore on the behavior of the different materials at planetary conditions (e.g., Seager et al. 2007; Chabrier et al. 2009; Grasset et al. 2009; Mordasini et al. 2012; Spiegel et al. 2014; Jontof-Hutter 2019). Theoretical M–R relations can be inferred from interior models that rely on equations of state (EOSs), which relate the density and pressure (and usually also the temperature) of a given composition. For simplicity, small terrestrial planets are often assumed to have constant densities and are thus expected to behave as R ∝ M1/3 (e.g., Spiegel et al. 2014). On the other hand, in massive giant planets that are hydrogen-helium (H-He) dominated in composition, the gravitational pressure is high enough for the materials to be pressure ionized, and the electron degeneracy pressure becomes substantial (e.g., Helled et al. 2020). This causes the radius to decrease with increasing mass, with R ∝ M−1/3. Of course, the planetary radius does not only depend on its mass, but also on other factors such as stellar irradiation or the planetary age. Today, there are enough planets with mass and radius measurements to statistically infer an observed M–R relation. The observed M–R relations of exoplanets have been investigated in several studies and are often fit by a broken power law. The breakpoints are particularly interesting as they represent the transitions between different planetary types. For example, Weiss et al. (2013) inferred a mass-radius-incident-flux relation. Based solely on the visual inspection of the M–R and mass-density (M–ρ) diagrams, they found a transition point at 150 Mθ. They inferred relations R ∝ M0.53F−0.03 for M < 150 M⊕ and R ∝ M−0.04F0.09 for M > 150 M⊕, where F is the instellation flux. They suggested that for small planets the mass is the most important parameter for predicting the planetary radius, whereas for giant planets the incident flux is more important. Below 150 M⊕ only 35 planets were available at the time, and the result for this region is less robust. A different approach was taken by Hatzes & Rauer (2015). In this study, the transition for giant planets was explored. They used observations to infer the mass–density (M–ρ) relation. A minimum in density at ≃95 M⊕ (0.3 MJ) and a maximum at ≃1.9 × 104 M⊕ (60 MJ) were inferred, suggesting that these values correspond to the transition into gas giant planets. This study also suggested there is no separation between brown dwarfs and giant planets as they display similar behaviors. The mass-density relation of giant planets was determined to be ρ = 0.78M1.15. Chen & Kipping (2017) presented an elaborate Markov chain Monte Carlo method to analyze the M–R relations of objects ranging from dwarf planets to stars. They inferred four distinct regions in their work: Terran worlds, Neptunian worlds, Jovian worlds, and stars. The corresponding transition points were fit and identified as 2.04, 132, and 2.66 ×104 M⊕, where the latter two values correspond to 0.41 MJ and 0.08 M⊕. The M–R relations were identified as R ∝ M0.28 for Terran worlds, R ∝ M0.59 for Neptunian worlds, R ∝ M−0.04 for Jovian worlds and R ∝ M0.88 for stars. The results were obtained by a solely data-driven analysis rather than being derived from any physical assumptions. At the time of the study, only a few objects at ~ 1 M⊕ had been observed, and therefore the transition point between the Terran and Neptunian worlds relied on a small number of planets. Bashi et al. (2017) fit the M–R relations of two distinct regions using a total least-squares approach, where the transition point was assumed to be unknown. It was found that the transition occurs at ~124 M⊕ and 12.1 R⊕. The data were best fit by the relations R ∝ M0.55 for small planets and R ∝ M0.01 for large planets. In this study, the transition point was also fit instead of being imposed by some prior assumption. Otegi et al. (2020) presented an updated catalog of exoplan-ets, for which robust measurements of both radius and mass are available, based on the NASA Exoplanet Archive catalog1. They focused on finding the transition between rocky planets and those with a substantial gas envelope, and therefore only planets with masses up to 120 M⊕ were considered. When displaying the planets in the M–R plane, two distinct regions were identified: the rocky and the volatile-rich populations. Because they overlap in mass and radius, the populations were separated by the pure-water composition line to distinguish volatile-rich planets from terrestrial ones. For the two distinct groups, the following (M–R) relations were inferred: R = 1.03 M0.29 for the rocky population and R = 0.70 M0.63 for the volatile-rich population. Edmondson et al. (2023) showed that a discontinuous M–R relation, as well as a temperature dependence for giant planets, results in a good fit to the M–R measurements. Similarly to Otegi et al. (2020) they separate the rocky and the icy planets with a pure-ice EOS, while finding the transition of icy planets to gas giants at 115 M⊕. For the rocky planets, they fit a relation of R ∝ M0.34 and for the icy planets it is R ∝ M0.55. When also considering the equilibrium temperature of the gas giants they find that R ∝ M0.00T0.35. This suggests that for the giant planets, the radius only depends on the temperature. Recently, Mousavi-Sadr et al. (2023) used a machine-learning approach to analyze the exoplanet population. By applying various clustering algorithms, they identified the transition between small and giant planets at masses of 52.48 M⊕ and sizes of 8.13 R⊕. For the small planets, the M–R relation was found to be R ∝ M0.50. They also showed that the radii of giant planets are positively correlated with the stellar mass. In this work, we investigated the M–R relations of exoplanets and their transitions using the PlanetS catalog2. We used a solely statistical approach to determine the breakpoints in the relation used to define the different planetary regimes and determine the distinct dependencies. We also analyzed the mass-density and radius-density (R-ρ) relations and investigated their validity in separating different planetary types. 2 Methods In this work, we used the data from the PlanetS catalog. An earlier version was presented in Otegi et al. (2020); since then the catalog has been extended with additional discoveries and planets with masses up to 30 MJ. There was also an update on the planetary masses, and some planetary parameters were reanalyzed. The catalog only includes planets with relative measurement uncertainties on the mass and radius smaller than 25% and 8%. Since the updated catalog contains many more planets, it is more reliable and allows for additional analyses. The data we used were downloaded in July 2023 and contain the mass and radius measurements of 688 exoplanets. Figure 1 shows how the planets are distributed in the M–R plane. Similarly to previous work our approach assumed a power-law relation between two planetary variables: for example, M and R. The first step was to transform the variables into the log-log plane to use a linear regression method. We further assumed that there is an unknown number of break points in the linear relation between the two log-transformed variables, that is, that a piece-wise linear function describes the data. For a single breakpoint, the two-segmented piece-wise linear relation can be parametrized as (e.g., Muggeo 2003) $ \alpha \xi \beta \xi \psi \xi \psi \rm $ (1) where c is the intercept, α the slope of the first segment, β the difference between the slopes of the two segments, ψ the breakpoint, and ξ the independent variable. H(x) is the Heaviside step function, defined as H(x) = 1 if x ≥ 0 and H(x) = 0 otherwise. For n breakpoints, Eq. (1) can be generalized as $ \alpha _ \xi \sum \limits _ ^ \beta _ \left \xi \psi _ \right \left \xi \psi _ \right $ (2) Because of the Heaviside step function, Eqs. (1) and (2) are nonlinear, and ordinary linear regression methods cannot fit the parameters. Furthermore, the data used in this work has significant uncertainties on the dependent and independent variables, -which need to be considered. Therefore, in order to determine the piece-wise linear regression fit, we used orthogonal distance regression (ODR; as implemented in the Python package SciPy). We treated the number of breakpoints n as an additional free parameter. To determine the number of breakpoints that yield the best fit to the data, we fit piece-wise linear functions with zero to four breakpoints. After fitting, we compared the models by calculating the Bayesian information criterion (BIC) in the form that is commonly used for linear regression: $ \rm \ln \left \over \sum \limits _ \left _ \hat _ \right ^ \right \ln $ (3) where xi are the data points, $ \hat _ $ the model predictions, and n, k are the number of data points and model parameters. For the M–R relation, ξ ≡ log(M[M⊕]) and y ≡ log(R[R⊕]), for the M–ρ relation ξ ≡ log(M[M⊕]) and y ≡ log ρ [g/cm3], and for the R-p relation ξ ≡ log(R[R⊕]) and y ≡ log ρ [g/cm3]. The planetary bulk density was calculated using $ \rho \over \pi ^ $ . Since all fitting variables required the data to be transformed, the uncertainties had to be propagated. For transforming the mass-radius measurements into their logarithms, the error propagation is simply σξ = ∂ (log x/∂x) σx = σx/x ln 10), with x the measured mass or radius. When the measurement uncertainties were asymmetric, we used their arithmetic mean as σx. For vector-valued functions f of potentially covariant parameters, the more general form of the error propagation has to be used (e.g., Tellinghuisen 2001): $ \sigma _ ^ \bf ^ \bf $ (4) where g is the gradient of f whose ith element is ∂f/∂xi, and V is the covariance matrix. In our case, this applied to the uncertainties of the density and of the power-law parameters that were calculated from the piece-wise linear fit. \| \| \| --- \| \| \| Fig. 1 Mass-radius data from the PlanetS catalog and the derived relation. Data from PlanetS catalog is displayed in purple. The best-fit mass–radius relation (Eq. (5)) is represented by the solid black line and the shaded region shows the 1σ-uncertainty. The breakpoints are shown as the dashed lines. \| 3 Results In this section, we first present our results for the M–R, M–ρ and R–ρ relations in Sects. 3.1–3.3. We then compare our results to previous studies in Sect. 3.4. 3.1 The mass-radius relation By comparing the BIC of piece-wise linear models with zero to four breakpoints, we determined that two breakpoints provided the best fit to the M–R distribution of the exoplanets from the PlanetS catalog. This led to the following M–R relation: $ \left\{\matrix \pm ^ \pm \hfill & \pm \hfill \cr \pm ^ \pm \hfill & \pm \pm \hfill \cr \pm ^ \pm \hfill & \pm \hfill \cr \right $ (5) where R and M are in Earth units. The piece-wise linear fit with two breakpoints is shown together with the data in Fig. 1, and the fit parameters (see Eq. (2)) are listed in Table 1. The M–R fit with two breakpoints is split into three different regimes (segments). These breakpoints correspond to small planets (M < 4.37 M⊕), intermediate-mass planets (4.37 M⊕ < M < 127 M⊕), and giant planets (M > 127 M⊕). The first breakpoint has a higher relative uncertainty than the second, with a transition mass of M1 = (4.37 ± 0.72) M⊕. Between the first and second segments, the gradient changes less than the transition from the second to the third segment, making it harder to identify the breakpoint. The fit is particularly suitable for determining both breakpoints and describing the data in the first and second segments. The description of the data in the third segment (the giant planets) is relatively uncertain. These planets show a large scatter, making it difficult to find a well-fitting gradient. The first regime of exoplanets corresponds to planets with masses below 4.4 M⊕ and roughly follows the relation of R ∝ M0.27. These planets are most likely “rocky worlds” with compositions similar to the Earth’s. If terrestrial planets can be approximated as constant-density homogeneous spheres, they would follow the simple relation R ∝ M1/3, which is very similar to what we find. The scatter around this relation in the actual data comes from the differentiated structures of planets and the diversity in their bulk densities due to, for instance, rocks-to-metals ratios and the possible existence of lighter elements such as water (e.g., Seager et al. 2007; Weiss et al. 2013; Zeng et al. 2016). The change in the slope at ~4.4 M⊕ defines the transition to the intermediate-mass planets, which could also have non-negligible H-He envelopes. Our fit to the data implies that the maximal mass of “rocky” exoplanets and possibly of naked planetary cores is somewhere between 4 and 5 M⊕. This also implies that the minimum mass to accrete a substantial amount of volatile elements is about 4 to 5 M⊕. The transition region is about half of the theoretical mass limit of about 10 M⊕ for rocky exoplanets (e.g., Seager et al. 2007; Fortney et al. 2007; Charbonneau et al. 20091). The intermediate-mass planets between 4.4 M⊕ and 127 M⊕ correspond to planets with H-He envelopes that still have a large heavy-element mass fraction. Since the mass range is large, the diversity of the envelope mass fractions varies significantly and can range from very thin atmospheres to rather gaseous envelopes (e.g., Weiss et al. 2013; Hatzes & Rauer 2015; Ulmer-Moll et al. 2019). These planets have the steepest M–R relation following R ∝ M0.67. An increase in mass results in a significantly larger radius, corresponding mainly to a larger envelope composed of volatile elements. The transition to the gas giants occurs around 127 M⊕ and is where the planets start to be dominated by the H-He envelope. Interestingly, this transition mass is consistent with the suggested transition mass to giant planets based on recent giant planet formation models (Helled 2023). As expected from their H-He dominated composition, for the giant planets we find that the radius is nearly independent of mass (R ∝ M−0.06). For high-mass objects consisting of a degenerate electron gas, we expect a relation of R ∝ M−1/3. In the giant planets, the gas is not completely degenerate, leading to a slightly compressible gas and a deviation from the expected relation. At the same time, we also observe a large scatter in radius due to stellar irradiation, different planetary ages, and metallicities that can strongly affect the radii of gas giants (e.g., Thorngren et al. 2016; Teske et al. 2019; Müller et al. 2020; Müller & Helled 2023). Table 1 Results for the parameters in Eq. (2) from fitting the exoplane-tary mass (log(M[M⊕])) and radius (log(R[R⊕])) data. 3.2 The mass-density relation In this subsection, we present a fit to the mass-density (M–ρ) relation using the planets from the PlanetS catalog. As in Sect. 3.1, we first determined the best-fit number of breakpoints. Similar to the M–R relation, we found that two breakpoints provided the best fit to the data. The fitting function was therefore given by Eq. (2)), with ξ = log(M[M⊕]) and y = log(ρ[g cm−3]) and n = 2. The piece-wise linear function with two breakpoints yielded the following power-law M–ρ relation: $ \rho \left\{\matrix \pm ^ \pm \hfill & \pm \hfill \cr \pm ^ \pm \hfill & \pm \pm \hfill \cr \pm \over ^ ^ \pm \hfill & \pm \hfill \cr \right $ (6) where ρ is in g cm−3 and M in M⊕. The inferred M–ρ fit is shown in the top panel of Fig. 2 together with the data. Table 2 lists the values of the fitting parameters. We find that the first breakpoint in the M–ρ relation is similar, albeit slightly lower than that found in the M–R relation. The second breakpoint, the transition to the giant planets, is at a significantly higher mass (183 M⊕) compared to what we found previously (127 M⊕). As discussed above, terrestrial planets can be approximated by a constant density, which is roughly consistent with our findings. The planets in the second and third segments have a large scatter in the M–ρ plane, implying that the planetary composition is rather diverse. Consequently, there is also a larger uncertainty in the power-law of the M–ρ density for the intermediate and giant planets. To compare the M–ρ to the M–R relation (as derived in Sect. 3.1), we converted the density ρ(M) from Eq. (6) to a radius using $ \left \pi \rho \over \right ^ $ . Since the M–ρ and M–R relations were fit with the same data, they should yield similar results, although some deviations are expected. The results are shown in the bottom panel of Fig. 2. Indeed, we find that the M–ρ and M–R relations have very similar behaviors, and the relation derived from the M–ρ fit lies well inside the uncertainty of the fit to the M–R distribution. The fact that the power laws were similar while being found separately suggests that our approach yields consistent results. As noted earlier, the largest difference between the two fits is the transition mass between the intermediate and giant planets. Using the M–ρ relation yields a significantly higher transition mass. \| \| \| --- \| \| \| Fig. 2 Mass–density and mass–radius data from the PlanetS catalog and the derived relations. Top: mass and density data of exoplanets from PlanetS catalog. The mass–density relation is represented by a solid black line. The dashed lines show the position of the breakpoints. The light shaded areas are the corresponding 1σ uncertainties. Bottom: mass versus radius distribution with best fit for relation and breakpoints in black, as in Fig. 1. The orange solid and dashed lines show the mass-radius relation and the corresponding breakpoints derived from the fit to the mass-density distribution. \| Table 2 Results for the parameters in Eq. (2) with n = 2 breakpoints from fitting the exoplanetary mass (log(M[M⊕])) and density (log(ρ[g cm−3])) data. 3.3 The radius-density relation When fitting the M–R and the M–ρ relations, the three different regimes were defined by a transition mass. Alternatively, is also possible to search for transitions in the R–ρ relation. Here, we calculated the mean density from the measured M and R (see Sect. 2) and attempted to find a piece-wise linear function that describes the R–ρ relation. The data are shown in the top panel of Fig. 3. Qualitatively, three regimes can be identified. For the smallest planets, the density seems nearly independent of radius. Then, there is a breakpoint where the density decreases steeply with increasing radius. The giant planets (around 13 R⊕) show a strong dispersion in density. This is similar to what we already observed in the M–R relation. For the giant planets, the bulk density can vary greatly due to their age, instellation flux, and composition. The intermediate and the giant planets start to overlap around 7 R⊕, and the density appears almost uncorre-lated with the radius. Therefore, when fitting the R–ρ relation, we excluded planets larger than 7 R⊕. We found that the best model (with the lowest BIC) to describe the R–ρ relation uses one breakpoint. This is unlike the two breakpoints for the M–R and M–ρ relations. However, it is somewhat expected since we excluded the giant planets. The resulting R–ρ relation is $ \rho \left\{\matrix \pm ^ \pm \hfill & \pm \hfill \cr \pm ^ \pm \hfill & \pm \hfill \cr \right $ (7) where ρ and R are in g cm−3 and R⊕, respectively. The R–ρ best fit and the data are shown in the bottom panel of Fig. 3. The values of the parameters for the piece-wise linear function with one breakpoint (see Eq. (1)) are listed in Table 3. A notable result is the breakpoint at around 1.6 R⊕. Its relative uncertainty of 3% is significantly lower than for the mass threshold between the small and intermediate planets derived from the M–R relation (14%) or the M–ρ relation (20%). This shows that it is beneficial to consider the radius when distinguishing between different planetary types (Rogers 2015; Lozovsky et al. 2018). Similar to the results for the M–ρ relation (see Sect. 3.2), the relative uncertainty of the R–ρ in the first segment is rather high, but it is consistent with a constant density approximation. Lozovsky et al. (2018) found threshold radii above which a certain composition is unlikely. For purely rocky planets, they found a threshold radius of $ _ ^ _ \oplus $ . Larger planets must consist at least partly of lighter elements, such as H and He. This is consistent with our result of a breakpoint at 1.64 R⊕. However, they only distinguished between super-Earths and mini-Neptunes at ≃3R⊕, because planets with a larger radius have a substantial H-He atmosphere (at least 2% mass fraction). In contrast, based on our data no distinction can be made there. Our result of the radius breakpoint at 1.64 R⊕ also coincides with the position of the radius valley around 1.5–2 R⊕. The radius valley is a bimodal feature in the occurrence rate of planets as a function of their radii, which manifests itself as a scarcity of planets with R ≃ 1.5–2R⊕. It has been observed for planets with short periods (e.g., Fulton et al. 2017) and is often used for the distinction between super-Earths (below the valley) and mini-Neptunes (above the valley). Several previous studies have shown how photoevaporation or core-powered mass loss can lead to the depletion of the gaseous envelopes of planets at such radii (Chen & Rogers 2016; Owen & Wu 2017; Venturini et al. 2020) and therefore explain the radius valley. In particular, Kubyshkina & Fossati (2022) suggested that the M–R relation of intermediate planets is shaped by their thermal evolution and hydrodynamic escape. Additionally, it has been suggested that the planets at the upper edge of the radius valley are helium-rich (Malsky et al. 2023). This suggests that due to evaporation of the gaseous envelope for masses with R ≃ 1.3 R⊕ planets are naked rocky cores, while around 2.6 R⊕ they sustain at least part of their H-He envelope. As an alternative, it has also been suggested that the bimodal radius distribution of planets smaller than about 4 R⊕ is due to different compositions of rocky super-Earths and ice- or water-rich mini-Neptunes (Zeng et al. 2019; Venturini et al. 2020; Izidoro et al. 2021, 2022). \| \| \| --- \| \| \| Fig. 3 Radius–density data from the PlanetS catalog and the derived relation. Top: radius-density distribution of all exoplanets from the PlanetS Catalog. Bottom: the exoplanets from the PlanetS Catalog smaller than 7 R⊕ are displayed together with the mass–density relation (solid black line) and the corresponding breakpoints (dashed line). The light-shaded areas are the 1σ-uncertainties. \| Table 3 Results for the parameters in Eq. (1) from fitting the exoplane-tary radius (log(R[R⊕])) and density (log(ρ[g cm−3])) data. 3.4 Comparison with previous studies A comparison of our results with previous studies is presented in Table 4. To facilitate the comparison, the relations and breakpoints were converted to Earth units (M⊕ and R⊕). For the M–R relation, we use the results from Sect. 3.1. Also, the mass-density relation from Hatzes & Rauer (2015) was converted to a M–R relation. From Edmondson et al. (2023) we chose the M–R relation for the giant planets instead of their mass-radius-temperature relation. Overall, it is clear that there is a rather good agreement between the results of the various studies despite the use of different methods and data. The M–R relations from the different studies are shown in Fig. 4 together with the data from the PlanetS catalog. It can be seen that the relations from Weiss et al. (2013) and Bashi et al. (2017) underestimate the radii of the smallest planets and are not a good fit. This is because they only use one breakpoint in the M–R relation, corresponding to the transition to giant planets. The relation by Chen & Kipping (2017) does not fit the planets around 10 M⊕ very well because the location of the transition from small to intermediate-mass planets is underestimated. Hatzes & Rauer (2015) did not fit planets below 95 M⊕ at all. The relation by Otegi et al. (2020) remains a good fit for the dataset. The main difference compared to our results is the transition mass from small to intermediate planets. Otegi et al. (2020) defined the transition with the water-composition line, while we used a statistical approach to find this transition. The benefit of our approach is that it does not rely on a priori assumptions or theoretical models to determine the transition (and the associated uncertainties in, e.g., the EOS). Perhaps surprisingly, despite the spread of the planets in the M–R diagram, our dataset yielded a rather small uncertainty in the transition masses. Similarly, Edmondson et al. (2023) used a pure-ice EOS to mark the transition between small and intermediate planets, which leads to a good description of the smallest planets and intermediate planets. However, from ~30– 90 M⊕ their relation for icy planets significantly under-predicts the radii of the planets in the PlanetS catalog, leading to a poor fit. Compared to all the previous small-planet M–R fits listed, our uncertainty on the power-law index is slightly higher. This is likely because the small planets in our updated data have diverse radii. For the giant exoplanets, all the relations can qualitatively describe the M–R relation, although they are quite different and can even have a different sign of the gradient. The data in this regime show a strong dispersion, which leads to rather large uncertainties in the fit relation. Interestingly, using the M–ρ relation to determine the transition to the giant planets yielded a significantly larger mass (183 M⊕) than both the M–R fit (127 M⊕) and results from previous studies (95 to 150 M⊕). Table 4 Results for the mass-radius relations and corresponding breakpoints (transitions) from different studies and this work. \| \| \| --- \| \| \| Fig. 4 Comparison of M–R relations from different studies with exo-planetary data from the PlanetS catalog. \| 4 Discussion and conclusions In this work, we used the updated PlanetS catalog to infer M–R, M–ρ, and R–ρ relations and determined the transitions between different planetary types. While the presented analysis provides insight into the different planet regimes, it was simplified and did not consider all the subtleties related to exoplanetary data. First, we treated the data as one unit, although it is clear that the dataset is inhomogeneous and combines different observational methods with different biases. The effects of observational bias, for the most part, have not been considered. Other parameters affect the M–R relation that was not investigated in this work. For example, for giant planets, stellar age and irradiation are important. Giant planets are massive enough that their self-gravity causes them to contract over long timescales (~1 Gyr; e.g., Hubbard 1977; Burrows et al. 2001), and therefore their radius is expected to be correlated with their age. Additionally, high instellation fluxes inflate the radii of warm giant planets (e.g., Guillot et al. 2006; Fortney et al. 2007; Fortney & Nettelmann 2010; Thorngren et al. 2016; Müller & Helled 2023). This effect was included in Weiss et al. (2013) and Edmondson et al. (2023), where a third parameter (instellation flux or equilibrium temperature) was added to better fit the M–R of the giant planets. Recently, there have also been attempts to move beyond the two-parametric M–R relationship. For example, Kanodia et al. (2023) presented a framework to characterize exoplanets using up to four simultaneous parameters. In the future, such approaches may better constrain the transition from small to intermediate planets from observational data by considering additional parameters. Out of the over 5000 detected exoplanets, only 688 have robust enough mass and radius measurements to be included in the PlanetS catalog. While this means that only a fraction of the currently detected exoplanets were used in this work, the results are also more robust, since only planets with low mass and radius uncertainties are included. More accurate data are needed to analyze the whole parameter space occupied by exoplanets. The key results from our study can be summarized as follows: Our analysis yielded a small-to-intermediate transition mass of (4.37 ± 0.72) M⊕. Small planets below the transition mass follow R ∝ M0.27. These are “rocky worlds” with different bulk compositions. The transition to the intermediate mass could imply a maximal mass of ~4.4 M⊕ of “rocky” exoplanets and naked planetary cores; The transition from rocky to volatile-rich planets can also be defined in terms of the radius. By fitting the radius-density relation, we found that the transition occurs around 1.64 ± 0.05 R⊕. The transition in radius has a significantly lower relative uncertainty than the one in mass. Furthermore, the transition radius is consistent with the radius valley around 1.5 to 2 M⊕; Intermediate-mass planets ranging from about 4.4 M⊕ to 127 M⊕ behave as R ∝ M0.67. They correspond to planets with H-He envelopes. The transition to giant planets occurs at (127 ± 7) M⊕ and corresponds to planets that are H-He-rich; Using the M–ρ relation to find the transition to the giant planets yielded a significantly higher transition mass of (183 ± 7) M⊕; The radii of giant planets are nearly independent of their masses, and the mass-radius relation in this regime follows R ∝ M–0.06; Overall, planets of different compositions and structures can have the same mass and radius. This leads to an intrinsic degeneracy of the mass-radius distribution of exoplanets. Ongoing and future observations on the ground and in space will improve our understanding of exoplanets. The James Webb Space Telescope (Gardner et al. 2006) and the Ariel mission (Tinetti et al. 2018) will enable us to characterize the atmospheres of transiting planets, providing information about their chemical compositions. High-resolution spectroscopy from current (e.g., SPIROU; Artigau et al. 2014, CARMENES; Quirrenbach et al. 2016) and future (e.g., NIRPS; Bouchy et al. 2017; Wildi et al. 2017, CRIRES+; Kaeufl et al. 2004; Dorn et al. 2014, 2023) ground-based telescopes will provide further improvements with accurate radial-velocity measurements and atmospheric characterizations. More exoplanets detected via direct imaging, for example by SPHERE at the Very Large Telescope (Beuzit et al. 2019), will facilitate studies of the properties of planets on wide orbits. Also, the upcoming PLATO mission (Rauer et al. 2014) will detect and characterize small terrestrial planets as well as intermediate-mass and giant planets. Theoretical studies to understand the key physical processes that shape the exoplanetary populations are also being developed, and we hope to be able to connect the properties of exoplanets with their origin and evolution. 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B. 2016, ApJ, 819, 127 [Google Scholar] Zeng, L., Jacobsen, S. B., Sasselov, D. D., et al. 2019, PNAS, 116, 9723 [Google Scholar] All Tables Table 1 Results for the parameters in Eq. (2) from fitting the exoplane-tary mass (log(M[M⊕])) and radius (log(R[R⊕])) data. In the text Table 2 Results for the parameters in Eq. (2) with n = 2 breakpoints from fitting the exoplanetary mass (log(M[M⊕])) and density (log(ρ[g cm−3])) data. In the text Table 3 Results for the parameters in Eq. (1) from fitting the exoplane-tary radius (log(R[R⊕])) and density (log(ρ[g cm−3])) data. In the text Table 4 Results for the mass-radius relations and corresponding breakpoints (transitions) from different studies and this work. In the text All Figures \| \| \| --- \| \| \| Fig. 1 Mass-radius data from the PlanetS catalog and the derived relation. Data from PlanetS catalog is displayed in purple. The best-fit mass–radius relation (Eq. (5)) is represented by the solid black line and the shaded region shows the 1σ-uncertainty. The breakpoints are shown as the dashed lines. \| \| In the text \| \| \| \| \| --- \| \| \| Fig. 2 Mass–density and mass–radius data from the PlanetS catalog and the derived relations. Top: mass and density data of exoplanets from PlanetS catalog. The mass–density relation is represented by a solid black line. The dashed lines show the position of the breakpoints. The light shaded areas are the corresponding 1σ uncertainties. Bottom: mass versus radius distribution with best fit for relation and breakpoints in black, as in Fig. 1. The orange solid and dashed lines show the mass-radius relation and the corresponding breakpoints derived from the fit to the mass-density distribution. \| \| In the text \| \| \| \| \| --- \| \| \| Fig. 3 Radius–density data from the PlanetS catalog and the derived relation. Top: radius-density distribution of all exoplanets from the PlanetS Catalog. Bottom: the exoplanets from the PlanetS Catalog smaller than 7 R⊕ are displayed together with the mass–density relation (solid black line) and the corresponding breakpoints (dashed line). The light-shaded areas are the 1σ-uncertainties. \| \| In the text \| \| \| \| \| --- \| \| \| Fig. 4 Comparison of M–R relations from different studies with exo-planetary data from the PlanetS catalog. \| \| In the text \| \| Current usage metrics show cumulative count of Article Views (full-text article views including HTML views, PDF and ePub downloads, according to the available data) and Abstracts Views on Vision4Press platform. Data correspond to usage on the plateform after 2015. The current usage metrics is available 48-96 hours after online publication and is updated daily on week days. Initial download of the metrics may take a while. Click anywhere to hide the fullscreen overlay
5869	https://forallx.openlogicproject.org/forallxyyc-f21.pdf	forallx CALGARY An Introduction to Formal Logic P. D. Magnus Tim Button with additions by J. Robert Loftis Robert Trueman remixed and revised by Aaron Thomas-Bolduc Richard Zach Fall 2021+ forall x: Calgary An Introduction to Formal Logic By P. D. Magnus Tim Button with additions by J. Robert Loftis Robert Trueman remixed and revised by Aaron Thomas-Bolduc Richard Zach Fall 2021+ This book is based on forallx: Cambridge, by Tim Button (University College London), used under a CC BY 4.0 license, which is based in turn on forallx, by P.D. Magnus (University at Albany, State Univer-sity of New York), used under a CC BY 4.0 license, and was remixed, revised, & expanded by Aaron Thomas-Bolduc & Richard Zach (Uni-versity of Calgary). It includes additional material from forallx by P. D. Magnus and Metatheory by Tim Button, used under a CC BY 4.0 license, from forallx: Lorain County Remix, by Cathal Woods and J. Robert Loftis, and from A Modal Logic Primer by Robert Trueman, used with permission. This work is licensed under a Creative Commons Attribution 4.0 license. You are free to copy and redistribute the material in any medium or format, and remix, transform, and build upon the material for any purpose, even commer-cially, under the following terms: ⊲You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use. ⊲You may not apply legal terms or technological measures that legally restrict others from doing anything the license permits. The L A T EX source for this book is available on GitHub and PDFs at forallx.openlogicproject.org. This version is revision fcdb224 (2023-07-22). The preparation of this textbook was made possible by a grant from the Taylor Institute for Teaching and Learning. Cover design by Mark Lyall. Contents Preface vi I Key notions of logic 1 1 Arguments 2 2 The scope of logic 7 3 Other logical notions 18 II Truth-functional logic 26 4 First steps to symbolization 27 5 Connectives 32 6 Sentences of TFL 49 7 Ambiguity 56 8 Use and mention 62 III Truth tables 68 9 Characteristic truth tables 69 10 Truth-functional connectives 72 11 Complete truth tables 77 12 Semantic concepts 85 iii CONTENTS iv 13 Truth table shortcuts 96 14 Partial truth tables 101 IV Natural deduction for TFL 108 15 The very idea of natural deduction 109 16 Basic rules for TFL 112 17 Constructing proofs 141 18 Additional rules for TFL 161 19 Proof-theoretic concepts 169 20 Derived rules 173 21 Soundness and completeness 181 V First-order logic 190 22 Building blocks of FOL 191 23 Sentences with one quantifier 200 24 Multiple generality 214 25 Identity 229 26 Sentences of FOL 236 27 Definite descriptions 244 28 Ambiguity 254 VI Interpretations 258 29 Extensionality 259 30 Truth in FOL 267 31 Semantic concepts 278 32 Using interpretations 280 33 Reasoning about interpretations 288 CONTENTS v VII Natural deduction for FOL 292 34 Basic rules for FOL 293 35 Proofs with quantifiers 308 36 Conversion of quantifiers 315 37 Rules for identity 318 38 Derived rules 322 39 Proofs and semantics 324 VIII Modal logic 328 40 Introducing modal logic 329 41 Natural deduction for ML 333 42 Semantics for ML 346 IX Metatheory 359 43 Normal forms 360 44 Functional completeness 366 45 Proving equivalences 372 46 Soundness 381 Appendices 390 A Symbolic notation 390 B Alternative proof systems 395 C Quick reference 402 Glossary 412 Preface As the title indicates, this is a textbook on formal logic. For-mal logic concerns the study of a certain kind of language which, like any language, can serve to express states of affairs. It is a formal language, i.e., its expressions (such as sentences) are de-fined formally. This makes it a very useful language for being very precise about the states of affairs its sentences describe. In particular, in formal logic it is impossible to be ambiguous. The study of these languages centres on the relationship of entailment between sentences, i.e., which sentences follow from which other sentences. Entailment is central because by understanding it bet-ter we can tell when some states of affairs must obtain provided some other states of affairs obtain. But entailment is not the only important notion. We will also consider the relationship of be-ing satisfiable, i.e., of not being mutually contradictory. These notions can be defined semantically, using precise definitions of entailment based on interpretations of the language—or proof-theoretically, using formal systems of deduction. Formal logic is of course a central sub-discipline of philoso-phy, where the logical relationship of assumptions to conclusions reached from them is important. Philosophers investigate the consequences of definitions and assumptions and evaluate these definitions and assumptions on the basis of their consequences. It is also important in mathematics and computer science. In mathematics, formal languages are used to describe not “every-vi PREFACE vii day” states of affairs, but mathematical states of affairs. Mathe-maticians are also interested in the consequences of definitions and assumptions, and for them it is equally important to estab-lish these consequences (which they call “theorems”) using com-pletely precise and rigorous methods. Formal logic provides such methods. In computer science, formal logic is applied to describe the state and behaviours of computational systems, e.g., circuits, programs, databases, etc. Methods of formal logic can likewise be used to establish consequences of such descriptions, such as whether a circuit is error-free, whether a program does what it’s intended to do, whether a database is consistent or if something is true of the data in it. The book is divided into nine parts. Part I introduces the topic and notions of logic in an informal way, without introduc-ing a formal language yet. Parts II–IV concern truth-functional languages. In it, sentences are formed from basic sentences using a number of connectives (‘or’, ‘and’, ‘not’, ‘if . . . then’) which just combine sentences into more complicated ones. We discuss log-ical notions such as entailment in two ways: semantically, using the method of truth tables (in Part III) and proof-theoretically, us-ing a system of formal derivations (in Part IV). Parts V–VII deal with a more complicated language, that of first-order logic. It in-cludes, in addition to the connectives of truth-functional logic, also names, predicates, identity, and the so-called quantifiers. These additional elements of the language make it much more expressive than the truth-functional language, and we’ll spend a fair amount of time investigating just how much one can ex-press in it. Again, logical notions for the language of first-order logic are defined semantically, using interpretations, and proof-theoretically, using a more complex version of the formal deriva-tion system introduced in Part IV. Part VIII discusses the exten-sion of TFL by non-truth-functional operators for possibility and necessity: modal logic. Part IX covers two advanced topics: that of conjunctive and disjunctive normal forms and the functional completeness of the truth-functional connectives, and the sound-ness of natural deduction for TFL. PREFACE viii In the appendices you’ll find a discussion of alternative no-tations for the languages we discuss in this text, of alternative derivation systems, and a quick reference listing most of the im-portant rules and definitions. The central terms are listed in a glossary at the very end. Credits This book is based on a text originally written by P. D. Magnus in the version revised and expanded by Tim Button. It also includes some material (mainly exercises) by J. Robert Loftis. The material in Part VIII is based on notes by Robert Trueman (but rewritten to use Fitch’s original natural deduction rules for modal logic), and the material in chapters 43, 44, and 46 on two chapters from Tim Button’s open text Metatheory. Aaron Thomas-Bolduc and Richard Zach have combined elements of these texts into the present version, changed some of the terminology and examples, rewritten some sections, and added material of their own. In particular, Richard Zach rewrote chapters 1 and 2, and added chapters 7, 17, 28, and 45. As of the Fall 2019 edition, the part on FOL uses the syntax more common in advanced texts (such as those based on the Open Logic Project) where ar-guments to predicate symbols are enclosed in parentheses (i.e., ‘R(a,b)’ instead of ‘Rab’). The resulting text is licensed under a Creative Commons Attribution 4.0 license. There are several other “remixes” of forall x, including translations of this version. Notes for instructors The material in this book is suitable for a semester-long introduction to formal logic. I cover parts I–VII plus chapters 43, 44, and 45 in 12 weeks, although I leave out partial truth tables and derived inference rules. The most recent version of this book is available in PDF at forallx.openlogicproject.org, but changes frequently. The CC BY license gives you the right to download and distribute the book yourself. In order to ensure that all your students have the same version of the book throughout the term you’re using it, you should do so: upload the PDF you decide to use to your LMS PREFACE ix rather than merely give your students the link. You are also free to have the PDFs printed by your bookstore, but some bookstores will be able to purchase and stock the softcover books available on Amazon. Note that solutions to many exercises in the book are available at the above site as well (to everyone, including your students). The syntax and proof systems (except those for modal logic) are supported by Graham Leach-Krouse’s free, online logic teach-ing software application Carnap (carnap.io). This allows for sub-mission and automated marking of exercises such as symboliza-tion, truth tables, and natural deduction proofs. Instructors on carnap.io will be able to find samples of additional exercises they may wish to adapt or assign as-is. PART I Key notions of logic 1 CHAPTER 1 Arguments Logic has many uses, as mention in the Preface. What we will be focussing on here, is its use in evaluating arguments; sorting the good from the bad. In everyday language, we sometimes use the word ‘argument’ to talk about belligerent shouting matches. If you and a friend have an argument in this sense, things are not going well between the two of you. Logic is not concerned with such teeth-gnashing and hair-pulling. They are not arguments, in our sense; they are just disagreements. An argument, as we will understand it, is something more like this: Either the butler or the gardener did it. The butler didn’t do it. ∴The gardener did it. We have here a series of sentences. The three dots on the third line of the argument are read ‘therefore.’ They indicate that the final sentence expresses the conclusion of the argument. The two sentences before that are the premises of the argument. If you be-lieve the premises, and you think the conclusion follows from the premises—that the argument, as we will say, is valid—then this (perhaps) provides you with a reason to believe the conclusion. 2 CHAPTER 1. ARGUMENTS 3 This is the sort of thing that logicians are interested in. We will say that an argument is any collection of premises, together with a conclusion. This Part discusses some basic logical notions that apply to arguments in a natural language like English. It is important to begin with a clear understanding of what arguments are and of what it means for an argument to be valid. Later we will represent English-language arguments in a formal language. In the example just given, we used individual sentences to express both of the argument’s premises, and we used a third sentence to express the argument’s conclusion. Many arguments are expressed in this way, but a single sentence can contain a complete argument. Consider: The butler has an alibi; so they cannot have done it. This argument has one premise followed by a conclusion. Many arguments start with premises, and end with a conclu-sion, but not all of them. The argument with which this section began might equally have been presented with the conclusion at the beginning, like so: The gardener did it. After all, it was either the butler or the gardener. And the butler didn’t do it. Equally, it might have been presented with the conclusion in the middle: The butler didn’t do it. Accordingly, it was the gar-dener, given that it was either the gardener or the butler. When approaching an argument, we want to know whether or not the conclusion follows from the premises. So the first thing to do is to separate out the conclusion from the premises. As a guide, these words are often used to indicate an argument’s conclusion: so, therefore, hence, thus, accordingly, consequently CHAPTER 1. ARGUMENTS 4 For this reason, they are sometimes called conclusion indica-tor words. By contrast, these expressions are premise indicator words, as they often indicate that we are dealing with a premise, rather than a conclusion: since, because, given that But in analysing an argument, there is no substitute for a good nose. 1.1 Sentences To be perfectly general, we can define an argument as a series of sentences. The sentences at the beginning of the series are premises. The final sentence in the series is the conclusion. If the premises are true and the argument is a good one, then you have a reason to accept the conclusion. In logic, we are only interested in sentences that can figure as a premise or conclusion of an argument, i.e., sentences that can be true or false. So we will restrict ourselves to sentences of this sort, and define a sentence as a sentence that can be true or false. You should not confuse the idea of a sentence that can be true or false with the difference between fact and opinion. Often, sentences in logic will express things that would count as facts— such as ‘Rudolf Carnap was born in Ronsdorf’ or ‘Simone de Beauvoir liked taking walks’. They can also express things that you might think of as matters of opinion—such as, ‘Rhubarb is tasty’. In other words, a sentence is not disqualified from being part of an argument because we don’t know if it is true or false, or because its truth or falsity is a matter of opinion. If it is the kind of sentence that could be true or false it can play the role of premise or conclusion. CHAPTER 1. ARGUMENTS 5 Also, there are things that would count as ‘sentences’ in a linguistics or grammar course that we will not count as sentences in logic. Questions In a grammar class, ‘Are you sleepy yet?’ would count as an interrogative sentence. Although you might be sleepy or you might be alert, the question itself is neither true nor false. For this reason, questions will not count as sentences in logic. Suppose you answer the question: ‘I am not sleepy.’ This is either true or false, and so it is a sentence in the logical sense. Generally, questions will not count as sentences, but answers will. ‘What is this course about?’ is not a sentence (in our sense). ‘No one knows what this course is about’ is a sentence. Imperatives Commands are often phrased as imperatives like ‘Wake up!’, ‘Sit up straight’, and so on. In a grammar class, these would count as imperative sentences. Although it might be good for you to sit up straight or it might not, the command is neither true nor false. Note, however, that commands are not always phrased as imperatives. ‘You will respect my authority’ is either true or false—either you will or you will not—and so it counts as a sentence in the logical sense. Exclamations ‘Ouch!’ is sometimes called an exclamatory sen-tence, but it is neither true nor false. We will treat ‘Ouch, I hurt my toe!’ as meaning the same thing as ‘I hurt my toe.’ The ‘ouch’ does not add anything that could be true or false. Practice exercises At the end of some chapters, there are exercises that review and explore the material covered in the chapter. There is no substitute for actually working through some problems, because learning logic is more about developing a way of thinking than it is about memorizing facts. CHAPTER 1. ARGUMENTS 6 So here’s the first exercise. Highlight the phrase which ex-presses the conclusion of each of these arguments: 1. It is sunny. So I should take my sunglasses. 2. It must have been sunny. I did wear my sunglasses, after all. 3. No one but you has had their hands in the cookie-jar. And the scene of the crime is littered with cookie-crumbs. You’re the culprit! 4. Miss Scarlett and Professor Plum were in the study at the time of the murder. Reverend Green had the candlestick in the ballroom, and we know that there is no blood on his hands. Hence Colonel Mustard did it in the kitchen with the lead pipe. Recall, after all, that the gun had not been fired. CHAPTER 2 The scope of logic 2.1 Consequence and validity In §1, we talked about arguments, i.e., a collection of sentences (the premises), followed by a single sentence (the conclusion). We said that some words, such as “therefore,” indicate which sen-tence is supposed to be the conclusion. “Therefore,” of course, suggests that there is a connection between the premises and the conclusion, namely that the conclusion follows from, or is a conse-quence of, the premises. This notion of consequence is one of the primary things logic is concerned with. One might even say that logic is the science of what follows from what. Logic develops theories and tools that tell us when a sentence follows from some others. What about the main argument discussed in §1? Either the butler or the gardener did it. The butler didn’t do it. ∴The gardener did it. We don’t have any context for what the sentences in this argu-ment refer to. Perhaps you suspect that “did it” here means “was 7 CHAPTER 2. THE SCOPE OF LOGIC 8 the perpetrator” of some unspecified crime. You might imagine that the argument occurs in a mystery novel or TV show, per-haps spoken by a detective working through the evidence. But even without having any of this information, you probably agree that the argument is a good one in the sense that whatever the premises refer to, if they are both true, the conclusion cannot but be true as well. If the first premise is true, i.e., it’s true that “the butler did it or the gardener did it,” then at least one of them “did it,” whatever that means. And if the second premise is true, then the butler did not “do it.” That leaves only one option: “the gardener did it” must be true. Here, the conclusion follows from the premises. We call arguments that have this property v alid. By way of contrast, consider the following argument: If the driver did it, the maid didn’t do it. The maid didn’t do it. ∴The driver did it. We still have no idea what is being talked about here. But, again, you probably agree that this argument is different from the pre-vious one in an important respect. If the premises are true, it is not guaranteed that the conclusion is also true. The premises of this argument do not rule out, by themselves, that someone other than the maid or the driver “did it.” So there is a case where both premises are true, and yet the driver didn’t do it, i.e., the conclu-sion is not true. In this second argument, the conclusion does not follow from the premises. If, like in this argument, the conclusion does not follow from the premises, we say it is inv alid. 2.2 Cases and types of validity How did we determine that the second argument is invalid? We pointed to a case in which the premises are true and in which the conclusion is not. This was the scenario where neither the driver nor the maid did it, but some third person did. Let’s call such a CHAPTER 2. THE SCOPE OF LOGIC 9 case a counterexample to the argument. If there is a counterex-ample to an argument, the conclusion cannot be a consequence of the premises. For the conclusion to be a consequence of the premises, the truth of the premises must guarantee the truth of the conclusion. If there is a counterexample, the truth of the premises does not guarantee the truth of the conclusion. As logicians, we want to be able to determine when the conclusion of an argument follows from the premises. And the conclusion is a consequence of the premises if there is no counterexample—no case where the premises are all true but the conclusion is not. This motivates a definition: A sentence A is a consequence of sentences B1, ..., Bn if and only if there is no case where B1, . . . , Bn are all true and A is not true. (We then also say that A follows from B1, . . . , Bn or that B1, . . . , Bn entail A.) This “definition” is incomplete: it does not tell us what a “case” is or what it means to be “true in a case.” So far we’ve only seen an example: a hypothetical scenario involving three people. Of the three people in the scenario—a driver, a maid, and some third person—the driver and maid didn’t do it, but the third person did. In this scenario, as described, the driver didn’t do it, and so it is a case in which the sentence “the driver did it” is not true. The premises of our second argument are true, but the conclusion is not true: the scenario is a counterexample. We said that arguments where the conclusion is a conse-quence of the premises are called valid, and those where the con-clusion isn’t a consequence of the premises are invalid. Since we now have at least a first stab at a definition of “consequence,” we’ll record this: An argument is v alid if and only if the conclusion is a consequence of the premises. CHAPTER 2. THE SCOPE OF LOGIC 10 An argument is inv alid if and only if it is not valid, i.e., it has a counterexample. Logicians are in the business of making the notion of “case” more precise, and investigating which arguments are valid when “case” is made precise in one way or another. If we take “case” to mean “hypothetical scenario” like the counterexample to the sec-ond argument, it’s clear that the first argument counts as valid. If we imagine a scenario in which either the butler or the gardener did it, and also the butler didn’t do it, we are automatically imag-ining a scenario in which the gardener did it. So any hypothetical scenario in which the premises of our first argument are true au-tomatically makes the conclusion of our first argument true. This makes the first argument valid. Making “case” more specific by interpreting it as “hypothet-ical scenario” is an advance. But it is not the end of the story. The first problem is that we don’t know what to count as a hy-pothetical scenario. Are they limited by the laws of physics? By what is conceivable, in a very general sense? What answers we give to these questions determine which arguments we count as valid. Suppose the answer to the first question is “yes.” Consider the following argument: The spaceship Rocinante took six hours to reach Jupiter from T ycho space station. ∴The distance between T ycho space station and Jupiter is less than 14 billion kilometers. A counterexample to this argument would be a scenario in which the Rocinante makes a trip of over 14 billion kilometers in 6 hours, exceeding the speed of light. Since such a scenario is incompat-ible with the laws of physics, there is no such scenario if hypo-thetical scenarios have to conform to the laws of physics. If hypo-thetical scenarios are not limited by the laws of physics, however, CHAPTER 2. THE SCOPE OF LOGIC 11 there is a counterexample: a scenario where the Rocinante travels faster than the speed of light. Suppose the answer to the second question is “yes,” and con-sider another argument: Priya is an ophthalmologist. ∴Priya is an eye doctor. If we’re allowing only conceivable scenarios, this is also a valid argument. If you imagine Priya being an ophthalmologist, you thereby imagine Priya being an eye doctor. That’s just what “oph-thalmologist” and “eye doctor” mean. A scenario where Priya is an ophthalmologist but not an eye doctor is ruled out by the con-ceptual connection between these words. Depending on what kinds of cases we consider as potential counterexamples, then, we arrive at different notions of conse-quence and validity. We might call an argument nomologically v alid if there are no counterexamples that don’t violate the laws of nature, and an argument conceptually v alid if there are no counterexamples that don’t violate conceptual connections be-tween words. For both of these notions of validity, aspects of the world (e.g., what the laws of nature are) and aspects of the mean-ing of the sentences in the argument (e.g., that “ophthalmologist” just means a kind of eye doctor) figure into whether an argument is valid. 2.3 Formal validity One distinguishing feature of logical consequence, however, is that it should not depend on the content of the premises and conclusion, but only on their logical form. In other words, as logicians we want to develop a theory that can make finer-grained distinctions still. For instance, both Priya is either an ophthalmologist or a dentist. Priya isn’t a dentist. CHAPTER 2. THE SCOPE OF LOGIC 12 ∴Priya is an eye doctor. and Priya is either an ophthalmologist or a dentist. Priya isn’t a dentist. ∴Priya is an ophthalmologist. are valid arguments. But while the validity of the first depends on the content (i.e., the meaning of “ophthalmologist” and “eye doc-tor”), the second does not. The second argument is formally v alid. We can describe the “form” of this argument as a pattern, something like this: A is either an X or a Y . A isn’t a Y . ∴A is an X . Here, A, X , and Y are placeholders for appropriate expressions that, when substituted for A, X , and Y , turn the pattern into an argument consisting of sentences. For instance, Mei is either a mathematician or a botanist. Mei isn’t a botanist. ∴Mei is a mathematician. is an argument of the same form, but the first argument above is not: we would have to replace Y by different expressions (once by “ophthalmologist” and once by “eye doctor”) to obtain it from the pattern. Moreover, the first argument is not formally valid. Its form is this: A is either an X or a Y . A isn’t a Y . ∴A is a Z. In this pattern we can replace X by “ophthalmologist” and Z by “eye doctor” to obtain the original argument. But here is another argument of the same form: CHAPTER 2. THE SCOPE OF LOGIC 13 Mei is either a mathematician or a botanist. Mei isn’t a botanist. ∴Mei is an acrobat. This argument is clearly not valid, since we can imagine a math-ematician named Mei who is not an acrobat. Our strategy as logicians will be to come up with a notion of “case” on which an argument turns out to be valid if it is formally valid. Clearly such a notion of “case” will have to violate not just some laws of nature but some laws of English. Since the first ar-gument is invalid in this sense, we must allow as counterexample a case where Priya is an ophthalmologist but not an eye doctor. That case is not a conceivable situation: it is ruled out by the meanings of “ophthalmologist” and “eye doctor.” When we consider cases of various kinds in order to evaluate the validity of an argument, we will make a few assumptions. The first assumption is that every case makes every sentence true or not true—at least, every sentence in the argument under consid-eration. That means first of all that any imagined scenario which leaves it undetermined if a sentence in our argument is true will not be considered as a potential counterexample. For instance, a scenario where Priya is a dentist but not an ophthalmologist will count as a case to be considered in the first few arguments in this section, but not as a case to be considered in the last two: it doesn’t tell us if Mei is a mathematician, a botanist, or an ac-robat. If a case doesn’t make a sentence true, we say it makes it false. We’ll thus assume that cases make sentences true or false but never both.1 1Even if these assumptions seem common-sensical to you, they are contro-versial among philosophers of logic. First of all, there are logicians who want to consider cases where sentences are neither true nor false, but have some kind of intermediate level of truth. More controversially, some philosophers think we should allow for the possibility of sentences to be both true and false at the same time. There are systems of logic in which sentences can be neither true nor false, or both, but we will not discuss them in this book. CHAPTER 2. THE SCOPE OF LOGIC 14 2.4 Sound arguments Before we go on and execute this strategy, a few clarifications. Arguments in our sense, as conclusions which (supposedly) fol-low from premises, are of course used all the time in everyday and scientific discourse. When they are, arguments are given to support or even prove their conclusions. Now, if an argument is valid, it will support its conclusion, but only if its premises are all true. Validity rules out the possibility that the premises are true and the conclusion is not true at the same time. It does not, by itself, rule out the possibility that the conclusion is not true, pe-riod. In other words, it is perfectly possibly for a valid argument to have a conclusion that isn’t true! Consider this example: Oranges are either fruit or musical instruments. Oranges are not fruit. ∴Oranges are musical instruments. The conclusion of this argument is ridiculous. Nevertheless, it follows from the premises. If both premises are true, then the conclusion just has to be true. So the argument is valid. Conversely, having true premises and a true conclusion is not enough to make an argument valid. Consider this example: London is in England. Beijing is in China. ∴Paris is in France. The premises and conclusion of this argument are, as a matter of fact, all true, but the argument is invalid. If Paris were to declare independence from the rest of France, then the conclusion would no longer be true, even though both of the premises would remain true. Thus, there is a case where the premises of this argument are true without the conclusion being true. So the argument is invalid. The important thing to remember is that validity is not about the actual truth or falsity of the sentences in the argument. It is CHAPTER 2. THE SCOPE OF LOGIC 15 about whether it is possible for all the premises to be true and the conclusion to be not true at the same time (in some hypothetical case). What is in fact the case has no special role to play; and what the facts are does not determine whether an argument is valid or not.2 Nothing about the way things are can by itself de-termine if an argument is valid. It is often said that logic doesn’t care about feelings. Actually, it doesn’t care about facts, either. When we use an argument to prove that its conclusion is true, then, we need two things. First, we need the argument to be valid; i.e., we need the conclusion to follow from the premises. But we also need the premises to be true. We will say that an argument is sound if and only if it is both valid and all of its premises are true. The flip side of this is that when you want to rebut an argu-ment, you have two options: you can show that (one or more of) the premises are not true, or you can show that the argument is not valid. Logic, however, will only help you with the latter! 2.5 Inductive arguments Many good arguments are invalid. Consider this one: Every winter so far, it has snowed in Calgary. ∴It will snow in Calgary this coming winter. This argument generalises from observations about many (past) cases to a conclusion about all (future) cases. Such arguments are called inductive arguments. Nevertheless, the argument is invalid. Even if it has snowed in Calgary every winter thus far, it remains possible that Calgary will stay dry all through the coming winter. In fact, even if it will henceforth snow every winter in Calgary, we could still imagine a case in which this year is the first year it doesn’t snow all winter. And that hypothetical scenario 2Well, there is one case where it does: if the premises are in fact true and the conclusion is in fact not true, then we live in a counterexample; so the argument is invalid. CHAPTER 2. THE SCOPE OF LOGIC 16 is a case where the premises of the argument are true but the conclusion is not, making the argument invalid. The point of all this is that inductive arguments—even good inductive arguments—are not (deductively) valid. They are not watertight. Unlikely though it might be, it is possible for their con-clusion to be false, even when all of their premises are true. In this book, we will set aside (entirely) the question of what makes for a good inductive argument. Our interest is simply in sorting the (deductively) valid arguments from the invalid ones. So: we are interested in whether or not a conclusion follows from some premises. Don’t, though, say that the premises infer the conclusion. Entailment is a relation between premises and conclusions; inference is something we do. So if you want to mention inference when the conclusion follows from the premises, you could say that one may infer the conclusion from the premises. Practice exercises A. Which of the following arguments are valid? Which are in-valid? 1. Socrates is a man. 2. All men are carrots. ∴Socrates is a carrot. 1. Either Abe Lincoln was born in Illinois or he was once president. 2. Abe Lincoln was never president. ∴Abe Lincoln was born in Illinois. 1. If I pull the trigger, Abe Lincoln will die. 2. I do not pull the trigger. ∴Abe Lincoln will not die. 1. Abe Lincoln was either from France or from Luxembourg. 2. Abe Lincoln was not from Luxembourg. CHAPTER 2. THE SCOPE OF LOGIC 17 ∴Abe Lincoln was from France. 1. If the world ends today, then I will not need to get up to-morrow morning. 2. I will need to get up tomorrow morning. ∴The world will not end today. 1. Joe is now 19 years old. 2. Joe is now 87 years old. ∴Bob is now 20 years old. B. Could there be: 1. A valid argument that has one false premise and one true premise? 2. A valid argument that has only false premises? 3. A valid argument with only false premises and a false con-clusion? 4. An invalid argument that can be made valid by the addition of a new premise? 5. A valid argument that can be made invalid by the addition of a new premise? In each case: if so, give an example; if not, explain why not. CHAPTER 3 Other logical notions In §2, we introduced the ideas of consequence and of valid argu-ment. This is one of the most important ideas in logic. In this section, we will introduce some similarly important ideas. They all rely, as did validity, on the idea that sentences are true (or not) in cases. For the rest of this section, we’ll take cases in the sense of conceivable scenario, i.e., in the sense in which we used them to define conceptual validity. The points we made about different kinds of validity can be made about our new notions along similar lines: if we use a different idea of what counts as a “case” we will get different notions. And as logicians we will, eventually, consider a more permissive definition of case than we do here. 3.1 Joint possibility Consider these two sentences: B1. Jane’s only brother is shorter than her. B2. Jane’s only brother is taller than her. 18 CHAPTER 3. OTHER LOGICAL NOTIONS 19 Logic alone cannot tell us which, if either, of these sentences is true. Yet we can say that if the first sentence (B1) is true, then the second sentence (B2) must be false. Similarly, if B2 is true, then B1 must be false. There is no possible scenario where both sentences are true together. These sentences are incompatible with each other, they cannot all be true at the same time. This motivates the following definition: Sentences are jointly possible if and only if there is a case where they are all true together. B1 and B2 are jointly impossible, while, say, the following two sentences are jointly possible: B1. Jane’s only brother is shorter than her. B2. Jane’s only brother is younger than her. We can ask about the joint possibility of any number of sen-tences. For example, consider the following four sentences: G1. There are at least four giraffes at the wild animal park. G2. There are exactly seven gorillas at the wild animal park. G3. There are not more than two Martians at the wild animal park. G4. Every giraffe at the wild animal park is a Martian. G1 and G4 together entail that there are at least four Martian giraffes at the park. This conflicts with G3, which implies that there are no more than two Martian giraffes there. So the sen-tences G1–G4 are jointly impossible. They cannot all be true together. (Note that the sentences G1, G3 and G4 are jointly im-possible. But if sentences are already jointly impossible, adding an extra sentence to the mix cannot make them jointly possible!) CHAPTER 3. OTHER LOGICAL NOTIONS 20 3.2 Necessary truths, necessary falsehoods, and contingency In assessing arguments for validity, we care about what would be true if the premises were true, but some sentences just must be true. Consider these sentences: 1. It is raining. 2. Either it is raining here, or it is not. 3. It is both raining here and not raining here. In order to know if sentence 1 is true, you would need to look outside or check the weather channel. It might be true; it might be false. A sentence which is capable of being true and capa-ble of being false (in different circumstances, of course) is called contingent. Sentence 2 is different. You do not need to look outside to know that it is true. Regardless of what the weather is like, it is either raining or it is not. That is a necessary truth. Equally, you do not need to check the weather to determine whether or not sentence 3 is true. It must be false, simply as a matter of logic. It might be raining here and not raining across town; it might be raining now but stop raining even as you finish this sentence; but it is impossible for it to be both raining and not raining in the same place and at the same time. So, whatever the world is like, it is not both raining here and not raining here. It is a necessary falsehood. Something might always be true and still be contingent. For instance, if there never were a time when the universe contained fewer than seven things, then the sentence ‘At least seven things exist’ would always be true. Yet the sentence is contingent: the world could have been much, much smaller than it is, and then the sentence would have been false. CHAPTER 3. OTHER LOGICAL NOTIONS 21 3.3 Necessary equivalence We can also ask about the logical relations between two sentences. For example: John went to the store after he washed the dishes. John washed the dishes before he went to the store. These two sentences are both contingent, since John might not have gone to the store or washed dishes at all. Yet they must have the same truth-value. If either of the sentences is true, then they both are; if either of the sentences is false, then they both are. When two sentences have the same truth value in every case, we say that they are necessarily equiv alent. Summary of logical notions ⊲An argument is v alid if there is no case where the premises are all true and the conclusion is not; it is inv alid other-wise. ⊲A necessary truth is a sentence that is true in every case. ⊲A necessary falsehood is a sentence that is false in every case. ⊲A contingent sentence is neither a necessary truth nor a necessary falsehood; a sentence that is true in some case and false in some other case. ⊲T wo sentences are necessarily equiv alent if, in every case, they are both true or both false. ⊲A collection of sentences is jointly possible if there is a case where they are all true together; it is jointly impos-sible otherwise. CHAPTER 3. OTHER LOGICAL NOTIONS 22 Practice exercises A. For each of the following: Is it a necessary truth, a necessary falsehood, or contingent? 1. Caesar crossed the Rubicon. 2. Someone once crossed the Rubicon. 3. No one has ever crossed the Rubicon. 4. If Caesar crossed the Rubicon, then someone has. 5. Even though Caesar crossed the Rubicon, no one has ever crossed the Rubicon. 6. If anyone has ever crossed the Rubicon, it was Caesar. B. For each of the following: Is it a necessary truth, a necessary falsehood, or contingent? 1. Elephants dissolve in water. 2. Wood is a light, durable substance useful for building things. 3. If wood were a good building material, it would be useful for building things. 4. I live in a three-story building that is two stories tall. 5. If gerbils were mammals, they would nurse their young. C. Which of the following pairs of sentences are necessarily equiv-alent? 1. Elephants dissolve in water. If you put an elephant in water, it will disintegrate. 2. All mammals dissolve in water. If you put an elephant in water, it will disintegrate. 3. George Bush was the 43rd president. Barack Obama is the 44th president. 4. Barack Obama is the 44th president. Barack Obama was president immediately after the 43rd president. 5. Elephants dissolve in water. All mammals dissolve in water. CHAPTER 3. OTHER LOGICAL NOTIONS 23 D. Which of the following pairs of sentences are necessarily equiv-alent? 1. Thelonious Monk played piano. John Coltrane played tenor sax. 2. Thelonious Monk played gigs with John Coltrane. John Coltrane played gigs with Thelonious Monk. 3. All professional piano players have big hands. Piano player Bud Powell had big hands. 4. Bud Powell suffered from severe mental illness. All piano players suffer from severe mental illness. 5. John Coltrane was deeply religious. John Coltrane viewed music as an expression of spirituality. E. Consider the following sentences: G1 There are at least four giraffes at the wild animal park. G2 There are exactly seven gorillas at the wild animal park. G3 There are not more than two Martians at the wild animal park. G4 Every giraffe at the wild animal park is a Martian. Now consider each of the following collections of sentences. Which are jointly possible? Which are jointly impossible? 1. Sentences G2, G3, and G4 2. Sentences G1, G3, and G4 3. Sentences G1, G2, and G4 4. Sentences G1, G2, and G3 F. Consider the following sentences. M1 All people are mortal. M2 Socrates is a person. M3 Socrates will never die. CHAPTER 3. OTHER LOGICAL NOTIONS 24 M4 Socrates is mortal. Which combinations of sentences are jointly possible? Mark each “possible” or “impossible.” 1. Sentences M1, M2, and M3 2. Sentences M2, M3, and M4 3. Sentences M2 and M3 4. Sentences M1 and M4 5. Sentences M1, M2, M3, and M4 G. Which of the following are possible? For each, if it is possible, give an example. If it is not possible, explain why. 1. A valid argument that has one false premise and one true premise 2. A valid argument that has a false conclusion 3. A valid argument, the conclusion of which is a necessary falsehood 4. An invalid argument, the conclusion of which is a necessary truth 5. A necessary truth that is contingent 6. T wo necessarily equivalent sentences, both of which are necessary truths 7. T wo necessarily equivalent sentences, one of which is a nec-essary truth and one of which is contingent 8. T wo necessarily equivalent sentences that together are jointly impossible 9. A jointly possible collection of sentences that contains a necessary falsehood 10. A jointly impossible set of sentences that contains a neces-sary truth H. Which of the following are possible? For each, if it is possible, give an example. If it is not possible, explain why. 1. A valid argument, whose premises are all necessary truths, and whose conclusion is contingent CHAPTER 3. OTHER LOGICAL NOTIONS 25 2. A valid argument with true premises and a false conclusion 3. A jointly possible collection of sentences that contains two sentences that are not necessarily equivalent 4. A jointly possible collection of sentences, all of which are contingent 5. A false necessary truth 6. A valid argument with false premises 7. A necessarily equivalent pair of sentences that are not jointly possible 8. A necessary truth that is also a necessary falsehood 9. A jointly possible collection of sentences that are all neces-sary falsehoods PART II Truth-functional logic 26 CHAPTER 4 First steps to symbolization 4.1 Validity in virtue of form Consider this argument: It is raining outside. If it is raining outside, then Jenny is miserable. ∴Jenny is miserable. and another argument: Jenny is an anarcho-syndicalist. If Jenny is an anarcho-syndicalist, then Dipan is an avid reader of Tolstoy. ∴Dipan is an avid reader of Tolstoy. Both arguments are valid, and there is a straightforward sense in which we can say that they share a common structure. We might express the structure thus: A If A, then C ∴C 27 CHAPTER 4. FIRST STEPS TO SYMBOLIZATION 28 This looks like an excellent argument structure. Indeed, surely any argument with this structure will be valid, and this is not the only good argument structure. Consider an argument like: Jenny is either happy or sad. Jenny is not happy. ∴Jenny is sad. Again, this is a valid argument. The structure here is something like: A or B not-A ∴B A superb structure! Here is another example: It’s not the case that Jim both studied hard and acted in lots of plays. Jim studied hard. ∴Jim did not act in lots of plays. This valid argument has a structure which we might represent thus: not-(A and B) A ∴not-B These examples illustrate an important idea, which we might de-scribe as validity in virtue of form. The validity of the arguments just considered has nothing very much to do with the meanings of English expressions like ‘Jenny is miserable’, ‘Dipan is an avid reader of Tolstoy’, or ‘Jim acted in lots of plays’. If it has to do with meanings at all, it is with the meanings of phrases like ‘and’, ‘or’, ‘not,’ and ‘if . . . , then . . . ’. In Parts II–IV, we are going to develop a formal language which allows us to symbolize many arguments in such a way as to show that they are valid in virtue of their form. That language will be truth-functional logic, or TFL. CHAPTER 4. FIRST STEPS TO SYMBOLIZATION 29 4.2 Validity for special reasons There are plenty of arguments that are valid, but not for reasons relating to their form. Take an example: Juanita is a vixen. ∴Juanita is a fox. It is impossible for the premise to be true and the conclusion false. So the argument is valid. However, the validity is not related to the form of the argument. Here is an invalid argument with the same form: Juanita is a vixen. ∴Juanita is a cathedral. This might suggest that the validity of the first argument is keyed to the meaning of the words ‘vixen’ and ‘fox’. But, whether or not that is right, it is not simply the shape of the argument that makes it valid. Equally, consider the argument: The sculpture is green all over. ∴The sculpture is not red all over. Again, it seems impossible for the premise to be true and the conclusion false, for nothing can be both green all over and red all over. So the argument is valid, but here is an invalid argument with the same form: The sculpture is green all over. ∴The sculpture is not shiny all over. The argument is invalid, since it is possible to be green all over and shiny all over. (One might paint their nails with an elegant shiny green varnish.) Plausibly, the validity of the first argument is keyed to the way that colours (or colour-words) interact, but, whether or not that is right, it is not simply the shape of the argu-ment that makes it valid. The important moral can be stated as follows. At best, TFL will help us to understand arguments that are valid due to their form. CHAPTER 4. FIRST STEPS TO SYMBOLIZATION 30 4.3 Atomic sentences We started isolating the form of an argument, in §4.1, by replac-ing subsentences of sentences with individual letters. Thus in the first example of this section, ‘it is raining outside’ is a subsen-tence of ‘If it is raining outside, then Jenny is miserable’, and we replaced this subsentence with ‘A’. Our artificial language, TFL, pursues this idea absolutely ruthlessly. We start with some sentence letters. These will be the basic building blocks out of which more complex sentences are built. We will use single uppercase letters as sentence letters of TFL. There are only twenty-six letters of the alphabet, but there is no limit to the number of sentence letters that we might want to consider. By adding subscripts to letters, we obtain new sentence letters. So, here are five different sentence letters of TFL: A,P,P1,P2,A234 We will use sentence letters to represent, or symbolize, certain English sentences. To do this, we provide a symbolization key, such as the following: A: It is raining outside C : Jenny is miserable In doing this, we are not fixing this symbolization once and for all. We are just saying that, for the time being, we will think of the sentence letter of TFL, ‘A’, as symbolizing the English sen-tence ‘It is raining outside’, and the sentence letter of TFL, ‘C ’, as symbolizing the English sentence ‘Jenny is miserable’. Later, when we are dealing with different sentences or different argu-ments, we can provide a new symbolization key; as it might be: A: Jenny is an anarcho-syndicalist C : Dipan is an avid reader of Tolstoy It is important to understand that whatever structure an English sentence might have is lost when it is symbolized by a sentence CHAPTER 4. FIRST STEPS TO SYMBOLIZATION 31 letter of TFL. From the point of view of TFL, a sentence letter is just a letter. It can be used to build more complex sentences, but it cannot be taken apart. CHAPTER 5 Connectives In the previous chapter, we considered symbolizing fairly basic English sentences with sentence letters of TFL. This leaves us wanting to deal with the English expressions ‘and’, ‘or’, ‘not’, and so forth. These are connectives—they can be used to form new sentences out of old ones. In TFL, we will make use of logical connectives to build complex sentences from atomic components. There are five logical connectives in TFL. This table summarizes them, and they are explained throughout this section. symbol what it is called rough meaning ¬ negation ‘It is not the case that. . .’ ∧ conjunction ‘Both. . . and . . .’ ∨ disjunction ‘Either. . . or . . .’ → conditional ‘If . . . then . . .’ ↔ biconditional ‘. . . if and only if . . .’ These are not the only connectives of English of interest. Oth-ers are, e.g., ‘unless’, ‘neither . . . nor . . . ’, and ‘because’. We will see that the first two can be expressed by the connectives we will discuss, while the last cannot. ‘Because’, in contrast to the others, is not truth functional. 32 CHAPTER 5. CONNECTIVES 33 5.1 Negation Consider how we might symbolize these sentences: 1. Mary is in Barcelona. 2. It is not the case that Mary is in Barcelona. 3. Mary is not in Barcelona. In order to symbolize sentence 1, we will need a sentence letter. We might offer this symbolization key: B: Mary is in Barcelona. Since sentence 2 is obviously related to sentence 1, we will not want to symbolize it with a completely different sentence letter. Roughly, sentence 2 means something like ‘It is not the case that B’. In order to symbolize this, we need a symbol for nega-tion. We will use ‘¬’. Now we can symbolize sentence 2 with ‘¬B’. Sentence 3 also contains the word ‘not’, and it is obviously equivalent to sentence 2. As such, we can also symbolize it with ‘¬B’. A sentence can be symbolized as ¬A if it can be para-phrased in English as ‘It is not the case that...’. It will help to offer a few more examples: 4. The widget can be replaced. 5. The widget is irreplaceable. 6. The widget is not irreplaceable. Let us use the following representation key: R: The widget is replaceable Sentence 4 can now be symbolized by ‘R’. Moving on to sentence 5: saying the widget is irreplaceable means that it is not the case CHAPTER 5. CONNECTIVES 34 that the widget is replaceable. So even though sentence 5 does not contain the word ‘not’, we will symbolize it as follows: ‘¬R’. Sentence 6 can be paraphrased as ‘It is not the case that the widget is irreplaceable.’ Which can again be paraphrased as ‘It is not the case that it is not the case that the widget is replace-able’. So we might symbolize this English sentence with the TFL sentence ‘¬¬R’. But some care is needed when handling negations. Consider: 7. Jane is happy. 8. Jane is unhappy. If we let the TFL-sentence ‘H ’ symbolize ‘Jane is happy’, then we can symbolize sentence 7 as ‘H ’. However, it would be a mistake to symbolize sentence 8 with ‘¬H ’. If Jane is unhappy, then she is not happy; but sentence 8 does not mean the same thing as ‘It is not the case that Jane is happy’. Jane might be neither happy nor unhappy; she might be in a state of blank indifference. In order to symbolize sentence 8, then, we would need a new sentence letter of TFL. 5.2 Conjunction Consider these sentences: 9. Adam is athletic. 10. Barbara is athletic. 11. Adam is athletic, and also Barbara is athletic. We will need separate sentence letters of TFL to symbolize sen-tences 9 and 10; perhaps A: Adam is athletic. B: Barbara is athletic. Sentence 9 can now be symbolized as ‘A’, and sentence 10 can be symbolized as ‘B’. Sentence 11 roughly says ‘A and B’. We CHAPTER 5. CONNECTIVES 35 need another symbol, to deal with ‘and’. We will use ‘∧’. Thus we will symbolize it as ‘(A ∧B)’. This connective is called con-junction. We also say that ‘A’ and ‘B’ are the two conjuncts of the conjunction ‘(A ∧B)’. Notice that we make no attempt to symbolize the word ‘also’ in sentence 11. Words like ‘both’ and ‘also’ function to draw our attention to the fact that two things are being conjoined. Maybe they affect the emphasis of a sentence, but we will not (and can-not) symbolize such things in TFL. Some more examples will bring out this point: 12. Barbara is athletic and energetic. 13. Barbara and Adam are both athletic. 14. Although Barbara is energetic, she is not athletic. 15. Adam is athletic, but Barbara is more athletic than him. Sentence 12 is obviously a conjunction. The sentence says two things (about Barbara). In English, it is permissible to refer to Barbara only once. It might be tempting to think that we need to symbolize sentence 12 with something along the lines of ‘B and energetic’. This would be a mistake. Once we symbolize part of a sentence as ‘B’, any further structure is lost, as ‘B’ is a sentence letter of TFL. Conversely, ‘energetic’ is not an English sentence at all. What we are aiming for is something like ‘B and Barbara is energetic’. So we need to add another sentence letter to the symbolization key. Let ‘E’ symbolize ‘Barbara is energetic’. Now the entire sentence can be symbolized as ‘(B ∧E)’. Sentence 13 says one thing about two different subjects. It says of both Barbara and Adam that they are athletic, even though in English we use the word ‘athletic’ only once. The sentence can be paraphrased as ‘Barbara is athletic, and Adam is athletic’. We can symbolize this in TFL as ‘(B ∧A)’, using the same symbol-ization key that we have been using. Sentence 14 is slightly more complicated. The word ‘al-though’ sets up a contrast between the first part of the sentence and the second part. Nevertheless, the sentence tells us both CHAPTER 5. CONNECTIVES 36 that Barbara is energetic and that she is not athletic. In order to make each of the conjuncts a sentence letter, we need to replace ‘she’ with ‘Barbara’. So we can paraphrase sentence 14 as, ‘Both Barbara is energetic, and Barbara is not athletic’. The second conjunct contains a negation, so we paraphrase further: ‘Both Barbara is energetic and it is not the case that Barbara is athletic’. Now we can symbolize this with the TFL sentence ‘(E ∧¬B)’. Note that we have lost all sorts of nuance in this symbolization. There is a distinct difference in tone between sentence 14 and ‘Both Barbara is energetic and it is not the case that Barbara is athletic’. TFL does not (and cannot) preserve these nuances. Sentence 15 raises similar issues. There is a contrastive struc-ture, but this is not something that TFL can deal with. So we can paraphrase the sentence as ‘Both Adam is athletic, and Barbara is more athletic than Adam’. (Notice that we once again replace the pronoun ‘him’ with ‘Adam’.) How should we deal with the second conjunct? We already have the sentence letter ‘A’, which is being used to symbolize ‘Adam is athletic’, and the sentence ‘B’ which is being used to symbolize ‘Barbara is athletic’; but neither of these concerns their relative athleticity. So, to sym-bolize the entire sentence, we need a new sentence letter. Let the TFL sentence ‘R’ symbolize the English sentence ‘Barbara is more athletic than Adam’. Now we can symbolize sentence 15 by ‘(A ∧R)’. A sentence can be symbolized as (A ∧B) if it can be paraphrased in English as ‘Both. . . , and. . . ’, or as ‘..., but ...’, or as ‘although . . . , . . . ’. You might be wondering why we put brackets around the con-junctions. The reason can be brought out by thinking about how negation interacts with conjunction. Consider: 16. It’s not the case that you will get both soup and salad. 17. You will not get soup but you will get salad. CHAPTER 5. CONNECTIVES 37 Sentence 16 can be paraphrased as ‘It is not the case that: both you will get soup and you will get salad’. Using this symbolization key: S1: You will get soup. S2: You will get salad. we would symbolize ‘both you will get soup and you will get salad’ as ‘(S1 ∧S2)’. To symbolize sentence 16, then, we simply negate the whole sentence, thus: ‘¬(S1 ∧S2)’. Sentence 17 is a conjunction: you will not get soup, and you will get salad. ‘You will not get soup’ is symbolized by ‘¬S1’. So to symbolize sentence 17 itself, we offer ‘(¬S1 ∧S2)’. These English sentences are very different, and their symbol-izations differ accordingly. In one of them, the entire conjunction is negated. In the other, just one conjunct is negated. Brackets help us to keep track of things like the scope of the negation. 5.3 Disjunction Consider these sentences: 18. Either Fatima will play videogames, or she will watch movies. 19. Either Fatima or Omar will play videogames. For these sentences we can use this symbolization key: F : Fatima will play videogames. O: Omar will play videogames. M : Fatima will watch movies. However, we will again need to introduce a new symbol. Sen-tence 18 is symbolized by ‘(F ∨M )’. The connective is called disjunction. We also say that ‘F ’ and ‘M ’ are the disjuncts of the disjunction ‘(F ∨M )’. CHAPTER 5. CONNECTIVES 38 Sentence 19 is only slightly more complicated. There are two subjects, but the English sentence only gives the verb once. How-ever, we can paraphrase sentence 19 as ‘Either Fatima will play videogames, or Omar will play videogames’. Now we can obvi-ously symbolize it by ‘(F ∨O)’ again. A sentence can be symbolized as (A ∨B) if it can be paraphrased in English as ‘Either. . . , or. . . .’ Sometimes in English, the word ‘or’ is used in a way that excludes the possibility that both disjuncts are true. This is called an exclusive or. An exclusive or is clearly intended when it says, on a restaurant menu, ‘Entrees come with either soup or salad’: you may have soup; you may have salad; but, if you want both soup and salad, then you have to pay extra. At other times, the word ‘or’ allows for the possibility that both disjuncts might be true. This is probably the case with sen-tence 19, above. Fatima might play videogames alone, Omar might play videogames alone, or they might both play. Sen-tence 19 merely says that at least one of them plays videogames. This is an inclusive or. The TFL symbol ‘∨’ always symbolizes an inclusive or. It will also help to see how negation interacts with disjunction. Consider: 20. Either you will not have soup, or you will not have salad. 21. You will have neither soup nor salad. 22. You get either soup or salad, but not both. Using the same symbolization key as before, sentence 20 can be paraphrased in this way: ‘Either it is not the case that you get soup, or it is not the case that you get salad’. To symbolize this in TFL, we need both disjunction and negation. ‘It is not the case that you get soup’ is symbolized by ‘¬S1’. ‘It is not the case that you get salad’ is symbolized by ‘¬S2’. So sentence 20 itself is symbolized by ‘(¬S1 ∨¬S2)’. CHAPTER 5. CONNECTIVES 39 Sentence 21 also requires negation. It can be paraphrased as, ‘It is not the case that: either you get soup or you get salad’. Since this negates the entire disjunction, we symbolize sentence 21 with ‘¬(S1 ∨S2)’. Sentence 22 is an exclusive or. We can break the sentence into two parts. The first part says that you get one or the other. We symbolize this as ‘(S1 ∨S2)’. The second part says that you do not get both. We can paraphrase this as: ‘It is not the case both that you get soup and that you get salad’. Using both negation and conjunction, we symbolize this with ‘¬(S1∧S2)’. Now we just need to put the two parts together. As we saw above, ‘but’ can usually be symbolized with ‘∧’. So sentence 22 can be symbolized as ‘((S1 ∨S2) ∧¬(S1 ∧S2))’. This last example shows something important. Although the TFL symbol ‘∨’ always symbolizes inclusive or, we can symbolize an exclusive or in TFL. We just have to use a few other symbols as well. 5.4 Conditional Consider these sentences: 23. If Jean is in Paris, then Jean is in France. 24. Jean is in France only if Jean is in Paris. Let’s use the following symbolization key: P : Jean is in Paris. F : Jean is in France Sentence 23 is roughly of this form: ‘if P , then F ’. We will use the symbol ‘→’ to symbolize this ‘if. . . , then. . . ’ structure. So we symbolize sentence 23 by ‘(P →F )’. The connective is called the conditional. Here, ‘P ’ is called the antecedent of the conditional ‘(P →F )’, and ‘F ’ is called the consequent. CHAPTER 5. CONNECTIVES 40 Sentence 24 is also a conditional. Since the word ‘if’ appears in the second half of the sentence, it might be tempting to sym-bolize this in the same way as sentence 23. That would be a mistake. Your knowledge of geography tells you that sentence 23 is unproblematically true: there is no way for Jean to be in Paris that doesn’t involve Jean being in France. But sentence 24 is not so straightforward: were Jean in Dieppe, Lyon, or Toulouse, Jean would be in France without being in Paris, thereby render-ing sentence 24 false. Since geography alone dictates the truth of sentence 23, whereas travel plans (say) are needed to know the truth of sentence 24, they must mean different things. In fact, sentence 24 can be paraphrased as ‘If Jean is in France, then Jean is in Paris’. So we can symbolize it by ‘(F → P )’. A sentence can be symbolized as (A →B) if it can be paraphrased in English as ‘If A, then B’ or ‘A only if B’. In fact, the conditional can represent many English expressions. Consider: 25. For Jean to be in Paris, it is necessary that Jean be in France. 26. It is a necessary condition on Jean’s being in Paris that she be in France. 27. For Jean to be in France, it is sufficient that Jean be in Paris. 28. It is a sufficient condition on Jean’s being in France that she be in Paris. If we think about it, all four of these sentences mean the same as ‘If Jean is in Paris, then Jean is in France’. So they can all be symbolized by ‘(P →F )’. It is important to bear in mind that the connective ‘→’ tells us only that, if the antecedent is true, then the consequent is true. It says nothing about a causal connection between two events (for example). In fact, we lose a huge amount when we use ‘→’ to CHAPTER 5. CONNECTIVES 41 symbolize English conditionals. We will return to this in §10.3 and §12.5. 5.5 Biconditional Consider these sentences: 29. Laika is a dog only if she is a mammal. 30. Laika is a dog if she is a mammal. 31. Laika is a dog if and only if she is a mammal. We will use the following symbolization key: D: Laika is a dog M : Laika is a mammal For reasons discussed above, sentence 29 can be symbolized by ‘(D →M )’. Sentence 31 says something stronger than either 29 or 30. It can be paraphrased as ‘Laika is a dog if Laika is a mammal, and Laika is a dog only if Laika is a mammal’. This is just the conjunction of sentences 29 and 30. So we can symbolize it as ‘(D →M ) ∧(M →D)’. We call this a biconditional, because it amounts to stating both directions of the conditional. We could treat every biconditional this way. So, just as we do not need a new TFL symbol to deal with exclusive or, we do not really need a new TFL symbol to deal with biconditionals. Because the biconditional occurs so often, however, we will use the symbol ‘↔’ for it. We can then symbolize sentence 31 with the TFL sentence ‘(D ↔M )’. The expression ‘if and only if’ occurs a lot especially in phi-losophy, mathematics, and logic. For brevity, we can abbreviate it with the snappier word ‘iff’. We will follow this practice. So ‘if’ with only one ‘f’ is the English conditional. But ‘iff’ with two ‘f’s is the English biconditional. Armed with this we can say: CHAPTER 5. CONNECTIVES 42 A sentence can be symbolized as (A ↔B) if it can be paraphrased in English as ‘A iff B’; that is, as ‘A if and only if B’. A word of caution. Ordinary speakers of English often use ‘if ..., then. . . ’ when they really mean to use something more like ‘... if and only if . . . ’. Perhaps your parents told you, when you were a child: ‘if you don’t eat your greens, you won’t get any dessert’. Suppose you ate your greens, but that your parents refused to give you any dessert, on the grounds that they were only committed to the conditional (roughly ‘if you get dessert, then you will have eaten your greens’), rather than the bicondi-tional (roughly, ‘you get dessert iff you eat your greens’). Well, a tantrum would rightly ensue. So, be aware of this when inter-preting people; but in your own writing, make sure you use the biconditional iff you mean to. 5.6 Unless We have now introduced all of the connectives of TFL. We can use them together to symbolize many kinds of sentences. An especially difficult case is when we use the English-language con-nective ‘unless’: 32. Unless you wear a jacket, you will catch a cold. 33. You will catch a cold unless you wear a jacket. These two sentences are clearly equivalent. To symbolize them, we will use the symbolization key: J : You will wear a jacket D: You will catch a cold Both sentences mean that if you do not wear a jacket, then you will catch a cold. With this in mind, we might symbolize them as ‘(¬ J →D)’. CHAPTER 5. CONNECTIVES 43 Equally, both sentences mean that if you do not catch a cold, then you must have worn a jacket. With this in mind, we might symbolize them as ‘(¬D →J )’. Equally, both sentences mean that either you will wear a jacket or you will catch a cold. With this in mind, we might symbolize them as ‘( J ∨D)’. All three are correct symbolizations. Indeed, in chapter 12 we will see that all three symbolizations are equivalent in TFL. If a sentence can be paraphrased as ‘Unless A, B,’ then it can be symbolized as ‘(A∨B)’. Again, though, there is a little complication. ‘Unless’ can be symbolized as a conditional; but as we said above, people often use the conditional (on its own) when they mean to use the bi-conditional. Equally, ‘unless’ can be symbolized as a disjunction; but there are two kinds of disjunction (exclusive and inclusive). So it will not surprise you to discover that ordinary speakers of English often use ‘unless’ to mean something more like the bicon-ditional, or like exclusive disjunction. Suppose someone says: ‘I will go running unless it rains’. They probably mean something like ‘I will go running iff it does not rain’ (i.e., the biconditional), or ‘either I will go running or it will rain, but not both’ (i.e., ex-clusive disjunction). Again: be aware of this when interpreting what other people have said, but be precise in your writing. Practice exercises A. Using the symbolization key given, symbolize each English sentence in TFL. M : Those creatures are men in suits C : Those creatures are chimpanzees G: Those creatures are gorillas 1. Those creatures are not men in suits. CHAPTER 5. CONNECTIVES 44 2. Those creatures are men in suits, or they are not. 3. Those creatures are either gorillas or chimpanzees. 4. Those creatures are neither gorillas nor chimpanzees. 5. If those creatures are chimpanzees, then they are neither gorillas nor men in suits. 6. Unless those creatures are men in suits, they are either chimpanzees or they are gorillas. B. Using the symbolization key given, symbolize each English sentence in TFL. A: Mister Ace was murdered. B: The butler did it. C : The cook did it. D: The Duchess is lying. E: Mister Edge was murdered. F : The murder weapon was a frying pan. 1. Either Mister Ace or Mister Edge was murdered. 2. If Mister Ace was murdered, then the cook did it. 3. If Mister Edge was murdered, then the cook did not do it. 4. Either the butler did it, or the Duchess is lying. 5. The cook did it only if the Duchess is lying. 6. If the murder weapon was a frying pan, then the culprit must have been the cook. 7. If the murder weapon was not a frying pan, then the culprit was either the cook or the butler. 8. Mister Ace was murdered if and only if Mister Edge was not murdered. 9. The Duchess is lying, unless it was Mister Edge who was murdered. 10. If Mister Ace was murdered, he was done in with a frying pan. 11. Since the cook did it, the butler did not. 12. Of course the Duchess is lying! CHAPTER 5. CONNECTIVES 45 C. Using the symbolization key given, symbolize each English sentence in TFL. E1: Ava is an electrician E2: Harrison is an electrician F1: Ava is a firefighter F2: Harrison is a firefighter S1: Ava is satisfied with her career S2: Harrison is satisfied with his career 1. Ava and Harrison are both electricians. 2. If Ava is a firefighter, then she is satisfied with her career. 3. Ava is a firefighter, unless she is an electrician. 4. Harrison is an unsatisfied electrician. 5. Neither Ava nor Harrison is an electrician. 6. Both Ava and Harrison are electricians, but neither of them find it satisfying. 7. Harrison is satisfied only if he is a firefighter. 8. If Ava is not an electrician, then neither is Harrison, but if she is, then he is too. 9. Ava is satisfied with her career if and only if Harrison is not satisfied with his. 10. If Harrison is both an electrician and a firefighter, then he must be satisfied with his work. 11. It cannot be that Harrison is both an electrician and a fire-fighter. 12. Harrison and Ava are both firefighters if and only if neither of them is an electrician. D. Using the symbolization key given, symbolize each English-language sentence in TFL. J1: John Coltrane played tenor sax J2: John Coltrane played soprano sax J3: John Coltrane played tuba M1: Miles Davis played trumpet M2: Miles Davis played tuba CHAPTER 5. CONNECTIVES 46 1. John Coltrane played tenor and soprano sax. 2. Neither Miles Davis nor John Coltrane played tuba. 3. John Coltrane did not play both tenor sax and tuba. 4. John Coltrane did not play tenor sax unless he also played soprano sax. 5. John Coltrane did not play tuba, but Miles Davis did. 6. Miles Davis played trumpet only if he also played tuba. 7. If Miles Davis played trumpet, then John Coltrane played at least one of these three instruments: tenor sax, soprano sax, or tuba. 8. If John Coltrane played tuba then Miles Davis played nei-ther trumpet nor tuba. 9. Miles Davis and John Coltrane both played tuba if and only if Coltrane did not play tenor sax and Miles Davis did not play trumpet. E. Give a symbolization key and symbolize the following English sentences in TFL. 1. Alice and Bob are both spies. 2. If either Alice or Bob is a spy, then the code has been bro-ken. 3. If neither Alice nor Bob is a spy, then the code remains unbroken. 4. The German embassy will be in an uproar, unless someone has broken the code. 5. Either the code has been broken or it has not, but the Ger-man embassy will be in an uproar regardless. 6. Either Alice or Bob is a spy, but not both. F. Give a symbolization key and symbolize the following English sentences in TFL. 1. If there is food to be found in the pridelands, then Rafiki will talk about squashed bananas. 2. Rafiki will talk about squashed bananas unless Simba is alive. CHAPTER 5. CONNECTIVES 47 3. Rafiki will either talk about squashed bananas or he won’t, but there is food to be found in the pridelands regardless. 4. Scar will remain as king if and only if there is food to be found in the pridelands. 5. If Simba is alive, then Scar will not remain as king. G. For each argument, write a symbolization key and symbolize all of the sentences of the argument in TFL. 1. If Dorothy plays the piano in the morning, then Roger wakes up cranky. Dorothy plays piano in the morning un-less she is distracted. So if Roger does not wake up cranky, then Dorothy must be distracted. 2. It will either rain or snow on Tuesday. If it rains, Neville will be sad. If it snows, Neville will be cold. Therefore, Neville will either be sad or cold on Tuesday. 3. If Zoog remembered to do his chores, then things are clean but not neat. If he forgot, then things are neat but not clean. Therefore, things are either neat or clean; but not both. H. For each argument, write a symbolization key and symbolize the argument as well as possible in TFL. The part of the passage in italics is there to provide context for the argument, and doesn’t need to be symbolized. 1. It is going to rain soon. I know because my leg is hurting, and my leg hurts if it’s going to rain. 2. Spider-man tries to figure out the bad guy’s plan. If Doctor Oc-topus gets the uranium, he will blackmail the city. I am certain of this because if Doctor Octopus gets the uranium, he can make a dirty bomb, and if he can make a dirty bomb, he will blackmail the city. 3. A westerner tries to predict the policies of the Chinese government. If the Chinese government cannot solve the water shortages in Beijing, they will have to move the capital. They don’t want to move the capital. Therefore they must solve the CHAPTER 5. CONNECTIVES 48 water shortage. But the only way to solve the water short-age is to divert almost all the water from the Yangzi river northward. Therefore the Chinese government will go with the project to divert water from the south to the north. I. We symbolized an exclusive or using ‘∨’, ‘∧’, and ‘¬’. How could you symbolize an exclusive or using only two connectives? Is there any way to symbolize an exclusive or using only one connective? CHAPTER 6 Sentences of TFL The sentence ‘either apples are red, or berries are blue’ is a sen-tence of English, and the sentence ‘(A ∨B)’ is a sentence of TFL. Although we can identify sentences of English when we encounter them, we do not have a formal definition of ‘sentence of English’. But in this chapter, we will define exactly what will count as a sentence of TFL. This is one respect in which a formal language like TFL is more precise than a natural language like English. 6.1 Expressions We have seen that there are three kinds of symbols in TFL: Atomic sentences A,B,C,. . . ,Z with subscripts, as needed A1,B1,Z1,A2,A25, J375,. . . Connectives ¬,∧,∨,→,↔ Brackets ( , ) Define an expression of tfl as any string of symbols of TFL. So: write down any sequence of symbols of TFL, in any order, 49 CHAPTER 6. SENTENCES OF TFL 50 and you have an expression of TFL. 6.2 Sentences Given what we just said, ‘(A∧B)’ is an expression of TFL, and so is ‘¬)(∨()∧(¬¬())((B’. But the former is a sentence, and the latter is gibberish. We want some rules to tell us which TFL expressions are sentences. Obviously, individual sentence letters like ‘A’ and ‘G13’ should count as sentences. (We’ll also call them atomic sentences.) We can form further sentences out of these by using the various con-nectives. Using negation, we can get ‘¬A’ and ‘¬G13’. Using conjunction, we can get ‘(A ∧G13)’, ‘(G13 ∧A)’, ‘(A ∧A)’, and ‘(G13 ∧G13)’. We could also apply negation repeatedly to get sentences like ‘¬¬A’ or apply negation along with conjunction to get sentences like ‘¬(A ∧G13)’ and ‘¬(G13 ∧¬G13)’. The possible combinations are endless, even starting with just these two sen-tence letters, and there are infinitely many sentence letters. So there is no point in trying to list all the sentences one by one. Instead, we will describe the process by which sentences can be constructed. Consider negation: Given any sentence A of TFL, ¬Ais a sentence of TFL. (Why the funny fonts? We return to this in §8.3.) We can say similar things for each of the other connectives. For instance, if A and B are sentences of TFL, then (A ∧B) is a sentence of TFL. Providing clauses like this for all of the connectives, we arrive at the following formal definition for a sentence of tfl: CHAPTER 6. SENTENCES OF TFL 51 1. Every sentence letter is a sentence. 2. If A is a sentence, then ¬A is a sentence. 3. If Aand Bare sentences, then (A∧B) is a sentence. 4. If Aand Bare sentences, then (A∨B) is a sentence. 5. If A and B are sentences, then (A →B) is a sen-tence. 6. If A and B are sentences, then (A ↔B) is a sen-tence. 7. Nothing else is a sentence. Definitions like this are called inductive. Inductive definitions begin with some specifiable base elements, and then present ways to generate indefinitely many more elements by compounding together previously established ones. To give you a better idea of what an inductive definition is, we can give an inductive definition of the idea of an ancestor of mine. We specify a base clause. • My parents are ancestors of mine. and then offer further clauses like: • If x is an ancestor of mine, then x’s parents are ancestors of mine. • Nothing else is an ancestor of mine. Using this definition, we can easily check to see whether someone is my ancestor: just check whether she is the parent of the parent of...one of my parents. And the same is true for our inductive definition of sentences of TFL. Just as the inductive definition allows complex sentences to be built up from simpler parts, the definition allows us to decompose sentences into their simpler parts. Once we get down to sentence letters, then we know we are ok. CHAPTER 6. SENTENCES OF TFL 52 Let’s consider some examples. Suppose we want to know whether or not ‘¬¬¬D’ is a sentence of TFL. Looking at the second clause of the definition, we know that ‘¬¬¬D’ is a sentence if ‘¬¬D’ is a sentence. So now we need to ask whether or not ‘¬¬D’ is a sentence. Again looking at the second clause of the definition, ‘¬¬D’ is a sentence if ‘¬D’ is. So, ‘¬D’ is a sentence if ‘D’ is a sentence. Now ‘D’ is a sentence letter of TFL, so we know that ‘D’ is a sentence by the first clause of the definition. So for a compound sentence like ‘¬¬¬D’, we must apply the definition repeatedly. Eventually we arrive at the sentence letters from which the sentence is built up. Next, consider the example ‘¬(P ∧¬(¬Q ∨R))’. Looking at the second clause of the definition, this is a sentence if ‘(P ∧ ¬(¬Q ∨R))’ is, and this is a sentence if both ‘P ’ and ‘¬(¬Q ∨R)’ are sentences. The former is a sentence letter, and the latter is a sentence if ‘(¬Q ∨R)’ is a sentence. It is. Looking at the fourth clause of the definition, this is a sentence if both ‘¬Q’ and ‘R’ are sentences, and both are! Ultimately, every sentence is constructed nicely out of sen-tence letters. When we are dealing with a sentence other than a sentence letter, we can see that there must be some sentential connective that was introduced last, when constructing the sen-tence. We call that connective the main logical operator of the sentence. In the case of ‘¬¬¬D’, the main logical operator is the very first ‘¬’ sign. In the case of ‘(P ∧¬(¬Q ∨R))’, the main logical operator is ‘∧’. In the case of ‘((¬E ∨F ) →¬¬G)’, the main logical operator is ‘→’. As a general rule, you can find the main logical operator for a sentence by using the following method: • If the first symbol in the sentence is ‘¬’, then that is the main logical operator • Otherwise, start counting the brackets. For each open-bracket, i.e., ‘(’, add 1; for each closing-bracket, i.e., ‘)’, subtract 1. When your count is at exactly 1, the first oper-ator you hit (apart from a ‘¬’) is the main logical operator. CHAPTER 6. SENTENCES OF TFL 53 (Note: if you do use this method, then make sure to include all the brackets in the sentence, rather than omitting some as per the conventions of §6.3!) The inductive structure of sentences in TFL will be important when we consider the circumstances under which a particular sen-tence would be true or false. The sentence ‘¬¬¬D’ is true if and only if the sentence ‘¬¬D’ is false, and so on through the struc-ture of the sentence, until we arrive at the atomic components. We will return to this point in Part III. The inductive structure of sentences in TFL also allows us to give a formal definition of the scope of a negation (mentioned in §5.2). The scope of a ‘¬’ is the subsentence for which ‘¬’ is the main logical operator. Consider a sentence like: (P ∧(¬(R ∧B) ↔Q)) which was constructed by conjoining ‘P ’ with ‘(¬(R ∧B) ↔Q)’. This last sentence was constructed by placing a biconditional between ‘¬(R ∧B)’ and ‘Q’. The former of these sentences—a subsentence of our original sentence—is a sentence for which ‘¬’ is the main logical operator. So the scope of the negation is just ‘¬(R ∧B)’. More generally: The scope of a connective (in a sentence) is the subsen-tence for which that connective is the main logical opera-tor. 6.3 Bracketing conventions Strictly speaking, the brackets in ‘(Q ∧R)’ are an indispensable part of the sentence. Part of this is because we might use ‘(Q ∧R)’ as a subsentence in a more complicated sentence. For example, we might want to negate ‘(Q ∧R)’, obtaining ‘¬(Q ∧R)’. If we just had ‘Q ∧R’ without the brackets and put a negation in front of it, we would have ‘¬Q ∧R’. It is most natural to read this as CHAPTER 6. SENTENCES OF TFL 54 meaning the same thing as ‘(¬Q ∧R)’, but as we saw in §5.2, this is very different from ‘¬(Q ∧R)’. Strictly speaking, then, ‘Q ∧R’ is not a sentence. It is a mere expression. When working with TFL, however, it will make our lives eas-ier if we are sometimes a little less than strict. So, here are some convenient conventions. First, we allow ourselves to omit the outermost brackets of a sentence. Thus we allow ourselves to write ‘Q ∧R’ instead of the sentence ‘(Q ∧R)’. However, we must remember to put the brackets back in, when we want to embed the sentence into a more complicated sentence! Second, it can be a bit painful to stare at long sentences with many nested pairs of brackets. To make things a bit easier on the eyes, we will allow ourselves to use square brackets, ‘[’ and ‘]’, instead of rounded ones. So there is no logical difference between ‘(P ∨Q)’ and ‘[P ∨Q]’, for example. Combining these two conventions, we can rewrite the un-wieldy sentence (((H →I ) ∨(I →H )) ∧( J ∨K)) rather more clearly as follows: [︁ (H →I ) ∨(I →H ) ]︁ ∧( J ∨K) The scope of each connective is now much easier to pick out. Practice exercises A. For each of the following: (a) Is it a sentence of TFL, strictly speaking? (b) Is it a sentence of TFL, allowing for our relaxed bracketing conventions? 1. (A) 2. J374 ∨¬ J374 3. ¬¬¬¬F CHAPTER 6. SENTENCES OF TFL 55 4. ¬ ∧S 5. (G ∧¬G) 6. (A →(A ∧¬F )) ∨(D ↔E) 7. [(Z ↔S) →W ] ∧[ J ∨X ] 8. (F ↔¬D →J ) ∨(C ∧D) B. Are there any sentences of TFL that contain no sentence letters? Explain your answer. C. What is the scope of each connective in the sentence [︁ (H →I ) ∨(I →H ) ]︁ ∧( J ∨K) CHAPTER 7 Ambiguity In English, sentences can be ambiguous, i.e., they can have more than one meaning. There are many sources of ambiguity. One is lexical ambiguity: a sentence can contain words which have more than one meaning. For instance, ‘bank’ can mean the bank of a river, or a financial institution. So I might say that ‘I went to the bank’ when I took a stroll along the river, or when I went to deposit a check. Depending on the situation, a different meaning of ‘bank’ is intended, and so the sentence, when uttered in these different contexts, expresses different meanings. A different kind of ambiguity is structural ambiguity. This arises when a sentence can be interpreted in different ways, and depending on the interpretation, a different meaning is selected. A famous example due to Noam Chomsky is the following: Flying planes can be dangerous. There is one reading in which ‘flying’ is used as an adjective which modifies ‘planes’. In this sense, what’s claimed to be dangerous are airplanes which are in the process of flying. In another read-ing, ‘flying’ is a gerund: what’s claimed to be dangerous is the act of flying a plane. In the first case, you might use the sentence to warn someone who’s about to launch a hot air baloon. In the sec-ond case, you might use it to counsel someone against becoming a pilot. 56 CHAPTER 7. AMBIGUITY 57 When the sentence is uttered, usually only one meaning is intended. Which of the possible meanings an utterance of a sen-tence intends is determined by context, or sometimes by how it is uttered (which parts of the sentence are stressed, for instance). Often one interpretation is much more likely to be intended, and in that case it will even be difficult to “see” the unintended read-ing. This is often the reason why a joke works, as in this example from Groucho Marx: One morning I shot an elephant in my pajamas. How he got in my pajamas, I don’t know. Ambiguity is related to, but not the same as, vagueness. An adjective, for instance ‘rich’ or ‘tall,’ is v ague when it is not always possible to determine if it applies or not. For instance, a person who’s 6 ft 4 in (1.9 m) tall is pretty clearly tall, but a building that size is tiny. Here, context has a role to play in determining what the clear cases and clear non-cases are (‘tall for a person,’ ‘tall for a basketball player,’ ‘tall for a building’). Even when the context is clear, however, there will still be cases that fall in a middle range. In TFL, we generally aim to avoid ambiguity. We will try to give our symbolization keys in such a way that they do not use ambiguous words or disambiguate them if a word has different meanings. So, e.g., your symbolization key will need two differ-ent sentence letters for ‘Rebecca went to the (money) bank’ and ‘Rebecca went to the (river) bank.’ Vagueness is harder to avoid. Since we have stipulated that every case (and later, every valua-tion) must make every basic sentence (or sentence letter) either true or false and nothing in between, we cannot accommodate borderline cases in TFL. It is an important feature of sentences of TFL that they can-not be structurally ambiguous. Every sentence of TFL can be read in one, and only one, way. This feature of TFL is also a strength. If an English sentence is ambiguous, TFL can help us make clear what the different meanings are. Although we are CHAPTER 7. AMBIGUITY 58 pretty good at dealing with ambiguity in everyday conversation, avoiding it can sometimes be terribly important. Logic can then be usefully applied: it helps philosophers express their thoughts clearly, mathematicians to state their theorems rigorously, and software engineers to specify loop conditions, database queries, or verification criteria unambiguously. Stating things without ambiguity is of crucial importance in the law as well. Here, ambiguity can, without exaggeration, be a matter of life and death. Here is a famous example of where a death sentence hinged on the interpretation of an ambiguity in the law. Roger Casement (1864–1916) was a British diplomat who was famous in his time for publicizing human-rights violations in the Congo and Peru (for which he was knighted in 1911). He was also an Irish nationalist. In 1914–16, Casement secretly travelled to Germany, with which Britain was at war at the time, and tried to recruit Irish prisoners of war to fight against Britain and for Irish independence. Upon his return to Ireland, he was captured by the British and tried for high treason. The law under which Casement was tried is the Treason Act of 1351. That act specifies what counts as treason, and so the prosecution had to establish at trial that Casement’s actions met the criteria set forth in the Treason Act. The relevant passage stipulated that someone is guilty of treason if a man is adherent to the King’s enemies in his realm, giving to them aid and comfort in the realm, or elsewhere. Casement’s defense hinged on the last comma in this sentence, which is not present in the original French text of the law from 1351. It was not under dispute that Casement had been ‘adher-ent to the King’s enemies’, but the question was whether being adherent to the King’s enemies constituted treason only when it was done in the realm, or also when it was done abroad. The de-fense argued that the law was ambiguous. The claimed ambiguity hinged on whether ‘or elsewhere’ attaches only to ‘giving aid and CHAPTER 7. AMBIGUITY 59 comfort to the King’s enemies’ (the natural reading without the comma), or to both ‘being adherent to the King’s enemies’ and ‘giving aid and comfort to the King’s enemies’ (the natural read-ing with the comma). Although the former interpretation might seem far fetched, the argument in its favor was actually not unper-suasive. Nevertheless, the court decided that the passage should be read with the comma, so Casement’s antics in Germany were treasonous, and he was sentenced to death. Casement himself wrote that he was ‘hanged by a comma’. We can use TFL to symbolize both readings of the passage, and thus to provide a disambiguiation. First, we need a symbol-ization key: A: Casement was adherent to the King’s enemies in the realm. G: Casement gave aid and comfort to the King’s enemies in the realm. B: Casement was adherent to the King’s enemies abroad. H : Casement gave aid and comfort to the King’s enemies abroad. The interpretation according to which Casement’s behavior was not treasonous is this: A ∨(G ∨H ) The interpretation which got him executed, on the other hand, can be symbolized by: (A ∨B) ∨(G ∨H ) Remember that in the case we’re dealing with Casement, was adherent to the King’s enemies abroad (B is true), but not in the realm, and he did not give the King’s enemies aid or comfort in or outside the realm (A, G, and H are false). One common source of structural ambiguity in English arises from its lack of parentheses. For instance, if I say ‘I like movies that are not long and boring’, you will most likely think that what CHAPTER 7. AMBIGUITY 60 I dislike are movies that are long and boring. A less likely, but possible, interpretation is that I like movies that are both (a) not long and (b) boring. The first reading is more likely because who likes boring movies? But what about ‘I like dishes that are not sweet and flavorful’? Here, the more likely interpretation is that I like savory, flavorful dishes. (Of course, I could have said that better, e.g., ‘I like dishes that are not sweet, yet flavor-ful’.) Similar ambiguities result from the interaction of ‘and’ with ‘or’. For instance, suppose I ask you to send me a picture of a small and dangerous or stealthy animal. Would a leopard count? It’s stealthy, but not small. So it depends whether I’m looking for small animals that are dangerous or stealthy (leopard doesn’t count), or whether I’m after either a small, dangerous animal or a stealthy animal (of any size). These kinds of ambiguities are called scope ambiguities, since they depend on whether or not a connective is in the scope of another. For instance, the sentence, ‘Avengers: Endgame is not long and boring’ is ambiguous between: 1. Avengers: Endgame is not: both long and boring. 2. Avengers: Endgame is both: not long and boring. Sentence 2 is certainly false, since Avengers: Endgame is over three hours long. Whether you think 1 is true depends on if you think it is boring or not. We can use the symbolization key: B: Avengers: Endgame is boring. L: Avengers: Endgame is long. Sentence 1 can now be symbolized as ‘¬(L ∧B)’, whereas sen-tence 2 would be ‘¬L ∧B’. In the first case, the ‘∧’ is in the scope of ‘¬’, in the second case ‘¬’ is in the scope of ‘∧’. The sentence ‘Tai Lung is small and dangerous or stealthy’ is ambiguous between: 3. Tai Lung is either both small and dangerous or stealthy. 4. Tai Lung is both small and either dangerous or stealthy. CHAPTER 7. AMBIGUITY 61 We can use the following symbolization key: D: Tai Lung is dangerous. S: Tai Lung is small. T : Tai Lung is stealthy. The symbolization of sentence 3 is ‘(S ∧D) ∨T ’ and that of sentence 4 is ‘S ∧(D ∨T )’. In the first, ‘∧’ is in the scope of ‘∨’, and in the second ‘∨’ is in the scope of ‘∧’. Practice exercises A. The following sentences are ambiguous. Give symbolization keys for each and symbolize the different readings. 1. Haskell is a birder and enjoys watching cranes. 2. The zoo has lions or tigers and bears. 3. The flower is not red or fragrant. CHAPTER 8 Use and mention In this Part, we have talked a lot about sentences. So we should pause to explain an important, and very general, point. 8.1 Quotation conventions Consider these two sentences: • Justin Trudeau is the Prime Minister. • The expression ‘Justin Trudeau’ is composed of two upper-case letters and eleven lowercase letters When we want to talk about the Prime Minister, we use his name. When we want to talk about the Prime Minister’s name, we men-tion that name, which we do by putting it in quotation marks. There is a general point here. When we want to talk about things in the world, we just use words. When we want to talk about words, we typically have to mention those words. We need to indicate that we are mentioning them, rather than using them. To do this, some convention is needed. We can put them in quotation marks, or display them centrally in the page (say). So this sentence: 62 CHAPTER 8. USE AND MENTION 63 • ‘Justin Trudeau’ is the Prime Minister. says that some expression is the Prime Minister. That’s false. The man is the Prime Minister; his name isn’t. Conversely, this sen-tence: • Justin Trudeau is composed of two uppercase letters and eleven lowercase letters. also says something false: Justin Trudeau is a man, made of flesh rather than letters. One final example: • “ ‘Justin Trudeau’ ” is the name of ‘Justin Trudeau’. On the left-hand-side, here, we have the name of a name. On the right hand side, we have a name. Perhaps this kind of sentence only occurs in logic textbooks, but it is true nonetheless. Those are just general rules for quotation, and you should ob-serve them carefully in all your work! To be clear, the quotation-marks here do not indicate reported speech. They indicate that you are moving from talking about an object, to talking about a name of that object. 8.2 Object language and metalanguage These general quotation conventions are very important for us. After all, we are describing a formal language here, TFL, and so we must often mention expressions from TFL. When we talk about a language, the language that we are talking about is called the object language. The language that we use to talk about the object language is called the metalan-guage. For the most part, the object language in this chapter has been the formal language that we have been developing: TFL. The metalanguage is English. Not conversational English exactly, but English supplemented with some additional vocabulary to help us get along. CHAPTER 8. USE AND MENTION 64 Now, we have used uppercase letters as sentence letters of TFL: A,B,C,Z,A1,B4,A25, J375,. . . These are sentences of the object language (TFL). They are not sentences of English. So we must not say, for example: • D is a sentence letter of TFL. Obviously, we are trying to come out with an English sentence that says something about the object language (TFL), but ‘D’ is a sentence of TFL, and not part of English. So the preceding is gibberish, just like: • Schnee ist weiß is a German sentence. What we surely meant to say, in this case, is: • ‘Schnee ist weiß’ is a German sentence. Equally, what we meant to say above is just: • ‘D’ is a sentence letter of TFL. The general point is that, whenever we want to talk in English about some specific expression of TFL, we need to indicate that we are mentioning the expression, rather than using it. We can either deploy quotation marks, or we can adopt some similar convention, such as placing it centrally in the page. 8.3 Metavariables However, we do not just want to talk about specific expressions of TFL. We also want to be able to talk about any arbitrary sentence of TFL. Indeed, we had to do this in §6.2, when we presented the inductive definition of a sentence of TFL. We used uppercase script letters to do this, namely: A,B, C, D,. . . CHAPTER 8. USE AND MENTION 65 These symbols do not belong to TFL. Rather, they are part of our (augmented) metalanguage that we use to talk about any ex-pression of TFL. To explain why we need them, recall the second clause of the recursive definition of a sentence of TFL: 2. If A is a sentence, then ¬A is a sentence. This talks about arbitrary sentences. If we had instead offered: • If ‘A’ is a sentence, then ‘¬A’ is a sentence. this would not have allowed us to determine whether ‘¬B’ is a sentence. To emphasize: ‘A’ is a symbol (called a metav ariable) in augmented En-glish, which we use to talk about expressions of TFL. ‘A’ is a particular sentence letter of TFL. But this last example raises a further complication, concern-ing quotation conventions. We did not include any quotation marks in the second clause of our inductive definition. Should we have done so? The problem is that the expression on the right-hand-side of this rule, i.e., ‘¬A’, is not a sentence of English, since it contains ‘¬’. So we might try to write: 2′. If A is a sentence, then ‘¬A’ is a sentence. But this is no good: ‘¬A’ is not a TFL sentence, since ‘A’ is a symbol of (augmented) English rather than a symbol of TFL. What we really want to say is something like this: 2′′. If Ais a sentence, then the result of concatenating the sym-bol ‘¬’ with the sentence A is a sentence. This is impeccable, but rather long-winded. But we can avoid long-windedness by creating our own conventions. We can per-fectly well stipulate that an expression like ‘¬A’ should simply be CHAPTER 8. USE AND MENTION 66 read directly in terms of rules for concatenation. So, officially, the metalanguage expression ‘¬A’ simply abbreviates: the result of concatenating the symbol ‘¬’ with the sentence A and similarly, for expressions like ‘(A∧B)’, ‘(A∨B)’, etc. 8.4 Quotation conventions for arguments One of our main purposes for using TFL is to study arguments, and that will be our concern in part III. In English, the premises of an argument are often expressed by individual sentences, and the conclusion by a further sentence. Since we can symbolize English sentences, we can symbolize English arguments using TFL. Or rather, we can use TFL to symbolize each of the sentences used in an English argument. However, TFL itself has no way to flag some of them as the premises and another as the conclusion of an argument. (Contrast this with natural English, which uses words like ‘so’, ‘therefore’, etc., to mark that a sentence is the conclusion of an argument.) So, we need another bit of notation. Suppose we want to symbolize the premises of an argument with A 1, ..., A n and the conclusion with C. Then we will write: A 1,. . . ,A n ∴C The role of the symbol ‘∴’ is simply to indicate which sentences are the premises and which is the conclusion. Strictly, the symbol ‘∴’ will not be a part of the object lan-guage, but of the metalanguage. As such, one might think that we would need to put quote-marks around the TFL-sentences which flank it. That is a sensible thought, but adding these quote-marks would make things harder to read. Moreover—and as above— recall that we are stipulating some new conventions. So, we can simply stipulate that these quote-marks are unnecessary. That is, CHAPTER 8. USE AND MENTION 67 we can simply write: A,A →B ∴B without any quotation marks, to indicate an argument whose premises are (symbolized by) ‘A’ and ‘A →B’ and whose con-clusion is (symbolized by) ‘B’. PART III Truth tables 68 CHAPTER 9 Characteristic truth tables Any sentence of TFL is composed of sentence letters, possibly combined using sentential connectives. The truth value of the compound sentence depends only on the truth value of the sen-tence letters that comprise it. In order to know the truth value of ‘(D ∧E)’, for instance, you only need to know the truth value of ‘D’ and the truth value of ‘E’. We introduced five connectives in chapter 5. So we just need to explain how they map between truth values. For convenience, we abbreviate ‘True’ with ‘T’ and ‘False’ with ‘F’. (But, to be clear, the two truth values are True and False; the truth values are not letters!) Negation. For any sentence A: If A is true, then ¬A is false; and if ¬A is true, then A is false. We can summarize this in the characteristic truth table for negation: A ¬A T F F T 69 CHAPTER 9. CHARACTERISTIC TRUTH TABLES 70 Conjunction. For any sentences A and B, A∧B is true if and only if both A and B are true. We can summarize this in the characteristic truth table for conjunction: A B A∧B T T T T F F F T F F F F Note that conjunction is symmetrical. The truth value for A∧B is always the same as the truth value for B∧A. Disjunction. Recall that ‘∨’ always represents inclusive or. So, for any sentences A and B, A∨B is true if and only if either A or B is true. We can summarize this in the characteristic truth table for disjunction: A B A∨B T T T T F T F T T F F F Like conjunction, disjunction is symmetrical. Conditional. We’re just going to come clean and admit it: Con-ditionals are a mess in TFL. Exactly how much of a mess they are is philosophically contentious. We’ll discuss a few of the sub-tleties in §§10.3 and 12.5. For now, we are going to stipulate the following: A →B is false if and only if A is true and B is false. We can summarize this with a characteristic truth table for the conditional. CHAPTER 9. CHARACTERISTIC TRUTH TABLES 71 A B A →B T T T T F F F T T F F T The conditional is asymmetric. You cannot swap the antecedent and consequent without changing the meaning of the sentence; A →B and B →A have different truth tables. Biconditional. Since a biconditional is to be the same as the conjunction of the conditionals running in both directions, we will want the truth table for the biconditional to be: A B A ↔B T T T T F F F T F F F T Unsurprisingly, the biconditional is symmetrical. CHAPTER 10 Truth-functional connectives 10.1 The idea of truth-functionality Let’s introduce an important idea. A connective is truth-functional iff the truth value of a sentence with that connective as its main logical oper-ator is uniquely determined by the truth value(s) of the constituent sentence(s). Every connective in TFL is truth-functional. The truth value of a negation is uniquely determined by the truth value of the unnegated sentence. The truth value of a conjunction is uniquely determined by the truth value of both conjuncts. The truth value of a disjunction is uniquely determined by the truth value of both disjuncts, and so on. To determine the truth value of some TFL sentence, we only need to know the truth value of its components. 72 CHAPTER 10. TRUTH-FUNCTIONAL CONNECTIVES 73 This is what gives TFL its name: it is truth-functional logic. Many languages use connectives that are not truth-functional. In English, for example, we can form a new sentence from any simpler sentence by prefixing it with ‘It is necessarily the case that...’. The truth value of this new sentence is not fixed solely by the truth value of the original sentence. For consider two true sentences: 1. 2 + 2 = 4 2. Shostakovich wrote fifteen string quartets Whereas it is necessarily the case that 2 + 2 = 4, it is not nec-essarily the case that Shostakovich wrote fifteen string quartets. If Shostakovich had died earlier, he would have failed to finish Quartet no. 15; if he had lived longer, he might have written a few more. So ‘It is necessarily the case that. . . ’ is not truth-functional. 10.2 Symbolizing versus translating All of the connectives of TFL are truth-functional, but more than that: they really do nothing but map us between truth values. When we symbolize a sentence or an argument in TFL, we ignore everything besides the contribution that the truth values of a component might make to the truth value of the whole. There are subtleties to our ordinary claims that far outstrip their mere truth values. Sarcasm; poetry; snide implicature; emphasis; these are important parts of everyday discourse, but none of this is retained in TFL. As remarked in §5, TFL cannot capture the subtle differences between the following English sentences: 1. Dana is a logician and Dana is a nice person 2. Although Dana is a logician, Dana is a nice person 3. Dana is a logician despite being a nice person 4. Dana is a nice person, but also a logician 5. Dana’s being a logician notwithstanding, he is a nice person CHAPTER 10. TRUTH-FUNCTIONAL CONNECTIVES 74 All of the above sentences will be symbolized with the same TFL sentence, perhaps ‘L ∧N ’. Now, we keep saying that we use TFL sentences to symbolize English sentences. Many other textbooks talk about translating English sentences into TFL. However, a good translation should preserve certain facets of meaning, and—as we just saw—TFL just cannot do that. This is why we will speak of symbolizing En-glish sentences, rather than of translating them. This affects how we should understand our symbolization keys. Consider a key like: L: Dana is a logician. N : Dana is a nice person. Other textbooks will understand this as a stipulation that the TFL sentence ‘L’ should mean that Dana is a logician, and that the TFL sentence ‘N ’ should mean that Dana is a nice person. But TFL just is totally unequipped to deal with meaning. The preceding symbolization key is doing no more and no less than stipulating that the TFL sentence ‘L’ should take the same truth value as the English sentence ‘Dana is a logician’ (whatever that might be), and that the TFL sentence ‘N ’ should take the same truth value as the English sentence ‘Dana is a nice person’ (what-ever that might be). When we treat a TFL sentence as symbolizing an English sentence, we are stipulating that the TFL sentence is to take the same truth value as that English sentence. 10.3 Indicative versus subjunctive conditionals We want to bring home the point that TFL can only deal with truth functions by considering the case of the conditional. When we introduced the characteristic truth table for the material con-CHAPTER 10. TRUTH-FUNCTIONAL CONNECTIVES 75 ditional in §9, we did not say anything to justify it. Let’s now offer a justification, which follows Dorothy Edgington.1 Suppose that Lara has drawn some shapes on a piece of pa-per, and coloured some of them in. We have not seen them, but nevertheless claim: If any shape is grey, then that shape is also circular. As it happens, Lara has drawn the following: A C D In this case, our claim is surely true. Shapes C and D are not grey, and so can hardly present counterexamples to our claim. Shape A is grey, but fortunately it is also circular. So our claim has no counterexamples. It must be true. That means that each of the following instances of our claim must be true too: • If A is grey, then it is circular (true antecedent, true consequent) • If C is grey, then it is circular (false antecedent, true consequent) • If D is grey, then it is circular (false antecedent, false consequent) However, if Lara had drawn a fourth shape, thus: A B C D then our claim would have been false. So this claim must also be false: • If B is grey, then it is circular (true antecedent, false consequent) 1Dorothy Edgington, ‘Conditionals’, 2014, in the Stanford Encyclopedia of Philosophy ( CHAPTER 10. TRUTH-FUNCTIONAL CONNECTIVES 76 Now, recall that every connective of TFL has to be truth-functional. This means that merely the truth values of the an-tecedent and consequent must uniquely determine the truth value of the conditional as a whole. Thus, from the truth values of our four claims—which provide us with all possible combinations of truth and falsity in antecedent and consequent—we can read off the truth table for the material conditional. What this argument shows is that ‘→’ is the best candidate for a truth-functional conditional. Otherwise put, it is the best conditional that TFL can provide. But is it any good, as a surrogate for the conditionals we use in everyday language? Consider two sentences: 1. If Hillary Clinton had won the 2016 election, then she would have been the first woman president of the USA. 2. If Hillary Clinton had won the 2016 election, then she would have turned into a helium-filled balloon and floated away into the night sky. Sentence 1 is true; sentence 2 is false, but both have false an-tecedents and false consequents. (Hillary did not win; she did not become the first woman president of the US; and she did not fill with helium and float away.) So the truth value of the whole sentence is not uniquely determined by the truth value of the parts. The crucial point is that sentences 1 and 2 employ subjunctive conditionals, rather than indicative conditionals. They ask us to imagine something contrary to fact—after all, Hillary Clinton lost the 2016 election—and then ask us to evaluate what would have happened in that case. Such considerations simply cannot be tackled using ‘→’. We will say more about the difficulties with conditionals in §12.5. For now, we will content ourselves with the observation that ‘→’ is the only candidate for a truth-functional conditional for TFL, but that many English conditionals cannot be repre-sented adequately using ‘→’. TFL is an intrinsically limited lan-guage. CHAPTER 11 Complete truth tables So far, we have used symbolization keys to assign truth values to TFL sentences indirectly. For example, we might say that the TFL sentence ‘B’ is to be true iff Big Ben is in London. Since Big Ben is in London, this symbolisation would make ‘B’ true. But we can also assign truth values directly. We can simply stipulate that ‘B’ is to be true, or stipulate that it is to be false. Such stipulations are called valuations: A v aluation is any assignment of truth values to particu-lar sentence letters of TFL. The power of truth tables lies in the following. Each row of a truth table represents a possible valuation. The complete truth table represents all possible valuations. And the truth table provides us with a means to calculate the truth value of complex sentences, on each possible valuation. But all of this is easiest to explain by example. 77 CHAPTER 11. COMPLETE TRUTH TABLES 78 11.1 A worked example Consider the sentence ‘(H ∧I ) →H ’. There are four possible ways to assign True and False to the sentence letter ‘H ’ and ‘I ’— four valuations—which we can represent as follows: H I (H ∧I ) →H T T T F F T F F To calculate the truth value of the entire sentence ‘(H ∧I ) →H ’, we first copy the truth values for the sentence letters and write them underneath the letters in the sentence: H I (H ∧I ) →H T T T T T T F T F T F T F T F F F F F F Now consider the subsentence ‘(H ∧I )’. This is a conjunction, (A∧B), with ‘H ’ as Aand with ‘I ’ as B. The characteristic truth table for conjunction gives the truth conditions for any sentence of the form (A∧B), whatever A and B might be. It represents the point that a conjunction is true iff both conjuncts are true. In this case, our conjuncts are just ‘H ’ and ‘I ’. They are both true on (and only on) the first line of the truth table. Accordingly, we can calculate the truth value of the conjunction on all four rows. A ∧B H I (H ∧I ) →H T T T T T T T F T F F T F T F F T F F F F F F F CHAPTER 11. COMPLETE TRUTH TABLES 79 Now, the entire sentence that we are dealing with is a conditional, A →B, with ‘(H ∧I )’ as A and with ‘H ’ as B. On the second row, for example, ‘(H ∧I )’ is false and ‘H ’ is true. Since a con-ditional is true when the antecedent is false, we write a ‘T’ in the second row underneath the conditional symbol. We continue for the other three rows and get this: A →B H I (H ∧I ) →H T T T T T T F F T T F T F T F F F F T F The conditional is the main logical operator of the sentence, so the column of ‘T’s underneath the conditional tells us that the sentence ‘(H ∧I ) →H ’ is true regardless of the truth values of ‘H ’ and ‘I ’. They can be true or false in any combination, and the compound sentence still comes out true. Since we have con-sidered all four possible assignments of truth and falsity to ‘H ’ and ‘I ’—since, that is, we have considered all the different valua-tions—we can say that ‘(H ∧I ) →H ’ is true on every valuation. In this example, we have not repeated all of the entries in every column in every successive table. When actually writing truth tables on paper, however, it is impractical to erase whole columns or rewrite the whole table for every step. Although it is more crowded, the truth table can be written in this way: H I (H ∧I ) →H T T T T T T T T F T F F T T F T F F T T F F F F F F T F Most of the columns underneath the sentence are only there for bookkeeping purposes. The column that matters most is the col-umn underneath the main logical operator for the sentence, since CHAPTER 11. COMPLETE TRUTH TABLES 80 this tells you the truth value of the entire sentence. We have em-phasized this, by putting this column in bold. When you work through truth tables yourself, you should similarly emphasize it (perhaps by highlighting). 11.2 Building complete truth tables A complete truth table has a line for every possible assign-ment of True and False to the relevant sentence letters. Each line represents a valuation, and a complete truth table has a line for all the different valuations. The size of the complete truth table depends on the num-ber of different sentence letters in the table. A sentence that contains only one sentence letter requires only two rows, as in the characteristic truth table for negation. This is true even if the same letter is repeated many times, as in the sentence ‘[(C ↔C ) →C ] ∧¬(C →C )’. The complete truth table re-quires only two lines because there are only two possibilities: ‘C ’ can be true or it can be false. The truth table for this sentence looks like this: C [(C ↔C ) →C ] ∧¬ (C →C ) T T T T T T FF T T T F F T F F F FF F T F Looking at the column underneath the main logical operator, we see that the sentence is false on both rows of the table; i.e., the sentence is false regardless of whether ‘C ’ is true or false. It is false on every valuation. There will be four lines in the complete truth table for a sentence containing two sentence letters, as in the characteris-tic truth tables, or the truth table for ‘(H ∧I ) →H ’. There will be eight lines in the complete truth table for a sentence containing three sentence letters, e.g.: CHAPTER 11. COMPLETE TRUTH TABLES 81 M N P M ∧(N ∨P ) T T T T T T T T T T F T T T T F T F T T T F T T T F F T F F F F F T T F F T T T F T F F F T T F F F T F F F T T F F F F F F F F From this table, we know that the sentence ‘M ∧(N ∨P )’ can be true or false, depending on the truth values of ‘M ’, ‘N ’, and ‘P ’. A complete truth table for a sentence that contains four dif-ferent sentence letters requires 16 lines. Five letters, 32 lines. Six letters, 64 lines. And so on. To be perfectly general: If a complete truth table has n different sentence letters, then it must have 2n lines. In order to fill in the columns of a complete truth table, begin with the right-most sentence letter and alternate between ‘T’ and ‘F’. In the next column to the left, write two ‘T’s, write two ‘F’s, and repeat. For the third sentence letter, write four ‘T’s followed by four ‘F’s. This yields an eight line truth table like the one above. For a 16 line truth table, the next column of sentence letters should have eight ‘T’s followed by eight ‘F’s. For a 32 line table, the next column would have 16 ‘T’s followed by 16 ‘F’s, and so on. 11.3 More about brackets Consider these two sentences: ((A ∧B) ∧C ) (A ∧(B ∧C )) These are truth functionally equivalent. Consequently, it will never make any difference from the perspective of truth value— CHAPTER 11. COMPLETE TRUTH TABLES 82 which is all that TFL cares about (see §10)—which of the two sen-tences we assert (or deny). Even though the order of the brackets does not matter as to their truth, we should not just drop them. The expression A ∧B ∧C is ambiguous between the two sentences above. The same obser-vation holds for disjunctions. The following sentences are logi-cally equivalent: ((A ∨B) ∨C ) (A ∨(B ∨C )) But we should not simply write: A ∨B ∨C In fact, it is a specific fact about the characteristic truth table of ∨ and ∧that guarantees that any two conjunctions (or disjunctions) of the same sentences are truth functionally equivalent, however you place the brackets. This is only true of conjunctions and disjunc-tions, however. The following two sentences have different truth tables: ((A →B) →C ) (A →(B →C )) So if we were to write: A →B →C it would be dangerously ambiguous. Leaving out brackets in this case would be disastrous. Equally, these sentences have different truth tables: ((A ∨B) ∧C ) (A ∨(B ∧C )) CHAPTER 11. COMPLETE TRUTH TABLES 83 So if we were to write: A ∨B ∧C it would be dangerously ambiguous. Never write this. The moral is: never drop brackets (except the outermost ones). Practice exercises A. Offer complete truth tables for each of the following: 1. A →A 2. C →¬C 3. (A ↔B) ↔¬(A ↔¬B) 4. (A →B) ∨(B →A) 5. (A ∧B) →(B ∨A) 6. ¬(A ∨B) ↔(¬A ∧¬B) 7. [︁ (A ∧B) ∧¬(A ∧B) ]︁ ∧C 8. [(A ∧B) ∧C ] →B 9. ¬ [︁ (C ∨A) ∨B ]︁ B. Check all the claims made in §11.3, i.e., show that: 1. ‘((A∧B) ∧C )’ and ‘(A∧(B ∧C ))’ have the same truth table 2. ‘((A∨B) ∨C )’ and ‘(A∨(B ∨C ))’ have the same truth table 3. ‘((A ∨B) ∧C )’ and ‘(A ∨(B ∧C ))’ do not have the same truth table 4. ‘((A →B) →C )’ and ‘(A →(B →C ))’ do not have the same truth table Also, check whether: 5. ‘((A ↔B) ↔C )’ and ‘(A ↔(B ↔C ))’ have the same truth table C. Write complete truth tables for the following sentences and mark the column that represents the possible truth values for the whole sentence. CHAPTER 11. COMPLETE TRUTH TABLES 84 1. ¬(S ↔(P →S)) 2. ¬[(X ∧Y ) ∨(X ∨Y )] 3. (A →B) ↔(¬B ↔¬A) 4. [C ↔(D ∨E)] ∧¬C 5. ¬(G ∧(B ∧H )) ↔(G ∨(B ∨H )) D. Write complete truth tables for the following sentences and mark the column that represents the possible truth values for the whole sentence. 1. (D ∧¬D) →G 2. (¬P ∨¬M ) ↔M 3. ¬¬(¬A ∧¬B) 4. [(D ∧R) →I ] →¬(D ∨R) 5. ¬[(D ↔O) ↔A] →(¬D ∧O) If you want additional practice, you can construct truth tables for any of the sentences and arguments in the exercises for the previous chapter. CHAPTER 12 Semantic concepts In the previous section, we introduced the idea of a valuation and showed how to determine the truth value of any TFL sentence, on any valuation, using a truth table. In this section, we will introduce some related ideas, and show how to use truth tables to test whether or not they apply. 12.1 Tautologies and contradictions In §3, we explained necessary truth and necessary falsity. Both no-tions have surrogates in TFL. We will start with a surrogate for necessary truth. A is a tautology iff it is true on every valuation. We can use truth tables to decide whether a sentence is a tautology. If the sentence is true on every line of its complete truth table, then it is true on every valuation, so it is a tautology. In the example of §11, ‘(H ∧I ) →H ’ is a tautology. This is only, though, a surrogate for necessary truth. There are some necessary truths that we cannot adequately symbolize 85 CHAPTER 12. SEMANTIC CONCEPTS 86 in TFL. One example is ‘2+2 = 4’. This must be true, but if we try to symbolize it in TFL, the best we can offer is a sentence letter, and no sentence letter is a tautology. Still, if we can adequately symbolize some English sentence using a TFL sentence which is a tautology, then that English sentence expresses a necessary truth. We have a similar surrogate for necessary falsity: A is a contradiction (in TFL) iff it is false on every valuation. We can use truth tables to decide whether a sentence is a con-tradiction. If the sentence is false on every line of its complete truth table, then it is false on every valuation, so it is a contra-diction. In the example of §11, ‘[(C ↔C ) →C ] ∧¬(C →C )’ is a contradiction. 12.2 Equivalence Here is a similar useful notion: Aand B are equiv alent (in TFL) iff, for every valuation, their truth values agree, i.e., if there is no valuation in which they have opposite truth values. We have already made use of this notion, in effect, in §11.3; the point was that ‘(A ∧B) ∧C ’ and ‘A ∧(B ∧C )’ are equivalent. Again, it is easy to test for equivalence using truth tables. Con-sider the sentences ‘¬(P ∨Q)’ and ‘¬P ∧¬Q’. Are they equivalent? To find out, we construct a truth table. P Q ¬ (P ∨Q) ¬P ∧¬Q T T F T T T F T F F T T F F T T F F T F T F F T F F T T T F F F T F F T F F F T F TT F CHAPTER 12. SEMANTIC CONCEPTS 87 Look at the columns for the main logical operators; negation for the first sentence, conjunction for the second. Go through the rows in the table one by one and compare the truth values in the columns for the main logical operators. On the first three rows, both are false. On the final row, both are true. Since they match on every row, the two sentences are equivalent. 12.3 Satisfiability In §3, we said that sentences are jointly possible iff it is possible for all of them to be true at once. We can offer a surrogate for this notion too: A 1,A 2,. . . ,A n are jointly satisfiable (in TFL) iff there is some valuation which makes them all true. Derivatively, sentences are jointly unsatisfiable iff no val-uation makes them all true. Again, it is easy to test for joint satisfiability using truth tables. 12.4 Entailment and validity The following idea is closely related to that of joint satisfiability: The sentences A 1,A 2,. . . ,A n entail (in TFL) the sen-tence C iff no valuation of the relevant sentence letters makes all of A 1,A 2,. . . ,A n true and C false. Again, it is easy to test this with a truth table. To check whether ‘¬L →( J ∨L)’ and ‘¬L’ entail ‘ J ’, we simply need to check whether there is any valuation which makes both ‘¬L → ( J ∨L)’ and ‘¬L’ true whilst making ‘ J ’ false. So we use a truth table: CHAPTER 12. SEMANTIC CONCEPTS 88 J L ¬ L →( J ∨L) ¬ L J T T F T T T T T F T T T F T F T T T F T F T F T F T T F T T F T F F F T F F F F F T F F The only row on which both‘¬L →( J ∨L)’ and ‘¬L’ are true is the second row, and that is a row on which ‘ J ’ is also true. So ‘¬L →( J ∨L)’ and ‘¬L’ entail ‘ J ’. We now make an important observation: If A 1,A 2,. . . ,A n entail C in TFL, then A 1,A 2,. . . ,A n ∴C is valid. Here’s why. Suppose A 1,A 2,. . . ,A n entail C in TFL. Can it happen that the argument A 1,A 2,. . . ,A n ∴C is invalid? Then there would be a case which makes all of A 1,A 2,. . . ,A n true and C false, relative to some symbolization key. This case would gener-ate a valuation of the sentence letters occurring in A 1,A 2,. . . ,A n, and C: take the truth value of any sentence letter to be just the truth value of the corresponding sentence in the case in question. This valuation would also make A 1,A 2,. . . ,A n true and C false, since truth values are determined from the truth values of sen-tence letters by the truth tables of the connectives, and the truth values of the sentence letters are the same in the valuation as they are in the (imagined) case. But this is impossible, since we’ve as-sumed that A 1,A 2,. . . ,A n entail C in TFL, and so there is no valuation which makes all of A 1,A 2,. . . ,A n true and also makes C false. So it can’t happen that the argument A 1,A 2,. . . ,A n ∴C is invalid; consequently, it must be valid. In short, we have a way to test for the validity of English arguments. First, we symbolize them in TFL; then we test for entailment in TFL using truth tables. CHAPTER 12. SEMANTIC CONCEPTS 89 12.5 The limits of these tests This is an important milestone: a test for the validity of argu-ments! However, we should not get carried away just yet. It is important to understand the limits of our achievement. We will illustrate these limits with four examples. First, consider the argument: 1. Daisy has four legs. So Daisy has more than two legs. To symbolize this argument in TFL, we would have to use two different sentence letters—perhaps ‘F ’ and ‘T ’—for the premise and the conclusion respectively. Now, it is obvious that ‘F ’ does not entail ‘T ’. But the English argument is surely valid! Second, consider the sentence: 2. Jan is neither bald nor not-bald. To symbolize this sentence in TFL, we would offer something like ‘¬ J ∧¬¬ J ’. This a contradiction (check this with a truth-table), but sentence 2 does not itself seem like a contradiction; for we might have happily added ‘Jan is on the borderline of baldness’! Third, consider the following sentence: 3. It’s not the case that, if God exists, She answers malevolent prayers. Symbolizing this in TFL, we would offer something like ‘¬(G → M )’. Now, ‘¬(G →M )’ entails ‘G’ (again, check this with a truth table). So if we symbolize sentence 3 in TFL, it seems to entail that God exists. But that’s strange: surely even an atheist can accept sentence 3, without contradicting herself! One lesson of this is that the symbolization of 3 as ‘¬(G → M )’ shows that 3 does not express what we intend. Perhaps we should rephrase it as 4. If God exists, She does not answer malevolent prayers. CHAPTER 12. SEMANTIC CONCEPTS 90 and symbolize 4 as ‘G →¬M ’. Now, if atheists are right, and there is no God, then ‘G’ is false and so ‘G →¬M ’ is true, and the puzzle disappears. However, if ‘G’ is false, ‘G →M ’, i.e., ‘If God exists, She answers malevolent prayers’, is also true! In different ways, these four examples highlight some of the limits of working with a language (like TFL) that can only handle truth-functional connectives. Moreover, these limits give rise to some interesting questions in philosophical logic. The case of Jan’s baldness (or otherwise) raises the general question of what logic we should use when dealing with vague discourse. The case of the atheist raises the question of how to deal with the (so-called) paradoxes of the material conditional. Part of the purpose of this book is to equip you with the tools to explore these questions of philosophical logic. But we have to walk before we can run; we have to become proficient in using TFL, before we can adequately discuss its limits, and consider alternatives. 12.6 The double turnstile In what follow, we will use the notion of entailment rather a lot in this book. It will help us, then, to introduce a symbol that abbreviates it. Rather than saying that the TFL sentences A 1, A 2, ... and A n together entail C, we will abbreviate this by: A 1,A 2,. . . ,A n ⊨C The symbol ‘⊨’ is known as the double turnstile, since it looks like a turnstile with two horizontal beams. Let’s be clear. ‘⊨’ is not a symbol of TFL. Rather, it is a symbol of our metalanguage, augmented English (recall the dif-ference between object language and metalanguage from §8). So the metalanguage sentence: • P,P →Q ⊨Q is just an abbreviation for this metalanguage sentence: CHAPTER 12. SEMANTIC CONCEPTS 91 • The TFL sentences ‘P ’ and ‘P →Q’ entail ‘Q’ Note that there is no limit on the number of TFL sentences that can be mentioned before the symbol ‘⊨’. Indeed, we can even consider the limiting case: ⊨C This says that there is no valuation which makes all the sentences mentioned on the left side of ‘⊨’ true whilst making Cfalse. Since no sentences are mentioned on the left side of ‘⊨’ in this case, this just means that there is no valuation which makes C false. Oth-erwise put, it says that every valuation makes C true. Otherwise put, it says that C is a tautology. Equally, to say that A is a contradiction, we can write: A ⊨ For this says that no valuation makes A true. Sometimes, we will want to deny that there is a tautological entailment, and say something of this shape: it is not the case that A 1,. . . ,A n ⊨C In that case, we can just slash the turnstile through, and write: A 1,A 2,. . . ,A n ⊭C This means that some valuation makes all of A 1,. . . ,A n true whilst making Cfalse. Note that it does not follow that A 1,. . . ,A n ⊨¬C, for it is possible that some other valuation makes all of A 1,. . . ,A n true and makes C true. For instance, P ⊭Q but also P ⊭¬Q. 12.7 ‘⊨’ versus ‘→’ We now want to compare and contrast ‘⊨’ and ‘→’. Observe: A ⊨C iff no valuation of the sentence letters makes A true and C false. CHAPTER 12. SEMANTIC CONCEPTS 92 Observe: A →C is a tautology iff no valuation of the sen-tence letters makes A →C false. Since a conditional is true ex-cept when its antecedent is true and its consequent false, A →C is a tautology iff no valuation makes A true and C false. Combining these two observations, we see that A →C is a tautology iff A ⊨C. But there is a really, really important difference between ‘⊨’ and ‘→’: ‘→’ is a sentential connective of TFL. ‘⊨’ is a symbol of augmented English. Indeed, when ‘→’ is flanked with two TFL sentences, the re-sult is a longer TFL sentence. By contrast, when we use ‘⊨’, we form a metalinguistic sentence that mentions the surrounding TFL sentences. Practice exercises A. Revisit your answers to §11A. Determine which sentences were tautologies, which were contradictions, and which were nei-ther tautologies nor contradictions. B. Use truth tables to determine whether these sentences are jointly satisfiable, or jointly unsatisfiable: 1. A →A, ¬A →¬A, A ∧A, A ∨A 2. A ∨B, A →C , B →C 3. B ∧(C ∨A), A →B, ¬(B ∨C ) 4. A ↔(B ∨C ), C →¬A, A →¬B C. Use truth tables to determine whether each argument is valid or invalid. 1. A →A ∴A 2. A →(A ∧¬A) ∴¬A 3. A ∨(B →A) ∴¬A →¬B CHAPTER 12. SEMANTIC CONCEPTS 93 4. A ∨B,B ∨C,¬A ∴B ∧C 5. (B ∧A) →C, (C ∧A) →B ∴(C ∧B) →A D. Determine whether each sentence is a tautology, a contradic-tion, or a contingent sentence, using a complete truth table. 1. ¬B ∧B 2. ¬D ∨D 3. (A ∧B) ∨(B ∧A) 4. ¬[A →(B →A)] 5. A ↔[A →(B ∧¬B)] 6. [(A ∧B) ↔B] →(A →B) E. Determine whether each the following sentences are logically equivalent using complete truth tables. If the two sentences re-ally are logically equivalent, write “equivalent.” Otherwise write, “Not equivalent.” 1. A and ¬A 2. A ∧¬A and ¬B ↔B 3. [(A ∨B) ∨C ] and [A ∨(B ∨C )] 4. A ∨(B ∧C ) and (A ∨B) ∧(A ∨C ) 5. [A ∧(A ∨B)] →B and A →B F. Determine whether each the following sentences are logically equivalent using complete truth tables. If the two sentences really are equivalent, write “equivalent.” Otherwise write, “not equiva-lent.” 1. A →A and A ↔A 2. ¬(A →B) and ¬A →¬B 3. A ∨B and ¬A →B 4. (A →B) →C and A →(B →C ) 5. A ↔(B ↔C ) and A ∧(B ∧C ) G. Determine whether each collection of sentences is jointly sat-isfiable or jointly unsatisfiable using a complete truth table. CHAPTER 12. SEMANTIC CONCEPTS 94 1. A ∧¬B, ¬(A →B), B →A 2. A ∨B, A →¬A, B →¬B 3. ¬(¬A ∨B), A →¬C , A →(B →C ) 4. A →B, A ∧¬B 5. A →(B →C ), (A →B) →C , A →C H. Determine whether each collection of sentences is jointly sat-isfiable or jointly unsatisfiable, using a complete truth table. 1. ¬B, A →B, A 2. ¬(A ∨B), A ↔B, B →A 3. A ∨B, ¬B, ¬B →¬A 4. A ↔B, ¬B ∨¬A, A →B 5. (A ∨B) ∨C , ¬A ∨¬B, ¬C ∨¬B I. Determine whether each argument is valid or invalid, using a complete truth table. 1. A →B, B ∴A 2. A ↔B, B ↔C ∴A ↔C 3. A →B, A →C ∴B →C 4. A →B, B →A ∴A ↔B J. Determine whether each argument is valid or invalid, using a complete truth table. 1. A ∨ [︁ A →(A ↔A) ]︁ ∴A 2. A ∨B, B ∨C , ¬B ∴A ∧C 3. A →B, ¬A ∴¬B 4. A, B ∴¬(A →¬B) 5. ¬(A ∧B), A ∨B, A ↔B ∴C K. Answer each of the questions below and justify your answer. 1. Suppose that A and B are logically equivalent. What can you say about A ↔B? CHAPTER 12. SEMANTIC CONCEPTS 95 2. Suppose that (A ∧B) →C is neither a tautology nor a contradiction. What can you say about whether A,B ∴C is valid? 3. Suppose that A, B and C are jointly unsatisfiable. What can you say about (A∧B∧C)? 4. Suppose that Ais a contradiction. What can you say about whether A,B ⊨C? 5. Suppose that C is a tautology. What can you say about whether A,B ⊨C? 6. Suppose that A and B are logically equivalent. What can you say about (A∨B)? 7. Suppose that A and B are not logically equivalent. What can you say about (A∨B)? L. Consider the following principle: • Suppose A and B are logically equivalent. Suppose an ar-gument contains A (either as a premise, or as the conclu-sion). The validity of the argument would be unaffected, if we replaced A with B. Is this principle correct? Explain your answer. CHAPTER 13 Truth table shortcuts With practice, you will quickly become adept at filling out truth tables. In this section, we consider (and justify) some shortcuts which will help you along the way. 13.1 Working through truth tables You will quickly find that you do not need to copy the truth value of each sentence letter, but can simply refer back to them. So you can speed things up by writing: P Q (P ∨Q) ↔¬P T T T F F T F T F F F T T T T F F F F T You also know for sure that a disjunction is true whenever one of the disjuncts is true. So if you find a true disjunct, there is no need to work out the truth value of the other disjunct. Thus you might offer: 96 CHAPTER 13. TRUTH TABLE SHORTCUTS 97 P Q (¬P ∨¬Q) ∨¬P T T F F F F F T F F TT T F F T TT F F TT Equally, you know for sure that a conjunction is false whenever one of the conjuncts is false. So if you find a false conjunct, there is no need to work out the truth value of the other conjunct. Thus you might offer: P Q ¬ (P ∧¬Q) ∧¬P T T F F T F F F F T T F TT F F T F TT A similar short cut is available for conditionals. You immediately know that a conditional is true if either its consequent is true, or its antecedent is false. Thus you might present: P Q ((P →Q)→P ) →P T T T T F T F T T F T F F T F T So ‘((P →Q) →P ) →P ’ is a tautology. In fact, it is an instance of Peirce’s Law, named after Charles Sanders Peirce. 13.2 T esting for validity and entailment In §12, we saw how to use truth tables to test for validity. In that test, we look for bad lines: lines where the premises are all true and the conclusion is false. Now: • If the conclusion is true on a line, then that line is not bad. (And we don’t need to evaluate anything else on that line to confirm this.) CHAPTER 13. TRUTH TABLE SHORTCUTS 98 • If any premise is false on a line, then that line is not bad. (And we don’t need to evaluate anything else on that line to confirm this.) With this in mind, we can speed up our tests for validity quite considerably. Let’s consider how we might test the following: ¬L →( J ∨L),¬L ∴J The first thing we should do is evaluate the conclusion. If we find that the conclusion is true on some line, then that is not a bad line. So we can simply ignore the rest of the line. So, after our first stage, we are left with something like this: J L ¬L→( J ∨L) ¬L J T T T T F T F T ? ? F F F ? ? F where the blanks indicate that we won’t bother with any more investigation (since the line is not bad), and the question marks indicate that we need to keep digging. The easiest premise to evaluate is the second, so we do that next, and get: J L ¬L→( J ∨L) ¬L J T T T T F T F T F F F F ? T F Note that we no longer need to consider the third line on the table: it is certainly not bad, because some premise is false on that line. And finally, we complete the truth table: CHAPTER 13. TRUTH TABLE SHORTCUTS 99 J L ¬L→( J ∨L) ¬L J T T T T F T F T F F F F T F F T F The truth table has no bad lines, so the argument is valid. Any valuation which makes every premise true makes the conclusion true. It’s probably worth illustrating the tactic again. Consider this argument: A ∨B,¬(B ∧C ) ∴(A ∨¬C ) Again, we start by evaluating the conclusion. Since this is a dis-junction, it is true whenever either disjunct is true, so we can speed things along a bit. A B C A ∨B ¬(B ∧C ) (A ∨¬C ) T T T T T T F T T F T T T F F T F T T ? ? F F F T F TT F F T ? ? F F F F F TT We can now ignore all but the two lines where the sentence after the turnstile is false. Evaluating the two sentences on the left of the turnstile, we get: CHAPTER 13. TRUTH TABLE SHORTCUTS 100 A B C A ∨B ¬(B ∧C ) (A ∨¬C ) T T T T T T F T T F T T T F F T F T T T F T F F F T F TT F F T F F F F F F TT So the entailment holds! And our shortcuts saved us a lot of work. We have been discussing shortcuts in testing for validity. But exactly the same shortcuts can be used in testing for entailment. By employing a similar notion of bad lines, you can save yourself a huge amount of work. Practice exercises A. Using shortcuts, check whether each sentence is a tautology, a contradiction, or neither. 1. ¬B ∧B 2. ¬D ∨D 3. (A ∧B) ∨(B ∧A) 4. ¬[A →(B →A)] 5. A ↔[A →(B ∧¬B)] 6. ¬(A ∧B) ↔A 7. A →(B ∨C ) 8. (A ∧¬A) →(B ∨C ) 9. (B ∧D) ↔[A ↔(A ∨C )] CHAPTER 14 Partial truth tables Sometimes, we do not need to know what happens on every line of a truth table. Sometimes, just a line or two will do. Tautology. In order to show that a sentence is a tautology, we need to show that it is true on every valuation. That is to say, we need to know that it comes out true on every line of the truth table. So we need a complete truth table. To show that a sentence is not a tautology, however, we only need one line: a line on which the sentence is false. Therefore, in order to show that some sentence is not a tautology, it is enough to provide a single valuation—a single line of the truth table— which makes the sentence false. Suppose that we want to show that the sentence ‘(U ∧T ) → (S ∧W )’ is not a tautology. We set up a partial truth table: S T U W (U ∧T ) →(S ∧W ) F We have only left space for one line, rather than 16, since we are only looking for one line, on which the sentence is false (hence, also, the ‘F’). 101 CHAPTER 14. PARTIAL TRUTH TABLES 102 The main logical operator of the sentence is a conditional. In order for the conditional to be false, the antecedent must be true and the consequent must be false. So we fill these in on the table: S T U W (U ∧T ) →(S ∧W ) T F F In order for the ‘(U ∧T )’ to be true, both ‘U ’ and ‘T ’ must be true. S T U W (U ∧T ) →(S ∧W ) T T T T T F F Now we just need to make ‘(S ∧W )’ false. To do this, we need to make at least one of ‘S’ and ‘W ’ false. We can make both ‘S’ and ‘W ’ false if we want. All that matters is that the whole sentence turns out false on this line. Making an arbitrary decision, we finish the table in this way: S T U W (U ∧T ) →(S ∧W ) F T T F T T T F F F F We now have a partial truth table, which shows that ‘(U ∧T ) → (S ∧W )’ is not a tautology. Put otherwise, we have shown that there is a valuation which makes ‘(U ∧T ) →(S ∧W )’ false, namely, the valuation which makes ‘S’ false, ‘T ’ true, ‘U ’ true and ‘W ’ false. Contradictions. Showing that something is a contradiction in TFL requires a complete truth table: we need to show that there is no valuation which makes the sentence true; that is, we need to show that the sentence is false on every line of the truth table. However, to show that something is not a contradiction, all we need to do is find a valuation which makes the sentence true, and a single line of a truth table will suffice. We can illustrate this with the same example. S T U W (U ∧T ) →(S ∧W ) T CHAPTER 14. PARTIAL TRUTH TABLES 103 To make the sentence true, it will suffice to ensure that the an-tecedent is false. Since the antecedent is a conjunction, we can just make one of them false. Making an arbitrary choice, let’s make ‘U ’ false; we can then assign any truth value we like to the other sentence letters. S T U W (U ∧T ) →(S ∧W ) F T F F F F T T F F F Equivalence. To show that two sentences are equivalent, we must show that the sentences have the same truth value on every valuation. So this requires a complete truth table. To show that two sentences are not equivalent, we only need to show that there is a valuation on which they have different truth values. So this requires only a one-line partial truth table: make the table so that one sentence is true and the other false. Joint satisfiability. To show that some sentences are jointly satisfiable, we must show that there is a valuation which makes all of the sentences true, so this requires only a partial truth table with a single line. To show that some sentences are jointly unsatisfiable, we must show that there is no valuation which makes all of the sentence true. So this requires a complete truth table: You must show that on every row of the table at least one of the sentences is false. Validity and entailment. To show that an argument is valid, we must show that there is no valuation which makes all of the premises true and the conclusion false. So this requires a com-plete truth table. (Likewise for entailment.) To show that argument is invalid, we must show that there is a valuation which makes all of the premises true and the conclusion false. So this requires only a one-line partial truth table on which all of the premises are true and the conclusion is false. (Likewise for a failure of entailment.) This table summarizes what is required: CHAPTER 14. PARTIAL TRUTH TABLES 104 Y es No tautology? complete one-line partial contradiction? complete one-line partial equivalent? complete one-line partial satisfiable? one-line partial complete valid? complete one-line partial entailment? complete one-line partial Practice exercises A. Use complete or partial truth tables (as appropriate) to deter-mine whether these pairs of sentences are logically equivalent: 1. A, ¬A 2. A, A ∨A 3. A →A, A ↔A 4. A ∨¬B, A →B 5. A ∧¬A, ¬B ↔B 6. ¬(A ∧B), ¬A ∨¬B 7. ¬(A →B), ¬A →¬B 8. (A →B), (¬B →¬A) B. Use complete or partial truth tables (as appropriate) to de-termine whether these sentences are jointly satisfiable, or jointly unsatisfiable: 1. A ∧B, C →¬B, C 2. A →B, B →C , A, ¬C 3. A ∨B, B ∨C , C →¬A 4. A, B, C , ¬D, ¬E, F 5. A ∧(B ∨C ), ¬(A ∧C ), ¬(B ∧C ) 6. A →B, B →C , ¬(A →C ) C. Use complete or partial truth tables (as appropriate) to deter-mine whether each argument is valid or invalid: 1. A ∨ [︁ A →(A ↔A) ]︁ ∴A CHAPTER 14. PARTIAL TRUTH TABLES 105 2. A ↔¬(B ↔A) ∴A 3. A →B,B ∴A 4. A ∨B,B ∨C,¬B ∴A ∧C 5. A ↔B,B ↔C ∴A ↔C D. Determine whether each sentence is a tautology, a contradic-tion, or a contingent sentence. Justify your answer with a com-plete or partial truth table as appropriate. 1. A →¬A 2. A →(A ∧(A ∨B)) 3. (A →B) ↔(B →A) 4. A →¬(A ∧(A ∨B)) 5. ¬B →[(¬A ∧A) ∨B] 6. ¬(A ∨B) ↔(¬A ∧¬B) 7. [(A ∧B) ∧C ] →B 8. ¬ [︁ (C ∨A) ∨B ]︁ 9. [︁ (A ∧B) ∧¬(A ∧B) ]︁ ∧C 10. (A ∧B)] →[(A ∧C ) ∨(B ∧D)] E. Determine whether each sentence is a tautology, a contradic-tion, or a contingent sentence. Justify your answer with a com-plete or partial truth table as appropriate. 1. ¬(A ∨A) 2. (A →B) ∨(B →A) 3. [(A →B) →A] →A 4. ¬[(A →B) ∨(B →A)] 5. (A ∧B) ∨(A ∨B) 6. ¬(A ∧B) ↔A 7. A →(B ∨C ) 8. (A ∧¬A) →(B ∨C ) 9. (B ∧D) ↔[A ↔(A ∨C )] CHAPTER 14. PARTIAL TRUTH TABLES 106 10. ¬[(A →B) ∨(C →D)] F. Determine whether each the following pairs of sentences are logically equivalent using complete truth tables. Justify your an-swer with a complete or partial truth table as appropriate. 1. A and A ∨A 2. A and A ∧A 3. A ∨¬B and A →B 4. (A →B) and (¬B →¬A) 5. ¬(A ∧B) and ¬A ∨¬B 6. ((U →(X ∨X )) ∨U ) and ¬(X ∧(X ∧U )) 7. ((C ∧(N ↔C )) ↔C ) and (¬¬¬N →C ) 8. [(A ∨B) ∧C ] and [A ∨(B ∧C )] 9. ((L ∧C ) ∧I ) and L ∨C G. Determine whether each collection of sentences is jointly satis-fiable or jointly unsatisfiable. Justify your answer with a complete or partial truth table as appropriate. 1. A →A, ¬A →¬A, A ∧A, A ∨A 2. A →¬A, ¬A →A 3. A ∨B, A →C , B →C 4. A ∨B, A →C , B →C , ¬C 5. B ∧(C ∨A), A →B, ¬(B ∨C ) 6. (A ↔B) →B, B →¬(A ↔B), A ∨B 7. A ↔(B ∨C ), C →¬A, A →¬B 8. A ↔B, ¬B ∨¬A, A →B 9. A ↔B, A →C , B →D, ¬(C ∨D) 10. ¬(A ∧¬B), B →¬A, ¬B H. Determine whether each argument is valid or invalid. Justify your answer with a complete or partial truth table as appropriate. 1. A →(A ∧¬A) ∴¬A 2. A ∨B, A →B, B →A ∴A ↔B 3. A ∨(B →A) ∴¬A →¬B CHAPTER 14. PARTIAL TRUTH TABLES 107 4. A ∨B, A →B, B →A ∴A ∧B 5. (B ∧A) →C , (C ∧A) →B ∴(C ∧B) →A 6. ¬(¬A ∨¬B), A →¬C ∴A →(B →C ) 7. A ∧(B →C ), ¬C ∧(¬B →¬A) ∴C ∧¬C 8. A ∧B, ¬A →¬C , B →¬D ∴A ∨B 9. A →B ∴(A ∧B) ∨(¬A ∧¬B) 10. ¬A →B,¬B →C ,¬C →A ∴¬A →(¬B ∨¬C ) I. Determine whether each argument is valid or invalid. Justify your answer with a complete or partial truth table as appropriate. 1. A ↔¬(B ↔A) ∴A 2. A ∨B, B ∨C , ¬A ∴B ∧C 3. A →C , E →(D ∨B), B →¬D ∴(A ∨C ) ∨(B →(E ∧D)) 4. A ∨B, C →A, C →B ∴A →(B →C ) 5. A →B, ¬B ∨A ∴A ↔B PART IV Natural deduction for TFL 108 CHAPTER 15 The very idea of natural deduction Way back in §2, we said that an argument is valid iff there is no case in which all of the premises are true and the conclusion is false. In the case of TFL, this led us to develop truth tables. Each line of a complete truth table corresponds to a valuation. So, when faced with a TFL argument, we have a very direct way to assess whether there is a valuation on which the premises are true and the conclusion is false: just thrash through the truth table. However, truth tables may not give us much insight. Consider two arguments in TFL: P ∨Q,¬P ∴Q P →Q,P ∴Q Clearly, these are valid arguments. You can confirm that they are valid by constructing four-line truth tables, but we might say that they make use of different forms of reasoning. It might be nice to keep track of these different forms of inference. 109 CHAPTER 15. THE VERY IDEA OF NATURAL DEDUCTION 110 One aim of a natural deduction system is to show that particular arguments are valid, in a way that allows us to understand the reasoning that the arguments might involve. We begin with very basic rules of inference. These rules can be combined to offer more complicated arguments. Indeed, with just a small starter pack of rules of inference, we hope to capture all valid arguments. This is a very different way of thinking about arguments. With truth tables, we directly consider different ways to make sentences true or false. With natural deduction systems, we ma-nipulate sentences in accordance with rules that we have set down as good rules. The latter promises to give us a better insight—or at least, a different insight—into how arguments work. The move to natural deduction might be motivated by more than the search for insight. It might also be motivated by necessity. Consider: A1 →C1 ∴(A1 ∧(A2 ∧(A3 ∧(A4 ∧A5)))) → ((((C1 ∨C2) ∨C3) ∨C4) ∨C5) To test this argument for validity, you might use a 1,024-line truth table. If you do it correctly, then you will see that there is no line on which all the premises are true and on which the conclusion is false. So you will know that the argument is valid. (But, as just mentioned, there is a sense in which you will not know why the argument is valid.) But now consider: A1 →C1 ∴(A1 ∧(A2 ∧(A3 ∧(A4 ∧(A5 ∧ (A6 ∧(A7 ∧(A8 ∧(A9 ∧A10))))))))) → (((((((((C1 ∨C2) ∨C3) ∨C4) ∨C5) ∨ C6) ∨C7) ∨C8) ∨C9) ∨C10) This argument is also valid—as you can probably tell—but to test it requires a truth table with 220 = 1,048,576 lines. In principle, we can set a machine to grind through truth tables and report CHAPTER 15. THE VERY IDEA OF NATURAL DEDUCTION 111 back when it is finished. In practice, complicated arguments in TFL can become intractable if we use truth tables. When we get to first-order logic (FOL) (beginning in chapter 22), though, the problem gets dramatically worse. There is noth-ing like the truth table test for FOL. To assess whether or not an argument is valid, we have to reason about all interpretations, but, as we will see, there are infinitely many possible interpreta-tions. We cannot even in principle set a machine to grind through infinitely many possible interpretations and report back when it is finished: it will never finish. We either need to come up with some more efficient way of reasoning about all interpretations, or we need to look for something different. There are, indeed, systems that codify ways to reason about all possible interpretations. They were developed in the 1950s by Evert Beth and Jaakko Hintikka, but we will not follow this path. We will, instead, look to natural deduction. Rather than reasoning directly about all valuations (in the case of TFL), we will try to select a few basic rules of inference. Some of these will govern the behaviour of the sentential con-nectives. Others will govern the behaviour of the quantifiers and identity that are the hallmarks of FOL. The resulting system of rules will give us a new way to think about the validity of ar-guments. The modern development of natural deduction dates from simultaneous and unrelated papers by Gerhard Gentzen and Stanisław Jaśkowski (1934). However, the natural deduction sys-tem that we will consider is based largely around work by Frederic Fitch (first published in 1952). CHAPTER 16 Basic rules for TFL We will develop a natural deduction system. For each con-nective, there will be introduction rules, that allow us to prove a sentence that has that connective as the main logical operator, and elimination rules, that allow us to prove something given a sentence that has that connective as the main logical operator. 16.1 The idea of a formal proof A formal proof or derivation is a sequence of sentences, some of which are marked as being initial assumptions (or premises). The last line of the formal proof is the conclusion. (Henceforth, we will simply call these ‘proofs’ or ‘derivations’, but be aware that there are informal proofs too.) As an illustration, consider: ¬(A ∨B) ∴¬A ∧¬B We will start a proof by writing the premise: 1 ¬(A ∨B) 112 CHAPTER 16. BASIC RULES FOR TFL 113 Note that we have numbered the premise, since we will want to refer back to it. Indeed, every line of the proof is numbered, so that we can refer back to it. Note also that we have drawn a line underneath the premise. Everything written above the line is an assumption. Everything written below the line will either be something which follows from the assumptions, or it will be some new assumption. We are hoping to conclude ‘¬A ∧¬B’; so we are hoping ultimately to conclude our proof with n ¬A ∧¬B for some number n. It doesn’t matter what line number we end on, but we would obviously prefer a short proof to a long one. Similarly, suppose we wanted to consider: A ∨B,¬(A ∧C ),¬(B ∧¬D) ∴¬C ∨D The argument has three premises, so we start by writing them all down, numbered, and drawing a line under them: 1 A ∨B 2 ¬(A ∧C ) 3 ¬(B ∧¬D) and we are hoping to conclude with some line: n ¬C ∨D All that remains to do is to explain each of the rules that we can use along the way from premises to conclusion. The rules are broken down by our logical connectives. CHAPTER 16. BASIC RULES FOR TFL 114 16.2 Reiteration The very first rule is so breathtakingly obvious that it is surprising we bother with it at all. If you already have shown something in the course of a proof, the reiteration rule allows you to repeat it on a new line. For example: 4 A ∧B . . . . . . 10 A ∧B R 4 This indicates that we have written ‘A∧B’ on line 4. Now, at some later line—line 10, for example—we have decided that we want to repeat this. So we write it down again. We also add a citation which justifies what we have written. In this case, we write ‘R’, to indicate that we are using the reiteration rule, and we write ‘4’, to indicate that we have applied it to line 4. Here is a general expression of the rule: m A A R m The point is that, if any sentence A occurs on some line, then we can repeat A on later lines. Each line of our proof must be justified by some rule, and here we have ‘R m’. This means: Reiteration, applied to line m. T wo things need emphasizing. First ‘A’ is not a sentence of TFL. Rather, it a symbol in the metalanguage, which we use when we want to talk about any sentence of TFL (see §8). Second, and similarly, ‘m’ is not a symbol that will appear on a proof. Rather, it is a symbol in the metalanguage, which we use when we want to talk about any line number of a proof. In an actual proof, the CHAPTER 16. BASIC RULES FOR TFL 115 lines are numbered ‘1’, ‘2’, ‘3’, and so forth. But when we define the rule, we use variables like ‘m’ to underscore the point that the rule may be applied at any point. 16.3 Conjunction Suppose we want to show that Ludwig is both reactionary and libertarian. One obvious way to do this would be as follows: first we show that Ludwig is reactionary; then we show that Ludwig is libertarian; then we put these two demonstrations together, to obtain the conjunction. Our natural deduction system will capture this thought straightforwardly. In the example given, we might adopt the fol-lowing symbolization key: R: Ludwig is reactionary L: Ludwig is libertarian Perhaps we are working through a proof, and we have obtained ‘R’ on line 8 and ‘L’ on line 15. Then on any subsequent line we can obtain ‘R ∧L’ thus: 8 R 15 L R ∧L ∧I 8, 15 Note that every line of our proof must either be an assumption, or must be justified by some rule. We cite ‘∧I 8, 15’ here to indicate that the line is obtained by the rule of conjunction introduction (∧I) applied to lines 8 and 15. We could equally well obtain: 8 R 15 L L ∧R ∧I 15, 8 CHAPTER 16. BASIC RULES FOR TFL 116 with the citation reversed, to reflect the order of the conjuncts. More generally, here is our conjunction introduction rule: m A n B A∧B ∧I m, n To be clear, the statement of the rule is schematic. It is not itself a proof. ‘A’ and ‘B’ are not sentences of TFL. Rather, they are symbols in the metalanguage, which we use when we want to talk about any sentence of TFL (see §8). Similarly, ‘m’ and ‘n’ are not numerals that will appear in any actual proof. Rather, they are symbols in the metalanguage, which we use when we want to talk about any line number of any proof. In an actual proof, the lines are numbered ‘1’, ‘2’, ‘3’, and so forth, but when we define the rule, we use variables to emphasize that the rule may be applied at any point. The rule requires only that we have both conjuncts available to us somewhere earlier in the proof. They can be separated from one another, and they can appear in any order. The rule is called ‘conjunction introduction’ because it intro-duces the symbol ‘∧’ into our proof where it may have been ab-sent. Correspondingly, we have a rule that eliminates that sym-bol. Suppose you have shown that Ludwig is both reactionary and libertarian. You are entitled to conclude that Ludwig is re-actionary. Equally, you are entitled to conclude that Ludwig is libertarian. Putting this together, we obtain our conjunction elim-ination rule(s): m A∧B A ∧E m CHAPTER 16. BASIC RULES FOR TFL 117 and equally: m A∧B B ∧E m The point is simply that, when you have a conjunction on some line of a proof, you can obtain either of the conjuncts by ∧E. One point is worth emphasising: you can only apply this rule when conjunction is the main logical operator. So you cannot infer ‘D’ just from ‘C ∨(D ∧E)’! Even with just these two rules, we can start to see some of the power of our formal proof system. Consider: [(A ∨B) →(C ∨D)] ∧[(E ∨F ) →(G ∨H )] ∴[(E ∨F ) →(G ∨H )] ∧[(A ∨B) →(C ∨D)] The main logical operator in both the premise and conclusion of this argument is ‘∧’. In order to provide a proof, we begin by writing down the premise, which is our assumption. We draw a line below this: everything after this line must follow from our assumptions by (repeated applications of) our rules of inference. So the beginning of the proof looks like this: 1 [(A ∨B) →(C ∨D)] ∧[(E ∨F ) →(G ∨H )] From the premise, we can get each of the conjuncts by ∧E. The proof now looks like this: 1 [(A ∨B) →(C ∨D)] ∧[(E ∨F ) →(G ∨H )] 2 [(A ∨B) →(C ∨D)] ∧E 1 3 [(E ∨F ) →(G ∨H )] ∧E 1 So by applying the ∧I rule to lines 3 and 2 (in that order), we arrive at the desired conclusion. The finished proof looks like this: CHAPTER 16. BASIC RULES FOR TFL 118 1 [(A ∨B) →(C ∨D)] ∧[(E ∨F ) →(G ∨H )] 2 [(A ∨B) →(C ∨D)] ∧E 1 3 [(E ∨F ) →(G ∨H )] ∧E 1 4 [(E ∨F ) →(G ∨H )] ∧[(A ∨B) →(C ∨D)] ∧I 3, 2 This is a very simple proof, but it shows how we can chain rules of proof together into longer proofs. In passing, note that investi-gating this argument with a truth table would have required 256 lines; our formal proof required only four lines. It is worth giving another example. Back in §11.3, we noted that this argument is valid: A ∧(B ∧C ) ∴(A ∧B) ∧C To provide a proof corresponding to this argument, we start by writing: 1 A ∧(B ∧C ) From the premise, we can get each of the conjuncts by applying ∧E twice. We can then apply ∧E twice more, so our proof looks like: 1 A ∧(B ∧C ) 2 A ∧E 1 3 B ∧C ∧E 1 4 B ∧E 3 5 C ∧E 3 But now we can merrily reintroduce conjunctions in the order we wanted them, so that our final proof is: CHAPTER 16. BASIC RULES FOR TFL 119 1 A ∧(B ∧C ) 2 A ∧E 1 3 B ∧C ∧E 1 4 B ∧E 3 5 C ∧E 3 6 A ∧B ∧I 2, 4 7 (A ∧B) ∧C ∧I 6, 5 Recall that our official definition of sentences in TFL only allowed conjunctions with two conjuncts. The proof just given suggests that we could drop inner brackets in all of our proofs. However, this is not standard, and we will not do this. Instead, we will maintain our more austere bracketing conventions. (Though we will still allow ourselves to drop outermost brackets, for legibility.) Let’s give one final illustration. When using the ∧I rule, there is no requirement to apply it to different sentences. So, if we want, we can formally prove ‘A ∧A’ from ‘A’ thus: 1 A 2 A ∧A ∧I 1, 1 Simple, but effective. 16.4 Conditional Consider the following argument: If Jane is smart then she is fast. Jane is smart. ∴Jane is fast. CHAPTER 16. BASIC RULES FOR TFL 120 This argument is certainly valid, and it suggests a straightforward conditional elimination rule (→E): m A →B n A B →E m, n This rule is also sometimes called modus ponens. Again, this is an elimination rule, because it allows us to obtain a sentence that may not contain ‘→’, having started with a sentence that did contain ‘→’. Note that the conditional A →B and the an-tecedent A can be separated from one another in the proof, and they can appear in any order. However, in the citation for →E, we always cite the conditional first, followed by the antecedent. The rule for conditional introduction is also quite easy to motivate. The following argument should be valid: Ludwig is reactionary. Therefore if Ludwig is libertar-ian, then Ludwig is both reactionary and libertarian. If someone doubted that this was valid, we might try to convince them otherwise by explaining ourselves as follows: Assume that Ludwig is reactionary. Now, additionally assume that Ludwig is libertarian. Then by conjunc-tion introduction—which we just discussed—Ludwig is both reactionary and libertarian. Of course, that’s conditional on the assumption that Ludwig is libertar-ian. But this just means that, if Ludwig is libertarian, then he is both reactionary and libertarian. Transferred into natural deduction format, here is the pattern of reasoning that we just used. We started with one premise, ‘Ludwig is reactionary’, thus: CHAPTER 16. BASIC RULES FOR TFL 121 1 R The next thing we did is to make an additional assumption (‘Lud-wig is libertarian’), for the sake of argument. To indicate that we are no longer dealing merely with our original assumption (‘R’), but with some additional assumption, we continue our proof as follows: 1 R 2 L Note that we are not claiming, on line 2, to have proved ‘L’ from line 1, so we do not write in any justification for the additional assumption on line 2. We do, however, need to mark that it is an additional assumption. We do this by drawing a line under it (to indicate that it is an assumption) and by indenting it with a further vertical line (to indicate that it is additional). With this extra assumption in place, we are in a position to use ∧I. So we can continue our proof: 1 R 2 L 3 R ∧L ∧I 1, 2 So we have now shown that, on the additional assumption, ‘L’, we can obtain ‘R ∧L’. We can therefore conclude that, if ‘L’ obtains, then so does ‘R ∧L’. Or, to put it more briefly, we can conclude ‘L →(R ∧L)’: 1 R 2 L 3 R ∧L ∧I 1, 2 4 L →(R ∧L) →I 2–3 CHAPTER 16. BASIC RULES FOR TFL 122 Observe that we have dropped back to using one vertical line on the left. We have discharged the additional assumption, ‘L’, since the conditional itself follows just from our original assumption, ‘R’. The general pattern at work here is the following. We first make an additional assumption, A; and from that additional as-sumption, we prove B. In that case, we know the following: If A is true, then B is true. This is wrapped up in the rule for condi-tional introduction: i A j B A →B →I i–j There can be as many or as few lines as you like between lines i and j. It will help to offer a second illustration of →I in action. Sup-pose we want to consider the following: P →Q,Q →R ∴P →R We start by listing both of our premises. Then, since we want to arrive at a conditional (namely, ‘P →R’), we additionally assume the antecedent to that conditional. Thus our main proof starts: 1 P →Q 2 Q →R 3 P Note that we have made ‘P ’ available, by treating it as an addi-tional assumption, but now, we can use →E on the first premise. This will yield ‘Q’. We can then use →E on the second premise. CHAPTER 16. BASIC RULES FOR TFL 123 So, by assuming ‘P ’ we were able to prove ‘R’, so we apply the →I rule—discharging ‘P ’—and finish the proof. Putting all this together, we have: 1 P →Q 2 Q →R 3 P 4 Q →E 1, 3 5 R →E 2, 4 6 P →R →I 3–5 16.5 Additional assumptions and subproofs The rule →I invoked the idea of making additional assumptions. These need to be handled with some care. Consider this proof: 1 A 2 B 3 B R 2 4 B →B →I 2–3 This is perfectly in keeping with the rules we have laid down already, and it should not seem particularly strange. Since ‘B → B’ is a tautology, no particular premises should be required to prove it. But suppose we now tried to continue the proof as follows: CHAPTER 16. BASIC RULES FOR TFL 124 1 A 2 B 3 B R 2 4 B →B →I 2–3 5 B naughty attempt to invoke →E 4, 3 If we were allowed to do this, it would be a disaster. It would allow us to prove any sentence letter from any other sentence letter. However, if you tell me that Anne is fast (symbolized by ‘A’), we shouldn’t be able to conclude that Queen Boudica stood twenty-feet tall (symbolized by ‘B’)! We must be prohibited from doing this, but how are we to implement the prohibition? We can describe the process of making an additional assump-tion as one of performing a subproof : a subsidiary proof within the main proof. When we start a subproof, we draw another vertical line to indicate that we are no longer in the main proof. Then we write in the assumption upon which the subproof will be based. A subproof can be thought of as essentially posing this question: what could we show, if we also make this additional assumption? When we are working within the subproof, we can refer to the additional assumption that we made in introducing the subproof, and to anything that we obtained from our original assumptions. (After all, those original assumptions are still in effect.) At some point though, we will want to stop working with the additional assumption: we will want to return from the subproof to the main proof. To indicate that we have returned to the main proof, the vertical line for the subproof comes to an end. At this point, we say that the subproof is closed. Having closed a subproof, we have set aside the additional assumption, so it will be illegit-imate to draw upon anything that depends upon that additional assumption. Thus we stipulate: CHAPTER 16. BASIC RULES FOR TFL 125 To cite an individual line when applying a rule: 1. the line must come before the line where the rule is applied, but 2. not occur within a subproof that has been closed before the line where the rule is applied. This stipulation rules out the disastrous attempted proof above. The application of rule →E on line 5 requires that we cite two individual lines from earlier in the proof. One of these lines (namely, line 3) occurs within a subproof (lines 2–3). By line 5, where we have to cite line 3, the subproof has been closed. This is illegitimate: we are not allowed to cite line 3 on line 5. Closing a subproof is called discharging the assumptions of that subproof. So we can put the point this way: you cannot refer back to anything that was obtained using discharged assumptions. Subproofs, then, allow us to think about what we could show, if we made additional assumptions. The point to take away from this is not surprising—in the course of a proof, we have to keep very careful track of what assumptions we are making, at any given moment. Our proof system does this very graphically. (In-deed, that’s precisely why we have chosen to use this proof sys-tem.) Once we have started thinking about what we can show by making additional assumptions, nothing stops us from posing the question of what we could show if we were to make even more assumptions? This might motivate us to introduce a subproof within a subproof. Here is an example using only the rules which we have considered so far: CHAPTER 16. BASIC RULES FOR TFL 126 1 A 2 B 3 C 4 A ∧B ∧I 1, 2 5 C →(A ∧B) →I 3–4 6 B →(C →(A ∧B)) →I 2–5 Notice that the citation on line 4 refers back to the initial assump-tion (on line 1) and an assumption of a subproof (on line 2). This is perfectly in order, since neither assumption has been dis-charged at the time (i.e., by line 4). Again, though, we need to keep careful track of what we are assuming at any given moment. Suppose we tried to continue the proof as follows: 1 A 2 B 3 C 4 A ∧B ∧I 1, 2 5 C →(A ∧B) →I 3–4 6 B →(C →(A ∧B)) →I 2–5 7 C →(A ∧B) naughty attempt to invoke →I 3–4 This would be awful. If we tell you that Anne is smart, you should not be able to infer that, if Cath is smart (symbolized by ‘C ’) then both Anne is smart and Queen Boudica stood 20-feet tall! But this is just what such a proof would suggest, if it were permissible. CHAPTER 16. BASIC RULES FOR TFL 127 The essential problem is that the subproof that began with the assumption ‘C ’ depended crucially on the fact that we had assumed ‘B’ on line 2. By line 6, we have discharged the assump-tion ‘B’: we have stopped asking ourselves what we could show, if we also assumed ‘B’. So it is simply cheating, to try to help ourselves (on line 7) to the subproof that began with the assump-tion ‘C ’. Thus we stipulate, much as before, that a subproof can only be cited on a line if it does not occur within some other subproof which is already closed at that line. The attempted dis-astrous proof violates this stipulation. The subproof of lines 3–4 occurs within a subproof that ends on line 5. So it cannot be invoked on line 7. Here is one further case we have to exclude: 1 A 2 B 3 C 4 B ∧C ∧I 2, 3 5 C ∧E 4 6 B →C naughty attempt to invoke →I 2–5 Here we are trying to cite a subproof that begins on line 2 and ends on line 5—but the sentence on line 5 depends not only on the assumption on line 2, but also on one another assumption (line 3) which we have not discharged at the end of the subproof. The subproof started on line 3 is still open at line 3. But →I re-quires that the last line of the subproof only relies on the assump-tion of the subproof being cited, i.e., the subproof beginning on line 2 (and anything before it), and not on assumptions of any subproofs within it. In particular, the last line of the subproof cited must not itself lie within a nested subproof. CHAPTER 16. BASIC RULES FOR TFL 128 To cite a subproof when applying a rule: 1. the cited subproof must come entirely before the ap-plication of the rule where it is cited, 2. the cited subproof must not lie within some other closed subproof which is closed at the line it is cited, and 3. the last line of the cited subproof must not occur inside a nested subproof. One last point to emphasize how rules can be applied: where a rule requires you to cite an individual line, you cannot cite a subproof instead; and where it requires you to cite a subproof, you cannot cite an individual line instead. So for instance, this is incorrect: 1 A 2 B 3 C 4 B ∧C ∧I 2, 3 5 C ∧E 4 6 C naughty attempt to invoke R 3–5 7 B →C →I 2–6 Here, we have tried to justify C on line 6 by the reiteration rule, but we have cited the subproof on lines 3–5 with it. That subproof is closed and can in principle be cited on line 6. (For instance, we could use it to justify C →C by →I.) But the reiteration rule R re-quires you to cite an individual line, so citing the entire subproof CHAPTER 16. BASIC RULES FOR TFL 129 is inadmissible (even if that subproof contains the sentence C we want to reiterate). It is always permissible to open a subproof with any assump-tion. However, there is some strategy involved in picking a useful assumption. Starting a subproof with an arbitrary, wacky assump-tion would just waste lines of the proof. In order to obtain a con-ditional by →I, for instance, you must assume the antecedent of the conditional in a subproof. Equally, it is always permissible to close a subproof (and dis-charge its assumptions). However, it will not be helpful to do so until you have reached something useful. Once the subproof is closed, you can only cite the entire subproof in any justification. Those rules that call for a subproof or subproofs, in turn, require that the last line of the subproof is a sentence of some form or other. For instance, you are only allowed to cite a subproof for →I if the line you are justifying is of the form A →B, A is the assumption of your subproof, and B is the last line of your subproof. 16.6 Biconditional The rules for the biconditional will be like double-barrelled ver-sions of the rules for the conditional. In order to prove ‘F ↔G’, for instance, you must be able to prove ‘G’ on the assumption ‘F ’ and prove ‘F ’ on the assumption ‘G’. The biconditional introduction rule (↔I) therefore requires two subproofs. Schematically, the rule works like this: CHAPTER 16. BASIC RULES FOR TFL 130 i A j B k B l A A ↔B ↔I i–j, k–l There can be as many lines as you like between i and j, and as many lines as you like between k and l. Moreover, the subproofs can come in any order, and the second subproof does not need to come immediately after the first. The biconditional elimination rule (↔E) lets you do a bit more than the conditional rule. If you have the left-hand subsen-tence of the biconditional, you can obtain the right-hand subsen-tence. If you have the right-hand subsentence, you can obtain the left-hand subsentence. So we allow: m A ↔B n A B ↔E m, n and equally: m A ↔B n B A ↔E m, n Note that the biconditional, and the right or left half, can be separated from one another, and they can appear in any order. CHAPTER 16. BASIC RULES FOR TFL 131 However, in the citation for ↔E, we always cite the biconditional first. 16.7 Disjunction Suppose Ludwig is reactionary. Then Ludwig is either reac-tionary or libertarian. After all, to say that Ludwig is either reac-tionary or libertarian is to say something weaker than to say that Ludwig is reactionary. Let’s emphasize this point. Suppose Ludwig is reactionary. It follows that Ludwig is either reactionary or a kumquat. Equally, it follows that either Ludwig is reactionary or kumquats are the only fruit. Equally, it follows that either Ludwig is reactionary or God is dead. Many of these are strange inferences to draw, but there is nothing logically wrong with them (even if they maybe violate all sorts of implicit conversational norms). Armed with all this, we present the disjunction introduction rule(s): m A A∨B ∨I m and m A B∨A ∨I m Notice that B can be any sentence whatsoever, so the follow-ing is a perfectly acceptable proof: 1 M 2 M ∨([(A ↔B) →(C ∧D)] ↔[E ∧F ]) ∨I 1 CHAPTER 16. BASIC RULES FOR TFL 132 Using a truth table to show this would have taken 128 lines. The disjunction elimination rule is slightly trickier. Suppose that either Ludwig is reactionary or he is libertarian. What can you conclude? Not that Ludwig is reactionary; it might be that he is libertarian instead. Equally, not that Ludwig is libertarian; for he might merely be reactionary. Disjunctions, just by themselves, are hard to work with. But suppose that we could somehow show both of the fol-lowing: first, that Ludwig’s being reactionary entails that he is an Austrian economist; second, that Ludwig’s being libertarian entails that he is an Austrian economist. Then if we know that Ludwig is either reactionary or libertarian, then we know that, whichever he is, Ludwig is an Austrian economist. This insight can be expressed in the following rule, which is our disjunction elimination (∨E) rule: m A∨B i A j C k B l C C ∨E m, i–j, k–l This is obviously a bit clunkier to write down than our previ-ous rules, but the point is fairly simple. Suppose we have some disjunction, A∨B. Suppose we have two subproofs, showing us that C follows from the assumption that A, and that C follows from the assumption that B. Then we can infer C itself. As usual, there can be as many lines as you like between i and j, and as many lines as you like between k and l. Moreover, the subproofs and the disjunction can come in any order, and do not have to be adjacent. CHAPTER 16. BASIC RULES FOR TFL 133 Some examples might help illustrate this. Consider this argu-ment: (P ∧Q) ∨(P ∧R) ∴P An example proof might run thus: 1 (P ∧Q) ∨(P ∧R) 2 P ∧Q 3 P ∧E 2 4 P ∧R 5 P ∧E 4 6 P ∨E 1, 2–3, 4–5 Here is a slightly harder example. Consider: A ∧(B ∨C ) ∴(A ∧B) ∨(A ∧C ) Here is a proof corresponding to this argument: 1 A ∧(B ∨C ) 2 A ∧E 1 3 B ∨C ∧E 1 4 B 5 A ∧B ∧I 2, 4 6 (A ∧B) ∨(A ∧C ) ∨I 5 7 C 8 A ∧C ∧I 2, 7 9 (A ∧B) ∨(A ∧C ) ∨I 8 10 (A ∧B) ∨(A ∧C ) ∨E 3, 4–6, 7–9 CHAPTER 16. BASIC RULES FOR TFL 134 Don’t be alarmed if you think that you wouldn’t have been able to come up with this proof yourself. The ability to come up with novel proofs comes with practice, and we’ll cover some strategies for finding proofs in §17. The key question at this stage is whether, looking at the proof, you can see that it conforms to the rules that we have laid down. That just involves checking every line, and making sure that it is justified in accordance with the rules we have laid down. 16.8 Contradiction and negation We have only one connective left to deal with: negation. But to tackle it, we must connect negation with contradiction. An effective form of argument is to argue your opponent into contradicting themselves. At that point, you have them on the ropes. They have to give up at least one of their assumptions. We are going to make use of this idea in our proof system, by adding a new symbol, ‘⊥’, to our proofs. This should be read as something like ‘contradiction!’ or ‘reductio!’ or ‘but that’s ab-surd!’ The rule for introducing this symbol is that we can use it whenever we explicitly contradict ourselves, i.e., whenever we find both a sentence and its negation appearing in our proof: m ¬A n A ⊥ ¬E m, n It does not matter what order the sentence and its negation appear in, and they do not need to appear on adjacent lines. However, we always cite the line number of the negation first, followed by that of the sentence it is a negation of. There is obviously a tight link between contradiction and negation. The rule ¬E lets us proceed from two contradictory CHAPTER 16. BASIC RULES FOR TFL 135 sentences—Aand its negation ¬A—to an explicit contradition ⊥. We choose the label for a reason: it is the the most basic rule that lets us proceed from a premise containing a negation, i.e., ¬A, to a sentence not containing it, i.e., ⊥. So it is a rule that elimi-nates ¬. We have said that ‘⊥’ should be read as something like ‘con-tradiction!’ but this does not tell us much about the symbol. There are, roughly, three ways to approach the symbol. • We might regard ‘⊥’ as a new atomic sentence of TFL, but one which can only ever have the truth value False. • We might regard ‘⊥’ as an abbreviation for some canonical contradiction, such as ‘A ∧¬A’. This will have the same effect as the above—obviously, ‘A ∧¬A’ only ever has the truth value False—but it means that, officially, we do not need to add a new symbol to TFL. • We might regard ‘⊥’, not as a symbol of TFL, but as some-thing more like a punctuation mark that appears in our proofs. (It is on a par with the line numbers and the vertical lines, say.) There is something very philosophically attractive about the third option, but here we will officially adopt the first. ‘⊥’ is to be read as a sentence letter that is always false. This means that we can manipulate it, in our proofs, just like any other sentence. We still have to state a rule for negation introduction. The rule is very simple: if assuming something leads you to a con-tradiction, then the assumption must be wrong. This thought motivates the following rule: i A j ⊥ ¬A ¬I i–j CHAPTER 16. BASIC RULES FOR TFL 136 There can be as many lines between i and j as you like. To see this in practice, and interacting with negation, consider this proof: 1 D 2 ¬D 3 ⊥ ¬E 2, 1 4 ¬¬D ¬I 2–3 If the assumption that A is true leads to a contradiction, A cannot be true, i.e., it must be false, i.e., ¬A must be true. Of course, if the assumption that Ais false (i.e., the assumption that ¬A is true) leads to a contradiction, then A cannot be false, i.e., A must be true. So we can consider the following rule: i ¬A j ⊥ A IP i–j This rule is called indirect proof, since it allows us to prove A indirectly, by assuming its negation. Formally, the rule is very similar to ¬I, but A and ¬A have changed places. Since ¬A is not the conclusion of the rule, we are not introducing ¬, so IP is not a rule that introduces any connective. It also doesn’t elim-inate a connective, since it has no free-standing premises which contain ¬, only a subproof with an assumption of the form ¬A. By contrast, ¬E does have a premise of the form ¬A: that’s why ¬E eliminates ¬, but IP does not.1 1There are logicians who have qualms about IP, but not about ¬E. They are called “intuitionists.” Intuitionists don’t buy our basic assumption that every sentence has one of two truth values, true or false. They also think that ¬ works differently—for them, a proof of ⊥from A guarantees ¬A, but a proof CHAPTER 16. BASIC RULES FOR TFL 137 Using ¬I, we were able to give a proof of ¬¬D from D. Us-ing IP, we can go the other direction (with essentially the same proof). 1 ¬¬D 2 ¬D 3 ⊥ ¬E 1, 2 4 D IP 2–3 We need one last rule. It is a kind of elimination rule for ‘⊥’, and known as explosion.2 If we obtain a contradiction, symbol-ized by ‘⊥’, then we can infer whatever we like. How can this be motivated, as a rule of argumentation? Well, consider the En-glish rhetorical device ‘. . . and if that’s true, I’ll eat my hat’. Since contradictions simply cannot be true, if one is true then not only will I eat my hat, I’ll have it too. Here is the formal rule: m ⊥ A X m Note that A can be any sentence whatsoever. The explosion rule is a bit odd. It looks like A arrives in our proof like a bunny out of a hat. When trying to find proofs, it is very tempting to try to use it everywhere, since it seems so powerful. Resist this temptation: you can only apply it when you already have ⊥! And you get ⊥only when your assumptions are contradictory. Still, isn’t it odd that from a contradiction anything whatso-ever should follow? Not according to our notion of entailment of ⊥from ¬A does not guarantee that A, but only ¬¬A. So, for them, A and ¬¬A are not equivalent. 2The Latin name for this principle is ex contradictione quodlibet, “from con-tradiction, anything.” CHAPTER 16. BASIC RULES FOR TFL 138 and validity. For A entails B iff there is no valuation of the sen-tence letters which makes A true and B false at the same time. Now ⊥is a contradiction—it is never true, whatever the valuation of the sentence letters. Since there is no valuation which makes ⊥ true, there of course is also no valuation that makes ⊥true and B false! So according to our definition of entailment, ⊥⊨B, whatever B is. A contradiction entails anything.3 These are all of the basic rules for the proof system for TFL. Practice exercises A. The following two ‘proofs’ are incorrect. Explain the mistakes they make. 1 (¬L ∧A) ∨L 2 ¬L ∧A 3 ¬L ∧E 3 4 A ∧E 1 5 L 6 ⊥ ¬E 3, 5 7 A X 6 8 A ∨E 1, 2–4, 5–7 3There are some logicians who don’t buy this. They think that if A entails B, there must be some relevant connection between A and B—and there isn’t one between ⊥and some arbitrary sentence B. So these logicians develop other, “relevant” logics in which you aren’t allowed the explosion rule. CHAPTER 16. BASIC RULES FOR TFL 139 1 A ∧(B ∧C ) 2 (B ∨C ) →D 3 B ∧E 1 4 B ∨C ∨I 3 5 D →E 4, 2 B. The following three proofs are missing their citations (rule and line numbers). Add them, to turn them into bona fide proofs. Additionally, write down the argument that corresponds to each proof. 1 P ∧S 2 S →R 3 P 4 S 5 R 6 R ∨E 1 A →D 2 A ∧B 3 A 4 D 5 D ∨E 6 (A ∧B) →(D ∨E) 1 ¬L →( J ∨L) 2 ¬L 3 J ∨L 4 J 5 J ∧J 6 J 7 L 8 ⊥ 9 J 10 J C. Give a proof for each of the following arguments: 1. J →¬ J ∴¬ J CHAPTER 16. BASIC RULES FOR TFL 140 2. Q →(Q ∧¬Q) ∴¬Q 3. A →(B →C ) ∴(A ∧B) →C 4. K ∧L ∴K ↔L 5. (C ∧D) ∨E ∴E ∨D 6. A ↔B,B ↔C ∴A ↔C 7. ¬F →G,F →H ∴G ∨H 8. (Z ∧K) ∨(K ∧M ),K →D ∴D 9. P ∧(Q ∨R),P →¬R ∴Q ∨E 10. S ↔T ∴S ↔(T ∨S) 11. ¬(P →Q) ∴¬Q 12. ¬(P →Q) ∴P CHAPTER 17 Constructing proofs There is no simple recipe for finding proofs, and there is no sub-stitute for practice. Here, though, are some rules of thumb and strategies to keep in mind. 17.1 Working backward from what we want So you’re trying to find a proof of some conclusion C, which will be the last line of your proof. The first thing you do is look at C and ask what the introduction rule is for its main logical operator. This gives you an idea of what should happen before the last line of the proof. The justifications for the introduction rule require one or two other sentences above the last line, or one or two subproofs. Moreover, you can tell from C what those sentences are, or what the assumptions and conclusions of the subproof(s) are. Then you can write down those sentence or outline the subproof(s) above the last line, and treat those as your new goals. For example: If your conclusion is a conditional A →B, plan to use the →I rule. This requires starting a subproof in which you assume A. The subproof ought to end with B. Then, continue by 141 CHAPTER 17. CONSTRUCTING PROOFS 142 thinking about what you should do to get B inside that subproof, and how you can use the assumption A. If your goal is a conjunction, conditional, or negated sen-tence, you should start by working backward in this way. We’ll describe what you have to do in each of these cases in detail. Working backward from a conjunction If we want to prove A∧B, working backward means we should write A∧B at the bottom of our proof, and try to prove it using ∧I. At the top, we’ll write out the premises of the proof, if there are any. Then, at the bottom, we write the sentence we want to prove. If it is a conjunction, we’ll prove it using ∧I. 1 P 1 . . . k P k . . . n A . . . m B m + 1 A∧B ∧I n, m For ∧I, we need to prove A first, then prove B. For the last line, we have to cite the lines where we (will have) proved Aand B, and use ∧I. The parts of the proof labelled by the vertical · · · have to still be filled in. We’ll mark the line numbers m, n for now. When the proof is complete, these placeholders can be replaced by actual numbers. CHAPTER 17. CONSTRUCTING PROOFS 143 Working backward from a conditional If our goal is to prove a conditional A →B, we’ll have to use →I. This requires a subproof starting with Aand ending with B. We’ll set up our proof as follows: n A . . . m B m + 1 A →B →I n–m Again we’ll leave placeholders in the line number slots. We’ll record the last inference as →I, citing the subproof. Working backward from a negated sentence If we want to prove ¬A, we’ll have to use ¬I. n A . . . m ⊥ m + 1 ¬A ¬I n–m For ¬I, we have to start a subproof with assumption A; the last line of the subproof has to be ⊥. We’ll cite the subproof, and use ¬I as the rule. When working backward, continue to do so as long as you can. So if you’re working backward to prove A →B and have set up a subproof in which you want to prove B. Now look at B. If, say, it is a conjunction, work backward from it, and write down the two conjuncts inside your subproof. Etc. CHAPTER 17. CONSTRUCTING PROOFS 144 Working backward from a disjunction Of course, you can also work backward from a disjunction A∨B, if that is your goal. The ∨I rule requires that you have one of the disjuncts in order to infer A∨B. So to work backward, you pick a disjunct, infer A∨B from it, and then continue to look for a proof of the disjunct you picked: . . . n A n + 1 A∨B ∨I n However, you may not be able to prove the disjunct you picked. In that case you have to backtrack. When you can’t fill in the part labelled by the vertical · · · , delete everything, and try with the other disjunct: . . . n B n + 1 A∨B ∨I n Obviously, deleting everything and starting over is frustrating, so you should avoid it. If your goal is a disjunction, therefore, you should not start by working backward: try working forward first, and apply the ∨I strategy only when working forward (and working backward using ∧I, →I, and ¬I) no longer work. 17.2 Work forward from what you have Your proof may have premises. And if you’ve worked backward in order to prove a conditional or a negated sentence, you will have set up subproofs with an assumption, and be looking to prove a final sentence in the subproof. These premises and assumptions are sentences you can work forward from in order to fill in the CHAPTER 17. CONSTRUCTING PROOFS 145 missing steps in your proof. That means applying elimination rules for the main operators of these sentences. The form of the rules will tell you what you’ll have to do. Working forward from a conjunction To work forward from a sentence of the form A∧B, we use ∧E. That rule allows us to do two things: infer A, and infer B. So in a proof where we have A∧B, we can work forward by writing A and/or B immediately below the conjunction: n A∧B n + 1 A ∧E n n + 2 B ∧E n Usually it will be clear in the particular situation you’re in which one of Aor Byou’ll need. It doesn’t hurt, however, to write them both down. Working forward from a disjunction Working forward from a disjunction works a bit differently. To use a disjunction, we use the ∨E rule. In order to apply that rule, it is not enough to know what the disjuncts of the disjunction are that we want to use. We must also keep in mind what we want to prove. Suppose we want to prove C, and we have A∨B to work with. (That A∨B may be a premise of the proof, an assumption of a subproof, or something already proved.) In order to be able to apply the ∨E rule, we’ll have to set up two subproofs: CHAPTER 17. CONSTRUCTING PROOFS 146 n A∨B n + 1 A . . . m C m + 1 B . . . k C k + 1 C ∨E n, (n + 1)–m, (m + 1)–k The first subproof starts with the first disjunct, A, and ends with the sentence we’re looking for, C. The second subproof starts with the other disjunct, B, and also ends with the goal sen-tence C. Each of these subproofs have to be filled in further. We can then justify the goal sentence C by using ∨E, citing the line with A∨B and the two subproofs. Working forward from a conditional In order to use a conditional A →B, you also need the an-tecedent A in order to apply →E. So to work forward from a conditional, you will derive B, justify it by →E, and set up A as a new subgoal. n A →B . . . m A m + 1 B →E n, m CHAPTER 17. CONSTRUCTING PROOFS 147 Working forward from a negated sentence Finally, to use a negated sentence ¬A, you would apply ¬E. It requires, in addition to ¬A, also the corresponding sentence A without the negation. The sentence you’ll get is always the same: ⊥. So working forward from a negated sentence works especially well inside a subproof that you’ll want to use for ¬I (or IP). You work forward from ¬A if you already have ¬A and you want to prove ⊥. To do it, you set up A as a new subgoal. n ¬A . . . m A m + 1 ⊥ ¬E n, m 17.3 Strategies at work Suppose we want to show that the argument (A ∧B) ∨(A ∧C ) ∴ A ∧(B ∨C ) is valid. We start the proof by writing the premise and conclusion down. (On a piece of paper, you would want as much space as possible between them, so write the premises at the top of the sheet and the conclusion at the bottom.) 1 (A ∧B) ∨(A ∧C ) . . . n A ∧(B ∨C ) We now have two options: either work backward from the con-clusion, or work forward from the premise. We’ll pick the sec-ond strategy: we use the disjunction on line 1, and set up the subproofs we need for ∨E. The disjunction on line 1 has two dis-juncts, A ∧B and A ∧C . The goal sentence you want to prove is A ∧(B ∨C ). So in this case you have to set up two subproofs, CHAPTER 17. CONSTRUCTING PROOFS 148 one with assumption A ∧B and last line A ∧(B ∨C ), the other with assumption A ∧C and last line A ∧(B ∨C ). The justification for the conclusion on line n will be ∨E, citing the disjunction on line 1 and the two subproofs. So your proof now looks like this: 1 (A ∧B) ∨(A ∧C ) 2 A ∧B . . . n A ∧(B ∨C ) n + 1 A ∧C . . . m A ∧(B ∨C ) m + 1 A ∧(B ∨C ) ∨E 1, 2–n, (n + 1)–m You now have two separate tasks, namely to fill in each of the two subproofs. In the first subproof, we now work backward from the conclusion A ∧(B ∨C ). That is a conjunction, so inside the first subproof, you will have two separate subgoals: proving A, and proving B ∨C . These subgoals will let you justify line n using ∧I. Your proof now looks like this: CHAPTER 17. CONSTRUCTING PROOFS 149 1 (A ∧B) ∨(A ∧C ) 2 A ∧B . . . i A . . . n −1 B ∨C n A ∧(B ∨C ) ∧I i, n −1 n + 1 A ∧C . . . m A ∧(B ∨C ) m + 1 A ∧(B ∨C ) ∨E 1, 2–n, (n + 1)–m We immediately see that we can get line i from line 2 by ∧E. So line i is actually line 3, and can be justified with ∧E from line 2. The other subgoal B ∨C is a disjunction. We’ll apply the strategy for working backward from a disjunctions to line n −1. We have a choice of which disjunct to pick as a subgoal, B or C . Picking C wouldn’t work and we’ d end up having to backtrack. And you can already see that if you pick B as a subgoal, you could get that by working forward again from the conjunction A ∧B on line 2. So we can complete the first subproof as follows: CHAPTER 17. CONSTRUCTING PROOFS 150 1 (A ∧B) ∨(A ∧C ) 2 A ∧B 3 A ∧E 2 4 B ∧E 2 5 B ∨C ∨I 4 6 A ∧(B ∨C ) ∧I 3, 5 7 A ∧C . . . m A ∧(B ∨C ) m + 1 A ∧(B ∨C ) ∨E 1, 2–6, 7–m Like line 3, we get line 4 from 2 by ∧E. Line 5 is justified by ∨I from line 4, since we were working backward from a disjunction there. That’s it for the first subproof. The second subproof is almost exactly the same. We’ll leave it as an exercise. Remember that when we started, we had the option of work-ing forward from the premise, or working backward from the conclusion, and we picked the first option. The second option also leads to a proof, but it will look different. The first steps would be to work backward from the conclusion and set up two subgoals, A and B ∨C , and then work forward from the premise to prove them, e.g., CHAPTER 17. CONSTRUCTING PROOFS 151 1 (A ∧B) ∨(A ∧C ) 2 A ∧B . . . k A k + 1 A ∧C . . . n −1 A n A ∨E 1, 2–k, (k + 1)–(n −1) n + 1 A ∧B . . . l B ∨C l + 1 A ∧C . . . m −1 B ∨C m B ∨C ∨E 1, (n + 1)–l, (l + 1)–(m −1) m + 1 A ∧(B ∨C ) ∧I n, m We’ll leave you to fill in the missing pieces indicated by . . .. Let’s give another example to illustrate how to apply the strategies to deal with conditionals and negation. The sentence (A →B) →(¬B →¬A) is a tautology. Let’s see if we can find a proof of it, from no premises, using the strategies. We first write the sentence at the bottom of a sheet of paper. Since working for-ward is not an option (there is nothing to work forward from), we work backward, and set up a subproof to establish the sentence we want, (A →B) →(¬B →¬A), using →I. Its assumption CHAPTER 17. CONSTRUCTING PROOFS 152 must be the antecedent of the conditional we want to prove, i.e., A →B, and its last line the consequent, ¬B →¬A. 1 A →B . . . n ¬B →¬A n + 1 (A →B) →(¬B →¬A) →I 1–n The new goal, ¬B →¬A is itself a conditional, so working back-ward we set up another subproof: 1 A →B 2 ¬B . . . n −1 ¬A n ¬B →¬A →I 2–(n −1) n + 1 (A →B) →(¬B →¬A) →I 1–n From ¬A we again work backward. To do this, look at the ¬I rule. It requires a subproof with A as assumption, and ⊥as its last line. So the proof is now: CHAPTER 17. CONSTRUCTING PROOFS 153 1 A →B 2 ¬B 3 A . . . n −2 ⊥ n −1 ¬A ¬I 3–(n −2) n ¬B →¬A →I 2–(n −1) n + 1 (A →B) →(¬B →¬A) →I 1–n Now our goal is to prove ⊥. We said above, when discussing how to work forward from a negated sentence, that the ¬E rule allows you to prove ⊥, which is our goal in the innermost subproof. So we look for a negated sentence which we can work forward from: that would be ¬B on line 2. That means we have to derive B inside the subproof, since ¬E requires not just ¬B (which we have already), but also B. And B, in turn, we get by working forward from A →B, since →E will allow us to justify the consequent of that conditional, B, by →E. The rule →E also requires the antecedent A of the conditional, but that is also already available (on line 3). So we finish with: CHAPTER 17. CONSTRUCTING PROOFS 154 1 A →B 2 ¬B 3 A 4 B →E 1, 3 5 ⊥ ¬E 2, 4 6 ¬A ¬I 3–5 7 ¬B →¬A →I 2–6 8 (A →B) →(¬B →¬A) →I 1–7 17.4 Working forward from ⊥ When applying the strategies, you will sometimes find yourself in a situation where you can justify ⊥. Using the explosion rule, this would allow you to justify anything. So ⊥works like a wild-card in proofs. For instance, suppose you want to give a proof of the argument A ∨B,¬A ∴B. You set up your proof, writing the premises A ∨B and ¬A at the top on lines 1 and 2, and the conclusion B at the bottom of the page. B has no main connec-tive, so you can’t work backward from it. Instead, you must work forward from A ∨B: That requires two subproofs, like so: CHAPTER 17. CONSTRUCTING PROOFS 155 1 A ∨B 2 ¬A 3 A . . . m B m + 1 B . . . k B k + 1 B ∨E 1, 3–m, (m + 1)–k Notice that you have ¬A on line 2 and A as the assumption of your first subproof. That gives you ⊥using ¬E, and from ⊥you get the conclusion B of the first subroof using X. Recall that you can repeat a sentence you already have by using the reiteration rule R. So our proof would be: 1 A ∨B 2 ¬A 3 A 4 ⊥ ¬E 2, 3 5 B X 4 6 B 7 B R 6 8 B ∨E 1, 3–5, 6–7 CHAPTER 17. CONSTRUCTING PROOFS 156 17.5 Proceed indirectly In very many cases, the strategies of working forward and back-ward will eventually pan out. But there are cases where they do not work. If you cannot find a way to show Adirectly using those, use IP instead. To do this, set up a subproof in which you assume ¬A and look for a proof of ⊥inside that subproof. n ¬A . . . m ⊥ m + 1 A IP n–m Here, we have to start a subproof with assumption ¬A; the last line of the subproof has to be ⊥. We’ll cite the subproof, and use IP as the rule. In the subproof, we now have an additional assumption (on line n) to work with. Suppose we used the indirect proof strategy, or we’re in some other situation where we’re looking for a proof of ⊥. What’s a good candidate? Of course the obvious candidate would be to use a negated sentence, since (as we saw above) ¬E always yields ⊥. If you set up a proof as above, trying to prove A using IP, you will have ¬A as the assumption of your subproof—so working forward from it to justify ⊥inside your subproof, you would next set up A as a goal inside your subproof. If you are using this IP strategy, you will find yourself in the following situation: n ¬A . . . m −1 A m ⊥ ¬E n, m −1 m + 1 A IP n–m CHAPTER 17. CONSTRUCTING PROOFS 157 This looks weird: We wanted to prove Aand the strategies failed us; so we used IP as a last resort. And now we find ourselves in the same situation: we are again looking for a proof of A. But notice that we are now inside a subproof, and in that subproof we have an additional assumption (¬A) to work with which we didn’t have before. Let’s look at an example. 17.6 Indirect proof of excluded middle The sentence A ∨¬A is a tautology, and so should have a proof even without any premises. But working backward fails us: to get A ∨¬A using ∨I we would have to prove either A or ¬A—again, from no premises. Neither of these is a tautology, so we won’t be able to prove either. Working forward doesn’t work either, since there is nothing to work forward from. So, the only option is indirect proof. 1 ¬(A ∨¬A) . . . m ⊥ m + 1 A ∨¬A IP 1–m Now we do have something to work forward from: the assumption ¬(A ∨¬A). To use it, we justify ⊥by ¬E, citing the assumption on line 1, and also the corresponding unnegated sentence A∨¬A, yet to be proved. 1 ¬(A ∨¬A) . . . m −1 A ∨¬A m ⊥ ¬E 1, m −1 m + 1 A ∨¬A IP 1–m CHAPTER 17. CONSTRUCTING PROOFS 158 At the outset, working backward to prove A ∨¬A by ∨I did not work. But we are now in a different situation: we want to prove A ∨¬A inside a subproof. In general, when dealing with new goals we should go back and start with the basic strategies. In this case, we should first try to work backward from the disjunction A ∨¬A, i.e., we have to pick a disjunct and try to prove it. Let’s pick ¬A. This would let us justify A ∨¬A on line m −1 using ∨I. Then working backward from ¬A, we start another subproof in order to justify ¬A using ¬I. That subproof must have A as the assumption and ⊥as its last line. 1 ¬(A ∨¬A) 2 A . . . m −3 ⊥ m −2 ¬A ¬I 2–(m −3) m −1 A ∨¬A ∨I m −2 m ⊥ ¬E 1, m −1 m + 1 A ∨¬A IP 1–m Inside this new subproof, we again need to justify ⊥. The best way to do this is to work forward from a negated sentence; ¬(A ∨¬A) on line 1 is the only negated sentence we can use. The corre-sponding unnegated sentence, A ∨¬A, however, directly follows from A (which we have on line 2) by ∨I. Our complete proof is: CHAPTER 17. CONSTRUCTING PROOFS 159 1 ¬(A ∨¬A) 2 A 3 A ∨¬A ∨I 2 4 ⊥ ¬E 1, 3 5 ¬A ¬I 2–4 6 A ∨¬A ∨I 5 7 ⊥ ¬E 1, 6 8 A ∨¬A IP 1–7 Practice exercises A. Use the strategies to find proofs for each of the following ar-guments: 1. A →B,A →C ∴A →(B ∧C ) 2. (A ∧B) →C ∴A →(B →C ) 3. A →(B →C ) ∴(A →B) →(A →C ) 4. A ∨(B ∧C ) ∴(A ∨B) ∧(A ∨C ) 5. (A ∧B) ∨(A ∧C ) ∴A ∧(B ∨C ) 6. A ∨B,A →C,B →D ∴C ∨D 7. ¬A ∨¬B ∴¬(A ∧B) 8. A ∧¬B ∴¬(A →B) B. Formulate strategies for working backward and forward from A ↔B. C. Use the strategies to find proofs for each of the following sen-tences: 1. ¬A →(A →⊥) 2. ¬(A ∧¬A) 3. [(A →C ) ∧(B →C )] →[(A ∨B) →C ] 4. ¬(A →B) →(A ∧¬B) CHAPTER 17. CONSTRUCTING PROOFS 160 5. (¬A ∨B) →(A →B) Since these should be proofs of sentences from no premises, you will start with the respective sentence at the bottom of the proof, which will have no premises. D. Use the strategies to find proofs for each one of the following arguments and sentences: 1. ¬¬A →A 2. ¬A →¬B ∴B →A 3. A →B ∴¬A ∨B 4. ¬(A ∧B) →(¬A ∨¬B) 5. A →(B ∨C ) ∴(A →B) ∨(A →C ) 6. (A →B) ∨(B →A) 7. ((A →B) →A) →A These all will require the IP strategy. The last three especially are quite hard! CHAPTER 18 Additional rules for TFL In §16, we introduced the basic rules of our proof system for TFL. In this section, we will add some additional rules to our system. Our extended proof system is a bit easier to work with. (However, in §20 we will see that they are not strictly speaking necessary.) 18.1 Disjunctive syllogism Here is a very natural argument form. Elizabeth is either in Massachusetts or in DC. She is not in DC. So, she is in Massachusetts. This is called disjunctive syllogism. We add it to our proof system as follows: m A∨B n ¬A B DS m, n 161 CHAPTER 18. ADDITIONAL RULES FOR TFL 162 and m A∨B n ¬B A DS m, n As usual, the disjunction and the negation of one disjunct may occur in either order and need not be adjacent. However, we always cite the disjunction first. 18.2 Modus tollens Another useful pattern of inference is embodied in the following argument: If Hilary has won the election, then she is in the White House. She is not in the White House. So she has not won the election. This inference pattern is called modus tollens. The corresponding rule is: m A →B n ¬B ¬A MT m, n As usual, the premises may occur in either order, but we al-ways cite the conditional first. CHAPTER 18. ADDITIONAL RULES FOR TFL 163 18.3 Double-negation elimination Another useful rule is double-negation elimination. It does exactly what it says on the tin: m ¬¬A A DNE m The justification for this is that, in natural language, double-negations tend to cancel out. That said, you should be aware that context and emphasis can prevent them from doing so. Consider: ‘Jane is not not happy’. Arguably, one cannot infer ‘Jane is happy’, since the first sen-tence should be understood as meaning the same as ‘Jane is not unhappy’. This is compatible with ‘Jane is in a state of profound indifference’. As usual, moving to TFL forces us to sacrifice cer-tain nuances of English expressions. 18.4 Excluded middle Suppose that we can show that if it’s sunny outside, then Bill will have brought an umbrella (for fear of burning). Suppose we can also show that, if it’s not sunny outside, then Bill will have brought an umbrella (for fear of rain). Well, there is no third way for the weather to be. So, whatever the weather, Bill will have brought an umbrella. This line of thinking motivates the following rule: CHAPTER 18. ADDITIONAL RULES FOR TFL 164 i A j B k ¬A l B B LEM i–j, k–l The rule is sometimes called the law of excluded middle, since it encapsulates the idea that A may be true or ¬A may be true, but there is no middle way where neither is true.1 There can be as many lines as you like between i and j, and as many lines as you like between k and l. Moreover, the subproofs can come in any order, and the second subproof does not need to come immediately after the first. To see the rule in action, consider: P ∴(P ∧D) ∨(P ∧¬D) Here is a proof corresponding with the argument: 1You may sometimes find logicians or philosophers talking about “tertium non datur.” That’s the same principle as excluded middle; it means “no third way.” Logicians who have qualms about indirect proof also have qualms about LEM. CHAPTER 18. ADDITIONAL RULES FOR TFL 165 1 P 2 D 3 P ∧D ∧I 1, 2 4 (P ∧D) ∨(P ∧¬D) ∨I 3 5 ¬D 6 P ∧¬D ∧I 1, 5 7 (P ∧D) ∨(P ∧¬D) ∨I 6 8 (P ∧D) ∨(P ∧¬D) LEM 2–4, 5–7 Here is another example: 1 A →¬A 2 A 3 ¬A →E 1, 2 4 ¬A 5 ¬A R 4 6 ¬A LEM 2–3, 4–5 18.5 De Morgan Rules Our final additional rules are called De Morgan’s Laws (named after Augustus De Morgan). The shape of the rules should be familiar from truth tables. The first De Morgan rule is: CHAPTER 18. ADDITIONAL RULES FOR TFL 166 m ¬(A∧B) ¬A∨¬B DeM m The second De Morgan is the reverse of the first: m ¬A∨¬B ¬(A∧B) DeM m The third De Morgan rule is the dual of the first: m ¬(A∨B) ¬A∧¬B DeM m And the fourth is the reverse of the third: m ¬A∧¬B ¬(A∨B) DeM m These are all of the additional rules of our proof system for TFL. Practice exercises A. The following proofs are missing their citations (rule and line numbers). Add them wherever they are required: CHAPTER 18. ADDITIONAL RULES FOR TFL 167 1. 1 W →¬B 2 A ∧W 3 B ∨( J ∧K) 4 W 5 ¬B 6 J ∧K 7 K 2. 1 L ↔¬O 2 L ∨¬O 3 ¬L 4 ¬O 5 L 6 ⊥ 7 ¬¬L 8 L CHAPTER 18. ADDITIONAL RULES FOR TFL 168 3. 1 Z →(C ∧¬N ) 2 ¬Z →(N ∧¬C ) 3 ¬(N ∨C ) 4 ¬N ∧¬C 5 ¬N 6 ¬C 7 Z 8 C ∧¬N 9 C 10 ⊥ 11 ¬Z 12 N ∧¬C 13 N 14 ⊥ 15 ¬¬(N ∨C ) 16 N ∨C B. Give a proof for each of these arguments: 1. E ∨F , F ∨G, ¬F ∴E ∧G 2. M ∨(N →M ) ∴¬M →¬N 3. (M ∨N ) ∧(O ∨P ), N →P , ¬P ∴M ∧O 4. (X ∧Y ) ∨(X ∧Z), ¬(X ∧D), D ∨M ∴M CHAPTER 19 Proof-theoretic concepts In this chapter we will introduce some new vocabulary. The fol-lowing expression: A 1,A 2,. . . ,A n ⊢C means that there is some proof which ends with C whose undis-charged assumptions are among A 1,A 2,. . . ,A n. When we want to say that it is not the case that there is some proof which ends with C from A 1, A 2, . . . , A n, we write: A 1,A 2,. . . ,A n ⊬C The symbol ‘⊢’ is called the single turnstile. We want to em-phasize that this is not the double turnstile symbol (‘⊨’) that we introduced in chapter 12 to symbolize entailment. The single turnstile, ‘⊢’, concerns the existence of proofs; the double turn-stile, ‘⊨’, concerns the existence of valuations (or interpretations, when used for FOL). They are very different notions. Armed with our ‘⊢’ symbol, we can introduce some more ter-minology. To say that there is a proof of Awith no undischarged assumptions, we write: ⊢A. In this case, we say that A is a theorem. 169 CHAPTER 19. PROOF-THEORETIC CONCEPTS 170 A is a theorem iff ⊢A To illustrate this, suppose we want to show that ‘¬(A ∧¬A)’ is a theorem. So we need a proof of ‘¬(A ∧¬A)’ which has no undischarged assumptions. However, since we want to prove a sentence whose main logical operator is a negation, we will want to start with a subproof within which we assume ‘A ∧¬A’, and show that this assumption leads to contradiction. All told, then, the proof looks like this: 1 A ∧¬A 2 A ∧E 1 3 ¬A ∧E 1 4 ⊥ ¬E 3, 2 5 ¬(A ∧¬A) ¬I 1–4 We have therefore proved ‘¬(A ∧¬A)’ on no (undischarged) as-sumptions. This particular theorem is an instance of what is sometimes called the Law of Non-Contradiction. To show that something is a theorem, you just have to find a suitable proof. It is typically much harder to show that something is not a theorem. To do this, you would have to demonstrate, not just that certain proof strategies fail, but that no proof is possible. Even if you fail in trying to prove a sentence in a thousand differ-ent ways, perhaps the proof is just too long and complex for you to make out. Perhaps you just didn’t try hard enough. Here is another new bit of terminology: T wo sentences Aand Bare prov ably equiv alent iff each can be proved from the other; i.e., both A ⊢B and B ⊢A. As in the case of showing that a sentence is a theorem, it is relatively easy to show that two sentences are provably equivalent: CHAPTER 19. PROOF-THEORETIC CONCEPTS 171 it just requires a pair of proofs. Showing that sentences are not provably equivalent would be much harder: it is just as hard as showing that a sentence is not a theorem. Here is a third, related, bit of terminology: The sentences A 1,A 2,. . . ,A n are jointly inconsistent iff the contradiction ⊥can be proved from them, i.e., A 1,A 2,. . . ,A n ⊢⊥. If they are not inconsistent, we call them jointly consistent. It is easy to show that some sentences are inconsistent: you just need to prove the contradiction ⊥, assuming all the sentences as premises. Showing that some sentences are not inconsistent is much harder. It would require more than just providing a proof or two; it would require showing that no proof of a certain kind is possible. This table summarizes whether one or two proofs suffice, or whether we must reason about all possible proofs. Y es No theorem? one proof all possible proofs inconsistent? one proof all possible proofs equivalent? two proofs all possible proofs consistent? all possible proofs one proof Practice exercises A. Show that each of the following sentences is a theorem: 1. O →O 2. N ∨¬N 3. J ↔[ J ∨(L ∧¬L)] 4. ((A →B) →A) →A B. Provide proofs to show each of the following: CHAPTER 19. PROOF-THEORETIC CONCEPTS 172 1. C →(E ∧G),¬C →G ⊢G 2. M ∧(¬N →¬M ) ⊢(N ∧M ) ∨¬M 3. (Z ∧K) ↔(Y ∧M ),D ∧(D →M ) ⊢Y →Z 4. (W ∨X ) ∨(Y ∨Z),X →Y,¬Z ⊢W ∨Y C. Show that each of the following pairs of sentences are provably equivalent: 1. R ↔E, E ↔R 2. G, ¬¬¬¬G 3. T →S, ¬S →¬T 4. U →I , ¬(U ∧¬I ) 5. ¬(C →D),C ∧¬D 6. ¬G ↔H , ¬(G ↔H ) D. If you know that A ⊢B, what can you say about (A∧C) ⊢B? What about (A∨C) ⊢B? Explain your answers. E. In this chapter, we claimed that it is just as hard to show that two sentences are not provably equivalent, as it is to show that a sentence is not a theorem. Why did we claim this? (Hint: think of a sentence that would be a theorem iff Aand B were provably equivalent.) CHAPTER 20 Derived rules In this section, we will see why we introduced the rules of our proof system in two separate batches. In particular, we want to show that the additional rules of §18 are not strictly speaking necessary, but can be derived from the basic rules of §16. 20.1 Derivation of Reiteration To illustrate what it means to derive a rule from other rules, first consider reiteration. It is a basic rule of our system, but it is also not necessary. Suppose you have some sentence on some line of your deduction: m A You now want to repeat yourself, on some line k. You could just invoke the rule R. But equally well, you can do this with other basic rules of §16: m A k A∧A ∧I m, m k + 1 A ∧E k 173 CHAPTER 20. DERIVED RULES 174 To be clear: this is not a proof. Rather, it is a proof scheme. After all, it uses a variable, ‘A’, rather than a sentence of TFL, but the point is simple: Whatever sentences of TFL we plugged in for ‘A’, and whatever lines we were working on, we could produce a bona fide proof. So you can think of this as a recipe for producing proofs. Indeed, it is a recipe which shows us the following: anything we can prove using the rule R, we can prove (with one more line) using just the basic rules of §16 without R. That is what it means to say that the rule R can be derived from the other basic rules: anything that can be justified using R can be justified using only the other basic rules. 20.2 Derivation of Disjunctive Syllogism Suppose that you are in a proof, and you have something of this form: m A∨B n ¬A You now want, on line k, to prove B. You can do this with the rule of DS, introduced in §18, but equally well, you can do this with the basic rules of §16: CHAPTER 20. DERIVED RULES 175 m A∨B n ¬A k A k + 1 ⊥ ¬E n, k k + 2 B X k + 1 k + 3 B k + 4 B R k + 3 k + 5 B ∨E m, k–k + 2, k + 3–k + 4 So the DS rule, again, can be derived from our more basic rules. Adding it to our system did not make any new proofs possible. Anytime you use the DS rule, you could always take a few extra lines and prove the same thing using only our basic rules. It is a derived rule. 20.3 Derivation of Modus T ollens Suppose you have the following in your proof: m A →B n ¬B You now want, on line k, to prove ¬A. You can do this with the rule of MT, introduced in §18. Equally well, you can do this with the basic rules of §16: CHAPTER 20. DERIVED RULES 176 m A →B n ¬B k A k + 1 B →E m, k k + 2 ⊥ ¬E n, k + 1 k + 3 ¬A ¬I k–k + 2 Again, the rule of MT can be derived from the basic rules of §16. 20.4 Derivation of Double-Negation Elimination Consider the following deduction scheme: m ¬¬A k ¬A k + 1 ⊥ ¬E m, k k + 2 A IP k–k + 1 Again, we can derive the DNE rule from the basic rules of §16. 20.5 Derivation of Excluded Middle Suppose you want to prove something using the LEM rule, i.e., you have in your proof CHAPTER 20. DERIVED RULES 177 m A n B k ¬A l B You now want, on line l + 1, to prove B. The rule LEM from §18 would allow you to do it. But can you do this with the basic rules of §16? One option is to first prove A∨¬A, and then apply ∨E, i.e., proof by cases: m A n B k ¬A l B . . . i A∨¬A i + 1 B ∨E i, m–n, k–l (We gave a proof of A∨¬A using only our basic rules in §17.6.) Here is another way that is a bit more complicated than the ones before. What you have to do is embed your two subproofs inside another subproof. The assumption of the subproof will be ¬B, and the last line will be ⊥. Thus, the complete subproof is the kind you need to conclude B using IP. Inside the proof, you’ d have to do a bit more work to get ⊥: CHAPTER 20. DERIVED RULES 178 m ¬B m + 1 A . . . n B n + 1 ⊥ ¬E m, n n + 2 ¬A . . . l B l + 1 ⊥ ¬E m, l l + 2 ¬A ¬I (m + 1)–(n + 1) l + 3 ¬¬A ¬I (n + 2)–(l + 1) l + 4 ⊥ ¬E l + 3, l + 2 l + 5 B IP m–(l + 4) Note that because we add an assumption at the top and additional conclusions inside the subproofs, the line numbers change. You may have to stare at this for a while before you understand what’s going on. 20.6 Derivation of De Morgan rules Here is a demonstration of how we could derive the first De Mor-gan rule: CHAPTER 20. DERIVED RULES 179 m ¬(A∧B) k A k + 1 B k + 2 A∧B ∧I k, k + 1 k + 3 ⊥ ¬E m, k + 2 k + 4 ¬B ¬I (k + 1)–(k + 3) k + 5 ¬A∨¬B ∨I k + 4 k + 6 ¬A k + 7 ¬A∨¬B ∨I k + 6 k + 8 ¬A∨¬B LEM k–(k + 5), (k + 6)–(k + 7) Here is a demonstration of how we could derive the second De Morgan rule: m ¬A∨¬B k A∧B k + 1 A ∧E k k + 2 B ∧E k k + 3 ¬A k + 4 ⊥ ¬E k + 3, k + 1 k + 5 ¬B k + 6 ⊥ ¬E k + 5, k + 2 k + 7 ⊥ ∨E m, (k + 3)–(k + 4), (k + 5)–(k + 6) k + 8 ¬(A∧B) ¬I k–(k + 7) CHAPTER 20. DERIVED RULES 180 Similar demonstrations can be offered explaining how we could derive the third and fourth De Morgan rules. These are left as exercises. Practice exercises A. Provide proof schemes that justify the addition of the third and fourth De Morgan rules as derived rules. B. The proofs you offered in response to the practice exercises of §§18–19 used derived rules. Replace the use of derived rules, in such proofs, with only basic rules. You will find some ‘repetition’ in the resulting proofs; in such cases, offer a streamlined proof using only basic rules. (This will give you a sense, both of the power of derived rules, and of how all the rules interact.) C. Give a proof of A∨¬A. Then give a proof that uses only the basic rules. D. Show that if you had LEM as a basic rule, you could justify IP as a derived rule. That is, suppose you had the proof: m ¬A . . . n ⊥ How could you use it to prove A without using IP but with using LEM as well as all the other basic rules? E. Give a proof of the first De Morgan rule, but using only the basic rules, in particular, without using LEM. (Of course, you can combine the proof using LEM with the proof of LEM. Try to find a proof directly.) CHAPTER 21 Soundness and completeness In §19, we saw that we could use derivations to test for the same concepts we used truth tables to test for. Not only could we use derivations to prove that an argument is valid, we could also use them to test if a sentence is a tautology or a pair of sentences are equivalent. We also started using the single turnstile the same way we used the double turnstile. If we could prove that A was a tautology using a truth table, we wrote ⊨A, and if we could prove it using a derivation, we wrote ⊢A. You may have wondered at that point if the two kinds of turn-stiles always worked the same way. If you can show that A is a tautology using truth tables, can you also always show that it is a theorem using a derivation? Is the reverse true? Are these things also true for valid arguments and pairs of equivalent sentences? As it turns out, the answer to all these questions and many more like them is yes. We can show this by defining all these concepts separately and then proving them equivalent. That is, we imag-181 CHAPTER 21. SOUNDNESS AND COMPLETENESS 182 ine that we actually have two notions of validity, valid⊨and valid⊢ and then show that the two concepts always work the same way. To begin with, we need to define all of our logical concepts separately for truth tables and derivations. A lot of this work has already been done. We handled all of the truth table definitions in §12. We have also already given proof-theoretic definitions for theorems and pairs of logically equivalent sentences. The other definitions follow naturally. For most logical properties we can devise a test using derivations, and those that we cannot test for directly can be defined in terms of the concepts that we can define. For instance, we defined a theorem as a sentence that can be derived without any premises (p. 170). Since the negation of a contradiction is a tautology, we can define an inconsistent sentence in tfl as a sentence whose negation can be derived without any premises.1 The syntactic definition of a contingent sentence is a little different. We don’t have any practical, finite method for proving that a sentence is contingent using deriva-tions, the way we did using truth tables. So we have to content ourselves with defining “contingent sentence” negatively. A sen-tence is proof-theoretically contingent in tfl if it is neither a theorem nor an inconsistent sentence. A collection of sentences is inconsistent in tfl if and only if one can derive the contradiction ⊥from them. Consistency, on the other hand, is like contingency, in that we do not have a practical finite method to test for it directly. So again, we have to define a term negatively. A collection of sentences is consistent in tfl if and only if they are not inconsistent. Finally, an argument is prov ably v alid in tfl if and only if there is a derivation of its conclusion from its premises. All of these definitions are given in Table 21.1. All of our concepts have now been defined both semantically (using valuations and truth tables) and proof-theoretically (on the basis of natural deduction). How can we establish that these 1Note that ¬A is a theorem iff A ⊢⊥. CHAPTER 21. SOUNDNESS AND COMPLETENESS 183 Concept Truth table (semantic) definition Proof-theoretic (syntactic) definition Tautology/ theorem A sentence whose truth table only has Ts under the main connective A sentence that can be derived without any premises. Contradiction/ inconsistent sentence A sentence whose truth table only has Fs under the main connective A sentence whose negation can be derived without any premises Contingent sentence A sentence whose truth table contains both Ts and Fs under the main connective A sentence that is not a theorem or contradiction Equivalent sentences The columns under the main connectives are identical. The sentences can be derived from each other Unsatisfiable/ inconsistent sentences Sentences which do not have a single line in their truth table where they are all true. Sentences from which one can derive the contradiction ⊥ Satisfiable/ Consistent sentences Sentences which have at least one line in their truth table where they are all true. Sentences from which one cannot derive the contradiction ⊥ Valid argument An argument whose truth table has no lines where there are all Ts under main connectives for the premises and an F under the main connective for the conclusion. An argument where one can derive the conclusion from the premises Table 21.1: Two ways to define logical concepts. CHAPTER 21. SOUNDNESS AND COMPLETENESS 184 definitions always work the same way? A full proof here goes well beyond the scope of this book. However, we can sketch what it would be like. We will focus on showing the two notions of va-lidity to be equivalent. From that the other concepts will follow quickly. The proof will have to go in two directions. First we will have to show that things which are proof-theoretically valid will also be semantically valid. In other words, everything that we can prove using derivations could also be proven using truth ta-bles. Put symbolically, we want to show that valid⊢implies valid⊨. Afterwards, we will need to show things in the other directions, valid⊨implies valid⊢ This argument from ⊢to ⊨is the problem of soundness. A proof system is sound if there are no derivations of arguments that can be shown invalid by truth tables. Demonstrating that the proof system is sound would require showing that any possible proof is the proof of a valid argument. It would not be enough simply to succeed when trying to prove many valid arguments and to fail when trying to prove invalid ones. The proof that we will sketch depends on the fact that we initially defined a sentence of TFL using an inductive definition (see p. 51). We could have also used inductive definitions to de-fine a proper proof in TFL and a proper truth table (although we didn’t.) If we had these definitions, we could then use an inductive proof to show the soundness of TFL. An inductive proof works the same way as an inductive definition. With the inductive defi-nition, we identified a group of base elements that were stipulated to be examples of the thing we were trying to define. In the case of a TFL sentence, the base class was the set of sentence letters A, B, C , . . . . We just announced that these were sentences. The second step of an inductive definition is to say that anything that is built up from your base class using certain rules also counts as an example of the thing you are defining. In the case of a defini-tion of a sentence, the rules corresponded to the five sentential connectives (see p. 51). Once you have established an inductive definition, you can use that definition to show that all the mem-bers of the class you have defined have a certain property. You CHAPTER 21. SOUNDNESS AND COMPLETENESS 185 simply prove that the property is true of the members of the base class, and then you prove that the rules for extending the base class don’t change the property. This is what it means to give an inductive proof. Even though we don’t have an inductive definition of a proof in TFL, we can sketch how an inductive proof of the soundness of TFL would go. Imagine a base class of one-line proofs, one for each of our eleven rules of inference. The members of this class would look like this A,B ⊢A∧B; A∧B ⊢A; A∨B,¬A ⊢ B ... etc. Since some rules have a couple different forms, we would have to have add some members to this base class, for instance A∧B ⊢B Notice that these are all statements in the metalanguage. The proof that TFL is sound is not a part of TFL, because TFL does not have the power to talk about itself. You can use truth tables to prove to yourself that each of these one-line proofs in this base class is valid⊨. For instance the proof A,B ⊢A∧Bcorresponds to a truth table that shows A,B ⊨A∧B This establishes the first part of our inductive proof. The next step is to show that adding lines to any proof will never change a valid⊨proof into an invalid⊨one. We would need to do this for each of our eleven basic rules of inference. So, for instance, for ∧I we need to show that for any proof that shows that A 1, . . . , A n ⊢B, adding a line where we use ∧I to infer C∧D, where C ∧D can be legitimately inferred from A 1, ..., A n, B, would result in a valid proof. But wait, if we can legitimately derive C∧D at the end of the given proof, then C and D must be already available in the proof. They are either already among A 1, ... , A n, B, or can be legitimately derived from them. As such, any truth table line in which the premises are true must be a truth table line in which C and D are true. According to the characteristic truth table for ∧, this means that C ∧D is also true on that line. Therefore, C∧D validly follows from the premises. This means that using the ∧I rule to extend a valid proof produces another valid proof. In order to show that the proof system is sound, we would need to show this for the other inference rules. Since the derived CHAPTER 21. SOUNDNESS AND COMPLETENESS 186 rules are consequences of the basic rules, it would suffice to pro-vide similar arguments for the 11 other basic rules. You can find the details of this proof worked out in chapter 46. So we have shown that A ⊢B implies A ⊨B. What about the other direction, that is, why think that every argument that can be shown valid using truth tables can also be proven using a derivation? This is the problem of completeness. A proof system has the property of completeness if and only if there is a derivation of every semantically valid argument. Proving that a system is com-plete is generally harder than proving that it is sound. Proving that a system is sound amounts to showing that all of the rules of your proof system work the way they are supposed to. Showing that a system is complete means showing that you have included all the rules you need, that you haven’t left any out. Showing this is beyond the scope of this book. The important point is that, happily, the proof system for TFL is both sound and complete. This is not the case for all proof systems or all formal languages. Because it is true of TFL, we can choose to give proofs or give truth tables—whichever is easier for the task at hand. Now that we know that the truth table method is interchange-able with the method of derivations, you can chose which method you want to use for any given problem. Students often prefer to use truth tables, because they can be produced purely mechani-cally, and that seems ‘easier’. However, we have already seen that truth tables become impossibly large after just a few sentence let-ters. On the other hand, there are a couple situations where using proofs simply isn’t possible. We syntactically defined a contingent sentence as a sentence that couldn’t be proven to be a tautology or a contradiction. There is no practical way to prove this kind of negative statement. We will never know if there isn’t some proof out there that a statement is a contradiction and we just haven’t found it yet. We have nothing to do in this situation but resort to truth tables. Similarly, we can use derivations to prove two sentences equivalent, but what if we want to prove that they are not equivalent? We have no way of proving that we will never find CHAPTER 21. SOUNDNESS AND COMPLETENESS 187 Logical property T o prove it present T o prove it absent Being a theorem Derive the sentence Find a false line in the truth table for the sentence Being a contradiction Derive the negation of the sentence Find a true line in the truth table for the sentence Contingency Find a false line and a true line in the truth table for the sentence Prove the sentence or its negation Equivalence Derive each sentence from the other Find a line in the truth tables for the sentence where they have different values Consistency Find a line in truth table for the sentence where they all are true Derive a contradiction from the sentences Validity Derive the conclusion from the premises Find a line in the truth table where the premises are true and the conclusion false. Table 21.2: When to provide a truth table and when to provide a proof. the relevant proof. So we have to fall back on truth tables again. Table 21.2 summarizes when it is best to give proofs and when it is best to give truth tables. Practice exercises A. Use either a derivation or a truth table for each of the follow-ing. 1. Show that A →[((B ∧C ) ∨D) →A] is a theorem. CHAPTER 21. SOUNDNESS AND COMPLETENESS 188 2. Show that A →(A →B) is not a theorem. 3. Show that the sentence A →¬A is not a contradiction. 4. Show that the sentence A ↔¬A is a contradiction. 5. Show that the sentence ¬(W →( J ∨J )) is contingent. 6. Show that the sentence ¬(X ∨(Y ∨Z)) ∨(X ∨(Y ∨Z)) is not contingent. 7. Show that the sentence B →¬S is equivalent to the sen-tence ¬¬B →¬S. 8. Show that the sentence ¬(X ∨O) is not equivalent to the sentence X ∧O. 9. Show that the sentences ¬(A ∨B), C , C →A are jointly inconsistent. 10. Show that the sentences ¬(A ∨B), ¬B, B →A are jointly consistent. 11. Show that ¬(A ∨(B ∨C )) ∴¬C is valid. 12. Show that ¬(A ∧(B ∨C )) ∴¬C is invalid. B. Use either a derivation or a truth table for each of the follow-ing. 1. Show that A →(B →A) is a theorem. 2. Show that ¬(((N ↔Q) ∨Q) ∨N ) is not a theorem. 3. Show that Z ∨(¬Z ↔Z) is contingent. 4. Show that (L ↔((N →N ) →L)) ∨H is not contingent. 5. Show that (A ↔A) ∧(B ∧¬B) is a contradiction. 6. Show that (B ↔(C ∨B)) is not a contradiction. CHAPTER 21. SOUNDNESS AND COMPLETENESS 189 7. Show that ((¬X ↔X ) ∨X ) is equivalent to X . 8. Show that F ∧(K ∧R) is not equivalent to (F ↔(K ↔R)). 9. Show that the sentences ¬(W →W ), (W ↔W ) ∧W , E ∨(W →¬(E ∧W )) are jointly inconsistent. 10. Show that the sentences ¬R ∨C , (C ∧R) →¬R, (¬(R ∨ R) →R) are jointly consistent. 11. Show that ¬¬(C ↔¬C ), ((G ∨C ) ∨G) ∴((G →C ) ∧G) is valid. 12. Show that ¬¬L, (C →¬L) →C ) ∴¬C is invalid. PART V First-order logic 190 CHAPTER 22 Building blocks of FOL 22.1 The need to decompose sentences Consider the following argument, which is obviously valid in En-glish: Willard is a logician. All logicians wear funny hats. ∴Willard wears a funny hat. To symbolize it in TFL, we might offer a symbolization key: L: Willard is a logician. A: All logicians wear funny hats. F : Willard wears a funny hat. And the argument itself becomes: L,A ∴F But the truth-table test will now indicate that this is invalid. What has gone wrong? 191 CHAPTER 22. BUILDING BLOCKS OF FOL 192 The problem is not that we have made a mistake while sym-bolizing the argument. This is the best symbolization we can give in TFL. The problem lies with TFL itself. ‘All logicians wear funny hats’ is about both logicians and hat-wearing. By not re-taining this structure in our symbolization, we lose the connec-tion between Willard’s being a logician and Willard’s wearing a hat. The basic units of TFL are sentence letters, and TFL cannot decompose these. To symbolize arguments like the preceding one, we will have to develop a new logical language which will allow us to split the atom. We will call this language first-order logic, or FOL. The details of FOL will be explained throughout this chapter, but here is the basic idea for splitting the atom. First, we have names. In FOL, we indicate these with lowercase italic letters. For instance, we might let ‘b’ stand for Bertie, or let ‘i’ stand for Willard. Second, we have predicates. English predicates are expres-sions like ‘ is a dog’ or ‘ is a logician’. These are not complete sentences by themselves. In order to make a complete sentence, we need to fill in the gap. We need to say something like ‘Bertie is a dog’ or ‘Willard is a logician’. In FOL, we in-dicate predicates with uppercase italic letters. For instance, we might let the FOL predicate ‘D’ symbolize the English predicate ‘ is a dog’. Then the expression ‘D(b)’ will be a sentence in FOL, which symbolizes the English sentence ‘Bertie is a dog’. Equally, we might let the FOL predicate ‘L’ symbolize the En-glish predicate ‘ is a logician’. Then the expression ‘L(i)’ will symbolize the English sentence ‘Willard is a logician’. Third, we have quantifiers. For instance, ‘∃’ will roughly con-vey ‘There is at least one . . . ’. So we might symbolize the English sentence ‘there is a dog’ with the FOL sentence ‘∃x D(x)’, which we might read aloud as ‘there is at least one thing, x, such that x is a dog’. That is the general idea, but FOL is significantly more subtle than TFL, so we will come at it slowly. CHAPTER 22. BUILDING BLOCKS OF FOL 193 22.2 Names In English, a singular term is a word or phrase that refers to a specific person, place, or thing. The word ‘dog’ is not a singular term, because there are a great many dogs. The phrase ‘Bertie’ is a singular term, because it refers to a specific terrier. Likewise, the phrase ‘Philip’s dog Bertie’ is a singular term, because it refers to a specific little terrier. Proper names are a particularly important kind of singular term. These are expressions that pick out individuals without describing them. The name ‘Emerson’ is a proper name, and the name alone does not tell you anything about Emerson. Of course, some names are traditionally given to boys and others are traditionally given to girls. If ‘Hilary’ is used as a singular term, you might guess that it refers to a woman. You might, though, be guessing wrongly. Indeed, the name does not necessarily mean that the person referred to is even a person: Hilary might be a giraffe, for all you could tell just from the name. In FOL, our names are lower-case letters ‘a’ through to ‘r’. We can add subscripts if we want to use some letter more than once. So here are some singular terms in FOL: a,b,c,. . . ,r,a1, f32, j390,m12 These should be thought of along the lines of proper names in English, but with one difference. ‘Tim Button’ is a proper name, but there are several people with this name. (Equally, there are at least two people with the name ‘P. D. Magnus’.) We live with this kind of ambiguity in English, allowing context to individuate the fact that ‘Tim Button’ refers to an author of this book, and not some other Tim. In FOL, we do not tolerate any such ambi-guity. Each name must pick out exactly one thing. (However, two different names may pick out the same thing.) As with TFL, we can provide symbolization keys. These indi-cate, temporarily, what a name will pick out. So we might offer: e: Elsa CHAPTER 22. BUILDING BLOCKS OF FOL 194 g: Gregor m: Marybeth 22.3 Predicates The simplest predicates are properties of individuals. They are things you can say about an object. Here are some examples of English predicates: is a dog is a member of Monty Python A piano fell on In general, you can think about predicates as things which com-bine with singular terms to make sentences. Conversely, you can start with sentences and make predicates out of them by remov-ing terms. Consider the sentence, ‘Vinnie borrowed the family car from Nunzio.’ By removing a singular term, we can obtain any of three different predicates: borrowed the family car from Nunzio Vinnie borrowed from Nunzio Vinnie borrowed the family car from In FOL, predicates are capital letters ‘A’ through ‘Z’, with or without subscripts. We might write a symbolization key for pred-icates thus: A(x): x is angry H (x): x is happy (Why the subscripts on the gaps? We will return to this in §24.) If we combine our two symbolization keys, we can start to symbolize some English sentences that use these names and pred-icates in combination. For example, consider the English sen-tences: 1. Elsa is angry. CHAPTER 22. BUILDING BLOCKS OF FOL 195 2. Gregor and Marybeth are angry. 3. If Elsa is angry, then so are Gregor and Marybeth. Sentence 1 is straightforward: we symbolize it by ‘A(e)’. Sentence 2: this is a conjunction of two simpler sentences. The simple sentences can be symbolized just by ‘A(g)’ and ‘A(m)’. Then we help ourselves to our resources from TFL, and symbolize the entire sentence by ‘A(g)∧A(m)’. This illustrates an important point: FOL has all of the truth-functional connectives of TFL. Sentence 3: this is a conditional, whose antecedent is sen-tence 1 and whose consequent is sentence 2, so we can symbolize this with ‘A(e) →(A(g) ∧A(m))’. 22.4 Quantifiers We are now ready to introduce quantifiers. Consider these sen-tences: 4. Everyone is happy. 5. Someone is angry. It might be tempting to symbolize sentence 4 as ‘H (e) ∧H (g) ∧ H (m)’. Yet this would only say that Elsa, Gregor, and Marybeth are happy. We want to say that everyone is happy, even those with no names. In order to do this, we introduce the ‘∀’ symbol. This is called the universal quantifier. A quantifier must always be followed by a v ariable. In FOL, variables are italic lowercase letters ‘s’ through ‘z’, with or with-out subscripts. So we might symbolize sentence 4 as ‘∀x H (x)’. The variable ‘x’ is serving as a kind of placeholder. The ex-pression ‘∀x’ intuitively means that you can pick anyone and put them in as ‘x’. The subsequent ‘H (x)’ indicates, of that thing you picked out, that it is happy. It should be pointed out that there is no special reason to use ‘x’ rather than some other variable. The sentences ‘∀x H (x)’, CHAPTER 22. BUILDING BLOCKS OF FOL 196 ‘∀y H (y)’, ‘∀z H (z)’, and ‘∀x5H (x5)’ use different variables, but they will all be logically equivalent. To symbolize sentence 5, we introduce another new symbol: the existential quantifier, ‘∃’. Like the universal quantifier, the existential quantifier requires a variable. Sentence 5 can be symbolized by ‘∃x A(x)’. Whereas ‘∀x A(x)’ is read naturally as ‘for all x, x is angry’, ‘∃x A(x)’ is read naturally as ‘there is some-thing, x, such that x is angry’. Once again, the variable is a kind of placeholder; we could just as easily have symbolized sentence 5 by ‘∃z A(z)’, ‘∃w256 A(w256)’, or whatever. Some more examples will help. Consider these further sen-tences: 6. No one is angry. 7. There is someone who is not happy. 8. Not everyone is happy. Sentence 6 can be paraphrased as, ‘It is not the case that some-one is angry’. We can then symbolize it using negation and an existential quantifier: ‘¬∃x A(x)’. Yet sentence 6 could also be paraphrased as, ‘Everyone is not angry’. With this in mind, it can be symbolized using negation and a universal quantifier: ‘∀x ¬A(x)’. Both of these are acceptable symbolizations. Indeed, it will transpire that, in general, ∀x ¬A is logically equivalent to ¬∃x A. (Notice that we have here returned to the practice of us-ing ‘A’ as a metavariable, from §8.) Symbolizing a sentence one way, rather than the other, might seem more ‘natural’ in some contexts, but it is not much more than a matter of taste. Sentence 7 is most naturally paraphrased as, ‘There is some x, such that x is not happy’. This then becomes ‘∃x ¬H (x)’. Of course, we could equally have written ‘¬∀x H (x)’, which we would naturally read as ‘it is not the case that everyone is happy’. That too would be a perfectly adequate symbolization of sen-tence 8. CHAPTER 22. BUILDING BLOCKS OF FOL 197 22.5 Domains Given the symbolization key we have been using, ‘∀x H (x)’ sym-bolizes ‘Everyone is happy’. Who is included in this everyone? When we use sentences like this in English, we usually do not mean everyone now alive on the Earth. We certainly do not mean everyone who was ever alive or who will ever live. We usually mean something more modest: everyone now in the building, everyone enrolled in the ballet class, or whatever. In order to eliminate this ambiguity, we will need to specify a domain. The domain is the collection of things that we are talking about. So if we want to talk about people in Chicago, we define the domain to be people in Chicago. We write this at the beginning of the symbolization key, like this: domain: people in Chicago The quantifiers range over the domain. Given this domain, ‘∀x’ is to be read roughly as ‘Every person in Chicago is such that...’ and ‘∃x’ is to be read roughly as ‘Some person in Chicago is such that...’. In FOL, the domain must always include at least one thing. Moreover, in English we can legitimately infer ‘something is an-gry’ from ‘Gregor is angry’. In FOL, then, we will want to be able to infer ‘∃x A(x)’ from ‘A(g)’. So we will insist that each name must pick out exactly one thing in the domain. If we want to name people in places beside Chicago, then we need to include those people in the domain. A domain must have at least one member. Every name must pick out exactly one member of the domain, but a member of the domain may be picked out by one name, many names, or none at all. Even allowing for a domain with just one member can pro-duce some strange results. Suppose we have this as a symboliza-tion key: CHAPTER 22. BUILDING BLOCKS OF FOL 198 domain: the Eiffel Tower P (x): x is in Paris. The sentence ∀x P (x) might be paraphrased in English as ‘Ev-erything is in Paris.’ Yet that would be misleading. It means that everything in the domain is in Paris. This domain contains only the Eiffel Tower, so with this symbolization key ∀x P (x) just means that the Eiffel Tower is in Paris. Non-referring terms In FOL, each name must pick out exactly one member of the domain. A name cannot refer to more than one thing—it is a singular term. Each name must still pick out something. This is connected to a classic philosophical problem: the so-called prob-lem of non-referring terms. Medieval philosophers typically used sentences about the chimera to exemplify this problem. Chimera is a mythological creature; it does not really exist. Consider these two sentences: 9. Chimera is angry. 10. Chimera is not angry. It is tempting just to define a name to mean ‘chimera.’ The sym-bolization key would look like this: domain: creatures on Earth A(x): x is angry. c: chimera We could then symbolize sentence 9 as A(c) and sentence 10 as ¬A(c). Problems will arise when we ask whether these sentences are true or false. One option is to say that sentence 9 is not true, because there is no chimera. If sentence 9 is false because it talks about a non-existent thing, then sentence 10 is false for the same reason. Yet CHAPTER 22. BUILDING BLOCKS OF FOL 199 this would mean that A(c) and ¬A(c) would both be false. Given the truth conditions for negation, this cannot be the case. Since we cannot say that they are both false, what should we do? Another option is to say that sentence 9 is meaningless because it talks about a non-existent thing. So A(c) would be a meaningful expression in FOL for some interpretations but not for others. Yet this would make our formal language hostage to particular interpretations. Since we are interested in logical form, we want to consider the logical force of a sentence like A(c) apart from any particular interpretation. If A(c) were sometimes meaningful and sometimes meaningless, we could not do that. This is the problem of non-referring terms, and we will return to it later (see p. 244.) The important point for now is that each name of FOL must refer to something in the domain, although the domain can contain any things we like. If we want to symbolize arguments about mythological creatures, then we must define a domain that includes them. This option is important if we want to consider the logic of stories. We can symbolize a sentence like ‘Sherlock Holmes lived at 221B Baker Street’ by including fictional characters like Sherlock Holmes in our domain. CHAPTER 23 Sentences with one quantifier We now have all of the pieces of FOL. Symbolizing more com-plicated sentences is just a matter of knowing how to combine predicates, names, quantifiers, and connectives. There is a knack to this, and there is no substitute for practice. 23.1 Common quantifier phrases Consider these sentences: 1. Every coin in my pocket is a quarter. 2. Some coin on the table is a dime. 3. Not all the coins on the table are dimes. 4. None of the coins in my pocket are dimes. In providing a symbolization key, we need to specify a domain. Since we are talking about coins in my pocket and on the table, the domain must at least contain all of those coins. Since we are not talking about anything besides coins, we let the domain be all coins. Since we are not talking about any specific coins, we do not need to deal with any names. So here is our key: 200 CHAPTER 23. SENTENCES WITH ONE QUANTIFIER 201 domain: all coins P (x): x is in my pocket T (x): x is on the table Q (x): x is a quarter D(x): x is a dime Sentence 1 is most naturally symbolized using a universal quanti-fier. The universal quantifier says something about everything in the domain, not just about the coins in my pocket. Sentence 1 can be paraphrased as ‘for any coin, if that coin is in my pocket then it is a quarter’. So we can symbolize it as ‘∀x(P (x) →Q (x))’. Since sentence 1 is about coins that are both in my pocket and that are quarters, it might be tempting to symbolize it using a conjunction. However, the sentence ‘∀x(P (x) ∧Q (x))’ would symbolize the sentence ‘every coin is both a quarter and in my pocket’. This obviously means something very different than sen-tence 1. And so we see: A sentence can be symbolized as ∀x(F(x) →G(x)) if it can be paraphrased in English as ‘every F is G’. Sentence 2 is most naturally symbolized using an existential quantifier. It can be paraphrased as ‘there is some coin which is both on the table and which is a dime’. So we can symbolize it as ‘∃x(T (x) ∧D(x))’. Notice that we needed to use a conditional with the universal quantifier, but we used a conjunction with the existential quanti-fier. Suppose we had instead written ‘∃x(T (x) →D(x))’. That would mean that there is some object in the domain of which ‘(T (x) →D(x))’ is true. Recall that, in TFL, A →B is logically equivalent (in TFL) to ¬A∨B. This equivalence will also hold in FOL. So ‘∃x(T (x) →D(x))’ is true if there is some object in the domain, such that ‘(¬T (x) ∨D(x))’ is true of that object. That is, ‘∃x(T (x) →D(x))’ is true if some coin is either not on the table or is a dime. Of course there is a coin that is not on the table: there are coins in lots of other places. So it is very easy CHAPTER 23. SENTENCES WITH ONE QUANTIFIER 202 for ‘∃x(T (x) →D(x))’ to be true. A conditional will usually be the natural connective to use with a universal quantifier, but a conditional within the scope of an existential quantifier tends to say something very weak indeed. As a general rule of thumb, do not put conditionals in the scope of existential quantifiers unless you are sure that you need one. A sentence can be symbolized as ∃x(F(x) ∧G(x)) if it can be paraphrased in English as ‘some F is G’. Sentence 3 can be paraphrased as, ‘It is not the case that every coin on the table is a dime’. So we can symbolize it by ‘¬∀x(T (x) →D(x))’. You might look at sentence 3 and para-phrase it instead as, ‘Some coin on the table is not a dime’. You would then symbolize it by ‘∃x(T (x) ∧¬D(x))’. Although it is probably not immediately obvious yet, these two sentences are logically equivalent. (This is due to the logical equivalence be-tween ¬∀x A and ∃x¬A, mentioned in §22, along with the equiv-alence between ¬(A →B) and A∧¬B.) Sentence 4 can be paraphrased as, ‘It is not the case that there is some dime in my pocket’. This can be symbolized by ‘¬∃x(P (x) ∧D(x))’. It might also be paraphrased as, ‘Everything in my pocket is a non-dime’, and then could be symbolized by ‘∀x(P (x) →¬D(x))’. Again the two symbolizations are logically equivalent; both are correct symbolizations of sentence 4. A sentence that can be paraphrased as ‘no F is G’ can be symbolized as ¬∃x(F(x) ∧G(x)) and also as ∀x(F(x) → ¬G(x)). Finally, consider ‘only’, as in: 5. Only dimes are on the table. How should we symbolize this? A good strategy is to consider when the sentence would be false. If we are saying that only dimes CHAPTER 23. SENTENCES WITH ONE QUANTIFIER 203 are on the table, we are excluding all the cases where something on the table is a non-dime. So we can symbolize the sentence the same way we would symbolize ‘No non-dimes are on the ta-ble.’ Remembering the lesson we just learned, and symbolizing ‘x is a non-dime’ as ‘¬D(x)’, the possible symbolizations are: ‘¬∃x(T (x) ∧¬D(x))’, or alternatively: ‘∀x(T (x) →¬¬D(x))’. Since double negations cancel out, the second is just as good as ‘∀x(T (x) →D(x))’. In other words, ‘Only dimes are on the ta-ble’ and ‘Everything on the table is a dime’ are symbolized the same way. A sentence that can be paraphrased as ‘only F s are Gs’ can be symbolized as ¬∃x(G(x) ∧¬F(x)) and also as ∀x(G(x) →F(x)). 23.2 Empty predicates In §22, we emphasized that a name must pick out exactly one object in the domain. However, a predicate need not apply to anything in the domain. A predicate that applies to nothing in the domain is called an empty predicate. This is worth exploring. Suppose we want to symbolize these two sentences: 6. Every monkey knows sign language. 7. Some monkey knows sign language. It is possible to write the symbolization key for these sentences in this way: domain: animals M (x): x is a monkey. S (x): x knows sign language. Sentence 6 can now be symbolized by ‘∀x(M (x) →S (x))’. Sen-tence 7 can be symbolized as ‘∃x(M (x) ∧S (x))’. It is tempting to say that sentence 6 entails sentence 7. That is, we might think that it is impossible that every monkey knows CHAPTER 23. SENTENCES WITH ONE QUANTIFIER 204 sign language unless some monkey knows sign language. But this would be a mistake. It is possible for the sentence ‘∀x(M (x) → S (x))’ to be true even though the sentence ‘∃x(M (x) ∧S (x))’ is false. How can this be? The answer comes from considering whether these sentences would be true or false if there were no monkeys. If there were no monkeys at all (in the domain), then ‘∀x(M (x) →S (x))’ would be vacuously true: take any monkey you like—it knows sign language! But if there were no monkeys at all (in the domain), then ‘∃x(M (x) ∧S (x))’ would be false. Another example will help to bring this home. Suppose we extend the above symbolization key, by adding: R(x): x is a refrigerator Now consider the sentence ‘∀x(R(x) →M (x))’. This symbol-izes ‘every refrigerator is a monkey’. This sentence is true, given our symbolization key, which is counterintuitive, since we (pre-sumably) do not want to say that there are a whole bunch of refrigerator monkeys. It is important to remember, though, that ‘∀x(R(x) →M (x))’ is true iff any member of the domain that is a refrigerator is a monkey. Since the domain is animals, there are no refrigerators in the domain. Again, then, the sentence is vacuously true. If you were actually dealing with the sentence ‘All refrigera-tors are monkeys’, then you would most likely want to include kitchen appliances in the domain. Then the predicate ‘R’ would not be empty and the sentence ‘∀x(R(x) →M (x))’ would be false. When F is an empty predicate, any sentence ∀x(F(x) → . . .) is vacuously true. CHAPTER 23. SENTENCES WITH ONE QUANTIFIER 205 23.3 Picking a domain The appropriate symbolization of an English language sentence in FOL will depend on the symbolization key. Choosing a key can be difficult. Suppose we want to symbolize the English sentence: 8. Every rose has a thorn. We might offer this symbolization key: R(x): x is a rose T (x): x has a thorn It is tempting to say that sentence 8 should be symbolized as ‘∀x(R(x) →T (x))’, but we have not yet chosen a domain. If the domain contains all roses, this would be a good symboliza-tion. Yet if the domain is merely things on my kitchen table, then ‘∀x(R(x) →T (x))’ would only come close to covering the fact that every rose on my kitchen table has a thorn. If there are no roses on my kitchen table, the sentence would be trivially true. This is not what we want. To symbolize sentence 8 adequately, we need to include all the roses in the domain, but now we have two options. First, we can restrict the domain to include all roses but only roses. Then sentence 8 can, if we like, be symbolized with ‘∀x T (x)’. This is true iff everything in the domain has a thorn; since the domain is just the roses, this is true iff every rose has a thorn. By restricting the domain, we have been able to symbolize our English sentence with a very short sentence of FOL. So this approach can save us trouble, if every sentence that we want to deal with is about roses. Second, we can let the domain contain things besides roses: rhododendrons, rats, rifles, whatevers; and we will certainly need to include a more expansive domain if we simultaneously want to symbolize sentences like: 9. Every cowboy sings a sad, sad song. CHAPTER 23. SENTENCES WITH ONE QUANTIFIER 206 Our domain must now include both all the roses (so that we can symbolize sentence 8) and all the cowboys (so that we can sym-bolize sentence 9). So we might offer the following symbolization key: domain: people and plants C (x): x is a cowboy S (x): x sings a sad, sad song R(x): x is a rose T (x): x has a thorn Now we will have to symbolize sentence 8 with ‘∀x(R(x) → T (x))’, since ‘∀x T (x)’ would symbolize the sentence ‘every per-son or plant has a thorn’. Similarly, we will have to symbolize sentence 9 with ‘∀x(C (x) →S (x))’. In general, the universal quantifier can be used to symbolize the English expression ‘everyone’ if the domain only contains people. If there are people and other things in the domain, then ‘everyone’ must be treated as ‘every person’. 23.4 The utility of paraphrase When symbolizing English sentences in FOL, it is important to understand the structure of the sentences you want to symbolize. What matters is the final symbolization in FOL, and sometimes you will be able to move from an English language sentence di-rectly to a sentence of FOL. Other times, it helps to paraphrase the sentence one or more times. Each successive paraphrase should move from the original sentence closer to something that you can easily symbolize directly in FOL. For the next several examples, we will use this symbolization key: domain: people B(x): x is a bassist R(x): x is a rock star CHAPTER 23. SENTENCES WITH ONE QUANTIFIER 207 k: Kim Deal Now consider these sentences: 10. If Kim Deal is a bassist, then she is a rock star. 11. If a person is a bassist, then she is a rock star. The same words appear as the consequent in sentences 10 and 11 (‘. . . she is a rock star’), but they mean very different things. To make this clear, it often helps to paraphrase the original sen-tences, removing pronouns. Sentence 10 can be paraphrased as, ‘If Kim Deal is a bassist, then Kim Deal is a rockstar’. This can obviously be symbolized as ‘B(k) →R(k)’. Sentence 11 must be paraphrased differently: ‘If a person is a bassist, then that person is a rock star’. This sentence is not about any particular person, so we need a variable. As an inter-mediate step, we can paraphrase this as, ‘For any person x, if x is a bassist, then x is a rockstar’. Now this can be symbolized as ‘∀x(B(x) →R(x))’. This is the same sentence we would have used to symbolize ‘Everyone who is a bassist is a rock star’. On reflection, that is surely true iff sentence 11 is true, as we would hope. Consider these further sentences: 12. If anyone is a bassist, then Kim Deal is a rock star. 13. If anyone is a bassist, then she is a rock star. The same words appear as the antecedent in sentences 12 and 13 (‘If anyone is a bassist. . .’), but it can be tricky to work out how to symbolize these two uses. Again, paraphrase will come to our aid. Sentence 12 can be paraphrased, ‘If there is at least one bassist, then Kim Deal is a rock star’. It is now clear that this is a conditional whose antecedent is a quantified expression; so we can symbolize the entire sentence with a conditional as the main logical operator: ‘∃xB(x) →R(k)’. CHAPTER 23. SENTENCES WITH ONE QUANTIFIER 208 Sentence 13 can be paraphrased, ‘For all people x, if x is a bassist, then x is a rock star’. Or, in more natural English, it can be paraphrased by ‘All bassists are rock stars’. It is best symbolized as ‘∀x(B(x) →R(x))’, just like sentence 11. The moral is that the English words ‘any’ and ‘anyone’ should typically be symbolized using quantifiers, and if you are having a hard time determining whether to use an existential or a uni-versal quantifier, try paraphrasing the sentence with an English sentence that uses words besides ‘any’ or ‘anyone’. 23.5 Quantifiers and scope Continuing the example, suppose we want to symbolize these sentences: 14. If everyone is a bassist, then Lars is a bassist. 15. Everyone is such that, if they are a bassist, then Lars is a bassist. To symbolize these sentences, we will have to add a new name to the symbolization key, namely: l: Lars Sentence 14 is a conditional, whose antecedent is ‘everyone is a bassist’, so we will symbolize it with ‘∀x B(x) →B(l)’. This sentence is necessarily true: if everyone is indeed a bassist, then take anyone you like—for example Lars—and he will be a bassist. Sentence 15, by contrast, might best be paraphrased by ‘every person x is such that, if x is a bassist, then Lars is a bassist’. This is symbolized by ‘∀x(B(x) →B(l))’. This sentence can well be false. For instance, Kim Deal is a bassist. So ‘B(k)’ is true. Suppose that Lars is not a bassist (say, he’s a drummer instead), so ‘B(l)’ is false. Accordingly, ‘B(k) →B(l)’ will be false, so ‘∀x(B(x) →B(l))’ will be false as well. In short, ‘∀xB(x) →B(l)’ and ‘∀x(B(x) →B(l))’ are very different sentences. We can explain the difference in terms of the CHAPTER 23. SENTENCES WITH ONE QUANTIFIER 209 scope of the quantifier. The scope of quantification is very much like the scope of negation, which we considered when discussing TFL, and it will help to explain it in this way. In the sentence ‘¬B(k) →B(l)’, the scope of ‘¬’ is just the antecedent of the conditional. We are saying something like: if ‘B(k)’ is false, then ‘B(l)’ is true. Similarly, in the sentence ‘∀xB(x) →B(l)’, the scope of ‘∀x’ is just the antecedent of the conditional. We are saying something like: if ‘B(x)’ is true of everything, then ‘B(l)’ is also true. In the sentence ‘¬(B(k) →B(l))’, the scope of ‘¬’ is the en-tire sentence. We are saying something like: ‘(B(k) →B(l))’ is false. Similarly, in the sentence ‘∀x(B(x) →B(l))’, the scope of ‘∀x’ is the entire sentence. We are saying something like: ‘(B(x) →B(l))’ is true of everything. The moral of the story is simple. When you are using condi-tionals, be very careful to make sure that you have sorted out the scope correctly. Ambiguous predicates Suppose we just want to symbolize this sentence: 16. Adina is a skilled surgeon. Let the domain be people, let K (x) mean ‘x is a skilled surgeon’, and let a mean Adina. Sentence 16 is simply K (a). Suppose instead that we want to symbolize this argument: The hospital will only hire a skilled surgeon. All sur-geons are greedy. Billy is a surgeon, but is not skilled. Therefore, Billy is greedy, but the hospital will not hire him. We need to distinguish being a skilled surgeon from merely being a surgeon. So we define this symbolization key: domain: people G (x): x is greedy CHAPTER 23. SENTENCES WITH ONE QUANTIFIER 210 H (x): The hospital will hire x R(x): x is a surgeon K (x): x is skilled b: Billy Now the argument can be symbolized in this way: ∀x [︁ ¬(R(x) ∧K (x)) →¬H (x) ]︁ ∀x(R(x) →G (x)) R(b) ∧¬K (b) ∴G (b) ∧¬H (b) Next suppose that we want to symbolize this argument: Carol is a skilled surgeon and a tennis player. There-fore, Carol is a skilled tennis player. If we start with the symbolization key we used for the previous argument, we could add a predicate (let T (x) mean ‘x is a tennis player’) and a name (let c mean Carol). Then the argument becomes: (R(c) ∧K (c)) ∧T (c) ∴T (c) ∧K (c) This symbolization is a disaster! It takes what in English is a terrible argument and symbolizes it as a valid argument in FOL. The problem is that there is a difference between being skilled as a surgeon and skilled as a tennis player. Symbolizing this argument correctly requires two separate predicates, one for each type of skill. If we let K1(x) mean ‘x is skilled as a surgeon’ and K2(x) mean ‘x is skilled as a tennis player,’ then we can symbolize the argument in this way: (R(c) ∧K1(c)) ∧T (c) ∴T (c) ∧K2(c) Like the English language argument it symbolizes, this is invalid. CHAPTER 23. SENTENCES WITH ONE QUANTIFIER 211 The moral of these examples is that you need to be careful of symbolizing predicates in an ambiguous way. Similar problems can arise with predicates like good, bad, big, and small. Just as skilled surgeons and skilled tennis players have different skills, big dogs, big mice, and big problems are big in different ways. Is it enough to have a predicate that means ‘x is a skilled surgeon’, rather than two predicates ‘x is skilled’ and ‘x is a sur-geon’? Sometimes. As sentence 16 shows, sometimes we do not need to distinguish between skilled surgeons and other surgeons. Must we always distinguish between different ways of being skilled, good, bad, or big? No. As the argument about Billy shows, sometimes we only need to talk about one kind of skill. If you are symbolizing an argument that is just about dogs, it is fine to define a predicate that means ‘x is big.’ If the domain includes dogs and mice, however, it is probably best to make the predicate mean ‘x is big for a dog.’ Practice exercises A. Here are the syllogistic figures identified by Aristotle and his successors, along with their medieval names: 1. Barbara. All G are F. All H are G. So: All H are F. 2. Celarent. No G are F. All H are G. So: No H are F. 3. Ferio. No G are F. Some H is G. So: Some H is not F. 4. Darii. All G are F. Some H is G. So: Some H is F. 5. Camestres. All F are G. No H are G. So: No H are F. 6. Cesare. No F are G. All H are G. So: No H are F. 7. Baroko. All F are G. Some H is not G. So: Some H is not F. 8. Festino. No F are G. Some H are G. So: Some H is not F. 9. Datisi. All G are F. Some G is H. So: Some H is F. 10. Disamis. Some G is F. All G are H. So: Some H is F. 11. Ferison. No G are F. Some G is H. So: Some H is not F. 12. Bokardo. Some G is not F. All G are H. So: Some H is not F. 13. Camenes. All F are G. No G are H. So: No H is F. CHAPTER 23. SENTENCES WITH ONE QUANTIFIER 212 14. Dimaris. Some F is G. All G are H. So: Some H is F. 15. Fresison. No F are G. Some G is H. So: Some H is not F. Symbolize each figure in FOL. B. Using the following symbolization key: domain: people K (x): x knows the combination to the safe S (x): x is a spy V (x): x is a vegetarian h: Hofthor i: Ingmar symbolize the following sentences in FOL: 1. Neither Hofthor nor Ingmar is a vegetarian. 2. No spy knows the combination to the safe. 3. No one knows the combination to the safe unless Ingmar does. 4. Hofthor is a spy, but no vegetarian is a spy. C. Using this symbolization key: domain: all animals A(x): x is an alligator M (x): x is a monkey R(x): x is a reptile Z (x): x lives at the zoo a: Amos b: Bouncer c: Cleo symbolize each of the following sentences in FOL: 1. Amos, Bouncer, and Cleo all live at the zoo. 2. Bouncer is a reptile, but not an alligator. 3. Some reptile lives at the zoo. 4. Every alligator is a reptile. CHAPTER 23. SENTENCES WITH ONE QUANTIFIER 213 5. Any animal that lives at the zoo is either a monkey or an alligator. 6. There are reptiles that are not alligators. 7. If any animal is an reptile, then Amos is. 8. If any animal is an alligator, then it is a reptile. D. For each argument, write a symbolization key and symbolize the argument in FOL. 1. Willard is a logician. All logicians wear funny hats. So Willard wears a funny hat. 2. Nothing on my desk escapes my attention. There is a com-puter on my desk. As such, there is a computer that does not escape my attention. 3. All my dreams are black and white. Old TV shows are in black and white. Therefore, some of my dreams are old TV shows. 4. Neither Holmes nor Watson has been to Australia. A per-son could have seen a kangaroo only if they had been to Australia or to a zoo. Although Watson has not seen a kangaroo, Holmes has. Therefore, Holmes has been to a zoo. 5. No one expects the Spanish Inquisition. No one knows the troubles I’ve seen. Therefore, anyone who expects the Spanish Inquisition knows the troubles I’ve seen. 6. All babies are illogical. Nobody who is illogical can man-age a crocodile. Berthold is a baby. Therefore, Berthold is unable to manage a crocodile. CHAPTER 24 Multiple generality So far, we have only considered sentences that require one-place predicates and one quantifier. The full power of FOL really comes out when we start to use many-place predicates and multiple quantifiers. For this insight, we largely have Gottlob Frege (1879) to thank, but also C. S. Peirce. 24.1 Many-placed predicates All of the predicates that we have considered so far concern prop-erties that objects might have. Those predicates have one gap in them, and to make a sentence, we simply need to slot in one term. They are one-place predicates. However, other predicates concern the relation between two things. Here are some examples of relational predicates in En-glish: loves is to the left of is in debt to 214 CHAPTER 24. MULTIPLE GENERALITY 215 These are two-place predicates. They need to be filled in with two terms in order to make a sentence. Conversely, if we start with an English sentence containing many singular terms, we can re-move two singular terms, to obtain different two-place predicates. Consider the sentence ‘Vinnie borrowed the family car from Nun-zio’. By deleting two singular terms, we can obtain any of three different two-place predicates Vinnie borrowed from borrowed the family car from borrowed from Nunzio and by removing all three singular terms, we obtain a three-place predicate: borrowed from Indeed, there is no in principle upper limit on the number of places that our predicates may contain. 24.2 Mind the gap(s)! We have used the same symbol, ‘ ’, to indicate a gap formed by deleting a term from a sentence. However, as Frege empha-sized, these are different gaps. To obtain a sentence, we can fill them in with the same term, but we can equally fill them in with different terms, and in various different orders. The following are three perfectly good sentences, obtained by filling in the gaps in ‘ loves ’ in different ways; but they all have distinc-tively different meanings: 1. Karl loves Imre. 2. Imre loves Karl. 3. Karl loves Karl. The point is that we need to keep track of the gaps in predicates, so that we can keep track of how we are filling them in. To keep CHAPTER 24. MULTIPLE GENERALITY 216 track of the gaps, we assign them variables. Suppose we want to symbolize the preceding sentences. Then I might start with the following representation key: domain: people i: Imre k: Karl L(x,y): x loves y Sentence 1 will be symbolized by ‘L(k,i)’, sentence 2 will be symbolized by ‘L(i,k)’, and sentence 3 will be symbolized by ‘L(k,k)’. Here are a few more sentences that we can symbolize with the same key: 4. Imre loves himself. 5. Karl loves Imre, but not vice versa. 6. Karl is loved by Imre. Sentence 4 can be paraphrased as ‘Imre loves Imre’, and so sym-bolized by ‘L(i,i)’. Sentence 5 is a conjunction. We can para-phrase it as ‘Karl loves Imre, and Imre does not love Karl’, and so symbolize it as ‘L(k,i) ∧¬L(i,k)’. Sentence 6 can be para-phrased by ‘Imre loves Karl’, and so symbolized as ‘L(i,k)’. In this last case, of course, we have lost the difference in tone between the active and passive voice; but we have at least preserved the truth conditions. But the relationship between ‘Imre loves Karl’ and ‘Karl is loved by Imre’ highlights something important. To see what, sup-pose we add another entry to our symbolization key: M (x,y): y loves x The entry for ‘M ’ uses exactly the same English word—‘loves’— as the entry for ‘L’. But the gaps have been swapped around! ( Just look closely at the subscripts.) And this matters. To explain: when we see a sentence like ‘L(k,i)’, we are being told to take the first name (i.e., ‘k’) and associate its value (i.e., Karl) with the gap labelled ‘x’, then take the second name (i.e., CHAPTER 24. MULTIPLE GENERALITY 217 ‘i’) and associate its value (i.e., Imre) with the gap labelled ‘y’, and so come up with: Karl loves Imre. The sentence ‘M (i,k)’ also tells us to take the first name (i.e., ‘i’) and plug its value into the gap labelled ‘x’, and take the second name (i.e., ‘k) and plug its value into the gap labelled ‘y’, and so come up with: Imre loves Karl. So, ‘L(i,k)’ and ‘M (k,i)’ both symbolize ‘Imre loves Karl’, whereas ‘L(k,i)’ and ‘M (i,k)’ both symbolize ‘Karl loves Imre’. Since love can be unrequited, these are different claims. One last example might be helpful. Suppose we add this to our symbolisation key: P (x,y): x prefers x to y Now the sentence ‘P (i,k)’ symbolizes ‘Imre prefers Imre to Karl’, and ‘P (k,i)’ symbolizes ‘Karl prefers Karl to Imre’. And note that we could have achieved the same effect, if we had instead specified: P (x,y): x prefers themselves to y In any case, the overall moral of this is simple. When dealing with predicates with more than one place, pay careful attention to the order of the gaps! 24.3 The order of quantifiers Consider the sentence ‘everyone loves someone’. This is poten-tially ambiguous. It might mean either of the following: 7. For every person x, there is some person that x loves. 8. There is some particular person whom every person loves. Sentence 7 can be symbolized by ‘∀x∃y L(x,y)’, and would be true of a love-triangle. For example, suppose that our domain of discourse is restricted to Imre, Juan and Karl. Suppose also that Karl loves Imre but not Juan, that Imre loves Juan but not Karl, and that Juan loves Karl but not Imre. Then sentence 7 is true. CHAPTER 24. MULTIPLE GENERALITY 218 Sentence 8 is symbolized by ‘∃y∀x L(x,y)’. Sentence 8 is not true in the situation just described. Again, suppose that our do-main of discourse is restricted to Imre, Juan and Karl. Then all of Juan, Imre and Karl must converge on (at least) one object of love. The point of the example is to illustrate that the order of the quantifiers matters a great deal. Indeed, to switch them around is called a quantifier shift fallacy. Here is an example, which comes up in various forms throughout the philosophical literature: For every person, there is some truth they cannot know. (∀∃) ∴There is some particular truth that no person can know. (∃∀) This argument form is obviously invalid. It’s just as bad as:1 Every dog has its day. (∀∃) ∴There is a day for all the dogs. (∃∀) The order of quantifiers is also important in definitions in mathematics. For instance, there is a big difference between pointwise and uniform continuity of functions: ⊲A function f is pointwise continuous if ∀𝜖∀x∀y∃𝛿( \|︁ \|︁x −y \|︁ \|︁< 𝛿→ \|︁ \|︁F (x) −f (y) \|︁ \|︁< 𝜖) ⊲A function f is uniformly continuous if ∀𝜖∃𝛿∀x∀y( \|︁ \|︁x −y \|︁ \|︁< 𝛿→ \|︁ \|︁F (x) −f (y) \|︁ \|︁< 𝜖) The moral is simple: take great care with the order of your quan-tifiers! 1Thanks to Rob Trueman for the example. CHAPTER 24. MULTIPLE GENERALITY 219 24.4 Stepping-stones to symbolization As we are starting to see, symbolization in FOL can become a bit tricky. So, when symbolizing a complex sentence, you should lay down several stepping-stones. As usual, the idea is best illustrated by example. Consider this symbolisation key: domain: people and dogs D(x): x is a dog F (x,y): x is a friend of y O (x,y): x owns y g: Geraldo Now let’s try to symbolize these sentences: 9. Geraldo is a dog owner. 10. Someone is a dog owner. 11. All of Geraldo’s friends are dog owners. 12. Every dog owner is a friend of a dog owner. 13. Every dog owner’s friend owns a dog of a friend. Sentence 9 can be paraphrased as, ‘There is a dog that Geraldo owns’. This can be symbolized by ‘∃x(D(x) ∧O (g,x))’. Sentence 10 can be paraphrased as, ‘There is some y such that y is a dog owner’. Dealing with part of this, we might write ‘∃y(y is a dog owner)’. Now the fragment we have left as ‘y is a dog owner’ is much like sentence 9, except that it is not specifi-cally about Geraldo. So we can symbolize sentence 10 by: ∃y∃x(D(x) ∧O (y,x)) We should pause to clarify something here. In working out how to symbolize the last sentence, we wrote down ‘∃y(y is a dog owner)’. To be very clear: this is neither an FOL sentence nor an English sentence: it uses bits of FOL (‘∃’, ‘y’) and bits of English (‘dog owner’). It is really just a stepping-stone on the way to symbolizing the entire English sentence with a sen-tence of FOL. You should regard it as a bit of rough-working-out, CHAPTER 24. MULTIPLE GENERALITY 220 on a par with the doodles that you might absent-mindedly draw in the margin of this book, whilst you are concentrating fiercely on some problem. Sentence 11 can be paraphrased as, ‘Every x who is a friend of Geraldo is a dog owner’. Using our stepping-stone tactic, we might write ∀x [︁ F (x,g) →x is a dog owner ]︁ Now the fragment that we have left to deal with, ‘x is a dog owner’, is structurally just like sentence 9. However, it would be a mistake for us simply to write ∀x [︁ F (x,g) →∃x(D(x) ∧O (x,x)) ]︁ for we would here have a clash of variables. The scope of the universal quantifier, ‘∀x’, is the entire conditional, so the ‘x’ in ‘D(x)’ should be governed by that, but ‘D(x)’ also falls under the scope of the existential quantifier ‘∃x’, so the ‘x’ in ‘D(x)’ should be governed by that. Now confusion reigns: which ‘x’ are we talking about? Suddenly the sentence becomes ambiguous (if it is even meaningful at all), and logicians hate ambiguity. The broad moral is that a single variable cannot serve two quantifier-masters simultaneously. To continue our symbolization, then, we must choose some different variable for our existential quantifier. What we want is something like: ∀x [︁ F (x,g) →∃z (D(z) ∧O (x,z)) ]︁ This adequately symbolizes sentence 11. Sentence 12 can be paraphrased as ‘For any x that is a dog owner, there is a dog owner who x is a friend of’. Using our stepping-stone tactic, this becomes ∀x [︁ x is a dog owner →∃y(y is a dog owner ∧F (x,y)) ]︁ Completing the symbolization, we end up with ∀x [︁ ∃z (D(z) ∧O (x,z)) →∃y (︁∃z (D(z) ∧O (y,z)) ∧F (x,y))︁]︁ CHAPTER 24. MULTIPLE GENERALITY 221 Note that we have used the same letter, ‘z’, in both the antecedent and the consequent of the conditional, but that these are gov-erned by two different quantifiers. This is ok: there is no clash here, because it is clear which quantifier that variable falls under. We might graphically represent the scope of the quantifiers thus: scope of ‘∀x’ ⏟ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏞⏞ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏟ ∀x [︁ scope of 1st ‘∃z’ ⏟ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏞⏞ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏟ ∃z (D(z) ∧O (x,z)) → scope of ‘∃y’ ⏟ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏞⏞ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏟ ∃y( scope of 2nd ‘∃z’ ⏟ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏞⏞ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏟ ∃z (D(z) ∧O (y,z)) ∧F (x,y)) ]︁ This shows that no variable is being forced to serve two masters simultaneously. Sentence 13 is the trickiest yet. First we paraphrase it as ‘For any x that is a friend of a dog owner, x owns a dog which is also owned by a friend of x’. Using our stepping-stone tactic, this becomes: ∀x [︁ x is a friend of a dog owner → x owns a dog which is owned by a friend of x ]︁ Breaking this down a bit more: ∀x [︁ ∃y(F (x,y) ∧y is a dog owner) → ∃y((D(y) ∧O (x,y)) ∧y is owned by a friend of x) ]︁ And a bit more: ∀x [︁ ∃y(F (x,y) ∧∃z (D(z) ∧O (y,z))) → ∃y((D(y) ∧O (x,y)) ∧∃z (F (z,x) ∧O (z,y))) ]︁ And we are done! There is one subtle issue we should briefly address. We para-phrased sentence 10 as ‘There is some y such that y is a dog CHAPTER 24. MULTIPLE GENERALITY 222 owner’. Now our domain includes people and dogs, and ‘some-one’ includes people, but (at least arguably) does not include dogs. To be more correct, we should have paraphrased sen-tence 10 as ‘There is some y such that y is a person and a dog owner’. A more accurate symbolization of ‘Someone is a dog owner’ would require that we add a predicate for ‘ is a per-son’ to our symbolization key: P (x): x is a person Then we can give a better symbolization of sentence 10: ∃y(P (y) ∧∃x(D(x) ∧O (y,x))) ‘Everyone’ and ‘no one’ have to be treated similarly: ‘Everyone is a friend of Geraldo’ and ‘No one is a friend of Geraldo’ would be symbolized, respectively, as ∀x(P (x) →F (x,g)) ∀x(P (x) →¬F (x,g)). Only ‘someone’, ‘everyone’, and ‘no one’ require this treatment. In particular, we do not need to explicitly state in the symbol-ization of sentence 9 that Geraldo is a person. Neither do we have to ensure in sentence 11 that Geraldo’s friend x is a person. Although it may be true that only people can own dogs or be friends with Geraldo, it is not part of what the sentences say, and so does not need to be taken into account when we symbolize them.2 2You might object: but the sentences also don’t say that dogs aren’t people. E.g., if we’re talking about the fictional world of Mickey Mouse, Goofy should be included in ‘everyone’, but Pluto should not be. That’s why we picked the predicate ‘ is a person’ and not ‘ is a human’: in that domain, Goofy would fall under both ‘ is a person’ and ‘ is a dog’, but Pluto would only fall under ‘ is a dog’. CHAPTER 24. MULTIPLE GENERALITY 223 24.5 Supressed quantifiers Logic can often help to get clear on the meanings of English claims, especially where the quantifiers are left implicit or their order is ambiguous or unclear. The clarity of expression and thinking afforded by FOL can give you a significant advantage in argument, as can be seen in the following takedown by British political philosopher Mary Astell (1666–1731) of her contempo-rary, the theologian William Nicholls. In Discourse IV: The Duty of Wives to their Husbands of his The Duty of Inferiors towards their Superiors, in Five Practical Discourses (London 1701), Nicholls ar-gued that women are naturally inferior to men. In the preface to the 3rd edition of her treatise Some Reflections upon Marriage, Occasion’d by the Duke and Duchess of Mazarine’s Case; which is also considered, Astell responded as follows: ’Tis true, thro’ Want of Learning, and of that Su-perior Genius which Men as Men lay claim to, she [Astell] was ignorant of the Natural Inferiority of our Sex, which our Masters lay down as a Self-Evident and Fundamental Truth. She saw nothing in the Rea-son of Things, to make this either a Principle or a Conclusion, but much to the contrary; it being Sedi-tion at least, if not Treason to assert it in this Reign. For if by the Natural Superiority of their Sex, they mean that every Man is by Nature superior to ev-ery Woman, which is the obvious meaning, and that which must be stuck to if they would speak Sense, it wou’ d be a Sin in any Woman to have Dominion over any Man, and the greatest Queen ought not to command but to obey her Footman, because no Mu-nicipal Laws can supersede or change the Law of Na-ture; so that if the Dominion of the Men be such, the Salique Law,3 as unjust as English Men have ever 3The Salique law was the common law of France which prohibited the crown be passed on to female heirs. CHAPTER 24. MULTIPLE GENERALITY 224 thought it, ought to take place over all the Earth, and the most glorious Reigns in the English, Danish, Castil-ian, and other Annals, were wicked Violations of the Law of Nature! If they mean that some Men are superior to some Women this is no great Discovery; had they turn’ d the Tables they might have seen that some Women are Superior to some Men. Or had they been pleased to remember their Oaths of Allegiance and Supremacy, they might have known that One Woman is superior to All the Men in these Nations, or else they have sworn to very little purpose.4 And it must not be suppos’ d, that their Reason and Religion wou’ d suffer them to take Oaths, contrary to the Laws of Nature and Reason of things.5 We can symbolize the different interpretations Astell offers of Nicholls’ claim that men are superior to women: He either meant that every man is superior to every woman, i.e., ∀x(M (x) →∀y(W (y) →S (x,y))) or that some men are superior to some women, ∃x(M (x) ∧∃y(W (y) ∧S (x,y))). The latter is true, but so is ∃y(W (y) ∧∃x(M (x) ∧S (y,x))) (some women are superior to some men), so that would be “no great discovery.” In fact, since the Queen is superior to all her subjects, it’s even true that some woman is superior to every man, i.e., ∃y(W (y) ∧∀x(M (x) →S (y,x))). 4In 1706, England was ruled by Queen Anne. 5Mary Astell, Reflections upon Marriage, 1706 Preface, iii–iv, and Mary Astell, Political Writings, ed. Patricia Springborg, Cambridge University Press, 1996, 9–10. CHAPTER 24. MULTIPLE GENERALITY 225 But this is incompatible with the “obvious meaning” of Nicholls’ claim, i.e., the first reading. So what Nicholls claims amounts to treason against the Queen! Practice exercises A. Using this symbolization key: domain: all animals A(x): x is an alligator M (x): x is a monkey R(x): x is a reptile Z (x): x lives at the zoo L(x,y): x loves y a: Amos b: Bouncer c: Cleo symbolize each of the following sentences in FOL: 1. If Cleo loves Bouncer, then Bouncer is a monkey. 2. If both Bouncer and Cleo are alligators, then Amos loves them both. 3. Cleo loves a reptile. 4. Bouncer loves all the monkeys that live at the zoo. 5. All the monkeys that Amos loves love him back. 6. Every monkey that Cleo loves is also loved by Amos. 7. There is a monkey that loves Bouncer, but sadly Bouncer does not reciprocate this love. B. Using the following symbolization key: domain: all animals D(x): x is a dog S (x): x likes samurai movies L(x,y): x is larger than y r: Rave CHAPTER 24. MULTIPLE GENERALITY 226 h: Shane d: Daisy symbolize the following sentences in FOL: 1. Rave is a dog who likes samurai movies. 2. Rave, Shane, and Daisy are all dogs. 3. Shane is larger than Rave, and Daisy is larger than Shane. 4. All dogs like samurai movies. 5. Only dogs like samurai movies. 6. There is a dog that is larger than Shane. 7. If there is a dog larger than Daisy, then there is a dog larger than Shane. 8. No animal that likes samurai movies is larger than Shane. 9. No dog is larger than Daisy. 10. Any animal that dislikes samurai movies is larger than Rave. 11. There is an animal that is between Rave and Shane in size. 12. There is no dog that is between Rave and Shane in size. 13. No dog is larger than itself. 14. Every dog is larger than some dog. 15. There is an animal that is smaller than every dog. 16. If there is an animal that is larger than any dog, then that animal does not like samurai movies. C. Using the symbolization key given, symbolize each English-language sentence into FOL. domain: candies C (x): x has chocolate in it M (x): x has marzipan in it S (x): x has sugar in it T (x): Boris has tried x B(x,y): x is better than y 1. Boris has never tried any candy. 2. Marzipan is always made with sugar. CHAPTER 24. MULTIPLE GENERALITY 227 3. Some candy is sugar-free. 4. The very best candy is chocolate. 5. No candy is better than itself. 6. Boris has never tried sugar-free chocolate. 7. Boris has tried marzipan and chocolate, but never together. 8. Any candy with chocolate is better than any candy without it. 9. Any candy with chocolate and marzipan is better than any candy that lacks both. D. Using the following symbolization key: domain: people and dishes at a potluck R(x): x has run out T (x): x is on the table F (x): x is food P (x): x is a person L(x,y): x likes y e: Eli f : Francesca g: the guacamole symbolize the following English sentences in FOL: 1. All the food is on the table. 2. If the guacamole has not run out, then it is on the table. 3. Everyone likes the guacamole. 4. If anyone likes the guacamole, then Eli does. 5. Francesca only likes the dishes that have run out. 6. Francesca likes no one, and no one likes Francesca. 7. Eli likes anyone who likes the guacamole. 8. Eli likes anyone who likes the people that he likes. 9. If there is a person on the table already, then all of the food must have run out. E. Using the following symbolization key: domain: people CHAPTER 24. MULTIPLE GENERALITY 228 D(x): x dances ballet F (x): x is female M (x): x is male C (x,y): x is a child of y S (x,y): x is a sibling of y e: Elmer j: Jane p: Patrick symbolize the following sentences in FOL: 1. All of Patrick’s children are ballet dancers. 2. Jane is Patrick’s daughter. 3. Patrick has a daughter. 4. Jane is an only child. 5. All of Patrick’s sons dance ballet. 6. Patrick has no sons. 7. Jane is Elmer’s niece. 8. Patrick is Elmer’s brother. 9. Patrick’s brothers have no children. 10. Jane is an aunt. 11. Everyone who dances ballet has a brother who also dances ballet. 12. Every woman who dances ballet is the child of someone who dances ballet. CHAPTER 25 Identity Consider this sentence: 1. Pavel owes money to everyone. Let the domain be people; this will allow us to symbolize ‘ev-eryone’ with a universal quantifier. Offering the symbolization key: O (x,y): x owes money to y p: Pavel we can symbolize sentence 1 by ‘∀x O (p,x)’. But this has a (per-haps) odd consequence. It requires that Pavel owes money to ev-ery member of the domain (whatever the domain may be). The domain certainly includes Pavel. So this entails that Pavel owes money to himself. And maybe we did not want to say that. Maybe we meant to leave it open if Pavel owes money to himself, some-thing we could have expressed more precisely by using either one of the following: 2. Pavel owes money to everyone else. 3. Pavel owes money to everyone other than Pavel. But we do not have any way for dealing with the italicized words yet. The solution is to add another symbol to FOL. 229 CHAPTER 25. IDENTITY 230 25.1 Adding identity The symbol ‘=’ will be a two-place predicate. Since it will have a special meaning, we shall write it a bit differently: we put it between two terms, rather than out front. (This should also be familiar; consider a mathematical equation like 1 2 = 0.5.) And the special meaning for ‘=’ is given by the fact that we always adopt the following symbolization key: x = y: x is identical to y This does not mean merely that the objects in question are in-distinguishable, or that all of the same things are true of them. Rather, it means that the objects in question are the very same object. To put this to use, suppose we want to symbolize this sen-tence: 4. Pavel is Mister Chekov. Let us add to our symbolization key: c: Mister Chekov Now sentence 4 can be symbolized as ‘p = c’. This tells us that the names ‘p’ and ‘c’ both name the same thing. We can also now deal with sentences 2 and 3. Both of these sentences can be paraphrased as ‘Everyone who is not Pavel is owed money by Pavel’. Paraphrasing some more, we get: ‘For all x, if x is not Pavel, then x is owed money by Pavel’. Now that we are armed with our new identity symbol, we can symbolize this as ‘∀x(¬x = p →O (p,x))’. This last sentence contains the formula ‘¬x = p’. That might look a bit strange, because the symbol that comes immediately after the ‘¬’ is a variable, rather than a predicate, but this is not a problem. We are simply negating the entire formula ‘x = p’. CHAPTER 25. IDENTITY 231 25.2 ‘Only’ and ‘except’ In addition to sentences that use the word ‘else’, and ‘other than’, identity is helpful when symbolizing some sentences that contain the words ‘only’, and ‘except’. Consider: 5. Only Pavel owes money to Hikaru. Let ‘h’ name Hikaru. Plausibly, sentence 5 is true if, and only if, both of the following conditions hold: 6. Pavel owes money to Hikaru. 7. No one who is not Pavel owes money to Hikaru. Sentence 7 can be symbolized by any one of: ¬∃x(¬x = p ∧O (x,h)), ∀x(¬x = p →¬O (x,h)), ∀x(O (x,h) →x = p). Thus, we can symbolize sentence 5 as the conjunction of one of the above with the symbolization of 6, ‘O (p,h)’, or more com-pactly using ‘↔’ as ‘∀x(O (x,h) ↔x = p)’. 8. Everyone except Pavel owes money to Hikaru. Sentence 8 can be treated similarly, although now of course Pavel does not owe Hikaru money. We can paraphrase it as ‘Ev-eryone who is not Pavel owes Hikaru money, and Pavel does not’. Consequently, it can be symbolized as ‘∀x(¬x = p → O (x,h)) ∧¬O (p,h)’, or more concisely, ‘∀x(¬x = p ↔O (x,h))’. Other locutions akin to ‘except’ such as ‘but’ or ‘besides’ (as used in ‘no one but Pavel’ or ‘someone besides Hikaru’) can be treated in similar ways. The above treatment of so-called “exceptives” is not uncon-tentious. Some linguists think that sentence 8 does not entail that Pavel doesn’t owe Hikaru money, and so the symbolization should just be ‘∀x(¬x = p →O (x,h))’. There are also uses of CHAPTER 25. IDENTITY 232 ‘except’ that clearly do not have that entailment, especially in mathematical writing. For instance, you may read in a calculus textbook that “the function f is defined everywhere except pos-sibly at a”. That means only that for every point x other than a, f is defined at x. It is not required that f is undefined at a; it’s left open whether f is or is not defined at a. 25.3 There are at least. . . We can also use identity to say how many things there are of a particular kind. For example, consider these sentences: 9. There is at least one apple. 10. There are at least two apples. 11. There are at least three apples. We will use the symbolization key: A(x): x is an apple Sentence 9 does not require identity. It can be adequately sym-bolized by ‘∃x A(x)’: There is an apple; perhaps many, but at least one. It might be tempting to also symbolize sentence 10 without identity. Yet consider the sentence ‘∃x∃y(A(x) ∧A(y))’. Roughly, this says that there is some apple x in the domain and some apple y in the domain. Since nothing precludes these from being one and the same apple, this would be true even if there were only one apple. In order to make sure that we are dealing with different apples, we need an identity predicate. Sentence 10 needs to say that the two apples that exist are not identical, so it can be symbolized by ‘∃x∃y((A(x) ∧A(y)) ∧¬x = y)’. Sentence 11 requires talking about three different apples. Now we need three existential quantifiers, and we need to make sure that each will pick out something different: ∃x∃y∃z [((A(x) ∧A(y)) ∧A(z)) ∧((¬x = y ∧¬y = z) ∧¬x = z)]. CHAPTER 25. IDENTITY 233 Note that it is not enough to use ‘¬x = y ∧¬y = z’ to symbolize ‘x, y, and z are all different.’ For that would be true if x and y were different, but x = z. In general, to say that x1, ..., xn are all different, we must have a conjunction of ¬xi = x j for every different pair i and j. 25.4 There are at most. . . Now consider these sentences: 12. There is at most one apple. 13. There are at most two apples. Sentence 12 can be paraphrased as, ‘It is not the case that there are at least two apples’. This is just the negation of sentence 10: ¬∃x∃y [(A(x) ∧A(y)) ∧¬x = y] But sentence 12 can also be approached in another way. It means that if you pick out an object and it’s an apple, and then you pick out an object and it’s also an apple, you must have picked out the same object both times. With this in mind, it can be symbolized by ∀x∀y [︁ (A(x) ∧A(y)) →x = y ]︁ The two sentences will turn out to be logically equivalent. Similarly, sentence 13 can be approached in two equivalent ways. It can be paraphrased as, ‘It is not the case that there are three or more distinct apples’, so we can offer: ¬∃x∃y∃z [︁ ((A(x) ∧A(y)) ∧A(z)) ∧((¬x = y ∧¬x = z) ∧¬y = z) ]︁ Alternatively we can read it as saying that if you pick out an apple, and an apple, and an apple, then you will have picked out (at least) one of these objects more than once. Thus: ∀x∀y∀z [︁ ((A(x) ∧A(y)) ∧A(z)) →((x = y ∨x = z) ∨y = z) ]︁ CHAPTER 25. IDENTITY 234 25.5 There are exactly. . . We can now consider statements of exact numerical quantity, like: 14. There is exactly one apple. 15. There are exactly two apples. 16. There are exactly three apples. Sentence 14 can be paraphrased as, ‘There is at least one apple and there is at most one apple’. This is just the conjunction of sentence 9 and sentence 12. So we can offer: ∃xA(x) ∧∀x∀y [︁ (A(x) ∧A(y)) →x = y ]︁ But it is perhaps more straightforward to paraphrase sentence 14 as, ‘There is a thing x which is an apple, and everything which is an apple is just x itself’. Thought of in this way, we offer: ∃x [︁ A(x) ∧∀y(A(y) →x = y) ]︁ Similarly, sentence 15 may be paraphrased as, ‘There are at least two apples, and there are at most two apples’. Thus we could offer ∃x∃y((A(x) ∧A(y)) ∧¬x = y) ∧ ∀x∀y∀z [︁ ((A(x) ∧A(y)) ∧A(z)) →((x = y ∨x = z) ∨y = z) ]︁ More efficiently, though, we can paraphrase it as ‘There are at least two different apples, and every apple is one of those two apples’. Then we offer: ∃x∃y [︁ ((A(x) ∧A(y)) ∧¬x = y) ∧∀z (A(z) →(x = z ∨y = z)) ]︁ Finally, consider these sentences: 17. There are exactly two things. 18. There are exactly two objects. CHAPTER 25. IDENTITY 235 It might be tempting to add a predicate to our symbolization key, to symbolize the English predicate ‘ is a thing’ or ‘ is an object’, but this is unnecessary. Words like ‘thing’ and ‘object’ do not sort wheat from chaff: they apply trivially to everything, which is to say, they apply trivially to every thing. So we can symbolize either sentence with either of the following: ∃x∃y ¬x = y ∧¬∃x∃y∃z ((¬x = y ∧¬y = z) ∧¬x = z) ∃x∃y [︁ ¬x = y ∧∀z (x = z ∨y = z) ]︁ Practice exercises A. Consider the sentence, 19. Every officer except Pavel owes money to Hikaru. Symbolize this sentence, using ‘F (x)’ for ‘ x is an officer’. Are you confident that your symbolization is true if, and only if, sentence 19 is true? What happens if every officer owes money to Hikaru, Pavel does not, but Pavel isn’t an officer? B. Explain why: • ‘∃x∀y(A(y) ↔x = y)’ is a good symbolization of ‘there is exactly one apple’. • ‘∃x∃y [︁ ¬x = y ∧∀z (A(z) ↔(x = z ∨y = z)) ]︁ ’ is a good symbolization of ‘there are exactly two apples’. CHAPTER 26 Sentences of FOL We know how to represent English sentences in FOL. The time has finally come to define the notion of a sentence of FOL. 26.1 Expressions There are six kinds of symbols in FOL: Predicates A,B,C,. . . ,Z, or with subscripts, as needed: A1,B1,Z1,A2,A25, J375,. . . Names a,b,c,. . . ,r, or with subscripts, as needed a1,b224,h7,m32,. . . Variables s,t,u,v,w,x,y,z, or with subscripts, as needed x1,y1,z1,x2,. . . Connectives ¬,∧,∨,→,↔ Brackets ( , ) Quantifiers ∀,∃ 236 CHAPTER 26. SENTENCES OF FOL 237 We define an expression of fol as any string of symbols of FOL. Take any of the symbols of FOL and write them down, in any order, and you have an expression. 26.2 T erms and formulas In §6, we went straight from the statement of the vocabulary of TFL to the definition of a sentence of TFL. In FOL, we will have to go via an intermediary stage: via the notion of a formula. The intuitive idea is that a formula is any sentence, or anything which can be turned into a sentence by adding quantifiers out front. But this intuitive idea will take some time to unpack. We start by defining the notion of a term. A term is any name or any variable. So, here are some terms: a,b,x,x1,x2,y,y254,z Next we need to define atomic formulas. 1. Any sentence letter is an atomic formula. 2. If R is an n-place predicate and t 1, t 2,. . . , t n are terms, then R(t 1, t 2,. . . , t n) is an atomic formula. 3. If t 1 and t 2 are terms, then t 1 = t 2 is an atomic formula. 4. Nothing else is an atomic formula. Note that we consider sentence letters also formulas of FOL, so every sentence of TFL is also a formula of FOL. The use of script letters here follows the conventions laid down in §8. So, ‘R’ is not itself a predicate of FOL. Rather, it is a symbol of our metalanguage (augmented English) that we use CHAPTER 26. SENTENCES OF FOL 238 to talk about any predicate of FOL. Similarly, ‘t 1’ is not a term of FOL, but a symbol of the metalanguage that we can use to talk about any term of FOL. So, where ‘F ’ is a one-place predicate, ‘G’ is a three-place predicate, and ‘S’ is a six-place predicate, here are some atomic formulas: D F (a) x = a G (x,a,y) a = b G (a,a,a) F (x) S (x1,x2,a,b,y,x1) Once we know what atomic formulas are, we can offer recursion clauses to define arbitrary formulas. The first few clauses are exactly analogous to those in the definition of ‘sentence of TFL’. 1. Every atomic formula is a formula. 2. If A is a formula, then ¬A is a formula. 3. If A and B are formulas, then (A∧B) is a formula. 4. If A and B are formulas, then (A∨B) is a formula. 5. If Aand Bare formulas, then (A →B) is a formula. 6. If Aand Bare formulas, then (A ↔B) is a formula. 7. If A is a formula and x is a variable, then ∀xA is a formula. 8. If A is a formula and x is a variable, then ∃xA is a formula. 9. Nothing else is a formula. So, assuming again that ‘F ’ is a one-place predicate, ‘G’ is a three-place predicate and ‘S’ is a six place-predicate, here are CHAPTER 26. SENTENCES OF FOL 239 some formulas you can build this way: F (x) G (a,y,z) S (y,z,y,a,y,x) (G (a,y,z) →S (y,z,y,a,y,x)) ∀z (G (a,y,z) →S (y,z,y,a,y,x)) F (x) ∧∀z (G (a,y,z) →S (y,z,y,a,y,x)) ∃y(F (x) ∧∀z (G (a,y,z) →S (y,z,y,a,y,x))) ∀x∃y(F (x) ∧∀z (G (a,y,z) →S (y,z,y,a,y,x))) We can now give a formal definition of scope, which incorpo-rates the definition of the scope of a quantifier. Here we follow the case of TFL, though we note that a logical operator can be either a connective or a quantifier: The main logical operator in a formula is the operator that was introduced most recently, when that formula was constructed using the recursion rules. The scope of a logical operator in a formula is the subfor-mula for which that operator is the main logical operator. So we can graphically illustrate the scope of the quantifiers in the preceding example thus: scope of ‘∀x’ ⏟ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏞⏞ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏟ ∀x scope of ‘∃y’ ⏟ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏞⏞ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏟ ∃y(F (x) ↔ scope of ‘∀z’ ⏟ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏞⏞ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏟ ∀z (G (a,y,z) →S (y,z,y,a,y,x))) CHAPTER 26. SENTENCES OF FOL 240 26.3 Sentences and free variables Recall that we are largely concerned in logic with assertoric sen-tences: sentences that can be either true or false. Many formulas are not sentences. Consider the following symbolization key: domain: people L(x,y): x loves y b: Boris Consider the atomic formula ‘L(z,z)’. All atomic formula are formulas, so ‘L(z,z)’ is a formula, but can it be true or false? You might think that it will be true just in case the person named by ‘z’ loves themself, in the same way that ‘L(b,b)’ is true just in case Boris (the person named by ‘b’) loves himself. However, ‘z’ is a variable, and does not name anyone or any thing. Of course, if we put an existential quantifier out front, obtain-ing ‘∃zL(z,z)’, then this would be true iff someone loves them-selves. Equally, if we wrote ‘∀zL(z,z)’, this would be true iff ev-eryone loves themselves. The point is that we need a quantifier to tell us how to deal with a variable. Let’s make this idea precise. An occurrence of a variable x is bound iff it falls within the scope of either ∀x or ∃x. An occurrence of a variable which is not bound is free. For example, consider the formula (∀x(E(x) ∨D(y)) →∃z (E(x) →L(z,x))) The scope of the universal quantifier ‘∀x’ is ‘∀x(E(x) ∨D(y))’, so the first ‘x’ is bound by the universal quantifier. However, the second and third occurrence of ‘x’ are free. Equally, the ‘y’ is free. The scope of the existential quantifier ‘∃z’ is ‘(E(x) →L(z,x))’, so ‘z’ is bound. Finally we can say the following. CHAPTER 26. SENTENCES OF FOL 241 A sentence of FOL is any formula of FOL that contains no free variables. 26.4 Bracketing conventions We will adopt the same notational conventions governing brack-ets that we did for TFL (see §6 and §11.3.) First, we may omit the outermost brackets of a formula. Second, we may use square brackets, ‘[’ and ‘]’, in place of brackets to increase the readability of formulas. Sentences of FOL used in our examples can become quite cumbersome, and so we also introduce a convention to deal with conjunctions and disjunctions of more than two sentences. We stipulate that A1 ∧A2 ∧· · · ∧An and A1 ∨A2 ∨· · · ∨An are to be interpreted as, respectively: (. . . (A1 ∧A2) ∧· · · ∧An) (. . . (A1 ∨A2) ∨· · · ∨An) In practice, this just means that you are allowed to leave out parentheses in long conjunctions and disjunctions. But remember that (unless they are the outermost parentheses of the sentence) you must still enclose the entire conjunction or disjunction in parentheses. Also, you cannot mix conjunctions and disjunctions with each other or with other connectives. So the following are still not allowed, and would be ambiguous if they were: A ∨B ∧C ∧D B ∨C →D 26.5 Superscripts on predicates Above, we said that an n-place predicate followed by n terms is an atomic formula. But there is a small issue with this definition: the symbols we use for predicates do not, themselves, indicate CHAPTER 26. SENTENCES OF FOL 242 how many places the predicate has. Indeed, in some places in this book, we have used the letter ‘G’ as a one-place predicate; in other places we have used it as a three-place predicate. So, unless we state explicitly whether we want to use ‘G’ as a one-place predicate or as a three place predicate, it is indeterminate whether ‘G (a)’ is an atomic formula or not. There is an easy way to avoid this, which many books adopt. Instead of saying that our predicates are just capital letters (with numerical subscripts as necessary), we could say that they are capital letters with numerical superscripts (and with numerical sub-scripts as necessary). The purpose of the superscript would be to say explicitly how many places the predicate has. On this ap-proach, ‘G 1’ would be a one-place predicate, and ‘G 3’ would be an (entirely different) three places predicate. They would need to have different entries in any symbolisation key. And ‘G 1(a)’ would be an atomic formula, whereas ‘G 3(a)’ would not; likewise ‘G 3(a,b,c)’ would be an atomic formula, and ‘G 1(a,b,c)’ would not. So, we could add superscripts to all our predicates. This would have the advantage of making certain things completely explicit. However, it would have the disadvantage of making our formulas much harder to read; the superscripts would distract the eye. So, we will not bother to make this change. Our predicates will remain without superscripts. (And, in practice, any book which includes superscripts almost immediately stops including them!) However, this leaves open a possibility of ambiguity. So, when any ambiguity could arise—in practice, very rarely—you should say, explicitly, how many places your predicate(s) have. Practice exercises A. Identify which variables are bound and which are free. 1. ∃x L(x,y) ∧∀y L(y,x) 2. ∀x A(x) ∧B(x) 3. ∀x(A(x) ∧B(x)) ∧∀y(C (x) ∧D(y)) CHAPTER 26. SENTENCES OF FOL 243 4. ∀x∃y [R(x,y) →( J (z) ∧K (x))] ∨R(y,x) 5. ∀x1(M (x2) ↔L(x2,x1)) ∧∃x2 L(x3,x2) CHAPTER 27 Definite descriptions Consider sentences like: 1. Nick is the traitor. 2. The traitor went to Cambridge. 3. The traitor is the deputy. These are definite descriptions: they are meant to pick out a unique object. They should be contrasted with indefinite descrip-tions, such as ‘Nick is a traitor’. They should equally be con-trasted with generics, such as ‘The whale is a mammal’ (when it’s inappropriate to ask which whale). The question we face is: How should we deal with definite descriptions in FOL? 27.1 Treating definite descriptions as terms One option would be to introduce new names whenever we come across a definite description. This is probably not a great idea. We know that the traitor—whoever it is—is indeed a traitor. We want to preserve that information in our symbolization. A second option would be to use a new definite description operator, such as ‘ 𝜄 ’. The idea would be to symbolize ‘the F ’ as 244 CHAPTER 27. DEFINITE DESCRIPTIONS 245 ‘ 𝜄 x F (x)’ (think ‘the x such that F (x)’); or to symbolize ‘the G’ as ‘ 𝜄 x G (x)’, etc. Expressions of the form 𝜄 xA(x) would then behave like names. If we were to follow this path, we could use the following symbolization key: domain: people T (x): x is a traitor D(x): x is a deputy C (x): x went to Cambridge n: Nick Then, we could symbolize sentence 1 with ‘n = 𝜄 x T (x)’, sentence 2 with ‘C ( 𝜄 x T (x))’, and sentence 3 with ‘ 𝜄 x T (x) = 𝜄 x D(x)’. However, it would be nice if we didn’t have to add a new symbol to FOL. And we might be able to make do without one. 27.2 Russell’s analysis Bertrand Russell offered an analysis of definite descriptions. Very briefly put, he observed that, when we say ‘the F ’ in the context of a definite description, our aim is to pick out the one and only thing that is F (in the appropriate context). Thus Russell analyzed the notion of a definite description as follows:1 the F is G iff there is at least one F , and there is at most one F , and every F is G Note a very important feature of this analysis: ‘the’ does not appear on the right-side of the equivalence. Russell is aiming to provide an understanding of definite descriptions in terms that do not presuppose them. Now, one might worry that we can say ‘the table is brown’ without implying that there is one and only one table in the uni-verse. But this is not (yet) a fantastic counterexample to Russell’s 1Bertrand Russell, ‘On Denoting’, 1905, Mind 14, pp. 479–93; also Russell, Introduction to Mathematical Philosophy, 1919, London: Allen and Unwin, ch. 16. CHAPTER 27. DEFINITE DESCRIPTIONS 246 analysis. The domain of discourse is likely to be restricted by context (e.g., to salient objects in my vicinity). If we accept Russell’s analysis of definite descriptions, then we can symbolize sentences of the form ‘the F is G’ using our strategy for numerical quantification in FOL. After all, we can deal with the three conjuncts on the right-hand side of Russell’s analysis as follows: ∃xF (x) ∧∀x∀y((F (x) ∧F (y)) →x = y) ∧∀x(F (x) →G (x)) In fact, we could express the same point rather more crisply, by recognizing that the first two conjuncts just amount to the claim that there is exactly one F , and that the last conjunct tells us that that object is G. So, equivalently, we could offer: ∃x [︁ (F (x) ∧∀y(F (y) →x = y)) ∧G (x) ]︁ Using these sorts of techniques, we can now symbolize sentences 1–3 without using any new-fangled fancy operator, such as ‘ 𝜄 ’. Sentence 1 is exactly like the examples we have just consid-ered. So we would symbolize it by ∃x [︁ T (x) ∧∀y(T (y) →x = y) ∧x = n ]︁ . Sentence 2 poses no problems either: ∃x [︁ T (x) ∧∀y(T (y) →x = y) ∧C (x) ]︁ . Sentence 3 is a little trickier, because it links two definite descrip-tions. But, deploying Russell’s analysis, it can be paraphrased by ‘there is exactly one traitor, x, and there is exactly one deputy, y, and x = y’. So we can symbolize it by: ∃x∃y (︁[︁ T (x) ∧∀z (T (z) →x = z) ]︁ ∧ [︁ D(y) ∧∀z (D(z) →y = z) ]︁ ∧x = y)︁ Note that the formula ‘x = y’ must fall within the scope of both quantifiers! CHAPTER 27. DEFINITE DESCRIPTIONS 247 27.3 Empty definite descriptions One of the nice features of Russell’s analysis is that it allows us to handle empty definite descriptions neatly. France has no king at present. Now, if we were to introduce a name, ‘k’, to name the present King of France, then everything would go wrong: remember from §22 that a name must always pick out some object in the domain, and whatever we choose as our domain, it will contain no present kings of France. Russell’s analysis neatly avoids this problem. Russell tells us to treat definite descriptions using predicates and quantifiers, in-stead of names. Since predicates can be empty (see §23), this means that no difficulty now arises when the definite description is empty. Indeed, Russell’s analysis helpfully highlights two ways to go wrong in a claim involving a definite description. To adapt an example from Stephen Neale (1990),2 suppose Alex claims: 4. I am dating the present king of France. Using the following symbolization key: a: Alex K (x): x is a present king of France D(x,y): x is dating y (Note that the symbolization key speaks of a present King of France, not the present King of France; i.e., it employs an indef-inite, rather than a definite, description.) Sentence 4 would be symbolized by ‘∃x [︁ (K (x) ∧∀y(K (y) →x = y)) ∧D(a,x) ]︁ ’. Now, this can be false in (at least) two ways, corresponding to these two different sentences: 5. There is no one who is both the present King of France and such that he and Alex are dating. 2Neale, Descriptions, 1990, Cambridge: MIT Press. CHAPTER 27. DEFINITE DESCRIPTIONS 248 6. There is a unique present King of France, but Alex is not dating him. Sentence 5 might be paraphrased by ‘It is not the case that: the present King of France and Alex are dating’. It will then be sym-bolized by ‘¬∃x [︁ (K (x) ∧∀y(K (y) →x = y)) ∧D(a,x) ]︁ ’. We might call this outer negation, since the negation governs the en-tire sentence. Note that the sentence is true if there is no present King of France. Sentence 6 can be symbolized by ‘∃x [︁ (K (x) ∧∀y(K (y) → x = y)) ∧¬D(a,x) ]︁ ’. We might call this inner negation, since the negation occurs within the scope of the definite description. Note that its truth requires that there is a present King of France, albeit one who is not dating Alex. 27.4 Possessives, ‘both’, ‘neither’ We can use Russell’s analysis of definite descriptions also to deal with singular possessive constructions in English. For instance, ‘Smith’s murderer’ means something like ‘the person who mur-dered Smith’, i.e., it is a disguised definite description. On Rus-sell’s analysis, the sentence 7. Smith’s murderer is insane. can be false in one of three ways. It can be false because the one person who murdered Smith is not, in fact, insane. But it can also be false if the definite description is empty, namely if either no-one murdered Smith (e.g., if Smith met with an unfortunate accident) or if more than one person murdered Smith. To symbolize sentences containing singular possessives such as ‘Smith’s murderer’ you should paraphrase them using an ex-plicit definite description, e.g., ‘The person who murdered Smith is insane’ and then symbolize it according to Russell’s analysis. In our case, we would use the symbolization key: Domain: people CHAPTER 27. DEFINITE DESCRIPTIONS 249 I (x): x is insane M (x,y): x murdered y s: Smith Our symbolization then reads, ‘∃x [︁ M (x,s) ∧∀y(M (y,s) →x = y) ∧I (x) ]︁ ’. T wo other determiners that we can extend Russell’s analysis to are ‘both’ and ‘neither’. To say ‘both F s are G’ is to say that there are exactly two F s, and each of them is G. To say that ‘neither F is G’, is to also say that there are exactly two F s, and neither of them is G. In FOL, the symbolizations would read, respectively, ∃x∃y [︁ F (x) ∧F (y) ∧¬x = y ∧ ∀z (F (z) →(x = z ∨y = z)) ∧G (x) ∧G (y) ]︁ ∃x∃y [︁ F (x) ∧F (y) ∧¬x = y ∧ ∀z (F (z) →(x = z ∨y = z)) ∧¬G (x) ∧¬G (y) ]︁ Compare these symbolizations with the symbolizations of ‘ex-actly two F s are Gs’ from section 25.5, i.e., of ‘there are exactly two things that are both F and G’: ∃x∃y [︁ (F (x) ∧G (x)) ∧(F (y) ∧G (y)) ∧¬x = y ∧ ∀z ((F (z) ∧G (z)) →(x = z ∨y = z)) ]︁ The difference between the symbolization of this and that of ‘both F s are Gs’ lies in the antecedent of the conditional. For ‘exactly two F s are Gs’, we only require that there are no F s that are also Gs other than x and y, whereas for ‘both F s are Gs’, there cannot be any F s, whether they are Gs or not, other than x and y. In other words, ‘both F s are Gs’ implies that exactly two F s are Gs. However, ‘exactly two F s are Gs’ does not imply that both F s are Gs (there might be a third F which isn’t a G). CHAPTER 27. DEFINITE DESCRIPTIONS 250 27.5 The adequacy of Russell’s analysis How good is Russell’s analysis of definite descriptions? This ques-tion has generated a substantial philosophical literature, but we will restrict ourselves to two observations. One worry focusses on Russell’s treatment of empty definite descriptions. If there are no F s, then on Russell’s analysis, both ‘the F is G’ and ‘the F is non-G’ are false. P. F. Strawson sug-gested that such sentences should not be regarded as false, ex-actly, but involve presupposition failure, and so need to be treated as neither true nor false.3 If we agree with Strawson here, we will need to revise our logic. For, in our logic, there are only two truth values (True and False), and every sentence is assigned exactly one of these truth values. But there is room to disagree with Strawson. Strawson is ap-pealing to some linguistic intuitions, but it is not clear that they are very robust. For example: isn’t it just false, not ‘gappy’, that Tim is dating the present King of France? Keith Donnellan raised a second sort of worry, which (very roughly) can be brought out by thinking about a case of mistaken identity.4 T wo men stand in the corner: a very tall man drinking what looks like a gin martini; and a very short man drinking what looks like a pint of water. Seeing them, Malika says: 8. The gin-drinker is very tall! Russell’s analysis will have us render Malika’s sentence as: 8′. There is exactly one gin-drinker [in the corner], and who-ever is a gin-drinker [in the corner] is very tall. Now suppose that the very tall man is actually drinking water from a martini glass; whereas the very short man is drinking a 3P. F. Strawson, ‘On Referring’, 1950, Mind 59, pp. 320–34. 4Keith Donnellan, ‘Reference and Definite Descriptions’, 1966, Philosophi-cal Review 77, pp. 281–304. CHAPTER 27. DEFINITE DESCRIPTIONS 251 pint of (neat) gin. By Russell’s analysis, Malika has said some-thing false, but don’t we want to say that Malika has said some-thing true? Again, one might wonder how clear our intuitions are on this case. We can all agree that Malika intended to pick out a partic-ular man, and say something true of him (that he was tall). On Russell’s analysis, she actually picked out a different man (the short one), and consequently said something false of him. But maybe advocates of Russell’s analysis only need to explain why Malika’s intentions were frustrated, and so why she said some-thing false. This is easy enough to do: Malika said something false because she had false beliefs about the men’s drinks; if Ma-lika’s beliefs about the drinks had been true, then she would have said something true.5 To say much more here would lead us into deep philosophical waters. That would be no bad thing, but for now it would distract us from the immediate purpose of learning formal logic. So, for now, we will stick with Russell’s analysis of definite descriptions, when it comes to putting things into FOL. It is certainly the best that we can offer, without significantly revising our logic, and it is quite defensible as an analysis. Practice exercises A. Using the following symbolization key: domain: people K (x): x knows the combination to the safe S (x): x is a spy V (x): x is a vegetarian T (x,y): x trusts y h: Hofthor 5Interested parties should read Saul Kripke, ‘Speaker Reference and Se-mantic Reference’, 1977, in French et al. (eds.), Contemporary Perspectives in the Philosophy of Language, Minneapolis: University of Minnesota Press, pp. 6–27. CHAPTER 27. DEFINITE DESCRIPTIONS 252 i: Ingmar symbolize the following sentences in FOL: 1. Hofthor trusts a vegetarian. 2. Everyone who trusts Ingmar trusts a vegetarian. 3. Everyone who trusts Ingmar trusts someone who trusts a vegetarian. 4. Only Ingmar knows the combination to the safe. 5. Ingmar trusts Hofthor, but no one else. 6. The person who knows the combination to the safe is a vegetarian. 7. The person who knows the combination to the safe is not a spy. B. Using the following symbolization key: domain: cards in a standard deck B(x): x is black. C (x): x is a club. D(x): x is a deuce. J (x): x is a jack. M (x): x is a man with an axe. O (x): x is one-eyed. W (x): x is wild. symbolize each sentence in FOL: 1. All clubs are black cards. 2. There are no wild cards. 3. There are at least two clubs. 4. There is more than one one-eyed jack. 5. There are at most two one-eyed jacks. 6. There are two black jacks. 7. There are four deuces. 8. The deuce of clubs is a black card. 9. One-eyed jacks and the man with the axe are wild. CHAPTER 27. DEFINITE DESCRIPTIONS 253 10. If the deuce of clubs is wild, then there is exactly one wild card. 11. The man with the axe is not a jack. 12. The deuce of clubs is not the man with the axe. C. Using the following symbolization key: domain: animals in the world B(x): x is in Farmer Brown’s field H (x): x is a horse P (x): x is a Pegasus W (x): x has wings symbolize the following sentences in FOL: 1. There are at least three horses in the world. 2. There are at least three animals in the world. 3. There is more than one horse in Farmer Brown’s field. 4. There are three horses in Farmer Brown’s field. 5. There is a single winged creature in Farmer Brown’s field; any other creatures in the field must be wingless. 6. The Pegasus is a winged horse. 7. The animal in Farmer Brown’s field is not a horse. 8. The horse in Farmer Brown’s field does not have wings. D. In this chapter, we symbolized ‘Nick is the traitor’ by ‘∃x(T (x) ∧∀y(T (y) →x = y) ∧x = n)’. Explain why these would be equally good symbolisations: • T (n) ∧∀y(T (y) →n = y) • ∀y(T (y) ↔y = n) CHAPTER 28 Ambiguity In chapter 7 we discussed the fact that sentences of English can be ambiguous, and pointed out that sentences of TFL are not. One important application of this fact is that the structural ambigu-ity of English sentences can often, and usefully, be straightened out using different symbolizations. One common source of am-biguity is scope ambiguity, where the English sentence does not make it clear which logical word is supposed to be in the scope of which other. Multiple interpretations are possible. In FOL, ev-ery connective and quantifier has a well-determined scope, and so whether or not one of them occurs in the scope of another in a given sentence of FOL is always determined. For instance, consider the English idiom, 1. Everything that glitters is not gold. If we think of this sentence as of the form ‘every F is not G’ where F (x) symbolizes ‘ x glitters’ and G (x) is ‘ x is gold’, we would symbolize it as: ∀x(F (x) →¬G (x)), in other words, we symbolize it the same way as we would ‘Noth-ing that glitters is gold’. But the idiom does not mean that! It means that one should not assume that just because something glitters, it is gold; not everything that appears valuable is in fact 254 CHAPTER 28. AMBIGUITY 255 valuable. To capture the actual meaning of the idiom, we would have to symbolize it instead as we would ‘Not everything that glitters is gold’, i.e., in the following way: ¬∀x(F (x) →G (x)) Compare this with the previous symbolization: again we see that the difference in the two meanings of the ambiguous sentence lies in whether the ‘¬’ is in the scope of the ‘∀’ (in the first symbol-ization) or ‘∀’ is in the scope of ‘¬’ (in the second). Of course we can alternatively symbolize the two readings using existential quantifiers as well: ¬∃x(F (x) ∧G (x)) ∃x(F (x) ∧¬G (x)) In chapter 23 we discussed how to symbolize sentences in-volving ‘only’. Consider the sentence: 2. Only young cats are playful. According to our schema, we would symbolize it this way: ∀x(P (x) →(Y (x) ∧C (x))) The meaning of this sentence of FOL is something like, ‘If an animal is playful, it is a young cat’. (Assuming that the domain is animals, of course.) This is probably not what’s intended in uttering sentence 2, however. It’s more likely that we want to say that old cats are not playful. In other words, what we mean to say is that if something is a cat and playful, it must be young. This would be symbolized as: ∀x((C (x) ∧P (x)) →Y (x)) There is even a third reading! Suppose we’re talking about young animals and their characteristics. And suppose you wanted to say that of all the young animals, only the cats are playful. You could symbolize this reading as: CHAPTER 28. AMBIGUITY 256 ∀x((Y (x) ∧P (x)) →C (x)) Each of the last two readings can be made salient in English by placing the stress appropriately. For instance, to suggest the last reading, you would say ‘Only young cats are playful’, and to get the other reading you would say ‘Only young cats are playful’. The very first reading can be indicate by stressing both ‘young’ and ‘cats’: ‘Only young cats are playful’ (but not old cats, or dogs of any age). In sections 24.3 and 24.5 we discussed the importance of the order of quantifiers. This is relevant here because, in English, the order of quantifiers is sometimes not completely determined. When both universal (‘all’) and existential (‘some’, ‘a’) quantifiers are involved, this can result in scope ambiguities. Consider: 3. Everyone went to see a movie. This sentence is ambiguous. In one interpretation, it means that there is a single movie that everyone went to see. In the other, it means that everyone went to see some movie or other, but not necessarily the same one. The two readings can be symbolized, respectively, by ∃x(M (x) ∧∀y(P (y) →S (y,x))) ∀y(P (y) →∃x(M (x) ∧S (y,x))) We assume here that the domain contains (at least) people and movies, and the symbolization key, P (y): y is a person, M (x): x is a movie S (y,x): y went to see x. In the first reading, we say that the existential quantifier has wide scope (and its scope contains the universal quantifier, which has narrow scope), and the other way round in the second. In chapter 27, we encountered another scope ambiguity, aris-ing from definite descriptions interacting with negation. Con-sider Russell’s own example: CHAPTER 28. AMBIGUITY 257 4. The King of France is not bald. If the definite description has wide scope, and we are interpreting the ‘not’ as an ‘inner’ negation (as we said before), sentence 4 is interpreted to assert the existence of a single King of France, to whom we are ascribing non-baldness. In this reading, it is symbolized as ‘∃x [︁ K (x) ∧∀y(K (y) →x = y)) ∧¬B(x) ]︁ ’. In the other reading, the ‘not’ denies the sentence ‘The King of France is bald’, and we would symbolize it as: ‘¬∃x [︁ K (x)∧∀y(K (y) →x = y)) ∧B(x) ]︁ ’. In the first case, we say that the definite description has wide scope and in the second that it has narrow scope. Practice exercises A. Each of the following sentences is ambiguous. Provide a sym-bolization key for each, and symbolize all readings. 1. No one likes a quitter. 2. CSI found only red hair at the scene. 3. Smith’s murderer hasn’t been arrested. B. Russell gave the following example in his paper ‘On Denoting’: I have heard of a touchy owner of a yacht to whom a guest, on first seeing it, remarked, ‘I thought your yacht was larger than it is’; and the owner replied, ‘No, my yacht is not larger than it is’. Explain what’s going on. PART VI Interpretations 258 CHAPTER 29 Extensionality Recall that TFL is a truth-functional language. Its connectives are all truth-functional, and all that we can do with TFL is key sentences to particular truth values. We can do this directly. For example, we might stipulate that the TFL sentence ‘P ’ is to be true. Alternatively, we can do this indirectly, offering a symbol-ization key, e.g.: P : Big Ben is in London. But recall from §10 that this is just a means of specifying ‘P ’s truth value; the symbolization key statement amounts to something like the following stipulation: • The TFL sentence ‘P ’ is true iff Big Ben is in London. And we emphasized in §10 that TFL cannot handle differences in meaning that go beyond mere differences in truth value. 29.1 Symbolizing versus translating FOL has some similar limitations. It gets beyond mere truth val-ues, since it enables us to split up sentences into terms, predicates and quantifiers. This enables us to consider what is true of some particular object, or of some or all objects. But that’s it. To unpack this a bit, consider this symbolization key: 259 CHAPTER 29. EXTENSIONALITY 260 C (x): x teaches Logic III in Calgary This stipulation does not carry the meaning of the English pred-icate across into our FOL predicate. We are simply stipulating something like this: • ‘C (x)’ and ‘ x teaches Logic III in Calgary’ are to be true of exactly the same things. So, in particular: • ‘C (x)’ is to be true of exactly those things which teach Logic III in Calgary (whatever those things might be). This is an indirect way of stipulating which things a predicate is true of. Alternatively, we can stipulate predicate extensions directly. For example, we can stipulate that ‘C (x)’ is to be true of Richard Zach, and Richard Zach alone. As it happens, this direct stipula-tion would have the same effect as the indirect stipulation, since Richard, and Richard alone, teaches Logic III in Calgary. Note, however, that the English predicates ‘ is Richard Zach’ and ‘ teaches Logic III in Calgary’ have very different meanings! The point is that FOL has no resources for dealing with nu-ances of meaning. When we interpret FOL, all we are considering is what the predicates are true of, regardless of whether we spec-ify these things directly or indirectly. The things a predicate is true of are known as the extension of that predicate. We say that FOL is an extensional language because FOL does not rep-resent differences of meaning between predicates that have the same extension. This is why we speak of symbolizing English sentences in FOL. It is doubtful that we are translating English into FOL, for trans-lation should preserve meaning. CHAPTER 29. EXTENSIONALITY 261 29.2 Extensions We can stipulate directly what predicates are to be true of. And our stipulations can be as arbitrary as we like. For example, we could stipulate that ‘H (x)’ should be true of, and only of, the following objects: Justin Trudeau the number 𝜋 every top-F key on every piano ever made Armed with this interpretation of ‘H (x)’, suppose we now add to our symbolization key: j: Justin Trudeau a: Angela Merkel p: the number 𝜋 Then ‘H (j)’ and ‘H (p)’ will both be true, on this interpretation, but ‘H (a)’ will be false, since Angela Merkel was not among the stipulated objects. This process of explicit stipulation is sometimes described as stipulating the extension of a predicate. Note that, in the stipula-tion we just gave, the objects we listed have nothing particularly in common. This doesn’t matter. Logic doesn’t care about what we humans (at a particular moment) think ‘naturally goes together’; to logic, all objects are on an equal footing. Any well-defined collection of objects is a potential extension of a one-place predicate. The example above shows one way of stipulating the extension of ‘H (x)’ by enumeration, i.e., we simply list the objects in the extension of ‘H (x)’. We can also stipulate the extension, as we have also already seen, by giving an English predicate, such as ‘ x teaches Logic III at Calgary’ or ‘ x is an even integer between 3 and 9’. The latter would specify an extension consisting of, and only of, 4, 6, and 8. Note that some predicates of English, such as ‘ x is a round square’, are not true of anything. In this case we say the ex-tension of the predicate is empty. We do allow empty extensions, CHAPTER 29. EXTENSIONALITY 262 and we can stipulate that the extension of ‘H (x)’ is to be empty simply by not listing any members. (It may be odd to consider collections of no things, but logic is odd this way sometimes.) 29.3 Many-place predicates All of this is quite easy to understand when it comes to one-place predicates, but it gets messier when we deal with two-place predicates. Consider a symbolization key like: L(x,y): x loves y Given what we said above, this symbolization key should be read as saying: • ‘L(x,y)’ and ‘ x loves y’ are to be true of exactly the same things. So, in particular: • ‘L(x,y)’ is to be true of x and y (in that order) iff x loves y. It is important that we insist upon the order here, since love— famously—is not always reciprocated. (Note that ‘x’ and ‘y’ on the right here are symbols of augmented English, and that they are being used. By contrast, ‘x’ and ‘y’ in ‘L(x,y)’ are symbols of FOL, and they are being mentioned.) That is an indirect stipulation. What about a direct stipula-tion? This is also tricky. If we simply list objects that fall un-der ‘L(x,y)’, we will not know whether they are the lover or the beloved (or both). We have to find a way to include the order in our explicit stipulation. To do this, we can specify that two-place predicates are true of pairs of objects, where the order of the pair is important. Thus we might stipulate that ‘B(x,y)’ is to be true of, and only of, the following pairs of objects: CHAPTER 29. EXTENSIONALITY 263 ⟨Lenin, Marx⟩ ⟨de Beauvoir, Sartre⟩ ⟨Sartre, de Beauvoir⟩ Here the angle-brackets keep us informed concerning order. Sup-pose we now add the following stipulations: l: Lenin m: Marx b: de Beauvoir r: Sartre Then ‘B(l,m)’ will be true, since ⟨Lenin, Marx⟩is in our explicit list, but ‘B(m,l)’ will be false, since ⟨Marx, Lenin⟩is not in our list. However, both ‘B(b,r)’ and ‘B(r,b)’ will be true, since both ⟨de Beauvoir, Sartre⟩and ⟨Sartre, de Beauvoir⟩are in our explicit list. To make these ideas more precise, we would need to develop some very elementary set theory. Set theory has formal apparatus which allows us to deal with extensions, ordered pairs, and so forth. However, set theory is not covered in this book. So I shall leave these ideas at an imprecise level. Nevertheless, the general idea should be clear. 29.4 Semantics for identity Identity is a special predicate of FOL. We write it a bit differently than other two-place predicates: ‘x = y’ instead of ‘I (x,y)’ (for example). More important, though, is that its interpretation is fixed, once and for all. If two names refer to the same object, then swapping one name for another will not change the truth value of any sentence. So, in particular, if ‘a’ and ‘b’ name the same object, then all of CHAPTER 29. EXTENSIONALITY 264 the following will be true: A(a) ↔A(b) B(a) ↔B(b) R(a,a) ↔R(b,b) R(a,a) ↔R(a,b) R(c,a) ↔R(c,b) ∀x R(x,a) ↔∀x R(x,b) Some philosophers have believed the reverse of this claim. That is, they have believed that when exactly the same sentences (not containing ‘=’) are true of a and b, then a and b are the very same object. This is a highly controversial philosophical claim— sometimes called the identity of indiscernibles—and our logic will not subscribe to it; we allow that exactly the same things might be true of two distinct objects. To bring this out, consider the following interpretation: domain: P. D. Magnus, Tim Button a: P. D. Magnus b: Tim Button • For every primitive predicate we care to consider, that pred-icate is true of nothing. Suppose ‘A’ is a one-place predicate; then ‘A(a)’ is false and ‘A(b)’ is false, so ‘A(a) ↔A(b)’ is true. Similarly, if ‘R’ is a two-place predicate, then ‘R(a,a)’ is false and ‘R(a,b)’ is false, so that ‘R(a,a) ↔R(a,b)’ is true. And so it goes: every atomic sentence not involving ‘=’ is false, so every biconditional linking such sentences is true. For all that, Tim Button and P. D. Magnus are two distinct people, not one and the same! 29.5 Interpretations We defined a v aluation in TFL as any assignment of truth and falsity to sentence letters. In FOL, we are going to define an CHAPTER 29. EXTENSIONALITY 265 interpretation as consisting of four things: • the specification of a domain • for each sentence letter we care to consider, a truth value • for each name that we care to consider, an assignment of exactly one object within the domain • for each predicate that we care to consider (apart from ‘=’), a specification of what things (in what order) the predicate is to be true of. We don’t need to specify anything for‘=’, since it has a fixed mean-ing, namely that of identity. Everything is identical to itself, and only to itself. The symbolization keys that we considered in Part V conse-quently give us one very convenient way to present an interpre-tation. We will continue to use them in this chapter. Following the discussion of §29.2, we now also allow extensions specified by enumerations on the right side, e.g., domain: heads of state, numbers H (x): Justin Trudeau, Angela Merkel, 𝜋 is a perfectly good way of specifying an interpretation, as is domain: 0, 1, 2 L(x,y): ⟨0,1⟩, ⟨0, 2⟩, ⟨1, 2⟩ We could have specified the same extension (on this particular do-main) by giving the English predicate ‘ x is less than y’. However, it is sometimes also convenient to present an inter-pretation diagrammatically. To illustrate (literally): suppose we want to consider just a single two-place predicate, ‘R(x,y)’. Then we can represent it just by drawing an arrow between two objects, and stipulate that ‘R(x,y)’ is to hold of x and y just in case there is an arrow running from x to y in our diagram. As an example, we might offer: CHAPTER 29. EXTENSIONALITY 266 1 2 3 4 This diagram could be used to describe an interpretation whose domain is the first four positive whole numbers, and which inter-prets ‘R(x,y)’ as being true of and only of: ⟨1, 2⟩, ⟨2, 3⟩, ⟨3, 4⟩, ⟨4, 1⟩, ⟨1, 3⟩ Equally we might offer this diagram: 1 2 3 4 The interpretation specified by this diagram can also be given by listing what’s in the domain and in the extension of ‘R(x,y)’: domain: 1, 2, 3, 4 R(x,y): ⟨1, 3⟩, ⟨3, 1⟩, ⟨3, 4⟩, ⟨1, 1⟩, ⟨3, 3⟩ If we wanted, we could make our diagrams more complex. For example, we could add names as labels for particular objects. Equally, to symbolize the extension of a one-place predicate, we might simply draw a circle around some particular objects and stipulate that the thus encircled objects (and only them) are to fall under the predicate ‘H (x)’, say. To specify multiple predicates we could use colored (or dashed, dotted) lines for arrows and circles. CHAPTER 30 Truth in FOL We have introduced you to interpretations. Since, among other things, they tell us which predicates are true of which objects, they will provide us with an account of the truth of atomic sentences. However, we now need to say, precisely, what it is for an arbitrary FOL sentence to be true or false in an interpretation. We know from §26 that there are three kinds of sentence in FOL: • atomic sentences • sentences whose main logical operator is a sentential con-nective • sentences whose main logical operator is a quantifier We need to explain truth for all three kinds of sentence. We will provide a completely general explanation in this sec-tion. However, to try to keep the explanation comprehensible, we will, at several points, use the following interpretation: domain: all people born before 2000 ce a: Aristotle b: Beyoncé P (x): x is a philosopher R(x,y): x was born before y This will be our go-to example in what follows. 267 CHAPTER 30. TRUTH IN FOL 268 30.1 Atomic sentences The truth of atomic sentences should be fairly straightforward. For sentence letters, the interpretation specifies if they are true or false. The sentence ‘P (a)’ should be true just in case ‘P (x)’ is true of ‘a’. Given our go-to interpretation, this is true iff Aristotle is a philosopher. Aristotle is a philosopher. So the sentence is true. Equally, ‘P (b)’ is false on our go-to interpretation. Likewise, on this interpretation, ‘R(a,b)’ is true iff the object named by ‘a’ was born before the object named by ‘b’. Well, Aristotle was born before Beyoncé. So ‘R(a,b)’ is true. Equally, ‘R(a,a)’ is false: Aristotle was not born before Aristotle. Dealing with atomic sentences, then, is very intuitive. When R is an n-place predicate and a1, a2, . . . , an are names, The sentence R(a1, a2,. . . , an) is true in an interpreta-tion iff R is true of the objects named by a1, a2, . . ., an (in that order) in that interpretation. Recall, though, that there is a special kind of atomic sentence: two names connected by an identity sign constitute an atomic sentence. This kind of atomic sentence is also easy to handle. Where a and b are any names, a = b is true in an interpretation iff a and b name the very same object in that interpretation So in our go-to interpretation, ‘a = b’ is false, since Aristotle is distinct from Beyoncé. 30.2 Sentential connectives We saw in §26 that FOL sentences can be built up from sim-pler ones using the truth-functional connectives that were familiar CHAPTER 30. TRUTH IN FOL 269 from TFL. The rules governing these truth-functional connectives are exactly the same as they were when we considered TFL. Here they are: A∧B is true in an interpretation iff both A is true and B is true in that interpretation. A∨B is true in an interpretation iff either A is true or B is true in that interpretation. ¬A is true in an interpretation iff A is false in that interpretation. A →B is true in an interpretation iff either A is false or B is true in that interpretation. A ↔B is true in an interpretation iff A has the same truth value as B in that interpretation. This presents the very same information as the characteristic truth tables for the connectives; it just does so in a slightly differ-ent way. Some examples will probably help to illustrate the idea. (Make sure you understand them!) On our go-to interpretation: • ‘a = a ∧P (a)’ is true • ‘R(a,b) ∧P (b)’ is false because, although ‘R(a,b)’ is true, ‘P (b)’ is false • ‘a = b ∨P (a)’ is true • ‘¬a = b’ is true • ‘P (a) ∧¬(a = b ∧R(a,b))’ is true, because ‘P (a)’ is true and ‘a = b’ is false Make sure you understand these examples. CHAPTER 30. TRUTH IN FOL 270 30.3 When the main logical operator is a quantifier The exciting innovation in FOL, though, is the use of quantifiers, but expressing the truth conditions for quantified sentences is a bit more fiddly than one might first expect. Here is a naïve first thought. We want to say that ‘∀x F (x)’ is true iff ‘F (x)’ is true of everything in the domain. This should not be too problematic: our interpretation will specify directly what ‘F (x)’ is true of. Unfortunately, this naïve thought is not general enough. For example, we want to be able to say that ‘∀x∃y L(x,y)’ is true just in case (speaking roughly) ‘∃y L(x,y)’ is true of everything in the domain. But our interpretation does not directly specify what ‘∃y L(x,y)’ is true of. Instead, whether or not this is true of something should follow just from the interpretation of the predicate ‘L’, the domain, and the meanings of the quantifiers. So here is a second naïve thought. We might try to say that ‘∀x∃y L(x,y)’ is to be true in an interpretation iff ∃y L(a,y) is true for every name a that we have included in our interpretation. Similarly, we might try to say that ∃y L(a,y) is true just in case L(a, b) is true for some name b that we have included in our interpretation. Unfortunately, this is not right either. To see this, observe that our go-to interpretation only interprets two names, ‘a’ and ‘b’. But the domain—all people born before the year 2000 ce— contains many more than two people. (And we have no intention of trying to correct for this by naming all of them!) So here is a third thought. (And this thought is not naïve, but correct.) Although it is not the case that we have named everyone, each person could have been given a name. So we should focus on this possibility of extending an interpretation by adding a new name. We will offer a few examples of how this might work, centering on our go-to interpretation, and we will then present the formal definition. CHAPTER 30. TRUTH IN FOL 271 In our go-to interpretation, ‘∃x R(b,x)’ should be true. After all, in the domain, there is certainly someone who was born after Beyoncé. Lady Gaga is one of those people. Indeed, if we were to extend our go-to interpretation—temporarily, mind—by adding the name ‘c’ to refer to Lady Gaga, then ‘R(b,c)’ would be true on this extended interpretation. This, surely, should suffice to make ‘∃x R(b,x)’ true on the original go-to interpretation. In our go-to interpretation, ‘∃x(P (x) ∧R(x,a))’ should also be true. After all, in the domain, there is certainly someone who was both a philosopher and born before Aristotle. Socrates is one such person. Indeed, if we were to extend our go-to interpretation by letting a new name, ‘c’, denote Socrates, then ‘P (c) ∧R(c,a)’ would be true on this extended interpretation. Again, this should surely suffice to make ‘∃x(P (x)∧R(x,a))’ true on the original go-to interpretation. In our go-to interpretation, ‘∀x∃y R(x,y)’ should be false. Af-ter all, consider the last person born in the year 1999. We don’t know who that was, but if we were to extend our go-to interpre-tation by letting a new name, ‘d’, denote that person, then we would not be able to find anyone else in the domain to denote with some further new name, perhaps ‘e’, in such a way that ‘R(d,e)’ would be true. Indeed, no matter whom we named with ‘e’, ‘R(d,e)’ would be false. This observation is surely sufficient to make ‘∃y R(d,y)’ false in our extended interpretation, which in turn is surely sufficient to make ‘∀x∃y R(x,y)’ false on the original go-to interpretation. If you have understood these three examples, that’s what mat-ters. It provides the basis for a formal definition of truth for quantified sentences. Strictly speaking, though, we still need to give that definition. The result, sadly, is a bit ugly, and requires a few new definitions. Brace yourself! Suppose that A is a formula containing at least one occur-rence of the variable x, and that x is free in A. We will write this thus: A(. . . x. . . x. . .) CHAPTER 30. TRUTH IN FOL 272 Suppose also that c is a name. Then we will write: A(. . . c. . . c. . .) for the formula we obtain by replacing every occurrence of x in A with c. The resulting formula is called a substitution instance of ∀xAand ∃xA. Also, c is called the instantiating name. So: ∃x(R(e,x) ↔F (x)) is a substitution instance of ∀y∃x(R(y,x) ↔F (x)) with the instantiating name ‘e’ and instantiated variable ‘y’. Our interpretation will include a specification of which names correspond to which objects in the domain. Take any object in the domain, say, d, and a name c which is not already assigned by the interpretation. If our interpretation is I, then we can con-sider the interpretation I[d/c] which is just like I except it also assigns the name c to the object d. Then we can say that d satisfies the formula A(. . . x. . . x. . . ) in the interpretation I if, and only if, A(. . . c. . . c. . . ) is true in I[d/c]. (If d satisfies A(. . . x. . . x. . . ) we also say that A(. . . x. . . x. . . ) is true of d.) The interpretation I[d/c] is just like the interpretation I except it also assigns the name c to the object d. An object d satisfies A(. . . x. . . x. . . ) in interpretation I iff A(. . . c. . . c. . . ) is true in I[d/c]. So, for instance, Socrates satisfies the formula P (x) since P (c) is true in the interpretation I[Socrates/c], i.e., the interpretation: domain: all people born before 2000 ce a: Aristotle b: Beyoncé c: Socrates CHAPTER 30. TRUTH IN FOL 273 P (x): x is a philosopher R(x,y): x was born before y Armed with this notation, the rough idea is as follows. The sentence ∀xA(. . . x. . . x. . .) will be true in I iff, for any ob-ject d in the domain, A(. . . c. . . c. . .) is true in I[d/c], i.e., no matter what object (in the domain) we name with c. In other words, ∀xA(. . . x. . . x. . .) is true iff every object in the domain satisfies A(. . . x. . . x. . .). Similarly, the sentence ∃xA will be true iff there is some object that satisifes A(. . . x. . . x. . .), i.e., A(. . . c. . . c. . .) is true in I[d/c] for some object d. ∀xA(. . . x. . . x. . .) is true in an interpretation iff every object in the domain satisfies A(. . . x. . . x. . .). ∃xA(. . . x. . . x. . .) is true in an interpretation iff at least one object in the domain satisfies A(. . . x. . . x. . .). To be clear: all this is doing is formalizing (very pedantically) the intuitive idea expressed on the previous page. The result is a bit ugly, and the final definition might look a bit opaque. Hopefully, though, the spirit of the idea is clear. 30.4 Satisfaction of formulas The concept of an object satisfying a formula with a free vari-able can also be extended to formulas with more than one free variable. If we have a formula A(x, y) with two free variables x and y, then we can say that a pair of objects ⟨a,b⟩satisfies A(x, y) iff A(c, d) is true in the interpretation extended by two names c and d, where c names a and d names b. So, for in-stance, ⟨Socrates,Plato⟩satisfies R(x,y) since R(c,d) is true in the interpretation: domain: all people born before 2000 ce a: Aristotle CHAPTER 30. TRUTH IN FOL 274 b: Beyoncé c: Socrates d: Plato P (x): x is a philosopher R(x,y): x was born before y For atomic formulas, the objects, pairs of objects, etc., that sat-isfy them are exactly the extension of the predicate given in the interpretation. But the notion of satisfaction also applies to non-atomic formulas, e.g., the formula P (x) ∧R(x,b) is satisfied by all philosophers born before Beyoncé. It even applies to formulas involving quantifiers, e.g., P (x) ∧¬∃y(P (y) ∧R(y,x)) is satisfied by all people who are philosophers and for whom it is true that no philosopher was born before them—in other words, it is true of the first philosopher. By considering formulas (possibly involving quantifiers) with two free variables, we can express relations for which we do not have dedicated predicate symbols in our interpretation or sym-bolization key. Consider the formula R(x,y). It expresses the re-lation ‘ x was born before y’, since that is how we have specified its extension. What happens if we switch the variables, i.e., consider ‘R(y,x)’? A pair of objects ⟨y,x⟩in the domain (i.e., a pair of people) satisfies R(y,x) if, and only if, the reverse pair ⟨x,y⟩satisfies R(x,y). In other words, R(y,x) expresses the rela-tion ‘ x was born after y’. Or suppose we add to our interpretation a predicate for ‘teacher of’. T (x,y): x was a teacher of y Then the formula ‘∃z (T (z,x) ∧T (z,y))’ is satisfied by x and y if, and only if, some person z was a teacher of both x and y, i.e., it expresses ‘ x and y have a teacher in common’. Similarly, ‘∀z (T (x,z) ↔T (y,z))’ expresses ‘ x and y taught the same people’. CHAPTER 30. TRUTH IN FOL 275 Practice exercises A. Consider the following interpretation: • The domain comprises only Corwin and Benedict • ‘A(x)’ is to be true of both Corwin and Benedict • ‘B(x)’ is to be true of Benedict only • ‘N (x)’ is to be true of no one • ‘c’ is to refer to Corwin Determine whether each of the following sentences is true or false in that interpretation: 1. B(c) 2. A(c) ↔¬N (c) 3. N (c) →(A(c) ∨B(c)) 4. ∀x A(x) 5. ∀x¬B(x) 6. ∃x(A(x) ∧B(x)) 7. ∃x(A(x) →N (x)) 8. ∀x(N (x) ∨¬N (x)) 9. ∃x B(x) →∀x A(x) B. Consider the following interpretation: • The domain comprises only Lemmy, Courtney and Eddy • ‘G (x)’ is to be true of Lemmy, Courtney and Eddy. • ‘H (x)’ is to be true of and only of Courtney • ‘M (x)’ is to be true of and only of Lemmy and Eddy • ‘c’ is to refer to Courtney • ‘e’ is to refer to Eddy Determine whether each of the following sentences is true or false in that interpretation: 1. H (c) 2. H (e) 3. M (c) ∨M (e) CHAPTER 30. TRUTH IN FOL 276 4. G (c) ∨¬G (c) 5. M (c) →G (c) 6. ∃x H (x) 7. ∀x H (x) 8. ∃x ¬M (x) 9. ∃x(H (x) ∧G (x)) 10. ∃x(M (x) ∧G (x)) 11. ∀x(H (x) ∨M (x)) 12. ∃x H (x) ∧∃x M (x) 13. ∀x(H (x) ↔¬M (x)) 14. ∃x G (x) ∧∃x¬G (x) 15. ∀x∃y(G (x) ∧H (y)) C. Following the diagram conventions introduced at the end of §29, consider the following interpretation: 1 2 3 4 5 Determine whether each of the following sentences is true or false in that interpretation: 1. ∃x R(x,x) 2. ∀x R(x,x) 3. ∃x∀y R(x,y) 4. ∃x∀y R(y,x) 5. ∀x∀y∀z ((R(x,y) ∧R(y,z)) →R(x,z)) 6. ∀x∀y∀z ((R(x,y) ∧R(x,z)) →R(y,z)) 7. ∃x∀y ¬R(x,y) 8. ∀x(∃y R(x,y) →∃y R(y,x)) 9. ∃x∃y(¬x = y ∧R(x,y) ∧R(y,x)) 10. ∃x∀y(R(x,y) ↔x = y) 11. ∃x∀y(R(y,x) ↔x = y) CHAPTER 30. TRUTH IN FOL 277 12. ∃x∃y(¬x = y ∧R(x,y) ∧∀z (R(z,x) ↔y = z)) CHAPTER 31 Semantic concepts Defining truth in FOL was quite fiddly. But now that we are done, we can define various other central logical notions. These defi-nitions will look very similar to those for TFL, from §12. How-ever, remember that they concern interpretations, rather than val-uations. We will use the symbol ‘⊨’ for FOL much as we did for TFL. So: A 1,A 2,. . . ,A n ⊨C means that there is no interpretation in which all of A 1, A 2, ..., A n are true and in which C is false. Derivatively, ⊨A means that A is true in every interpretation. The other logical notions also have corresponding definitions in FOL: ⊲An FOL sentence A is a v alidity iff A is true in every interpretation; i.e., ⊨A. ⊲Ais a contradiction iff Ais false in every interpretation; i.e., ⊨¬A. 278 CHAPTER 31. SEMANTIC CONCEPTS 279 ⊲A 1,A 2,. . . ,A n ∴Cis v alid in fol iff there is no interpreta-tion in which all of the premises are true and the conclusion is false; i.e., A 1,A 2,. . . ,A n ⊨C. It is inv alid in fol other-wise. ⊲T wo FOL sentences A and B are equiv alent iff they are true in exactly the same interpretations as each other; i.e., both A ⊨B and B ⊨A. ⊲The FOL sentences A 1, A 2, . . . , A n are jointly satisfiable iff some interpretation makes all of them true. They are jointly unsatisfiable iff there is no such interpretation. CHAPTER 32 Using interpretations 32.1 Validities and contradictions Suppose we want to show that ‘∃x A(x,x) →B(d)’ is not a valid-ity. This requires showing that the sentence is not true in every interpretation; i.e., that it is false in some interpretation. If we can provide just one interpretation in which the sentence is false, then we will have shown that the sentence is not a validity. In order for ‘∃x A(x,x) →B(d)’ to be false, the antecedent (‘∃x A(x,x)’) must be true, and the consequent (‘B(d)’) must be false. To construct such an interpretation, we start by specifying a domain. Keeping the domain small makes it easier to specify what the predicates will be true of, so we will start with a domain that has just one member. For concreteness, let’s say it is just the city of Paris. domain: Paris The name ‘d’ must refer to something in the domain, so we have no option but: d: Paris 280 CHAPTER 32. USING INTERPRETATIONS 281 Recall that we want ‘∃x A(x,x)’ to be true, so we want all mem-bers of the domain to be paired with themselves in the extension of ‘A’. We can just offer: A(x,y): x is identical with y Now ‘A(d,d)’ is true, so it is surely true that ‘∃x A(x,x)’. Next, we want ‘B(d)’ to be false, so the referent of ‘d’ must not be in the extension of ‘B’. We might simply offer: B(x): x is in Germany Now we have an interpretation where ‘∃x A(x,x)’ is true, but where ‘B(d)’ is false. So there is an interpretation where ‘∃x A(x,x) →B(d)’ is false. So ‘∃x A(x,x) →B(d)’ is not a validity. We can just as easily show that ‘∃xA(x,x) →B(d)’ is not a contradiction. We need only specify an interpretation in which ‘∃xA(x,x) →B(d)’ is true; i.e., an interpretation in which either ‘∃x A(x,x)’ is false or ‘B(d)’ is true. Here is one: domain: Paris d: Paris A(x,y): x is identical with y B(x): x is in France This shows that there is an interpretation where ‘∃xA(x,x) → B(d)’ is true. So ‘∃x A(x,x) →B(d)’ is not a contradiction. To show that A is not a validity, it suffices to find an inter-pretation where A is false. To show that Ais not a contradiction, it suffices to find an interpretation where A is true. 32.2 Logical equivalence Suppose we want to show that ‘∀x S (x)’ and ‘∃x S (x)’ are not logically equivalent. We need to construct an interpretation in CHAPTER 32. USING INTERPRETATIONS 282 which the two sentences have different truth values; we want one of them to be true and the other to be false. We start by specifying a domain. Again, we make the domain small so that we can specify extensions easily. In this case, we will need at least two objects. (If we chose a domain with only one member, the two sentences would end up with the same truth value. In order to see why, try constructing some partial interpretations with one-member domains.) For concreteness, let’s take: domain: Ornette Coleman, Miles Davis We can make ‘∃x S (x)’ true by including something in the exten-sion of ‘S’, and we can make ‘∀x S (x)’ false by leaving something out of the extension of ‘S’. For concreteness, let’s say: S (x): x plays saxophone Now ‘∃x S (x)’ is true, because ‘S (x)’ is true of Ornette Coleman. Slightly more precisely, extend our interpretation by allowing ‘c’ to name Ornette Coleman. ‘S (c)’ is true in this extended in-terpretation, so ‘∃x S (x)’ was true in the original interpretation. Similarly, ‘∀x S (x)’ is false, because ‘S (x)’ is false of Miles Davis. Slightly more precisely, extend our interpretation by allowing ‘d’ to name Miles Davis, and ‘S (d)’ is false in this extended interpre-tation, so ‘∀x S (x)’ was false in the original interpretation. We have provided a counter-interpretation to the claim that ‘∀x S (x)’ and ‘∃x S (x)’ are logically equivalent. To show that A and B are not logically equivalent, it suf-fices to find an interpretation where one is true and the other is false. 32.3 Validity, entailment and satisfiability To test for validity, entailment, or satisfiability, we typically need to produce interpretations that determine the truth value of sev-eral sentences simultaneously. CHAPTER 32. USING INTERPRETATIONS 283 Consider the following argument in FOL: ∃x(G (x) →G (a)) ∴∃x G (x) →G (a) To show that this is invalid, we must make the premise true and the conclusion false. The conclusion is a conditional, so to make it false, the antecedent must be true and the consequent must be false. Clearly, our domain must contain two objects. Let’s try: domain: Karl Marx, Ludwig von Mises G (x): x hated communism a: Karl Marx Given that Marx wrote The Communist Manifesto, ‘G (a)’ is plainly false in this interpretation. But von Mises famously hated com-munism, so ‘∃x G (x)’ is true in this interpretation. Hence ‘∃x G (x) →G (a)’ is false, as required. Does this interpretation make the premise true? Yes it does! Note that ‘G (a) →G (a)’ is true. (Indeed, it is a validity.) But then certainly ‘∃x(G (x) →G (a))’ is true, so the premise is true, and the conclusion is false, in this interpretation. The argument is therefore invalid. In passing, note that we have also shown that ‘∃x(G (x) → G (a))’ does not entail ‘∃x G (x) →G (a)’, i.e., that ∃x(G (x) → G (a)) ⊭∃xG (x) →G (a). Equally, we have shown that the sen-tences ‘∃x(G (x) →G (a))’ and ‘¬(∃x G (x) →G (a))’ are jointly satisfiable. Let’s consider a second example. Consider: ∀x∃y L(x,y) ∴∃y∀x L(x,y) Again, we want to show that this is invalid. To do this, we must make the premises true and the conclusion false. Here is a sug-gestion: domain: Canadian citizens currently in a domestic partnership with another Canadian citizen L(x,y): x is in a domestic partnership with y CHAPTER 32. USING INTERPRETATIONS 284 The premise is clearly true on this interpretation. Anyone in the domain is a Canadian citizen in a domestic partnership with some other Canadian citizen. That other citizen will also, then, be in the domain. So for everyone in the domain, there will be someone (else) in the domain with whom they are in a domestic partnership. Hence ‘∀x∃y L(x,y)’ is true. However, the conclu-sion is clearly false, for that would require that there is some single person who is in a domestic partnership with everyone in the domain, and there is no such person, so the argument is in-valid. We observe immediately that the sentences ‘∀x∃y L(x,y)’ and ‘¬∃y∀x L(x,y)’ are jointly satisfiable and that ‘∀x∃y L(x,y)’ does not entail ‘∃y∀x L(x,y)’. For our third example, we’ll mix things up a bit. In §29, we de-scribed how we can present some interpretations using diagrams. For example: 1 2 3 Using the conventions employed in §29, the domain of this inter-pretation is the first three positive whole numbers, and ‘R(x,y)’ is true of x and y just in case there is an arrow from x to y in our diagram. Here are some sentences that the interpretation makes true: • ‘∀x∃y R(y,x)’ • ‘∃x∀y R(x,y)’ witness 1 • ‘∃x∀y(R(y,x) ↔x = y)’ witness 1 • ‘∃x∃y∃z ((¬y = z ∧R(x,y)) ∧R(z,x))’ witness 2 • ‘∃x∀y ¬R(x,y)’ witness 3 • ‘∃x(∃y R(y,x) ∧¬∃y R(x,y))’ witness 3 This immediately shows that all of the preceding six sentences are jointly satisfiable. We can use this observation to generate CHAPTER 32. USING INTERPRETATIONS 285 invalid arguments, e.g.: ∀x∃y R(y,x),∃x∀y R(x,y) ∴∀x∃y R(x,y) ∃x∀y R(x,y),∃x∀y¬R(x,y) ∴¬∃x∃y∃z (¬y = z ∧(R(x,y) ∧R(z,x))) and many more besides. If some interpretation makes all of A 1,A 2,. . . ,A n true and C is false, then: • A 1,A 2,. . . ,A n ∴C is invalid; and • A 1,A 2,. . . ,A n ⊭C; and • And A 1,A 2,. . . ,A n,¬C are jointly satisfiable. An interpretation which refutes a claim—to logical truth, say, or to entailment—is called a counter-interpretation, or a counter-model. We’ll close this section, though, with a caution about the rela-tionship between (in)validity and (non)entailment. Recall FOL’s limitations: it is an extensional language; it ignores issues of vagueness; and it cannot handle cases of validity for ‘special reasons’. To take one illustration of these issues, consider this natural-language argument: Every fox is cute. ∴All vixens are cute. This is valid: necessarily every vixen is a fox, so it is impossible for the premise to be true and the conclusion false. Now, we might sensibly symbolize the argument as follows: ∀x(F (x) →C (x)) ∴∀x(V (x) →C (x)) However, it is easy to find counter-models which show that ∀x(F (x) →C (x)) ⊭∀x(V (x) →C (x)). (Exercise: find one.) So, it would be wrong to infer that the English argument is invalid, just because there is a counter-model to the relevant FOL-entailment. CHAPTER 32. USING INTERPRETATIONS 286 The general moral is this. If you want to infer from the ab-sence of an entailment in FOL to the invalidity of some English argument, then you need to argue that nothing important is lost in the way you have symbolized the English argument. Practice exercises A. Show that each of the following is neither a validity nor a contradiction: 1. D(a) ∧D(b) 2. ∃x T (x,h) 3. P (m) ∧¬∀x P (x) 4. ∀z J (z) ↔∃y J (y) 5. ∀x(W (x,m,n) ∨∃yL(x,y)) 6. ∃x(G (x) →∀y M (y)) 7. ∃x(x = h ∧x = i) B. Show that the following pairs of sentences are not logically equivalent. 1. J (a), K (a) 2. ∃x J (x), J (m) 3. ∀x R(x,x), ∃x R(x,x) 4. ∃x P (x) →Q (c), ∃x(P (x) →Q (c)) 5. ∀x(P (x) →¬Q (x)), ∃x(P (x) ∧¬Q (x)) 6. ∃x(P (x) ∧Q (x)), ∃x(P (x) →Q (x)) 7. ∀x(P (x) →Q (x)), ∀x(P (x) ∧Q (x)) 8. ∀x∃y R(x,y), ∃x∀y R(x,y) 9. ∀x∃y R(x,y), ∀x∃y R(y,x) C. Show that the following sentences are jointly satisfiable: 1. M (a),¬N (a),P (a),¬Q (a) 2. L(e,e),L(e,g),¬L(g,e),¬L(g,g) 3. ¬(M (a) ∧∃x A(x)),M (a) ∨F (a),∀x(F (x) →A(x)) 4. M (a) ∨M (b),M (a) →∀x¬M (x) CHAPTER 32. USING INTERPRETATIONS 287 5. ∀y G (y),∀x(G (x) →H (x)),∃y¬I (y) 6. ∃x(B(x) ∨A(x)),∀x¬C (x),∀x [︁ (A(x) ∧B(x)) →C (x) ]︁ 7. ∃x X (x),∃x Y (x),∀x(X (x) ↔¬Y (x)) 8. ∀x(P (x) ∨Q (x)),∃x¬(Q (x) ∧P (x)) 9. ∃z (N (z) ∧O (z,z)),∀x∀y(O (x,y) →O (y,x)) 10. ¬∃x∀y R(x,y),∀x∃y R(x,y) 11. ¬R(a,a), ∀x(x = a ∨R(x,a)) 12. ∀x∀y∀z [(x = y ∨y = z) ∨x = z], ∃x∃y ¬x = y 13. ∃x∃y((Z (x) ∧Z (y)) ∧x = y), ¬Z (d), d = e D. Show that the following arguments are invalid: 1. ∀x(A(x) →B(x)) ∴∃x B(x) 2. ∀x(R(x) →D(x)),∀x(R(x) →F (x)) ∴∃x(D(x) ∧F (x)) 3. ∃x(P (x) →Q (x)) ∴∃x P (x) 4. N (a) ∧N (b) ∧N (c) ∴∀x N (x) 5. R(d,e),∃x R(x,d) ∴R(e,d) 6. ∃x(E(x) ∧F (x)),∃x F (x) →∃x G (x) ∴∃x(E(x) ∧G (x)) 7. ∀x O (x,c),∀x O (c,x) ∴∀x O (x,x) 8. ∃x( J (x)∧K (x)),∃x¬K (x),∃x¬ J (x) ∴∃x(¬ J (x)∧¬K (x)) 9. L(a,b) →∀x L(x,b),∃x L(x,b) ∴L(b,b) 10. ∀x(D(x) →∃y T (y,x)) ∴∃y∃z ¬y = z CHAPTER 33 Reasoning about interpretations 33.1 Validities and contradictions We can show that a sentence is not a validity just by providing one carefully specified interpretation: an interpretation in which the sentence is false. To show that something is a validity, on the other hand, it would not be enough to construct ten, one hundred, or even a thousand interpretations in which the sentence is true. A sentence is only a validity if it is true in every interpretation, and there are infinitely many interpretations. We need to reason about all of them, and we cannot do this by dealing with them one by one! Sometimes, we can reason about all interpretations fairly eas-ily. For example, we can offer a relatively simple argument that ‘R(a,a) ∨¬R(a,a)’ is a validity: Any relevant interpretation will give ‘R(a,a)’ a truth value. If ‘R(a,a)’ is true in an interpretation, then 288 CHAPTER 33. REASONING ABOUT INTERPRETATIONS 289 ‘R(a,a) ∨¬R(a,a)’ is true in that interpretation. If ‘R(a,a)’ is false in an interpretation, then ¬R(a,a) is true, and so ‘R(a,a) ∨¬R(a,a)’ is true in that interpretation. These are the only alternatives. So ‘R(a,a) ∨¬R(a,a)’ is true in every interpretation. Therefore, it is a validity. This argument is valid, of course, and its conclusion is true. How-ever, it is not an argument in FOL. Rather, it is an argument in English about FOL: it is an argument in the metalanguage. Note another feature of the argument. Since the sentence in question contained no quantifiers, we did not need to think about how to interpret ‘a’ and ‘R’; the point was just that, however we interpreted them, ‘R(a,a)’ would have some truth value or other. (We could ultimately have given the same argument concerning TFL sentences.) Let’s have another example. The sentence ‘∀x(R(x,x) ∨ ¬R(x,x))’ should obviously be a validity. However, saying pre-cisely why is quite tricky. We cannot say that ‘R(x,x) ∨¬R(x,x)’ is true in every interpretation, since ‘R(x,x) ∨¬R(x,x)’ is not even a sentence of FOL (remember that ‘x’ is a variable, not a name). Instead, we should say something like this: Consider some arbitrary interpretation. ∀x(R(x,x) ∨ ¬R(x,x)) is true in our interpretation iff R(x,x) ∨ ¬R(x,x) is satisfied by every object of its domain. Consider some arbitrary member of the domain, which, for convenience, we will call Fred. Either Fred satisfies R(x,x) or it does not. If Fred sat-isfies ‘R(x,x)’, then Fred also satisfies ‘R(x,x) ∨ ¬R(x,x)’. If Fred does not satisfy ‘R(x,x)’, it does satisfy ‘¬R(x,x)’ and so also ‘R(x,x) ∨¬R(x,x)’.1 So either way, Fred satisfies ‘R(x,x) ∨¬R(x,x)’. Since there was nothing special about Fred—we might have 1We use here the fact that the truth conditions for connectives also apply to satisfaction: a satisfies A(x) ∨B(x) iff a satisfies A(x) or B(x), etc. CHAPTER 33. REASONING ABOUT INTERPRETATIONS 290 chosen any object—we see that every object in the do-main satisfies ‘R(x,x) ∨¬R(x,x)’. So ‘∀x(R(x,x) ∨ ¬R(x,x))’ is true in our interpretation. But we chose our interpretation arbitrarily, so ‘∀x(R(x,x) ∨ ¬R(x,x))’ is true in every interpretation. It is there-fore a validity. This is quite long-winded, but, as things stand, there is no alter-native. In order to show that a sentence is a validity, we must reason about all interpretations. 33.2 Other cases Similar points hold of other cases too. Thus, we must reason about all interpretations if we want to show: • that a sentence is a contradiction; for this requires that it is false in every interpretation. • that two sentences are logically equivalent; for this requires that they have the same truth value in every interpretation. • that some sentences are jointly unsatisfiable; for this re-quires that there is no interpretation in which all of those sentences are true together; i.e., that, in every interpreta-tion, at least one of those sentences is false. • that an argument is valid; for this requires that the conclu-sion is true in every interpretation where the premises are true. • that some sentences entail another sentence. The problem is that, with the tools available to you so far, rea-soning about all interpretations is a serious challenge! For a final example, here is a perfectly obvious entailment: ∀x(H (x) ∧J (x)) ⊨∀x H (x) After all, if everything is both H and J , then everything is H . But we can only establish the entailment by considering what must be CHAPTER 33. REASONING ABOUT INTERPRETATIONS 291 true in every interpretation in which the premise ∀x(H (x)∧J (x)) is true. To show this, we would have to reason as follows: Consider an arbitrary interpretation in which ‘∀x(H (x)∧J (x))’ is true. It follows that ‘H (x)∧J (x)’ is satisfied by every object in this interpretation. ‘H (x)’ will, then, also be satisfied by every object.2 So it must be that ‘∀x H (x)’ is true in the interpretation. We’ve assumed nothing about the interpretation ex-cept that it was one in which ‘∀x(H (x)∧J (x))’ is true. So any interpretation in which ‘∀x(H (x) ∧J (x))’ is true is one in which ‘∀x H (x)’ is true. Even for a simple entailment like this one, the reasoning is some-what complicated. For more complicated entailments, the rea-soning can be extremely torturous. The following table summarizes whether a single interpreta-tion or counter-interpretation suffices, or whether we must reason about all interpretations. Y es No validity? all interpretations one counter-interpretation contradiction? all interpretations one counter-interpretation equivalent? all interpretations one counter-interpretation satisfiable? one interpretation all interpretations valid? all interpretations one counter-interpretation entailment? all interpretations one counter-interpretation You might want to compare this table with the table at the end of §14. The key difference resides in the fact that TFL con-cerns truth tables, whereas FOL concerns interpretations. This difference is deeply important, since each truth-table only ever has finitely many lines, so that a complete truth table is a rela-tively tractable object. By contrast, there are infinitely many in-terpretations for any given sentence(s), so that reasoning about all interpretations can be a deeply tricky business. 2Here again we make use of the fact that any object that satisfies A(x) ∧ B(x) must satisfy both A(x) and B(x). PART VII Natural deduction for FOL 292 CHAPTER 34 Basic rules for FOL The language of FOL makes use of all of the connectives of TFL. So proofs in FOL will use all of the basic and derived rules from Part IV. We will also use the proof-theoretic notions (particularly, the symbol ‘⊢’) introduced there. However, we will also need some new basic rules to govern the quantifiers, and to govern the identity sign. 34.1 Universal elimination From the claim that everything is F , you can infer that any par-ticular thing is F . You name it; it’s F . So the following should be fine: 1 ∀x R(x,x,d) 2 R(a,a,d) ∀E 1 We obtained line 2 by dropping the universal quantifier and re-placing every instance of ‘x’ with ‘a’. Equally, the following should be allowed: 293 CHAPTER 34. BASIC RULES FOR FOL 294 1 ∀x R(x,x,d) 2 R(d,d,d) ∀E 1 We obtained line 2 here by dropping the universal quantifier and replacing every instance of ‘x’ with ‘d’. We could have done the same with any other name we wanted. This motivates the universal elimination rule (∀E): m ∀xA(. . . x. . . x. . .) A(. . . c. . . c. . .) ∀E m The notation here was introduced in §30. The point is that you can obtain any substitution instance of a universally quantified formula: replace every instance of the quantified variable with any name you like. We should emphasize that (as with every elimination rule) you can only apply the ∀E rule when the universal quantifier is the main logical operator. So the following is banned: 1 ∀x B(x) →B(k) 2 B(b) →B(k) naughy attempt to invoke ∀E 1 This is illegitimate, since ‘∀x’ is not the main logical operator in line 1. (If you need a reminder as to why this sort of inference should be banned, reread §23.) 34.2 Existential introduction From the claim that some particular thing is F , you can infer that something is F . So we ought to allow: 1 R(a,a,d) 2 ∃x R(a,a,x) ∃I 1 CHAPTER 34. BASIC RULES FOR FOL 295 Here, we have replaced the name ‘d’ with a variable ‘x’, and then existentially quantified over it. Equally, we would have allowed: 1 R(a,a,d) 2 ∃x R(x,x,d) ∃I 1 Here we have replaced both instances of the name ‘a’ with a variable, and then existentially generalized. But we do not need to replace both instances of a name with a variable: if Narcissus loves himself, then there is someone who loves Narcissus. So we also allow: 1 R(a,a,d) 2 ∃x R(x,a,d) ∃I 1 Here we have replaced one instance of the name ‘a’ with a vari-able, and then existentially generalized. These observations mo-tivate our introduction rule, although to explain it, we will need to introduce some new notation. Where A is a sentence containing the name c, we can emphasize this by writing ‘A(. . . c. . . c. . .)’. We will write ‘A(. . . x. . . c. . .)’ to indicate any formula obtained by replacing some or all of the instances of the name c with the variable x. Armed with this, our introduction rule is: m A(. . . c. . . c. . .) ∃xA(. . . x. . . c. . .) ∃I m 34.3 Empty domains The following proof combines our two new rules for quantifiers: CHAPTER 34. BASIC RULES FOR FOL 296 1 ∀x F (x) 2 F (a) ∀E 1 3 ∃x F (x) ∃I 2 Could this be a bad proof? If anything exists at all, then certainly we can infer that something is F , from the fact that everything is F . But what if nothing exists at all? Then it is surely vacuously true that everything is F ; however, it does not following that something is F , for there is nothing to be F . So if we claim that, as a matter of logic alone, ‘∃x F (x)’ follows from ‘∀x F (x)’, then we are claiming that, as a matter of logic alone, there is something rather than nothing. This might strike us as a bit odd. Actually, we are already committed to this oddity. In §22, we stipulated that domains in FOL must have at least one member. We then defined a validity (of FOL) as a sentence which is true in every interpretation. Since ‘∃x x = x’ will be true in every interpretation, this also had the effect of stipulating that it is a matter of logic that there is something rather than nothing. Since it is far from clear that logic should tell us that there must be something rather than nothing, we might well be cheat-ing a bit here. If we refuse to cheat, though, then we pay a high cost. Here are three things that we want to hold on to: • ∀x F (x) ⊢F (a): after all, that was ∀E. • F (a) ⊢∃x F (x): after all, that was ∃I. • the ability to copy-and-paste proofs together: after all, rea-soning works by putting lots of little steps together into rather big chains. If we get what we want on all three counts, then we have to coun-tenance that ∀x F (x) ⊢∃x F (x). So, if we get what we want on all three counts, the proof system alone tells us that there is some-thing rather than nothing. And if we refuse to accept that, then CHAPTER 34. BASIC RULES FOR FOL 297 we have to surrender one of the three things that we want to hold on to! Before we start thinking about which to surrender, we might want to ask how much of a cheat this is. Granted, it may make it harder to engage in theological debates about why there is some-thing rather than nothing. But the rest of the time, we will get along just fine. So maybe we should just regard our proof sys-tem (and FOL, more generally) as having a very slightly limited purview. If we ever want to allow for the possibility of nothing, then we will have to cast around for a more complicated proof system. But for as long as we are content to ignore that possibil-ity, our proof system is perfectly in order. (As, similarly, is the stipulation that every domain must contain at least one object.) 34.4 Universal introduction Suppose you had shown of each particular thing that it is F (and that there are no other things to consider). Then you would be justified in claiming that everything is F. This would motivate the following proof rule. If you had established each and every single substitution instance of ‘∀x F (x)’, then you can infer ‘∀x F (x)’. Unfortunately, that rule would be utterly unusable. To es-tablish each and every single substitution instance would require proving ‘F (a)’, ‘F (b)’, . . . , ‘F (j2)’, . . . , ‘F (r79002)’, ..., and so on. Indeed, since there are infinitely many names in FOL, this process would never come to an end. So we could never apply that rule. We need to be a bit more cunning in coming up with our rule for introducing universal quantification. A solution will be inspired by considering: ∀x F (x) ∴∀y F (y) This argument should obviously be valid. After all, alphabetical variation ought to be a matter of taste, and of no logical conse-quence. But how might our proof system reflect this? Suppose we begin a proof thus: CHAPTER 34. BASIC RULES FOR FOL 298 1 ∀x F (x) 2 F (a) ∀E 1 We have proved ‘F (a)’. And, of course, nothing stops us from using the same justification to prove ‘F (b)’, ‘F (c)’, ..., ‘F (j2)’, ..., ‘F (r79002), . . . , and so on until we run out of space, time, or patience. But reflecting on this, we see that there is a way to prove F c, for any name c. And if we can do it for any thing, we should surely be able to say that ‘F ’ is true of everything. This therefore justifies us in inferring ‘∀y F (y)’, thus: 1 ∀x F (x) 2 F (a) ∀E 1 3 ∀y F (y) ∀I 2 The crucial thought here is that ‘a’ was just some arbitrary name. There was nothing special about it—we might have chosen any other name—and still the proof would be fine. And this crucial thought motivates the universal introduction rule (∀I): m A(. . . c. . . c. . .) ∀xA(. . . x. . . x. . .) ∀I m c must not occur in any undischarged assumption A crucial aspect of this rule, though, is bound up in the con-straint. This constraint ensures that we are always reasoning at a sufficiently general level. To see the constraint in action, consider this terrible argument: Everyone loves Kylie Minogue; therefore everyone loves themselves. CHAPTER 34. BASIC RULES FOR FOL 299 We might symbolize this obviously invalid inference pattern as: ∀x L(x,k) ∴∀x L(x,x) Now, suppose we tried to offer a proof that vindicates this argu-ment: 1 ∀x L(x,k) 2 L(k,k) ∀E 1 3 ∀x L(x,x) naughty attempt to invoke ∀I 2 This is not allowed, because ‘k’ occurred already in an undis-charged assumption, namely, on line 1. The crucial point is that, if we have made any assumptions about the object we are working with, then we are not reasoning generally enough to license ∀I. Although the name may not occur in any undischarged as-sumption, it may occur in a discharged assumption. That is, it may occur in a subproof that we have already closed. For exam-ple, this is just fine: 1 G (d) 2 G (d) R 1 3 G (d) →G (d) →I 1–2 4 ∀z (G (z) →G (z)) ∀I 3 This tells us that ‘∀z (G (z) →G (z))’ is a theorem. And that is as it should be. We should emphasize one last point. As per the conventions of §30.3, the use of ∀I requires that we are replacing every instance of the name c in A(. . . c. . . c. . .) with the variable x. If we only replace some names and not others, we end up ‘proving’ silly things. For example, consider the argument: Everyone is as old as themselves; so everyone is as old as Judi Dench. CHAPTER 34. BASIC RULES FOR FOL 300 We might symbolize this as follows: ∀x O (x,x) ∴∀x O (x,d) But now suppose we tried to vindicate this terrible argument with the following: 1 ∀x O (x,x) 2 O (d,d) ∀E 1 3 ∀x O (x,d) naughty attempt to invoke ∀I 2 Fortunately, our rules do not allow for us to do this: the attempted proof is banned, since it doesn’t replace every occurrence of ‘d’ in line 2 with an ‘x’. 34.5 Existential elimination Suppose we know that something is F . The problem is that sim-ply knowing this does not tell us which thing is F . So it would seem that from ‘∃x F (x)’ we cannot immediately conclude ‘F (a)’, ‘F (e23)’, or any other substitution instance of the sentence. What can we do? Suppose we know that something is F , and that everything which is F is also G. In (almost) natural English, we might reason thus: Since something is F , there is some particular thing which is an F . We do not know anything about it, other than that it’s an F , but for convenience, let’s call it ‘Becky’. So: Becky is F . Since everything which is F is G, it follows that Becky is G. But since Becky is G, it follows that something is G. And nothing depended on which object, exactly, Becky was. So, something is G. CHAPTER 34. BASIC RULES FOR FOL 301 We might try to capture this reasoning pattern in a proof as fol-lows: 1 ∃x F (x) 2 ∀x(F (x) →G (x)) 3 F (b) 4 F (b) →G (b) ∀E 2 5 G (b) →E 4, 3 6 ∃x G (x) ∃I 5 7 ∃x G (x) ∃E 1, 3–6 Breaking this down: we started by writing down our assumptions. At line 3, we made an additional assumption: ‘F (b)’. This was just a substitution instance of ‘∃x F (x)’. On this assumption, we established ‘∃x G (x)’. Note that we had made no special assump-tions about the object named by ‘b’; we had only assumed that it satisfies ‘F (x)’. So nothing depends upon which object it is. And line 1 told us that something satisfies ‘F (x)’, so our reason-ing pattern was perfectly general. We can discharge the specific assumption ‘F (b)’, and simply infer ‘∃x G (x)’ on its own. Putting this together, we obtain the existential elimination rule (∃E): CHAPTER 34. BASIC RULES FOR FOL 302 m ∃xA(. . . x. . . x. . .) i A(. . . c. . . c. . .) j B B ∃E m, i–j c must not occur in any assumption undischarged before line i c must not occur in ∃xA(. . . x. . . x. . .) c must not occur in B As with universal introduction, the constraints are extremely important. To see why, consider the following terrible argument: Tim Button is a lecturer. Someone is not a lecturer. So Tim Button is both a lecturer and not a lecturer. We might symbolize this obviously invalid inference pattern as follows: L(b),∃x ¬L(x) ∴L(b) ∧¬L(b) Now, suppose we tried to offer a proof that vindicates this argu-ment: 1 L(b) 2 ∃x ¬L(x) 3 ¬L(b) 4 L(b) ∧¬L(b) ∧I 1, 3 5 L(b) ∧¬L(b) naughty attempt to invoke ∃E 2, 3–4 CHAPTER 34. BASIC RULES FOR FOL 303 The last line of the proof is not allowed. The name that we used in our substitution instance for ‘∃x ¬L(x)’ on line 3, namely ‘b’, occurs in line 4. This would be no better: 1 L(b) 2 ∃x ¬L(x) 3 ¬L(b) 4 L(b) ∧¬L(b) ∧I 1, 3 5 ∃x(L(x) ∧¬L(x)) ∃I 4 6 ∃x(L(x) ∧¬L(x)) naughty attempt to invoke ∃E 2, 3–5 The last line is still not allowed. For the name that we used in our substitution instance for ‘∃x ¬L(x)’, namely ‘b’, occurs in an undischarged assumption, namely line 1. The moral of the story is this. If you want to squeeze information out of an existential quantifier, choose a new name for your substitu-tion instance. That way, you can guarantee that you meet all the constraints on the rule for ∃E. Practice exercises A. Explain why these two ‘proofs’ are incorrect. Also, provide interpretations which would invalidate the fallacious argument forms the ‘proofs’ enshrine: 1 ∀x R(x,x) 2 R(a,a) ∀E 1 3 ∀y R(a,y) ∀I 2 4 ∀x ∀y R(x,y) ∀I 3 1 ∀x ∃y R(x,y) 2 ∃y R(a,y) ∀E 1 3 R(a,a) 4 ∃x R(x,x) ∃I 3 5 ∃x R(x,x) ∃E 2, 3–4 CHAPTER 34. BASIC RULES FOR FOL 304 B. The following three proofs are missing their citations (rule and line numbers). Add them, to turn them into bona fide proofs. 1. 1 ∀x∃y(R(x,y) ∨R(y,x)) 2 ∀x ¬R(m,x) 3 ∃y(R(m,y) ∨R(y,m)) 4 R(m,a) ∨R(a,m) 5 ¬R(m,a) 6 R(a,m) 7 ∃x R(x,m) 8 ∃x R(x,m) 2. 1 ∀x(∃y L(x,y) →∀z L(z,x)) 2 L(a,b) 3 ∃y L(a,y) →∀zL(z,a) 4 ∃y L(a,y) 5 ∀z L(z,a) 6 L(c,a) 7 ∃y L(c,y) →∀z L(z,c) 8 ∃y L(c,y) 9 ∀z L(z,c) 10 L(c,c) 11 ∀x L(x,x) CHAPTER 34. BASIC RULES FOR FOL 305 3. 1 ∀x( J (x) →K (x)) 2 ∃x ∀y L(x,y) 3 ∀x J (x) 4 ∀y L(a,y) 5 L(a,a) 6 J (a) 7 J (a) →K (a) 8 K (a) 9 K (a) ∧L(a,a) 10 ∃x(K (x) ∧L(x,x)) 11 ∃x(K (x) ∧L(x,x)) C. In §23 problem A, we considered fifteen syllogistic figures of Aristotelian logic. Provide proofs for each of the argument forms. NB: You will find it much easier if you symbolize (for example) ‘No F is G’ as ‘∀x(F (x) →¬G (x))’. D. Aristotle and his successors identified other syllogistic forms which depended upon ‘existential import’. Symbolize each of these argument forms in FOL and offer proofs. 1. Barbari. Something is H. All G are F. All H are G. So: Some H is F. 2. Celaront. Something is H. No G are F. All H are G. So: Some H is not F. 3. Cesaro. Something is H. No F are G. All H are G. So: Some H is not F. 4. Camestros. Something is H. All F are G. No H are G. So: Some H is not F. CHAPTER 34. BASIC RULES FOR FOL 306 5. Felapton. Something is G. No G are F. All G are H. So: Some H is not F. 6. Darapti. Something is G. All G are F. All G are H. So: Some H is F. 7. Calemos. Something is H. All F are G. No G are H. So: Some H is not F. 8. Fesapo. Something is G. No F is G. All G are H. So: Some H is not F. 9. Bamalip. Something is F. All F are G. All G are H. So: Some H are F. E. For each of the following claims, provide an FOL proof that shows it is true. 1. ⊢∀x F (x) →∀y(F (y) ∧F (y)) 2. ∀x(A(x) →B(x)),∃x A(x) ⊢∃x B(x) 3. ∀x(M (x) ↔N (x)),M (a) ∧∃x R(x,a) ⊢∃x N (x) 4. ∀x ∀y G (x,y) ⊢∃x G (x,x) 5. ⊢∀x R(x,x) →∃x ∃y R(x,y) 6. ⊢∀y ∃x(Q (y) →Q (x)) 7. N (a) →∀x(M (x) ↔M (a)),M (a),¬M (b) ⊢¬N (a) 8. ∀x ∀y(G (x,y) →G (y,x)) ⊢∀x∀y(G (x,y) ↔G (y,x)) 9. ∀x(¬M (x) ∨L(j,x)),∀x(B(x) → L(j,x)),∀x(M (x) ∨ B(x)) ⊢∀xL( j,x) F. Write a symbolization key for the following argument, symbol-ize it, and prove it: There is someone who likes everyone who likes every-one that she likes. Therefore, there is someone who likes herself. G. Show that each pair of sentences is provably equivalent. 1. ∀x(A(x) →¬B(x)), ¬∃x(A(x) ∧B(x)) 2. ∀x(¬A(x) →B(d)), ∀x A(x) ∨B(d) 3. ∃x P (x) →Q (c), ∀x(P (x) →Q (c)) CHAPTER 34. BASIC RULES FOR FOL 307 H. For each of the following pairs of sentences: If they are prov-ably equivalent, give proofs to show this. If they are not, construct an interpretation to show that they are not logically equivalent. 1. ∀x P (x) →Q (c),∀x(P (x) →Q (c)) 2. ∀x ∀y ∀z B(x,y,z),∀x B(x,x)x 3. ∀x ∀y D(x,y),∀y ∀x D(x,y) 4. ∃x ∀y D(x,y),∀y ∃x D(x,y) 5. ∀x(R(c,a) ↔R(x,a)),R(c,a) ↔∀x R(x,a) I. For each of the following arguments: If it is valid in FOL, give a proof. If it is invalid, construct an interpretation to show that it is invalid. 1. ∃y ∀x R(x,y) ∴∀x ∃y R(x,y) 2. ∀x ∃y R(x,y) ∴∃y ∀x R(x,y) 3. ∃x(P (x) ∧¬Q (x)) ∴∀x(P (x) →¬Q (x)) 4. ∀x(S (x) →T (a)),S (d) ∴T (a) 5. ∀x(A(x) →B(x)),∀x(B(x) →C (x)) ∴∀x(A(x) →C (x)) 6. ∃x(D(x) ∨E(x)),∀x(D(x) →F (x)) ∴∃x(D(x) ∧F (x)) 7. ∀x ∀y(R(x,y) ∨R(y,x)) ∴R(j, j) 8. ∃x ∃y(R(x,y) ∨R(y,x)) ∴R(j, j) 9. ∀x P (x) →∀x Q (x),∃x ¬P (x) ∴∃x ¬Q (x) 10. ∃x M (x) →∃x N (x), ¬∃x N (x) ∴∀x ¬M (x) CHAPTER 35 Proofs with quantifiers In §17 we discussed strategies for constructing proofs using the basic rules of natural deduction for TFL. The same principles apply to the rules for the quantifiers. If we want to prove a quan-tifier sentence ∀xA(x) or ∃xA(x), we can work backward by justifying the sentence we want by ∀I or ∃I and trying to find a proof of the corresponding premise of that rule. And to work forward from a quantified sentence, we apply ∀E or ∃E, as the case may be. Specifically, suppose you want to prove ∀xA(x). To do so using ∀I, we would need a proof of A(c) for some name c which does not occur in any undischarged assumption. To apply the corresponding strategy, i.e., to construct a proof of ∀xA(x) by working backward, is thus to write A(c) above it and then to continue to try to find a proof of that sentence. . . . n A(c) n + 1 ∀xA(x) ∀I n 308 CHAPTER 35. PROOFS WITH QUANTIFIERS 309 A(c) is obtained from A(x) by replacing every free occurrence of x in A(x) by c. For this to work, c must satisfy the special condition. We can ensure that it does by always picking a name that does not already occur in the proof constructed so far. (Of course, it will occur in the proof we end up constructing—just not in an assumption that is undischarged at line n + 1.) To work backward from a sentence ∃xA(x) we similarly write a sentence above it that can serve as a justification for an application of the ∃I rule, i.e., a sentence of the form A(c). . . . n A(c) n + 1 ∃xA(x) ∃I n This looks just like what we would do if we were working back-ward from a universally quantified sentence. The difference is that whereas for ∀I we have to pick a name c which does not occur in the proof (so far), for ∃I we may and in general must pick a name c which already occurs in the proof. Just like in the case of ∨I, it is often not clear which c will work out, and so to avoid having to backtrack you should work backward from exis-tentially quantified sentences only when all other strategies have been applied. By contrast, working forward from sentences ∃xA(x) gen-erally always works and you won’t have to backtrack. Working forward from an existentially quantified sentence takes into ac-count not just ∃xA(x) but also whatever sentence B you would like to prove. It requires that you set up a subproof above B, wherein B is the last line, and a substitution instance A(c) of ∃xA(x) as the assumption. In order to ensure that the condi-tion on c that governs ∃E is satisfied, chose a name c which does not already occur in the proof. CHAPTER 35. PROOFS WITH QUANTIFIERS 310 . . . m ∃xA(x) . . . n A(c) . . . k B k + 1 B ∃E m, n–k You’ll then continue with the goal of proving B, but now inside a subproof in which you have an additional sentence to work with, namely A(c). Lastly, working forward from ∀xA(x) means that you can always write down A(c) and justify it using ∀E, for any name c. Of course, you wouldn’t want to do that willy-nilly. Only certain names c will help in your task of proving whatever goal sentence you are working on. So, like working backward from ∃xA(x), you should work forward from ∀xA(x) only after all other strate-gies have been applied. Let’s consider as an example the argument ∀x(A(x) →B) ∴ ∃x A(x) →B. To start constructing a proof, we write the premise at the top and the conclusion at the bottom. 1 ∀x(A(x) →B) . . . n ∃x A(x) →B The strategies for connectives of TFL still apply, and you should apply them in the same order: first work backward from condi-tionals, negated sentences, conjunctions, and now also universal quantifiers, then forward from disjunctions and now existential CHAPTER 35. PROOFS WITH QUANTIFIERS 311 quantifiers, and only then try to apply →E, ¬E, ∨I, ∀E, or ∃I. In our case, that means, working backward from the conclusion: 1 ∀x(A(x) →B) 2 ∃x A(x) . . . n −1 B n ∃x A(x) →B →I 2–(n −1) Our next step should be to work forward from ∃x A(x) on line 2. For that, we have to pick a name not already in our proof. Since no names appear, we can pick any name, say ‘d’ 1 ∀x(A(x) →B) 2 ∃x A(x) 3 A(d) . . . n −2 B n −1 B ∃E 2, 3–(n −2) n ∃x A(x) →B →I 2–(n −1) Now we’ve exhausted our primary strategies, and it is time to work forward from the premise ∀x(A(x) →B). Applying ∀E means we can justify any instance of A(c) →B, regardless of what c we choose. Of course, we’ll do well to choose d, since that will give us A(d) →B. Then we can apply →E to justify B, finishing the proof. CHAPTER 35. PROOFS WITH QUANTIFIERS 312 1 ∀x(A(x) →B) 2 ∃x A(x) 3 A(d) 4 A(d) →B ∀E 1 5 B →E 4, 3 6 B ∃E 2, 3–5 7 ∃x A(x) →B →I 2–6 Now let’s construct a proof of the converse. We begin with 1 ∃x A(x) →B . . . n ∀x(A(x) →B) Note that the premise is a conditional, not an existentially quanti-fied sentence, so we should not (yet) work forward from it. Work-ing backward from the conclusion, ∀x(A(x) →B), leads us to look for a proof of A(d) →B: 1 ∃x A(x) →B . . . n −1 A(d) →B n ∀x(A(x) →B) ∀I n −1 And working backward from A(d) →B means we should set up a subproof with A(d) as an assumption and B as the last line: CHAPTER 35. PROOFS WITH QUANTIFIERS 313 1 ∃x A(x) →B 2 A(d) . . . n −2 B n −1 A(d) →B →I 2–(n −2) n ∀x(A(x) →B) ∀I n −1 Now we can work forward from the premise on line 1. That’s a conditional, and its consequent happens to be the sentence B we are trying to justify. So we should look for a proof of its antecedent, ∃x A(x). Of course, that is now readily available, by ∃I from line 2, and we’re done: 1 ∃x A(x) →B 2 A(d) 3 ∃x A(x) ∃I 2 4 B →E 1, 3 5 A(d) →B →I 2–4 6 ∀x(A(x) →B) ∀I 5 Practice exercises A. Use the strategies to find proofs for each of the following ar-guments and theorems: 1. A →∀x B(x) ∴∀x(A →B(x)) 2. ∃x(A →B(x)) ∴A →∃x B(x) 3. ∀x(A(x) ∧B(x)) ↔(∀x A(x) ∧∀x B(x)) 4. ∃x(A(x) ∨B(x)) ↔(∃x A(x) ∨∃x B(x)) 5. A ∨∀x B(x)) ∴∀x(A ∨B(x)) CHAPTER 35. PROOFS WITH QUANTIFIERS 314 6. ∀x(A(x) →B) ∴∃x A(x) →B 7. ∃x(A(x) →B) ∴∀x A(x) →B 8. ∀x(A(x) →∃y A(y)) Use only the basic rules of TFL in addition to the basic quantifier rules. B. Use the strategies to find proofs for each of the following ar-guments and theorems: 1. ∀x R(x,x) ∴∀x ∃y R(x,y) 2. ∀x ∀y ∀z [(R(x,y) ∧R(y,z)) →R(x,z)] ∴∀x ∀y [R(x,y) →∀z (R(y,z) →R(x,z))] 3. ∀x ∀y ∀z [(R(x,y) ∧R(y,z)) →R(x,z)], ∀x ∀y(R(x,y) →R(y,x)) ∴∀x ∀y ∀z [(R(x,y) ∧R(x,z)) →R(y,z)] 4. ∀x ∀y(R(x,y) →R(y,x)) ∴∀x ∀y ∀z [(R(x,y) ∧R(x,z)) →∃u(R(y,u) ∧R(z,u))] 5. ¬∃x ∀y(A(x,y) ↔¬A(y,y)) C. Use the strategies to find proofs for each of the following ar-guments and theorems: 1. ∀x A(x) →B ∴∃x(A(x) →B) 2. A →∃x B(x) ∴∃x(A →B(x)) 3. ∀x(A ∨B(x)) ∴A ∨∀x B(x)) 4. ∃x(A(x) →∀y A(y)) 5. ∃x(∃y A(y) →A(x)) These require the use of IP. Use only the basic rules of TFL in addition to the basic quantifier rules. CHAPTER 36 Conversion of quantifiers In this section, we will add some additional rules to the basic rules of the previous section. These govern the interaction of quantifiers and negation. In §22, we noted that ¬∃xA is logically equivalent to ∀x ¬A. We will add some rules to our proof system that govern this. In particular, we add: m ∀x¬A ¬∃xA CQ m and m ¬∃xA ∀x¬A CQ m Equally, we add: 315 CHAPTER 36. CONVERSION OF QUANTIFIERS 316 m ∃x¬A ¬∀xA CQ m and m ¬∀xA ∃x¬A CQ m Practice exercises A. Show in each case that the sentences are inconsistent: 1. S (a) →T (m),T (m) →S (a),T (m) ∧¬S (a) 2. ¬∃x R(x,a),∀x ∀y R(y,x) 3. ¬∃x ∃y L(x,y),L(a,a) 4. ∀x(P (x) →Q (x)),∀z (P (z) →R(z)),∀y P (y),¬Q (a) ∧ ¬R(b) B. Show that each pair of sentences is provably equivalent: 1. ∀x(A(x) →¬B(x)),¬∃x(A(x) ∧B(x)) 2. ∀x(¬A(x) →B(d)),∀x A(x) ∨B(d) C. In §23, we considered what happens when we move quanti-fiers ‘across’ various logical operators. Show that each pair of sentences is provably equivalent: 1. ∀x(F (x) ∧G (a)),∀x F (x) ∧G (a) 2. ∃x(F (x) ∨G (a)),∃x F (x) ∨G (a) 3. ∀x(G (a) →F (x)),G (a) →∀x F (x) 4. ∀x(F (x) →G (a)),∃x F (x) →G (a) 5. ∃x(G (a) →F (x)),G (a) →∃x F (x) 6. ∃x(F (x) →G (a)),∀x F (x) →G (a) CHAPTER 36. CONVERSION OF QUANTIFIERS 317 NB: the variable ‘x’ does not occur in ‘G (a)’. When all the quan-tifiers occur at the beginning of a sentence, that sentence is said to be in prenex normal form. These equivalences are sometimes called prenexing rules, since they give us a means for putting any sentence into prenex normal form. CHAPTER 37 Rules for identity In §29, we mentioned the philosophically contentious thesis of the identity of indiscernibles. This is the claim that objects which are indiscernible in every way are, in fact, identical to each other. It was also mentioned that we will not subscribe to this thesis. It follows that, no matter how much you learn about two objects, we cannot prove that they are identical. That is unless, of course, you learn that the two objects are, in fact, identical, but then the proof will hardly be very illuminating. The general point, though, is that no sentences which do not already contain the identity predicate could justify an inference to ‘a = b’. So our identity introduction rule cannot allow us to infer to an identity claim containing two different names. However, every object is identical to itself. No premises, then, are required in order to conclude that something is identical to itself. So this will be the identity introduction rule: c = c =I Notice that this rule does not require referring to any prior 318 CHAPTER 37. RULES FOR IDENTITY 319 lines of the proof. For any name c, you can write c = c on any line, with only the =I rule as justification. Our elimination rule is more fun. If you have established ‘a = b’, then anything that is true of the object named by ‘a’ must also be true of the object named by ‘b’. For any sentence with ‘a’ in it, you can replace some or all of the occurrences of ‘a’ with ‘b’ and produce an equivalent sentence. For example, from ‘R(a,a)’ and ‘a = b’, you are justified in inferring ‘R(a,b)’, ‘R(b,a)’ or ‘R(b,b)’. More generally: m a = b n A(. . . a. . . a. . .) A(. . . b. . . a. . .) =E m, n The notation here is as for ∃I. So A(. . . a. . . a. . .) is a for-mula containing the name a, and A(. . . b. . . a. . .) is a formula obtained by replacing one or more instances of the name a with the name b. Lines m and n can occur in either order, and do not need to be adjacent, but we always cite the statement of identity first. Symmetrically, we allow: m a = b n A(. . . b. . . b. . .) A(. . . a. . . b. . .) =E m, n This rule is sometimes called Leibniz’s Law, after Gottfried Leibniz. To see the rules in action, we will prove some quick results. First, we will prove that identity is symmetric: CHAPTER 37. RULES FOR IDENTITY 320 1 a = b 2 a = a =I 3 b = a =E 1, 2 4 a = b →b = a →I 1–3 5 ∀y(a = y →y = a) ∀I 4 6 ∀x ∀y(x = y →y = x) ∀I 5 We obtain line 3 by replacing one instance of ‘a’ in line 2 with an instance of ‘b’; this is justified given ‘a = b’. Second, we will prove that identity is transitive: 1 a = b ∧b = c 2 a = b ∧E 1 3 b = c ∧E 1 4 a = c =E 2, 3 5 (a = b ∧b = c) →a = c →I 1–4 6 ∀z ((a = b ∧b = z) →a = z) ∀I 5 7 ∀y ∀z ((a = y ∧y = z) →a = z) ∀I 6 8 ∀x ∀y∀z ((x = y ∧y = z) →x = z) ∀I 7 We obtain line 4 by replacing ‘b’ in line 3 with ‘a’; this is justified given ‘a = b’. Practice exercises A. For each of the following claims, provide an FOL proof that shows it is true. 1. P (a) ∨Q (b),Q (b) →b = c,¬P (a) ⊢Q (c) CHAPTER 37. RULES FOR IDENTITY 321 2. m = n ∨n = o,A(n) ⊢A(m) ∨A(o) 3. ∀x x = m,R(m,a) ⊢∃x R(x,x) 4. ∀x ∀y(R(x,y) →x = y) ⊢R(a,b) →R(b,a) 5. ¬∃x¬x = m ⊢∀x ∀y(P (x) →P (y)) 6. ∃x J (x),∃x ¬ J (x) ⊢∃x ∃y ¬x = y 7. ∀x(x = n ↔M (x)),∀x(O (x) ∨¬M (x)) ⊢O (n) 8. ∃x D(x),∀x(x = p ↔D(x)) ⊢D(p) 9. ∃x [︁ (K (x) ∧∀y(K (y) →x = y)) ∧B(x) ]︁ ,Kd ⊢B(d) 10. ⊢P (a) →∀x(P (x) ∨¬x = a) B. Show that the following are provably equivalent: • ∃x (︁[F (x) ∧∀y(F (y) →x = y)] ∧x = n)︁ • F (n) ∧∀y(F (y) →n = y) And hence that both have a decent claim to symbolize the En-glish sentence ‘Nick is the F ’. C. In §25, we claimed that the following are logically equivalent symbolizations of the English sentence ‘there is exactly one F ’: • ∃x F (x) ∧∀x ∀y [︁ (F (x) ∧F (y)) →x = y ]︁ • ∃x [︁ F (x) ∧∀y(F (y) →x = y) ]︁ • ∃x ∀y(F (y) ↔x = y) Show that they are all provably equivalent. (Hint: to show that three claims are provably equivalent, it suffices to show that the first proves the second, the second proves the third and the third proves the first; think about why.) D. Symbolize the following argument There is exactly one F . There is exactly one G. Noth-ing is both F and G. So: there are exactly two things that are either F or G. And offer a proof of it. CHAPTER 38 Derived rules As in the case of TFL, we first introduced some rules for FOL as basic (in §34), and then added some further rules for conversion of quantifiers (in §36). In fact, the CQ rules should be regarded as derived rules, for they can be derived from the basic rules of §34. (The point here is as in §20.) Here is a justification for the first CQ rule: 1 ∀x ¬A(x) 2 ∃x A(x) 3 A(c) 4 ¬A(c) ∀E 1 5 ⊥ ¬E 4, 3 6 ⊥ ∃E 2, 3–5 7 ¬∃x A(x) ¬I 2–6 Here is a justification of the third CQ rule: 322 CHAPTER 38. DERIVED RULES 323 1 ∃x ¬A(x) 2 ∀x A(x) 3 ¬A(c) 4 A(c) ∀E 2 5 ⊥ ¬E 3, 4 6 ⊥ ∃E 1, 3–5 7 ¬∀x A(x) ¬I 2–6 This explains why the CQ rules can be treated as derived. Similar justifications can be offered for the other two CQ rules. Practice exercises A. Offer proofs which justify the addition of the second and fourth CQ rules as derived rules. CHAPTER 39 Proofs and semantics We have used two different turnstiles in this book. This: A 1,A 2,. . . ,A n ⊢C means that there is some proof which ends with C and whose only undischarged assumptions are among A 1,A 2,. . . ,A n. This is a proof-theoretic notion. By contrast, this: A 1,A 2,. . . ,A n ⊨C means that no valuation (or interpretation) makes all of A 1,A 2,. . . ,A n true and C false. This concerns assignments of truth and falsity to sentences. It is a semantic notion. It cannot be emphasized enough that these are different no-tions. But we can emphasize it a bit more: They are different no-tions. Once you have internalised this point, continue reading. Although our semantic and proof-theoretic notions are differ-ent, there is a deep connection between them. To explain this connection,we will start by considering the relationship between validities and theorems. 324 CHAPTER 39. PROOFS AND SEMANTICS 325 To show that a sentence is a theorem, you need only produce a proof. Granted, it may be hard to produce a twenty line proof, but it is not so hard to check each line of the proof and confirm that it is legitimate; and if each line of the proof individually is legitimate, then the whole proof is legitimate. Showing that a sen-tence is a validity, though, requires reasoning about all possible interpretations. Given a choice between showing that a sentence is a theorem and showing that it is a validity, it would be easier to show that it is a theorem. Contrawise, to show that a sentence is not a theorem is hard. We would need to reason about all (possible) proofs. That is very difficult. However, to show that a sentence is not a validity, you need only construct an interpretation in which the sentence is false. Granted, it may be hard to come up with the interpretation; but once you have done so, it is relatively straightforward to check what truth value it assigns to a sentence. Given a choice between showing that a sentence is not a theorem and showing that it is not a validity, it would be easier to show that it is not a validity. Fortunately, a sentence is a theorem if and only if it is a validity. As a result, if we provide a proof of A on no assumptions, and thus show that A is a theorem, i.e., ⊢A, we can legitimately infer that A is a validity, i.e., ⊨A. Similarly, if we construct an interpretation in which A is false and thus show that it is not a validity, i.e., ⊭A, it follows that A is not a theorem, i.e., ⊬A. More generally, we have the following powerful result: A 1,A 2,. . . ,A n ⊢B iff A 1,A 2,. . . ,A n ⊨B This shows that, whilst provability and entailment are different notions, they are extensionally equivalent. As such: • An argument is valid iff the conclusion can be proved from the premises. • A sentence is a validity iff it is a theorem. • T wo sentences are equivalent iff they are provably equivalent. • Sentences are jointly satisfiable iff they are jointly consistent. CHAPTER 39. PROOFS AND SEMANTICS 326 For this reason, you can pick and choose when to think in terms of proofs and when to think in terms of valuations/interpretations, doing whichever is easier for a given task. The table on the next page summarizes which is (usually) easier. It is intuitive that provability and semantic entailment should agree. But—let us repeat this—do not be fooled by the similarity of the symbols ‘⊨’ and ‘⊢’. These two symbols have very differ-ent meanings. The fact that provability and semantic entailment agree is not an easy result to come by. In fact, demonstrating that provability and semantic entail-ment agree is, very decisively, the point at which introductory logic becomes intermediate logic. CHAPTER 39. PROOFS AND SEMANTICS 327 Y es No Is A a validity? give a proof which shows ⊢A give an interpretation in which A is false Is A a contradiction? give a proof which shows ⊢¬A give an interpretation in which A is true Are A and B equivalent? give two proofs, one for A ⊢B and one for B ⊢A give an interpretation in which Aand B have different truth values Are A 1,A 2,. . . ,A n jointly satisfiable? give an interpretation in which all of A 1,A 2,. . . ,A n are true prove a contradiction from assump-tions A 1,A 2,. . . ,A n Is A 1,A 2,. . . ,A n ∴C valid? give a proof with assumptions A 1,A 2,. . . ,A n and concluding with C give an interpretation in which each of A 1,A 2,. . . ,A n is true and Cis false PART VIII Modal logic 328 CHAPTER 40 Introducing modal logic Modal logic (ML) is the logic of modalities, ways in which a state-ment can be true. Necessity and possibility are two such modalities: a statement can be true, but it can also be necessarily true (true no matter how the world might have been). For instance, logical truths are not just true because of some accidental feature of the world, but true come what may. A possible statement may not actually be true, but it might have been true. We use □to express necessity, and ◇to express possibility. So you can read □A as It is necessarily the case that A, and ◇Aas It is possibly the case that A. There are lots of different kinds of necessity. It is humanly impossible for me to run at 100mph. Given the sorts of creatures that we are, no human can do that. But still, it isn’t physically impossible for me to run that fast. We haven’t got the technology to do it yet, but it is surely physically possible to swap my biological legs for robotic ones which could run at 100mph. By contrast, it is physically impossible for me to run faster than the speed of light. The laws of physics forbid any object from accelerating up to that speed. But even that isn’t logically impossible. It isn’t a contradiction to imagine that the laws of physics might have been different, and that they might have allowed objects to move faster 329 CHAPTER 40. INTRODUCING MODAL LOGIC 330 than light. Which kind of modality does ML deal with? All of them! ML is a very flexible tool. We start with a basic set of rules that govern □and ◇, and then add more rules to fit whatever kind of modality we are interested in. In fact, ML is so flexible that we do not even have to think of □and ◇as expressing necessity and possibility. We might instead read □as expressing provability, so that □A means It is provable that A, and ◇A means It is not refutable that A. Similarly, we can interpret □to mean S knows that A or S believes that A. Or we might read □as expressing moral obligation, so that □A means It is morally obligatory that A, and ◇A means It is morally permissible that A. All we would need to do is cook up the right rules for these different readings of □ and ◇. A modal formula is one that includes modal operators such as □and ◇. Depending on the interpretation we assign to □ and ◇, different modal formulas will be provable or valid. For instance, □A →A might say that “if A is necessary, it is true,” if □is interpreted as necessity. It might express “if A is known, then it is true,” if □expresses known truth. Under both these interpretations, □A →A is valid: All necessary propositions are true come what may, so are true in the actual world. And if a proposition is known to be true, it must be true (one can’t know something that’s false). However, when □is interpreted as “it is believed that” or “it ought to be the case that,” □A →A is not valid: We can believe false propositions. Not every proposition that ought to be true is in fact true, e.g., “Every murderer will be brought to justice.” This ought to be true, but it isn’t. We will consider different kinds of systems of ML. They differ in the rules of proof allowed, and in the semantics we use to de-fine our logical notions. The different systems we’ll consider are called K, T, S4, and S5. K is the basic system; everything that is valid or provable in K is also provable in the others. But there are some things that K does not prove, such as the formula □A →A for sentence letter A. So K is not an appropriate modal logic for necessity and possibility (where □A →A should be provable). CHAPTER 40. INTRODUCING MODAL LOGIC 331 This is provable in the system T, so T is more appropriate when dealing with necessity and possibiliity, but less apropriate when dealing with belief or obligation, since then □A →A should not (always) be provable. The perhaps best system of ML for neces-sity and possibility, and in any case the most widely accepted, is the strongest of the systems we consider, S5. 40.1 The Language of ML In order to do modal logic, we have to do two things. First, we want to learn how to prove things in ML. Second, we want to see how to construct interpretations for ML. But before we can do either of these things, we need to explain how to construct sentences in ML. The language of ML is an extension of TFL. We could have started with FOL, which would have given us Quantified Modal Logic (QML). QML is much more powerful than ML, but it is also much, much more complicated. So we are going to keep things simple, and start with TFL. Just like TFL, ML starts with an infinite stock of atoms. These are written as capital letters, with or without numerical sub-scripts: A, B, . . .A1, B1, . . . We then take all of the rules about how to make sentences from TFL, and add two more for □and ◇: (1) Every atom of ML is a sentence of ML. (2) If A is a sentence of ML, then ¬A is a sentence of ML. (3) If Aand B are sentences of ML, then (A∧B) is a sentence of ML. (4) If Aand B are sentences of ML, then (A∨B) is a sentence of ML. (5) If A and B are sentences of ML, then (A →B) is a sen-tence of ML. CHAPTER 40. INTRODUCING MODAL LOGIC 332 (6) If A and B are sentences of ML, then (A ↔B) is a sen-tence of ML. (7) If A is a sentence of ML, then □A is a sentence of ML. (8) If A is a sentence of ML, then ◇A is a sentence of ML. (9) Nothing else is a sentence of ML. Here are some examples of ML sentences: A, P ∨Q, □A, C ∨□D, □□(A →R), □◇(S ∧(Z ↔ (□W ∨◇Q))) CHAPTER 41 Natural deduction for ML Now that we know how to make sentences in ML, we can look at how to prove things in ML. We will use ⊢to express provabil-ity. So A 1,A 2,. . . A n ⊢C means that C can be proven from A 1,A 2,. . . A n. However, we will be looking at a number of dif-ferent systems of ML, and so it will be useful to add a subscript to indicate which system we are working with. So for example, if we want to say that we can prove C from A 1,A 2,. . . A n in sys-tem K, we will write: A 1,A 2,. . . A n ⊢K C. 41.1 System K We start with a particularly simple system called K, in honour of the philosopher and logician Saul Kripke. K includes all of the natural deduction rules from TFL, including the derived rules as well as the basic ones. K then adds a special kind of subproof, plus two new basic rules for □. 333 CHAPTER 41. NATURAL DEDUCTION FOR ML 334 The special kind of subproof looks like an ordinary subproof, except it has a □in its assumption line instead of a formula. We call them strict subproofs—they allow as to reason and prove things about alternate possibilities. What we can prove inside a strict subproof holds in any alternate possibility, in particular, in alter-nate possibilities where the assumptions in force in our proof may not hold. In a strict subproofs, all assumptions are disregarded, and we are not allowed to appeal to any lines outside the strict subproof (except as allowed by the modal rules given below). The □I rule allows us to derive a formula □A if we can de-rive A inside a strict subproof. It is our fundamental method of introducing □into proofs. The basic idea is simple enough: if A is a theorem, then □Ashould be a theorem too. (Remember that to call A a theorem is to say that we can prove A without relying on any undischarged assumptions.) Suppose we wanted to prove □(A →A). The first thing we need to do is prove that A →A is a theorem. You already know how to do that using TFL. You simply present a proof of A →A which doesn’t start with any premises, like this: 1 A 2 A R 1 3 A →A →I 1–2 But to apply □I, we need to have proven the formula inside a strict subproof. Since our proof of A →A makes use of no assumptions at all, this is possible. 1 □ 2 A 3 A R 2 4 A →A →I 2–3 5 □(A →A) □I 1–4 CHAPTER 41. NATURAL DEDUCTION FOR ML 335 m □ n A □A □I m–n No line above line m may be cited by any rule within the strict subproof begun at line m unless the rule explicitly allows it. It is essential to emphasize that in strict subproof you can-not use any rule which appeals to anything you proved outside of the strict subproof. There are exceptions, e.g., the □E rule be-low. These rules will explicitly state that they can be used inside strict subproofs and cite lines outside the strict subproof. This restriction is essential, otherwise we would get terrible results. For example, we could provide the following proof to vindicate A ∴□A: 1 A 2 □ 3 A incorrect use of R 1 4 □A □I 2–3 This is not a legitimate proof, because at line 3 we appealed to line 1, even though line 1 comes before the beginning of the strict subproof at line 2. We said above that a strict subproof allows us to reason about arbitrary alternate possible situations. What can be proved in a strict subproof holds in all alternate possible situtations, and so is necessary. This is the idea behind the □I rule. On the other hand, if we’ve assumed that something is necessary, we have therewith assumed that it is true in all alternate possbile situations. Hence, we have the rule □E: CHAPTER 41. NATURAL DEDUCTION FOR ML 336 m □A □ n A □E m □E can only be applied if line m (containing □A) lies out-side of the strict subproof in which line n falls, and this strict subproof is not itself part of a strict subproof not containing m. □E allows you to assert A inside a strict subproof if you have □A outside the strict subproof. The restriction means that you can only do this in the first strict subproof, you cannot apply the □E rule inside a nested strict subproof. So the following is not allowed: 1 □A 2 □ 3 □ 4 A incorrect use of □E 1 The incorrect use of □E on line 4 violates the condition, because although line 1 lies outside the strict subproof in which line 4 falls, the strict subproof containing line 4 lies inside the strict subproof beginning on line 2 which does not contain line 1. CHAPTER 41. NATURAL DEDUCTION FOR ML 337 Let’s begin with an example. 1 □A 2 □B 3 □ 4 A □E 1 5 B □E 2 6 A ∧B ∧I 4, 5 7 □(A ∧B) □I 3–7 We can also mix regular subproofs and strict subproofs: 1 □(A →B) 2 □A 3 □ 4 A □E 2 5 A →B □E 1 6 B →E 4, 5 7 □B 8 □A →□B →I 2–7 This is called the Distribution Rule, because it tells us that □‘dis-tributes’ over →. The rules □I and □E look simple enough, and indeed K is a very simple system! But K is more powerful than you might have thought. You can prove a fair few things in it. CHAPTER 41. NATURAL DEDUCTION FOR ML 338 41.2 Possibility In the last subsection, we looked at all of the basic rules for K. But you might have noticed that all of these rules were about necessity, □, and none of them were about possibility, ◇. That’s because we can define possibility in terms of necessity: ◇A =df ¬□¬A In other words, to say that A is possibly true, is to say that A is not necessarily false. As a result, it isn’t really essential to add a ◇, a special symbol for possibility, into system K. Still, the system will be much easier to use if we do, and so we will add the following definitional rules: m ¬□¬A ◇A Def◇m m ◇A ¬□¬A Def◇m Importantly, you should not think of these rules as any real addition to K: they just record the way that ◇is defined in terms of □. If we wanted, we could leave our rules for K here. But it will be helpful to add some Modal Conversion rules, which give us some more ways of flipping between □and ◇: CHAPTER 41. NATURAL DEDUCTION FOR ML 339 m ¬□A ◇¬A MC m m ◇¬A ¬□A MC m m ¬◇A □¬A MC m m □¬A ¬◇A MC m These Modal Conversion Rules are also no addition to the power of K, because they can be derived from the basic rules, along with the definition of ◇. In system K, using Def◇(or the modal conversion rules), one can prove ◇A ↔¬□¬A. When laying out system K, we started with □as our primitive modal symbol, and then defined ◇in terms of it. But if we had preferred, we could have started with ◇ as our primitive, and then defined □as follows: □A =df ¬◇¬A. There is, then, no sense in which necessity is somehow more fundamental than possibility. Necessity and possibility are exactly as fundamental as each other. 41.3 System T So far we have focussed on K, which is a very simple modal system. K is so weak that it will not even let you prove A from □A. But if we are thinking of □as expressing necessity, then we will want to be able to make this inference: if Ais necessarily true, then it must surely be true! CHAPTER 41. NATURAL DEDUCTION FOR ML 340 This leads us to a new system, T, which we get by adding the following rule to K: m □A n A RT m The line n on which rule RT is applied must not lie in a strict subproof that begins after line m. The restriction on rule T is in a way the opposite of the re-striction on □E: you can only use □E in a nested strict subproof, but you cannot use T in a nested strict subproof. We can prove things in T which we could not prove in K, e.g., □A →A. 41.4 System S4 T allows you to strip away the necessity boxes: from □A, you may infer A. But what if we wanted to add extra boxes? That is, what if we wanted to go from □Ato □□A? Well, that would be no problem, if we had proved □Aby applying □I to a strict subproof of A which itself does not use □E. In that case, A is a tautology, and by nesting the strict subproof inside another strict subproof and applying □I again, we can prove □□A. For example, we could CHAPTER 41. NATURAL DEDUCTION FOR ML 341 prove □□(P →P ) like this: 1 □ 2 □ 3 P 4 P R 3 5 P →P →I 3–4 6 □(P →P ) □I 2–5 7 □□(P →P ) □I 1–6 But what if we didn’t prove □A in this restricted way, but used □E inside the strict subproof of A. If we put that strict subproof inside another strict subproof, the requirement of rule □E to not cite a line containing □A which lies in another strict subproof that has not yet concluded, is violated. Or what if □A were just an assumption we started our proof with? Could we infer □□A then? Not in T, we couldn’t. And this might well strike you as a limitation of T, at least if we are reading □as expressing necessity. It seems intuitive that if Ais necessarily true, then it couldn’t have failed to be necessarily true. This leads us to another new system, S4, which we get by adding the following rule to T: CHAPTER 41. NATURAL DEDUCTION FOR ML 342 m □A □ n □A R4 m Note that R4 can only be applied if line m (containing □A) lies outside of the strict subproof in which line n falls, and this strict subproof is not itself part of a strict subproof not containing m. Rule R4 looks just like □E, except that instead of yielding A from □A it yields □A inside a strict subproof. The restriction is the same, however: R4 allows us to “import” □A into a strict subproof, but not into a strict subproof itself nested inside a strict subproof. However, if that is necessary, an additional application of R4 would have the same result. Now we can prove even more results. For instance: 1 □A 2 □ 3 □A R4 1 4 □□A □I 2–3 5 □A →□□A →I 1–6 Similarly, we can prove ◇◇A →◇A. This shows us that as well as letting us add extra boxes, S4 lets us delete extra diamonds: from ◇◇A, you can always infer ◇A. 41.5 System S5 In S4, we can always add a box in front of another box. But S4 does not automatically let us add a box in front of a diamond. CHAPTER 41. NATURAL DEDUCTION FOR ML 343 That is, S4 does not generally permit the inference from ◇A to □◇A. But again, that might strike you as a shortcoming, at least if you are reading □and ◇as expressing necessity and possibility. It seems intuitive that if A is possibly true, then it couldn’t have failed to be possibly true. This leads us to our final modal system, S5, which we get by adding the following rule to S4: m ¬□A □ n ¬□A R5 m Rule R5 can only be applied if line m (containing ¬□A) lies outside of the strict subproof in which line n falls, and this strict subproof is not itself part of a strict subproof not containing line m. This rule allows us to show, for instance, that ◇□A ⊢S5 □A: 1 ◇□A 2 ¬□¬□A Def◇1 3 ¬□A 4 □ 5 ¬□A R5 3 6 □¬□A □I 4–5 7 ⊥ ¬E 2, 6 8 □A IP 3–7 So, as well as adding boxes in front of diamonds, we can also delete diamonds in front of boxes. CHAPTER 41. NATURAL DEDUCTION FOR ML 344 We got S5 just by adding the rule R5 rule to S4. In fact, we could have added rule R5 to T alone, and leave out rule R4). Everything we can prove by rule R4 can also be proved using RT together with R5. For instance, here is a proof that shows □A ⊢S5 □□A without using R4: 1 □A 2 □¬□A 3 ¬□A RT 2 4 ⊥ ¬E 1, 3 5 ¬□¬□A ¬I 2–4 6 □ 7 ¬□A 8 □ 9 ¬□A R5 7 10 □¬□A □I 8–9 11 ¬□¬□A R5 5 12 ⊥ ¬E 10, 11 13 □A IP 7–12 14 □□A □I 6–13 S5 is strictly stronger than S4: there are things which can be proved in S5, but not in S4 (e.g., ◇□A →□A). The important point about S5 can be put like this: if you have a long string of boxes and diamonds, in any combination what-soever, you can delete all but the last of them. So for example, ◇□◇◇□□◇□A can be simplified down to just □A. CHAPTER 41. NATURAL DEDUCTION FOR ML 345 Practice exercises A. Provide proofs for the following: 1. □(A ∧B) ⊢K □A ∧□B 2. □A ∧□B ⊢K □(A ∧B) 3. □A ∨□B ⊢K □(A ∨B) 4. □(A ↔B) ⊢K □A ↔□B B. Provide proofs for the following (without using Modal Con-version!): 1. ¬□A ⊢K ◇¬A 2. ◇¬A ⊢K ¬□A 3. ¬◇A ⊢K □¬A 4. □¬A ⊢K ¬◇A C. Provide proofs of the following (and now feel free to use Modal Conversion!): 1. □(A →B),◇A ⊢K ◇B 2. □A ⊢K ¬◇¬A 3. ¬◇¬A ⊢K □A D. Provide proofs for the following: 1. P ⊢T ◇P 2. ⊢T (A ∧B) ∨(¬□A ∨¬□B) E. Provide proofs for the following: 1. □(□A →B),□(□B →C ),□A ⊢S4 □□C 2. □A ⊢S4 □(□A ∨B) 3. ◇◇A ⊢S4 ◇A F. Provide proofs in S5 for the following: 1. ¬□¬A,◇B ⊢S5 □(◇A ∧◇B) 2. A ⊢S5 □◇A 3. ◇◇A ⊢S5 ◇A CHAPTER 42 Semantics for ML So far, we have focussed on laying out various systems of Natural Deduction for ML. Now we will look at the semantics for ML. A semantics for a language is a method for assigning truth-values to the sentences in that language. So a semantics for ML is a method for assigning truth-values to the sentences of ML. 42.1 Interpretations of ML The big idea behind the semantics for ML is this. In ML, sen-tences are not just true or false, full stop. A sentence is true or false at a given possible world, and a single sentence may well be true at some worlds and false at others. We then say that □A is true iff A is true at every world, and ◇A is true iff A is true at some world. That’s the big idea, but we need to refine it and make it more precise. To do this, we need to introduce the idea of an interpre-tation of ML. The first thing you need to include in an interpreta-tion is a collection of possible worlds. Now, at this point you might well want to ask: What exactly is a possible world? The intuitive idea is that a possible world is another way that this world could 346 CHAPTER 42. SEMANTICS FOR ML 347 have been. But what exactly does that mean? This is an excellent philosophical question, and we will look at it in a lot of detail later. But we do not need to worry too much about it right now. As far as the formal logic goes, possible worlds can be anything you like. All that matters is that you supply each interpretation with a non-empty collection of things labelled possible worlds. Once you have chosen your collection of possible worlds, you need to find some way of determining which sentences of ML are true at which possible worlds. To do that, we need to intro-duce the notion of a valuation function. Those of you who have studied some maths will already be familiar with the general idea of a function. But for those of you who haven’t, a function is a mathematical entity which maps arguments to values. That might sound a little bit abstract, but some familiar examples will help. Take the function x+1. This is a function which takes in a number as argument, and then spits out the next number as value. So if you feed in the number 1 as an argument, the function x + 1 will spit out the number 2 as a value; if you feed in 2, it will spit out 3; if you feed in 3, it will spit out 4 . . . Or here is another example: the function x + y. This time, you have to feed two arguments into this function if you want it to return a value: if you feed in 2 and 3 as your arguments, it spits out 5; if you feed in 1003 and 2005, it spits out 3008; and so on. A valuation function for ML takes in a sentence and a world as its arguments, and then returns a truth-value as its value. So if 𝜈is a valuation function and w is a possible world, 𝜈w (A) is whatever truth-value 𝜈maps A and w to: if 𝜈w (A) = F , then A is false at world w on valuation 𝜈; if 𝜈w (A) = T , then A is true at world w on valuation 𝜈. These valuation functions are allowed to map any atomic sen-tence to any truth-value at any world. But there are rules about which truth-values more complex sentences get assigned at a world. Here are the rules for the connectives from TFL: (1) 𝜈w (¬A) = T iff: 𝜈w (A) = F (2) 𝜈w (A∧B) = T iff: 𝜈w (A) = T and 𝜈w (B) = T CHAPTER 42. SEMANTICS FOR ML 348 (3) 𝜈w (A∨B) = T iff: 𝜈w (A) = T or 𝜈w (B) = T , or both (4) 𝜈w (A →B) = T iff: 𝜈w (A) = F or 𝜈w (B) = T , or both (5) 𝜈w (A ↔B) = T iff: 𝜈w (A) = T and 𝜈w (B) = T , or 𝜈w (A) = F and 𝜈w (B) = F So far, these rules should all look very familiar. Essentially, they all work exactly like the truth-tables for TFL. The only difference is that these truth-table rules have to be applied over and over again, to one world at a time. But what are the rules for the new modal operators, □and ◇? The most obvious idea would be to give rules like these: 𝜈w (□A) = T iff ∀w′(𝜈w′(A) = T ) 𝜈w (◇A) = T iff ∃w′(𝜈w′(A) = T ) This is just the fancy formal way of writing out the idea that □A is true at w just in case A is true at every world, and ◇A is true at w just in case A is true at some world. However, while these rules are nice and simple, they turn out not to be quite as useful as we would like. As we mentioned, ML is meant to be a very flexible tool. It is meant to be a general framework for dealing with lots of different kinds of necessity. As a result, we want our semantic rules for □and ◇to be a bit less rigid. We can do this by introducing another new idea: accessibility relations. An accessibility relation, R, is a relation between possible worlds. Roughly, to say that Rw1w2 (in English: world w1 accesses world w2) is to say that w2 is possible relative to w1. In other words, by introducing accessibility relations, we open up the idea that a given world might be possible relative to some worlds but not others. This turns out to be a very fruitful idea when studying modal systems. We can now give the following semantic rules for □and ◇: (6) 𝜈w1(□A) = T iff ∀w2(Rw1w2 →𝜈w2(A) = T ) CHAPTER 42. SEMANTICS FOR ML 349 (7) 𝜈w1(◇A) = T iff ∃w2(Rw1w2 ∧𝜈w2(A) = T ) Or in plain English: □A is true in world w1 iff A is true in every world that is possible relative to w1; and ◇A is true in world w1 iff A is true in some world that is possible relative to w1. So, there we have it. An interpretation for ML consists of three things: a collection of possible worlds, W ; an accessibility relation, R; and a valuation function, 𝜈. The collection of ‘pos-sible worlds’ can really be a collection of anything you like. It really doesn’t matter, so long as W isn’t empty. (For many pur-poses, it is helpful just to take a collection of numbers to be your collection of worlds.) And for now, at least, R can be any rela-tion between the worlds in W that you like. It could be a relation which every world in W bears to every world in W , or one which no world bears to any world, or anything in between. And lastly, 𝜈can map any atomic sentence of ML to any truth-value at any world. All that matters is that it follows the rules (1)–(7) when it comes to the more complex sentences. Let’s look at an example. It is often helpful to present inter-pretations of ML as diagrams, like this: 1 2 A ¬A ¬B B Here is how to read the interpretation off from this diagram. It contains just two worlds, 1 and 2. The arrows between the worlds indicate the accessibility relation. So 1 and 2 both access 1, but neither 1 nor 2 accesses 2. The boxes at each world let us know which atomic sentences are true at each world: A is true at 1 but false at 2; B is false at 1 but true at 2. You may only write an atomic sentence or the negation of an atomic sentence into one of these boxes. We can figure out what truth-values the more CHAPTER 42. SEMANTICS FOR ML 350 complex sentences get at each world from that. For example, on this interpretation all of the following sentences are true at w1: A ∧¬B, B →A, ◇A, □¬B If you don’t like thinking diagrammatically, then you can also present an interpretation like this: W : 1,2 R: ⟨1,1⟩,⟨2,1⟩ 𝜈1(A) = T ,𝜈1(B) = F,𝜈2(A) = F,𝜈2(B) = T You will get the chance to cook up some interpretations of your own shortly, when we start looking at counter-interpretations. 42.2 A Semantics for System K We can now extend all of the semantic concepts of TFL to cover ML: ⊲A 1,A 2,. . . A n ∴C is modally v alid iff there is no world in any interpretation at which A 1,A 2,. . . A n are all true and C is false. ⊲A is a modal truth iff A is true at every world in every interpretation. ⊲Ais a modal contradiction iff Ais false at every world in every interpretation. ⊲Ais modally satisfiable iff Ais true at some world in some interpretation. (From now on we will drop the explicit ‘modal’ qualifications, since they can be taken as read.) CHAPTER 42. SEMANTICS FOR ML 351 We can also extend our use of ⊨. However, we need to add subscripts again, just as we did with ⊢. So, when we want to say that A 1,A 2,. . . A n ∴Cis valid, we will write: A 1,A 2,. . . A n ⊨K C. Let’s get more of a feel for this semantics by presenting some counter-interpretations. Consider the following (false) claim: ¬A ⊨K ¬◇A In order to present a counter-interpretation to this claim, we need to cook up an interpretation which makes ¬A true at some world w, and ¬◇A false at w. Here is one such interpretation, presented diagrammatically: 1 2 ¬A A It is easy to see that this will work as a counter-interpretation for our claim. First, ¬A is true at world 1. And second, ¬◇A is false at 1: A is true at 2, and 2 is accessible from 1. So there is some world in this interpretation where ¬A is true and ¬◇A is false, so it is not the case that ¬A ⊨K ¬◇A. Why did we choose the subscript K? Well, it turns out that there is an important relationship between system K and the def-inition of validity we have just given. In particular, we have the following two results: ⊲If A 1,A 2,. . . A n ⊢K C, then A 1,A 2,. . . A n ⊨K C ⊲If A 1,A 2,. . . A n ⊨K C, then A 1,A 2,. . . A n ⊢K C The first result is known as a soundness result, since it tells us that the rules of K are good, sound rules: if you can vindicate an argu-ment by giving a proof for it using system K, then that argument really is valid. The second result is known as a completeness result, since it tells us that the rules of K are broad enough to capture CHAPTER 42. SEMANTICS FOR ML 352 all of the valid arguments: if an argument is valid, then it will be possible to offer a proof in K which vindicates it. Now, it is one thing to state these results, quite another to prove them. However, we will not try to prove them here. But the idea behind the proof of soundness will perhaps make clearer how strict subproofs work. In a strict subproof, we are not allowed to make use of any information from outside the strict subproof, except what we im-port into the strict subproof using □E. If we’ve assumed or proved □A, by □E, we can used A inside a strict subproof. And in K, that is the only way to import a formula into a strict subproof. So everything that can be proved inside a strict subproof must fol-low from formulas A where outside the strict subproof we have □A. Let’s imagine that we are reasoning about what’s true in a possible world in some interpretation. If we know that □Ais true in that possible world, we know that A is true in all accessible worlds. So, everything proved inside a strict subproof is true in all accessible possible worlds. That is why □I is a sound rule. 42.3 A Semantics for System T A few moments ago, we said that system K is sound and com-plete. Where does that leave the other modal systems we looked at, namely T, S4 and S5? Well, they are all unsound, relative to the definition of validity we gave above. For example, all of these systems allow us to infer A from □A, even though □A ⊭K A. Does that mean that these systems are a waste of time? Not at all! These systems are only unsound relative to the definition of validity we gave above. (Or to use symbols, they are unsound relative to ⊨K.) So when we are dealing with these stronger modal systems, we just need to modify our definition of validity to fit. This is where accessibility relations come in really handy. When we introduced the idea of an accessibility relation, we said that it could be any relation between worlds that you like: you could have it relating every world to every world, no world to any CHAPTER 42. SEMANTICS FOR ML 353 world, or anything in between. That is how we were thinking of accessibility relations in our definition of ⊨K. But if we wanted, we could start putting some restrictions on the accessibility relation. In particular, we might insist that it has to be reflexive: ⊲∀wRww In English: every world accesses itself. Or in terms of relative possibility: every world is possible relative to itself. If we imposed this restriction, we could introduce a new consequence relation, ⊨T, as follows: A 1,A 2,. . . A n ⊨T C iff there is no world in any interpre-tation which has a reflexive accessibility relation, at which A 1,A 2,. . . A n are all true and C is false We have attached the T subscript to ⊨because it turns out that system T is sound and complete relative to this new definition of validity: ⊲If A 1,A 2,. . . A n ⊢T C, then A 1,A 2,. . . A n ⊨T C ⊲If A 1,A 2,. . . A n ⊨T C, then A 1,A 2,. . . A n ⊢T C As before, we will not try to prove these soundness and complete-ness results. However, it is relatively easy to see how insisting that the accessibility relation must be reflexive will vindicate the RT rule: m □A A RT m To see this, just imagine trying to cook up a counter-interpretation to this claim: □A ⊨T A CHAPTER 42. SEMANTICS FOR ML 354 We would need to construct a world, w, at which □A was true, but A was false. Now, if □A is true at w, then A must be true at every world w accesses. But since the accessibility relation is reflexive, w accesses w. So Amust be true at w. But now Amust be true and false at w. Contradiction! 42.4 A Semantics for S4 How else might we tweak our definition of validity? Well, we might also stipulate that the accessibility relation has to be tran-sitive: ⊲∀w1∀w2∀w3((Rw1w2 ∧Rw2w3) →Rw1w3) In English: if w1 accesses w2, and w2 accesses w3, then w1 ac-cesses w3. Or in terms of relative possibility: if w3 is possible relative to w2, and w2 is possible relative to w1, then w3 is possi-ble relative to w1. If we added this restriction on our accessibility relation, we could introduce a new consequence relation, ⊨S4, as follows: A 1,A 2,. . . A n ⊨S4 C iff there is no world in any interpreta-tion which has a reflexive and transitive accessibility relation, at which A 1,A 2,. . . A n are all true and C is false We have attached the S4 subscript to ⊨because it turns out that system S4 is sound and complete relative to this new defini-tion of validity: ⊲If A 1,A 2,. . . A n ⊢S4 C, then A 1,A 2,. . . A n ⊨S4 C ⊲If A 1,A 2,. . . A n ⊨S4 C, then A 1,A 2,. . . A n ⊢S4 C As before, we will not try to prove these soundness and complete-ness results. However, it is relatively easy to see how insisting that the accessibility relation must be transitive will vindicate the S4 rule: CHAPTER 42. SEMANTICS FOR ML 355 m □A □ □A R4 m The idea behind strict subproofs, remember, is that they are ways to prove things that must be true in all accessible worlds. So the R4 rule means that whenever □A is true, □A must also be true in every accessible world. In other words, we must have □A ⊨S4 □□A. To see this, just imagine trying to cook up a counter-interpretation to this claim: □A ⊨S4 □□A We would need to construct a world, w1, at which □A was true, but □□A was false. Now, if □□A is false at w1, then w1 must access some world, w2, at which □A is false. Equally, if □A is false at w2, then w2 must access some world, w3, at which A is false. We just said that w1 accesses w2, and w2 accesses w3. So since we are now insisting that the accessibility relation be transitive, w1 must access w3. And as □Ais true at w1, and w3 is accessible from w1, it follows that A must be true at w3. So A is true and false at w3. Contradiction! 42.5 A Semantics for S5 Let’s put one more restriction on the accessibility relation. This time, let’s insist that it must also be symmetric: ⊲∀w1∀w2(Rw1w2 →Rw2w1) In English: if w1 accesses w2, then w2 accesses w1. Or in terms of relative possibility: if w2 is possible relative to w1, then w1 is possible relative to w2. Logicians call a relation that is reflexive, CHAPTER 42. SEMANTICS FOR ML 356 symmetric, and transitive an equivalence relation. We can now define a new consequence relation, ⊨S5, as follows: A 1,A 2,. . . A n ⊨S5 C iff there is no world in any interpre-tation whose accessibility relation is an equivalence relation, at which A 1,A 2,. . . A n are all true and C is false We have attached the S5 subscript to ⊨because it turns out that system S5 is sound and complete relative to this new defini-tion of validity: ⊲If A 1,A 2,. . . A n ⊢S5 C, then A 1,A 2,. . . A n ⊨S5 C ⊲If A 1,A 2,. . . A n ⊨S5 C, then A 1,A 2,. . . A n ⊢S5 C As before, we will not try to prove these soundness and com-pleteness results here. However, it is relatively easy to see how insisting that the accessibility relation must be an equivalence relation will vindicate the R5 rule: m ¬□A □ ¬□A R5 m The rule says that if A is not necessary, i.e., false in some accessible world, it is also not necessary in any accessible possible world, i.e., we have ¬□A ⊢S5 □¬□A. To see this, just imagine trying to cook up a counter-interpretation to this claim: ¬□A ⊨S5 □¬□A We would need to construct a world, w1, at which ¬□Awas true, but □¬□A was false. Now, if ¬□A is true at w1, then w1 must access some world, w2, at which A is false. Equally, if □¬□A is CHAPTER 42. SEMANTICS FOR ML 357 false at w1, then w1 must access some world, w3, at which ¬□A is false. Since we are now insisting that the accessibility relation is an equivalence relation, and hence symmetric, we can infer that w3 accesses w1. Thus, w3 accesses w1, and w1 accesses w2. Again, since we are now insisting that the accessibility relation is an equivalence relation, and hence transitive, we can infer that w3 accesses w2. But earlier we said that ¬□Ais false at w3, which implies that A is true at every world which w3 accesses. So A is true and false at w2. Contradiction! In the definition of ⊨S5, we stipulated that the accessibility re-lation must be an equivalence relation. But it turns out that there is another way of getting a notion of validity fit for S5. Rather than stipulating that the accessibility relation be an equivalence relation, we can instead stipulate that it be a universal relation: ⊲∀w1∀w2Rw1w2 In English: every world accesses every world. Or in terms of relative possibility: every world is possible relative to every world. Using this restriction on the accessibility relation, we could have defined ⊨S5 like this: A 1,A 2,. . . A n ⊨S5 C iff there is no world in any interpre-tation which has a universal accessibility relation, at which A 1,A 2,. . . A n are all true and C is false. If we defined ⊨S5 like this, we would still get the same sound-ness and completeness results for S5. What does this tell us? Well, it means that if we are dealing with a notion of necessity according to which every world is possible relative to every world, then we should use S5. What is more, most philosophers assume that the notions of necessity that they are most concerned with, like logical necessity and metaphysical necessity, are of exactly this kind. So S5 is the modal system that most philosophers use most of the time. CHAPTER 42. SEMANTICS FOR ML 358 Practice exercises A. Present counter-interpretations to the following false claims: 1. ¬P ⊨K ¬◇P 2. □(P ∨Q) ⊨K □P ∨□Q 3. ⊨K ¬□(A ∧¬A) 4. □A ⊨K A B. Present counter-interpretations to the following false claims: 1. ◇A ⊨S4 □◇A 2. ◇A,□(◇A →B) ⊨S4 □B C. Present counter-interpretations to the following false claims: 1. □(M →O),◇M ⊨T O 2. □A ⊨T □□A Further reading Modal logic is a large subfield of logic. We have only scratched the surface. If you want to learn more about modal logic, here are some textbooks you might consult. ⊲Hughes, G. E., & Cresswell, M. J. (1996). A New Introduction to Modal Logic, Oxford: Routledge. ⊲Priest, G. (2008). An Introduction to Non-Classical Logic, 2nd ed., Cambridge: Cambridge University Press. ⊲Garson, J. W. (2013). Modal Logic for Philosophers, 2nd ed., Cambridge: Cambridge University Press. None of these authors formulate their modal proof systems in quite the way we did, but the closest formulation is given by Garson. PART IX Metatheory 359 CHAPTER 43 Normal forms 43.1 Disjunctive normal form Sometimes it is useful to consider sentences of a particularly sim-ple form. For instance, we might consider sentences in which ¬ only attaches to atomic sentences, or those which are combi-nations of atomic sentences and negated atomic sentences us-ing only ∧. A relatively general but still simple form is that where a sentence is a disjunction of conjunctions of atomic or negated atomic sentences. When such a sentence is constructed, we start with atomic sentences, then (perhaps) attach negations, then (perhaps) combine using ∧, and finally (perhaps) combine using ∨. Let’s say that a sentence is in disjunctive normal form iff it meets all of the following conditions: (dnf1) No connectives occur in the sentence other than nega-tions, conjunctions and disjunctions; (dnf2) Every occurrence of negation has minimal scope (i.e. any ‘¬’ is immediately followed by an atomic sentence); (dnf3) No disjunction occurs within the scope of any conjunc-tion. 360 CHAPTER 43. NORMAL FORMS 361 So, here are are some sentences in disjunctive normal form: A (A ∧¬B ∧C ) (A ∧B) ∨(A ∧¬B) (A ∧B) ∨(A ∧B ∧C ∧¬D ∧¬E) A ∨(C ∧¬P234 ∧P233 ∧Q) ∨¬B Note that we have here broken one of the maxims of this book and temporarily allowed ourselves to employ the relaxed bracketing-conventions that allow conjunctions and disjunctions to be of ar-bitrary length. These conventions make it easier to see when a sentence is in disjunctive normal form. We will continue to help ourselves to these relaxed conventions, without further comment. To further illustrate the idea of disjunctive normal form, we will introduce some more notation. We write ‘±A’ to indicate that A is an atomic sentence which may or may not be prefaced with an occurrence of negation. Then a sentence in disjunctive normal form has the following shape: (±A 1 ∧. . .∧±A i) ∨(±A i+1 ∧. . .∧±Aj ) ∨. . .∨(±A m+1 ∧. . .∧±A n) We now know what it is for a sentence to be in disjunctive normal form. The result that we are aiming at is: Disjunctive Normal Form Theorem. For any sentence, there is an equivalent sentence in disjunctive normal form. Henceforth, we will abbreviate ‘Disjunctive Normal Form’ by ‘DNF’. 43.2 Proof of DNF theorem via truth tables Our first proof of the DNF Theorem employs truth tables. We will first illustrate the technique for finding an equivalent sentence in DNF, and then turn this illustration into a rigorous proof. CHAPTER 43. NORMAL FORMS 362 Let’s suppose we have some sentence, S, which contains three atomic sentences, ‘A’, ‘B’ and ‘C ’. The very first thing to do is fill out a complete truth table for S. Maybe we end up with this: A B C S T T T T T T F F T F T T T F F F F T T F F T F F F F T T F F F T As it happens, S is true on four lines of its truth table, namely lines 1, 3, 7 and 8. Corresponding to each of those lines, we will write down four sentences, whose only connectives are negations and conjunctions, where every negation has minimal scope: 1. ‘A ∧B ∧C ’ which is true on line 1 (and only then) 2. ‘A ∧¬B ∧C ’ which is true on line 3 (and only then) 3. ‘¬A ∧¬B ∧C ’ which is true on line 7 (and only then) 4. ‘¬A ∧¬B ∧¬C ’ which is true on line 8 (and only then) We now combine all of these conjunctions using ∨, like so: (A ∧B ∧C ) ∨(A ∧¬B ∧C ) ∨(¬A ∧¬B ∧C ) ∨(¬A ∧¬B ∧¬C ) This gives us a sentence in DNF which is true on exactly those lines where one of the disjuncts is true, i.e. it is true on (and only on) lines 1, 3, 7, and 8. So this sentence has exactly the same truth table as S. So we have a sentence in DNF that is logically equivalent to S, which is exactly what we wanted! Now, the strategy that we just adopted did not depend on the specifics of S; it is perfectly general. Consequently, we can use it to obtain a simple proof of the DNF Theorem. Pick any arbitrary sentence, S, and let A 1,. . . ,A n be the atomic sentences that occur in S. To obtain a sentence in DNF CHAPTER 43. NORMAL FORMS 363 that is logically equivalent S, we consider S’s truth table. There are two cases to consider: 1. S is false on every line of its truth table. Then, S is a con-tradiction. In that case, the contradiction (A 1 ∧¬A 1) is in DNF and logically equivalent to S. 2. S is true on at least one line of its truth table. For each line i of the truth table, let B i be a conjunction of the form (±A 1 ∧. . . ∧±A n) where the following rules determine whether or not to in-clude a negation in front of each atomic sentence: A m is a conjunct of B i iff A m is true on line i ¬A m is a conjunct of B i iff A m is false on line i Given these rules, Bi is true on (and only on) line i of the truth table which considers all possible valuations of A 1,. . . ,A n (i.e. S’s truth table). Next, let i1, i2, . . . , im be the numbers of the lines of the truth table where S is true. Now let D be the sentence: B i1 ∨B i2 ∨. . . ∨B im Since S is true on at least one line of its truth table, D is indeed well-defined; and in the limiting case where Sis true on exactly one line of its truth table, D is just B i1, for some i1. By construction, D is in DNF. Moreover, by construction, for each line i of the truth table: S is true on line i of the truth table iff one of D’s disjuncts (namely, Bi) is true on, and only on, line i. Hence S and D have the same truth table, and so are logically equivalent. CHAPTER 43. NORMAL FORMS 364 These two cases are exhaustive and, either way, we have a sen-tence in DNF that is logically equivalent to S. So we have proved the DNF Theorem. Before we say any more, though, we should immediately flag that we are hereby returning to the austere definition of a (TFL) sentence, according to which we can assume that any conjunction has exactly two conjuncts, and any disjunction has exactly two disjuncts. 43.3 Conjunctive normal form So far in this chapter, we have discussed disjunctive normal form. It may not come as a surprise to hear that there is also such a thing as conjunctive normal form (CNF). The definition of CNF is exactly analogous to the definition of DNF. So, a sentence is in CNF iff it meets all of the following conditions: (cnf1) No connectives occur in the sentence other than nega-tions, conjunctions and disjunctions; (cnf2) Every occurrence of negation has minimal scope; (cnf3) No conjunction occurs within the scope of any disjunc-tion. Generally, then, a sentence in CNF looks like this (±A 1 ∨. . .∨±A i) ∧(±A i+1 ∨. . .∨±Aj ) ∧. . .∧(±A m+1 ∨. . .∨±A n) where each A k is an atomic sentence. We can now prove another normal form theorem: Conjunctive Normal Form Theorem. For any sentence, there is an equivalent sentence in conjunctive normal form. Given a TFL sentence, S, we begin by writing down the com-plete truth table for S. If S is true on every line of the truth table, then S and (A 1 ∨ ¬A 1) are logically equivalent. CHAPTER 43. NORMAL FORMS 365 If Sis false on at least one line of the truth table then, for every line on the truth table where S is false, write down a disjunction (±A 1 ∨. . . ∨±A n) which is false on (and only on) that line. Let C be the conjunction of all of these disjuncts; by construction, C is in CNF and S and C are logically equivalent. Practice exercises A. Consider the following sentences: 1. (A →¬B) 2. ¬(A ↔B) 3. (¬A ∨¬(A ∧B)) 4. (¬(A →B) ∧(A →C )) 5. (¬(A ∨B) ↔((¬C ∧¬A) →¬B)) 6. ((¬(A ∧¬B) →C ) ∧¬(A ∧D)) For each sentence, find an equivalent sentence in DNF and one in CNF. CHAPTER 44 Functional completeness Of our connectives, ¬ attaches to a single sentence, and the others all combine exactly two sentences. We may also introduce the idea of an n-place connective. For example, we could consider a three-place connective, ‘♥’, and stipulate that it is to have the following characteristic truth table: A B C ♥(A,B,C ) T T T F T T F T T F T T T F F F F T T F F T F T F F T F F F F F Probably this new connective would not correspond with any nat-ural English expression (at least not in the way that ‘∧’ corre-sponds with ‘and’). But a question arises: if we wanted to employ a connective with this characteristic truth table, must we add a new connective to TFL? Or can we get by with the connectives 366 CHAPTER 44. FUNCTIONAL COMPLETENESS 367 we already have (as we can for the connective ‘neither...nor’ for instance)? Let us make this question more precise. Say that some con-nectives are jointly functionally complete iff, for any possi-ble truth table, there is a sentence containing only those connec-tives with that truth table. The general point is, when we are armed with some jointly functionally complete connectives, no characteristic truth table lies beyond our grasp. And in fact, we are in luck. Functional Completeness Theorem. The connectives of TFL are jointly functionally complete. Indeed, the fol-lowing pairs of connectives are jointly functionally com-plete: 1. ‘¬’ and ‘∨’ 2. ‘¬’ and ‘∧’ 3. ‘¬’ and ‘→’ Given any truth table, we can use the method of proving the DNF Theorem (or the CNF Theorem) via truth tables from chap-ter 43, to write down a scheme which has the same truth table. For example, employing the truth table method for proving the DNF Theorem, we find that the following scheme has the same characteristic truth table as ♥(A,B,C ), above: (A ∧B ∧¬C ) ∨(A ∧¬B ∧C ) ∨(¬A ∧B ∧¬C ) It follows that the connectives of TFL are jointly functionally com-plete. We now prove each of the subsidiary results. Subsidiary Result 1: functional completeness of ‘¬’ and ‘∨’. Ob-serve that the scheme that we generate, using the truth table method of proving the DNF Theorem, will only contain the con-nectives ‘¬’, ‘∧’ and ‘∨’. So it suffices to show that there is an equivalent scheme which contains only ‘¬’ and ‘∨’. To demon-CHAPTER 44. FUNCTIONAL COMPLETENESS 368 strate this, we simply consider that (A∧B) and ¬(¬A∨¬B) are logically equivalent. Subsidiary Result 2: functional completeness of ‘¬’ and ‘∧’. Exactly as in Subsidiary Result 1, making use of the fact that (A∨B) and ¬(¬A∧¬B) are logically equivalent. Subsidiary Result 3: functional completeness of ‘¬’ and ‘→’. Ex-actly as in Subsidiary Result 1, making use of these equivalences instead: (A∨B) and (¬A →B) (A∧B) and ¬(A →¬B) Alternatively, we could simply rely upon one of the other two subsidiary results, and (repeatedly) invoke only one of these two equivalences. In short, there is never any need to add new connectives to TFL. Indeed, there is already some redundancy among the con-nectives we have: we could have made do with just two connec-tives, if we had been feeling really austere. 44.1 Individually functionally complete connectives In fact, some two-place connectives are individually functionally complete. These connectives are not standardly included in TFL, since they are rather cumbersome to use. But their existence shows that, if we had wanted to, we could have defined a truth-functional language that was functionally complete, which con-tained only a single primitive connective. The first such connective we will consider is ‘↑’, which has the following characteristic truth table. CHAPTER 44. FUNCTIONAL COMPLETENESS 369 A B A↑B T T F T F T F T T F F T This is often called ‘the Sheffer stroke’, after Henry Sheffer, who used it to show how to reduce the number of logical connectives in Russell and Whitehead’s Principia Mathematica.1 (In fact, Charles Sanders Peirce had anticipated Sheffer by about 30 years, but never published his results, and the Polish philosopher Erward Stamm published the same result two years before Sheffer.)2 It is quite common, as well, to call it ‘nand’, since its characteristic truth table is the negation of the truth table for ‘∧’. ‘↑’ is functionally complete all by itself. The functional completeness Theorem tells us that ‘¬’ and ‘∨’ are jointly functionally complete. So it suffices to show that, given any scheme which contains only those two connectives, we can rewrite it as an equivalent scheme which contains only ‘↑’. As in the proof of the subsidiary cases of the functional completeness Theorem, then, we simply apply the following equivalences: ¬A and (A↑A) (A∨B) and ((A↑A) ↑(B ↑B)) to the Subsidiary Result 1. Similarly, we can consider the connective ‘↓’: 1Sheffer, ‘A Set of Five Independent Postulates for Boolean Algebras, with application to logical constants,’ (1913, Transactions of the American Mathemati-cal Society 14.4) 2See Peirce, ‘A Boolian Algebra with One Constant’, which dates to c. 1880, Peirce’s Collected Papers, 4.264–5, and Stamm, “Beitrag zur Algebra der Logik,” Monatshefte für Mathematik und Physik 22 (1911): 137–49. CHAPTER 44. FUNCTIONAL COMPLETENESS 370 A B A↓B T T F T F F F T F F F T This is sometimes called the ‘Peirce arrow’ (Peirce himself called it ‘ampheck’). More often, though, it is called ‘nor’, since its characteristic truth table is the negation of ‘∨’, that is, of ‘neither ... nor . . . ’. ‘↓’ is functionally complete all by itself. As in the previous result for ↑, although invoking the equiva-lences: ¬A and (A↓A) (A∧B) and ((A↓A) ↓(B ↓B)) and Subsidiary Result 2. 44.2 Failures of functional completeness In fact, the only two-place connectives which are individually functionally complete are ‘↑’ and ‘↓’. But how would we show this? More generally, how can we show that some connectives are not jointly functionally complete? The obvious thing to do is to try to find some truth table which we cannot express, using just the given connectives. But there is a bit of an art to this. To make this concrete, let’s consider the question of whether ‘∨’ is functionally complete all by itself. After a little reflection, it should be clear that it is not. In particular, it should be clear that any scheme which only contains disjunctions cannot have the same truth table as negation, i.e.: CHAPTER 44. FUNCTIONAL COMPLETENESS 371 A ¬A T F F T The intuitive reason, why this should be so, is simple: the top line of the desired truth table needs to have the value False; but the top line of any truth table for a scheme which only contains ∨will always be True. The same is true for ∧, →, and ↔. ‘∨’, ‘∧’, ‘→’, and ‘↔’ are not functionally complete by themselves. In fact, the following is true: The only two-place connectives that are functionally com-plete by themselves are ‘↑’ and ‘↓’. This is of course harder to prove than for the primitive con-nectives. For instance, the “exclusive or” connective does not have a T in the first line of its characteristic truth table, and so the method used above no longer suffices to show that it cannot express all truth tables. It is also harder to show that, e.g., ‘↔’ and ‘¬’ together are not functionally complete. CHAPTER 45 Proving equivalences 45.1 Substitutability of equivalents Recall from §12.2 that P and Q are equivalent (in TFL) iff, for every valuation, their truth values agree. We have seen many examples of this and used both truth tables and natural dedcu-tion proofs to establish such equivalences. In chapter 43 we’ve even proved that ever sentence of TFL is equivalent to one in conjunctive and one in disjunctive normal form. If P and Q are equivalent, they always have the same truth value, either one en-tails the other, and from either one you can prove the other. Equivalent sentences are not the same, of course: the sen-tences ¬¬A and A may always have the same truth value, but the first starts with the ‘¬’ symbol while the second doesn’t. But you may wonder if it’s always true that we can replace one of a pair of equivalent sentences by the other, and the results will be equivalent, too. For instance, consider ¬¬A →B and A →B. The second results from the first by replacing ‘¬¬A’ by ‘A’. And these two sentences are also equivalent. This is a general fact, and it is not hard to see why it is true. In any valuation, we compute the truth value of a sentence “from 372 CHAPTER 45. PROVING EQUIVALENCES 373 the inside out.” So when it comes to determining the truth value of ‘¬¬A →B’, we first compute the truth value of ‘¬¬A’, and the truth value of the overall sentence then just depends on that truth value (true or false, as the case may be) and the rest of the sentence (the truth value of ‘B’ and the truth table for ‘→’). But since ‘¬¬A’ are equivalent, they always have the same truth value in a given valuation—hence, replacing ‘¬¬A’ by ‘A’ can-not change the truth value of the overall sentence. The same of course is true for any other sentence equivalent to ‘¬¬A’, say, ‘A ∧(A ∨A)’. To state the result in general, let’s use the notation R(P) to mean a sentence which contains the sentence P as a part. Then by R(Q) we mean the result of replacing the occurrence of P by the sentence Q. For instance, if P is the sentence letter ‘A’, Q is the sentence ‘¬¬A’, and R(P) is ‘A →B’, then R(Q) is ‘¬¬A →B’. If P and Q are equivalent, then so are R(P) and R(Q). It follows from this fact that any sentence of the form R(P) ↔R(Q), where P and Q are equivalent, is a tautology. However, the proofs in natural deduction will be wildly different for different R. (As an exercise, give proofs that show that ⊢(¬¬P →Q) ↔(P →Q) and ⊢(¬¬P ∧Q) ↔(P ∧Q) and compare the two.) Here is another fact: if two sentences P and Q are equivalent, and you replace some sentence letter in both P and Q by the same sentence R, the results are also equivalent. For instance, if you replace ‘A’ in both ‘A ∧B’ and ‘B ∧A’ by, say, ‘¬C ’, you get ‘¬C ∧B’ and ‘B ∧¬C ’, and those are equivalent. We can record this, too: CHAPTER 45. PROVING EQUIVALENCES 374 Equivalence is preserved under replacement of sentence letters, i.e., if P(A) and Q(A) both contain the sentence letter ‘A’ and are equivalent, then the sentences P(R) and Q(R) (resulting by replacing ‘A’ by R in both) are also equivalent. This means that once we have shown that two sentence are equivalent (e.g., ‘¬¬A’ and ‘A’, or ‘A ∧B’ and ‘B ∧A’) we know that all their common “instances” are also equivalent. Note that we do not immediately get this from a truth table or a natural deduction proof. E.g., a truth table that shows that ‘¬¬A’ and ‘A’ are equivalent does not also show that ‘¬¬(B →C )’ and ‘B →C ’ are equivalent: the former needs just 2 lines, the latter 4. 45.2 Chains of equivalences When you want to verify that two sentences are equivalent, you can of course do a truth table, or look for a formal proof. But there is a simpler method, based on the principle of substitutabil-ity of equivalents we just discussed: Armed with a small catalog of simple equivalences, replace parts of your first sentence by equiv-alent parts, and repeat until you reach your second sentence. This method of showing sentences equivalent is underwritten by the two facts from the previous section. The first fact tells us that if, say, ¬¬P and P are equivalent (for any sentence P), then replacing ¬¬P in a sentence by P results in an equivalent sen-tence. The second fact tells us that ¬¬P and P are always equiv-alent, for any sentence P. (A simple truth table shows that ‘¬¬A’ and ‘A’ are equivalent.) By the second fact we know that when-ever we replace ‘A’ in both ‘¬¬A’ and ‘A’ by some sentence P, we get equivalent results. In other words, from the fact that ‘¬¬A’ and ‘A’ are equivalent and the second fact, we can conclude that, for any sentence P, ¬¬P and P are equivalent. Let’s give an example. By De Morgan’s Laws, the following CHAPTER 45. PROVING EQUIVALENCES 375 pairs of sentences are equivalent: ¬(A ∧B) and (¬A ∨¬B) ¬(A ∨B) and (¬A ∧¬B) This can be verified by constructing two truth tables, or four natural deduction proofs that show: ¬(A ∧B) ⊢(¬A ∨¬B) (¬A ∨¬B) ⊢¬(A ∧B) ¬(A ∨B) ⊢(¬A ∧¬B) (¬A ∧¬B) ⊢¬(A ∨B) By the second fact, any pairs of sentences of the following forms are equivalent: ¬(P ∧Q) and (¬P ∨¬Q) ¬(P ∨Q) and (¬P ∧¬Q) Now consider the sentence ‘¬(R ∨(S ∧T ))’. We will find an equivalent sentence in which all ‘¬’ signs attach directly to sen-tence letters. In the first step, we consider this as a sentence of the form ¬(P ∨Q)—then P is the sentence ‘R’ and Q is ‘(S ∧T )’. Since ¬(P ∨Q) is equivalent to (¬P ∧¬Q) (by the second of De Morgan’s Laws) we can replace the entire sentence by (¬P ∧¬Q). In this case (where P is ‘R’ and Q is ‘(S ∧T )’) we obtain ‘(¬R ∧¬(S ∧T ))’. This new sentence contains as a part the sentence ‘¬(S ∧T )’. It is of the form ¬(P ∧Q), except now P is the sentence letter ‘S’ and Q is ‘T ’. By De Morgan’s Law (the first one this time), this is equivalent to (¬P ∨¬Q), or in this specific case, to ‘(¬S ∨¬T )’. So we can replace the part ‘¬(S ∧T )’ by ‘(¬S ∨¬T )’. This now results in the sentence ‘(¬R ∧(¬S ∨¬T ))’, in which the ‘¬’ symbols all attach directly to sentence letters. We’ve “pushed” the negations inwards as far as possible. We can record such a chain of equivalences by list-ing the individual steps, and recording, just as we do in natural CHAPTER 45. PROVING EQUIVALENCES 376 deduction, which basic equivalence we use in each case: ¬( R ∨(S ∧T ) ) (¬( R ∧¬ (S ∧T ) ) DeM (¬(R ∧¬( S ∧T ) ) (¬(R ∧(¬ S ∨¬ T ) ) DeM We’ve highlighted the sentence replaced, and those matching the Pand Q in De Morgan’s Laws for clarity, but this is not necessary, and we won’t keep doing it. In table 45.1 we’ve given a list of basic equivalences you can use for such chains of equivalences. The labels abbreviate the customary name for the respective logical laws: double negation (DN), De Morgan (DeM), commutativity (Comm), distributivity (Dist), associativity (Assoc), idempotence (Id), and absorption (Abs). 45.3 Finding equivalent normal forms In chapter 43 we showed that every sentence of TFL is equiv-alent to one in disjunctive normal form (DNF) and to one in conjunctive normal form (CNF). We did this by giving a method to construct a sentences in DNF or CNF equivalent to the origi-nal sentence by first constructing a truth table, and then “reading off” a sentence in DNF or CNF from the truth table. This method has two drawbacks. The first one is that the resulting sentences in DNF or CNF are not always the shortest ones. The second one is that the method itself becomes hard to apply when the sentence you start with contains more than a handful of sentence letters (since the truth table for a sentence with n sentence letters has 2n lines). We can use chains of equivalences as an alternative method: To find a sentence in DNF, we can successively apply basic equiv-CHAPTER 45. PROVING EQUIVALENCES 377 ¬¬P ⇔P (DN) (P →Q) ⇔(¬P ∨Q) (Cond) ¬(P →Q) ⇔(P ∧¬Q) (P ↔Q) ⇔((P →Q) ∧(Q →P)) (Bicond) ¬(P ∧Q) ⇔(¬P ∨¬Q) (DeM) ¬(P ∨Q) ⇔(¬P ∧¬Q) (P ∨Q) ⇔(Q ∨P) (Comm) (P ∧Q) ⇔(Q ∧P) (P ∧(Q ∨R)) ⇔((P ∧Q) ∨(P ∧R)) (Dist) (P ∨(Q ∧R)) ⇔((P ∨Q) ∧(P ∨R)) (P ∨(Q ∨R)) ⇔((P ∨Q) ∨R) (Assoc) (P ∧(Q ∧R)) ⇔((P ∧Q) ∧R) (P ∨P) ⇔P (Id) (P ∧P) ⇔P (P ∧(P ∨Q)) ⇔P (Abs) (P ∨(P ∧Q)) ⇔P (P ∧(Q ∨¬Q)) ⇔P (Simp) (P ∨(Q ∧¬Q)) ⇔P (P ∨(Q ∨¬Q)) ⇔(Q ∨¬Q) (P ∧(Q ∧¬Q)) ⇔(Q ∧¬Q) Table 45.1: Basic equivalences CHAPTER 45. PROVING EQUIVALENCES 378 alences until we have found an equivalent sentence that is in DNF. Recall the conditions a sentence in DNF must satisfy: (dnf1) No connectives occur in the sentence other than nega-tions, conjunctions and disjunctions; (dnf2) Every occurrence of negation has minimal scope (i.e., any ‘¬’ is immediately followed by an atomic sentence); (dnf3) No disjunction occurs within the scope of any conjunc-tion. Condition (dnf1) says that we must remove all ‘→’ and ‘↔’ sym-bols from a sentence. This is what the basic equivalences (Cond) and (Bicond) are good for. For instance, suppose we start with the sentence ¬(A ∧¬C ) ∧(¬A →¬B). We can get rid of the ‘→’ by using (Cond). In this case P is ‘¬A’ and Q is ‘¬B’. We get: ¬(A ∧¬C ) ∧(¬¬A ∨¬B) Cond The double negation can be removed, since ‘¬¬A’ is equivalent to ‘A’: ¬(A ∧¬C ) ∧(A ∨¬B) DN Now condition (dnf1) is satisfied: our sentence contains only ‘¬’, ‘∧’, and ‘∨’. Condition (dnf2) says that we must find a way to have all ‘¬’s apply immediately to sentence letters. But in the first conjunct it doesn’t. To ensure (dnf2) is satisfied, we use De Morgan’s Laws and the double negation (DN) law as many times as needed. (¬A ∨¬¬C ) ∧(A ∨¬B) DeM (¬A ∨C ) ∧(A ∨¬B) DN CHAPTER 45. PROVING EQUIVALENCES 379 The resulting sentence is now in CNF—it is a conjunction of dis-junctions of sentence letters and negated sentence letters. But we want a sentence in DNF, i.e., a sentence in which (dnf3) is satisfied: no ‘∨’ occurs in the scope of an ‘∧’. We use the distribu-tive laws (Dist) to ensure this. The last sentence is of the form P∧(Q ∨R), where P is ‘(¬A ∨C )’, Q is ‘A’, and R is ‘¬B’. By applying (Dist) once we get: ((¬A ∨C ) ∧A)) ∨((¬A ∨C ) ∧¬B) Dist This looks worse, but if we keep going, it’s going to look better! The two disjuncts almost look like we can apply (Dist) again, except the ‘∨’ is on the wrong side. This is what commutativity (Comm) is good for. let’s apply it to ‘(¬A ∨C ) ∧A’: (A ∧(¬A ∨C )) ∨((¬A ∨C ) ∧¬B) Comm We can apply (Dist) again to the resulting part, ‘A ∧(¬A ∨C )’: ((A ∧¬A) ∨(A ∧C )) ∨((¬A ∨C ) ∧¬B) Dist Now in the left half, no ‘∨’ is in the scope of an ‘∧’. Let’s apply the same principles to the right half: ((A ∧¬A) ∨(A ∧C )) ∨(¬B ∧(¬A ∨C )) Comm ((A ∧¬A) ∨(A ∧C )) ∨((¬B ∧¬A) ∨(¬B ∧C )) Dist Our sentence is now in DNF! But we can simplify it a bit: ‘(A ∧¬A)’ is a contradiction in TFL, i.e., it is always false. And if you combine something that’s always false using ‘∨’ with a sen-tence P, you get something equivalent to just P. This is the second of the simplification (Simp) rules. ((A ∧C ) ∨(A ∧¬A)) ∨((¬B ∧¬A) ∨(¬B ∧C )) Comm (A ∧C ) ∨((¬B ∧¬A) ∨(¬B ∧C )) Simp CHAPTER 45. PROVING EQUIVALENCES 380 The final result is still in DNF, but a bit simpler still. It is also much simpler than the DNF we would have obtained by the method of chapter 43. In fact, the sentence we started with could have been the Sof §43.2—it has exactly the truth table used as an example there. The DNF we found there (on p. 362), was (with all necessary brackets): ((((A∧B)∧C )∨((A∧¬B)∧C ))∨((¬A∧¬B)∧C ))∨((¬A∧¬B)∧¬C ). Practice exercises A. Consider the following sentences: 1. (A →¬B) 2. ¬(A ↔B) 3. (¬A ∨¬(A ∧B)) 4. (¬(A →B) ∧(A →C )) 5. (¬(A ∨B) ↔((¬C ∧¬A) →¬B)) 6. ((¬(A ∧¬B) →C ) ∧¬(A ∧D)) For each sentence, find an equivalent sentence in DNF and one in CNF by giving a chain of equivalences. Use (Id), (Abs), and (Simp) to simplify your sentences as much as possible. CHAPTER 46 Soundness In this chapter we relate TFL’s semantics to its natural deduction proof system (as defined in Part IV). We will prove that the formal proof system is safe: you can only prove sentences from premises from which they actually follow. Intuitively, a formal proof system is sound iff it does not allow you to prove any invalid arguments. This is obviously a highly desirable property. It tells us that our proof system will never lead us astray. Indeed, if our proof system were not sound, then we would not be able to trust our proofs. The aim of this chapter is to prove that our proof system is sound. Let’s make the idea more precise. We’ll abbreviate a list of sentences using the greek letter Γ (‘gamma’). A formal proof sys-tem is sound (relative to a given semantics) iff, whenever there is a formal proof of C from assumptions among Γ, then Γ gen-uinely entails C (given that semantics). Otherwise put, to prove that TFL’s proof system is sound, we need to prove the following Soundness Theorem. For any sentences Γ and C: if Γ ⊢ C, then Γ ⊨C To prove this, we will check each of the rules of TFL’s proof system individually. We want to show that no application of those rules ever leads us astray. Since a proof just involves repeated application of those rules, this will show that no proof ever leads 381 CHAPTER 46. SOUNDNESS 382 us astray. Or at least, that is the general idea. To begin with, we must make the idea of ‘leading us astray’ more precise. Say that a line of a proof is shiny iff the assump-tions on which that line depends entail the sentence on that line.1 To illustrate the idea, consider the following: 1 F →(G ∧H ) 2 F 3 G ∧H →E 1, 2 4 G ∧E 3 5 F →G →I 2–4 Line 1 is shiny iff F →(G ∧H ) ⊨F →(G ∧H ). You should be easily convinced that line 1 is, indeed, shiny! Similarly, line 4 is shiny iff F →(G ∧H ),F ⊨G. Again, it is easy to check that line 4 is shiny. As is every line in this TFL-proof. We want to show that this is no coincidence. That is, we want to prove: Shininess Lemma. Every line of every TFL-proof is shiny. Then we will know that we have never gone astray, on any line of a proof. Indeed, given the Shininess Lemma, it will be easy to prove the Soundness Theorem: Proof. Suppose Γ ⊢C. Then there is a TFL-proof, with C appearing on its last line, whose only undischarged assumptions are among Γ. The Shininess Lemma tells us that every line on every TFL-proof is shiny. So this last line is shiny, i.e. Γ ⊨C. QED It remains to prove the Shininess Lemma. To do this, we observe that every line of any TFL-proof is obtained by applying some rule. So what we want to show is that no application of a rule of TFL’s proof system will lead us astray. 1The word ‘shiny’ is not standard among logicians. CHAPTER 46. SOUNDNESS 383 More precisely, say that a rule of inference is rule-sound iff for all TFL-proofs, if we obtain a line on a TFL-proof by applying that rule, and every earlier line in the TFL-proof is shiny, then our new line is also shiny. What we need to show is that every rule in TFL’s proof system is rule-sound. We will do this in the next section. But having demonstrated the rule-soundness of every rule, the Shininess Lemma will follow immediately: Proof. Fix any line, line n, on any TFL-proof. The sentence written on line n must be obtained using a formal inference rule which is rule-sound. This is to say that, if every earlier line is shiny, then line n itself is shiny. Hence, by strong induction on the length of TFL-proofs, every line of every TFL-proof is shiny. QED Note that this proof appeals to a principle of strong induction on the length of TFL-proofs. This is the first time we have seen that principle, and you should pause to confirm that it is, indeed, justified. It remains to show that every rule is rule-sound. This is not difficult, but it is time-consuming, since we need to check each rule individually, and TFL’s proof system has plenty of rules! To speed up the process marginally, we will introduce a conve-nient abbreviation: ‘Δi’ (‘delta’) will abbreviate the assumptions (if any) on which line i depends in our TFL-proof (context will indicate which TFL-proof we have in mind). Introducing an assumption is rule-sound. If Ais introduced as an assumption on line n, then Ais among Δn, and so Δn ⊨A. ∧I is rule-sound. Proof. Consider any application of ∧I in any TFL-proof, i.e., something like: CHAPTER 46. SOUNDNESS 384 i A j B n A∧B ∧I i, j To show that ∧I is rule-sound, we assume that every line before line n is shiny; and we aim to show that line n is shiny, i.e. that Δn ⊨A∧B. So, let v be any valuation that makes all of Δn true. We first show that v makes A true. To prove this, note that all of Δi are among Δn. By hypothesis, line i is shiny. So any valuation that makes all of Δi true makes A true. Since v makes all of Δi true, it makes A true too. We can similarly see that v makes B true. So v makes A true and v makes B true. Consequently, v makes A∧Btrue. So any valuation that makes all of the sentences among Δn true also makes A∧B true. That is: line n is shiny. QED All of the remaining lemmas establishing rule-soundness will have, essentially, the same structure as this one did. ∧E is rule-sound. Proof. Assume that every line before line n on some TFL-proof is shiny, and that ∧E is used on line n. So the situation is: i A∧B n A ∧E i (or perhaps with B on line n instead; but similar reasoning will apply in that case). Let v be any valuation that makes all of Δn true. Note that all of Δi are among Δn. By hypothesis, line i is shiny. So any valuation that makes all of Δi true makes A∧B true. So v makes A∧Btrue, and hence makes Atrue. So Δn ⊨A. QED CHAPTER 46. SOUNDNESS 385 ∨I is rule-sound. We leave this as an exercise. ∨E is rule-sound. Proof. Assume that every line before line n on some TFL-proof is shiny, and that ∧E is used on line n. So the situation is: m A∨B i A j C k B l C n C ∨E m, i–j, k–l Let v be any valuation that makes all of Δn true. Note that all of Δm are among Δn. By hypothesis, line m is shiny. So any valuation that makes Δn true makes A∨B true. So in particular, v makes A∨B true, and hence either v makes Atrue, or v makes B true. We now reason through these two cases: Case 1: v makes A true. All of Δi are among Δn, with the possible exception of A. Since v makes all of Δn true, and also makes A true, v makes all of Δi true. Now, by assumption, line j is shiny; so Δj ⊨C. But the sentences Δi are just the sentences Δj , so Δi ⊨C. So, any valuation that makes all of Δi true makes C true. But v is just such a valuation. So v makes C true. Case 2: v makes B true. Reasoning in exactly the same way, con-sidering lines k and l, v makes C true. Either way, v makes C true. So Δn ⊨C. QED CHAPTER 46. SOUNDNESS 386 ¬E is rule-sound. Proof. Assume that every line before line n on some TFL-proof is shiny, and that ¬E is used on line n. So the situation is: i A j ¬A n ⊥ ¬E i, j Note that all of Δi and all of Δj are among Δn. By hypothesis, lines i and j are shiny. So any valuation which makes all of Δn true would have to make both A and ¬A true. But no valuation can do that. So no valuation makes all of Δn true. So Δn ⊨⊥, vacuously. QED X is rule-sound. We leave this as an exercise. ¬I is rule-sound. Proof. Assume that every line before line n on some TFL-proof is shiny, and that ¬I is used on line n. So the situation is: i A j ⊥ n ¬A ¬I i–j Let v be any valuation that makes all of Δn true. Note that all of Δn are among Δi, with the possible exception of A itself. By hypothesis, line j is shiny. But no valuation can make ‘⊥’ true, so no valuation can make all of Δj true. Since the sentences Δi are just the sentences Δj , no valuation can make all of Δi true. CHAPTER 46. SOUNDNESS 387 Since v makes all of Δn true, it must therefore make A false, and so make ¬A true. So Δn ⊨¬A. QED IP, →I, →E, ↔I, and ↔E are all rule-sound. We leave these as exercises. This establishes that all the basic rules of our proof system are rule-sound. Finally, we show: All of the derived rules of our proof system are rule-sound. Proof. Suppose that we used a derived rule to obtain some sentence, A, on line n of some TFL-proof, and that every earlier line is shiny. Every use of a derived rule can be replaced (at the cost of long-windedness) with multiple uses of basic rules. That is to say, we could have used basic rules to write A on some line n +k, without introducing any further assumptions. So, applying our individual results that all basic rules are rule-sound several times (k + 1 times, in fact), we can see that line n + k is shiny. Hence the derived rule is rule-sound. QED And that’s that! We have shown that every rule—basic or otherwise—is rule-sound, which is all that we required to establish the Shininess Lemma, and hence the Soundness Theorem. But it might help to round off this chapter if we repeat my informal explanation of what we have done. A formal proof is just a sequence—of arbitrary length—of applications of rules. We have shown that any application of any rule will not lead you astray. It follows (by induction)that no formal proof will lead you astray. That is: our proof system is sound. Practice exercises A. Complete the Lemmas left as exercises in this chapter. That is, show that the following are rule-sound: 1. ∨I. (Hint: this is similar to the case of ∧E.) CHAPTER 46. SOUNDNESS 388 2. X. (Hint: this is similar to the case of ¬E.) 3. →I. (Hint: this is similar to ∨E.) 4. →E. 5. IP. (Hint: this is similar to the case of ¬I.) Appendices 389 APPENDIX A Symbolic notation 1.1 Alternative nomenclature Truth-functional logic. TFL goes by other names. Sometimes it is called sentential logic, because it deals fundamentally with sentences. Sometimes it is called propositional logic, on the idea that it deals fundamentally with propositions. We have stuck with truth-functional logic, to emphasize the fact that it deals only with assignments of truth and falsity to sentences, and that its connec-tives are all truth-functional. First-order logic. FOL goes by other names. Sometimes it is called predicate logic, because it allows us to apply predicates to objects. Sometimes it is called quantified logic, because it makes use of quantifiers. Formulas. Some texts call formulas well-formed formulas. Since ‘well-formed formula’ is such a long and cumbersome phrase, they then abbreviate this as wff. This is both barbarous and un-necessary (such texts do not countenance ‘ill-formed formulas’). We have stuck with ‘formula’. 390 APPENDIX A. SYMBOLIC NOTATION 391 In §6, we defined sentences of TFL. These are also sometimes called ‘formulas’ (or ‘well-formed formulas’) since in TFL, unlike FOL, there is no distinction between a formula and a sentence. Valuations. Some texts call valuations truth-assignments, or truth-value assignments. n-place predicates. We have chosen to call predicates ‘one-place’, ‘two-place’, ‘three-place’, etc. Other texts respectively call them ‘monadic’, ‘dyadic’, ‘triadic’, etc. Still other texts call them ‘unary’, ‘binary’, ‘ternary’, etc. Names. In FOL, we have used ‘a’, ‘b’, ‘c’, for names. Some texts call these ‘constants’. Other texts do not mark any difference between names and variables in the syntax. Those texts focus simply on whether the symbol occurs bound or unbound. Domains. Some texts describe a domain as a ‘domain of dis-course’, or a ‘universe of discourse’. 1.2 Alternative symbols In the history of formal logic, different symbols have been used at different times and by different authors. Often, authors were forced to use notation that their printers could typeset. This ap-pendix presents some common symbols, so that you can recog-nize them if you encounter them in an article or in another book. Negation. T wo commonly used symbols are the hoe, ‘¬’, and the swung dash or tilde, ‘∼.’ In some more advanced formal sys-tems it is necessary to distinguish between two kinds of negation; the distinction is sometimes represented by using both ‘¬’ and ‘∼’. Older texts sometimes indicate negation by a line over the formula being negated, e.g., A ∧B. Some texts use ‘x ≠y’ to abbreviate ‘¬x = y’. APPENDIX A. SYMBOLIC NOTATION 392 Disjunction. The symbol ‘∨’ is typically used to symbolize in-clusive disjunction. One etymology is from the Latin word ‘vel’, meaning ‘or’. Conjunction. Conjunction is often symbolized with the amper-sand, ‘&’. The ampersand is a decorative form of the Latin word ‘et’, which means ‘and’. (Its etymology still lingers in certain fonts, particularly in italic fonts; thus an italic ampersand might appear as ‘&’.) This symbol is commonly used in natural English writing (e.g. ‘Smith & Sons’), and so even though it is a natural choice, many logicians use a different symbol to avoid confusion between the object and metalanguage: as a symbol in a formal system, the ampersand is not the English word ‘&’. The most common choice now is ‘∧’, which is a counterpart to the symbol used for disjunction. Sometimes a single dot, ‘•’, is used. In some older texts, there is no symbol for conjunction at all; ‘A and B’ is simply written ‘AB’. Material Conditional. There are two common symbols for the material conditional: the arrow, ‘→’, and the horseshoe, ‘⊃’. Material Biconditional. The double-headed arrow, ‘↔’, is used in systems that use the arrow to represent the material condi-tional. Systems that use the horseshoe for the conditional typi-cally use the triple bar, ‘≡’, for the biconditional. Quantifiers. The universal quantifier is typically symbolized as a rotated ‘A’, and the existential quantifier as a rotated, ‘E’. In some texts, there is no separate symbol for the universal quanti-fier. Instead, the variable is just written in parentheses in front of the formula that it binds. For example, they might write ‘(x)P (x)’ where we would write ‘∀x P (x)’. These alternative typographies are summarised below: APPENDIX A. SYMBOLIC NOTATION 393 negation ¬, ∼ conjunction ∧, &, • disjunction ∨ conditional →, ⊃ biconditional ↔, ≡ universal quantifier ∀x, (x) Polish notation This section briefly discusses sentential logic in Polish notation, a system of notation introduced in the late 1920s by the Polish logician Jan Łukasiewicz. Lower case letters are used as sentence letters. The capital letter N is used for negation. A is used for disjunction, K for conjunction, C for the conditional, E for the biconditional. (‘A’ is for alternation, another name for logical disjunction. ‘E’ is for equivalence.) In Polish notation, a binary connective is written before the two sentences that it connects. For example, the sentence A ∧B of TFL would be written Kab in Polish notation. The sentences ¬A →B and ¬(A →B) are very different; the main logical operator of the first is the conditional, but the main connective of the second is negation. In TFL, we show this by putting parentheses around the conditional in the second sen-tence. In Polish notation, parentheses are never required. The left-most connective is always the main connective. The first sen-tence would simply be written CN ab and the second N Cab. This feature of Polish notation means that it is possible to evaluate sentences simply by working through the symbols from right to left. If you were constructing a truth table for N Kab, for example, you would first consider the truth-values assigned to b and a, then consider their conjunction, and then negate the result. The general rule for what to evaluate next in TFL is not nearly so simple. In TFL, the truth table for ¬(A ∧B) requires looking at A and B, then looking in the middle of the sentence at the conjunction, and then at the beginning of the sentence at the APPENDIX A. SYMBOLIC NOTATION 394 negation. Because the order of operations can be specified more mechanically in Polish notation, variants of Polish notation are used as the internal structure for many computer programming languages. APPENDIX B Alternative proof systems In formulating our natural deduction system, we treated certain rules of natural deduction as basic, and others as derived. How-ever, we could equally well have taken various different rules as basic or derived. We will illustrate this point by considering some alternative treatments of disjunction, negation, and the quanti-fiers. We will also explain why we have made the choices that we have. 2.1 Alternative disjunction elimination Some systems take DS as their basic rule for disjunction elimina-tion. Such systems can then treat the ∨E rule as a derived rule. For they might offer the following proof scheme: 395 APPENDIX B. ALTERNATIVE PROOF SYSTEMS 396 m A∨B i A j C k B l C n A →C →I i–j n + 1 B →C →I k–l n + 2 ¬C n + 3 A n + 4 C →E n + 3, n n + 5 ⊥ ¬E n + 2, n + 4 n + 6 ¬A ¬I n + 3–n + 5 n + 7 B DS m, n + 6 n + 8 C →E n + 7, n + 1 n + 9 ⊥ ¬E n + 2, n + 8 n + 10 C IP n + 2–n + 9 So why did we choose to take ∨E as basic, rather than DS?1 Our reasoning is that DS involves the use of ‘¬’ in the statement of the rule. It is in some sense ‘cleaner’ for our disjunction elimination rule to avoid mentioning other connectives. 1P. D. Magnus’s original version of this book went the other way. APPENDIX B. ALTERNATIVE PROOF SYSTEMS 397 2.2 Alternative negation rules Some systems take the following rule as their basic negation in-troduction rule: m A n −1 B n ¬B ¬A ¬I m–n and a corresponding version of the rule we called IP as their basic negation elimination rule: m ¬A n −1 B n ¬B A ¬E m–n Using these two rules, we could we could have avoided all use of the symbol ‘⊥’ altogether.2 The resulting system would have had fewer rules than ours. Another way to deal with negation is to use either LEM or DNE as a basic rule and introduce IP as a derived rule. T ypically, in such a system the rules are given different names, too. E.g., sometimes what we call ¬E is called ⊥I, and what we call X is called ⊥E.3 So why did we chose our rules for negation and contradiction? Our first reason is that adding the symbol ‘⊥’ to our natural deduction system makes proofs considerably easier to work with. For instance, in our system it’s always clear what the conclusion 2Again, P. D. Magnus’s original version of this book went the other way. 3The version of this book due to Tim Button goes this route and replaces IP with LEM, which he calls TND, for “tertium non datur.” APPENDIX B. ALTERNATIVE PROOF SYSTEMS 398 of a subproof is: the sentence on the last line, e.g., ⊥in IP or ¬I. In ¬I and ¬E, subproofs have two conclusions, so you can’t check at one glance if an application of them is correct. Our second reason is that a lot of fascinating philosophical discussion has focussed on the acceptability or otherwise of indi-rect proof IP (equivalently, excluded middle, i.e., LEM, or double negation elimination DNE) and explosion (i.e., X). By treating these as separate rules in the proof system, you will be in a better position to engage with that philosophical discussion. In particu-lar: having invoked these rules explicitly, it would be much easier for us to know what a system which lacked these rules would look like. This discussion, and in fact the vast majority of mathematical study on applications of natural deduction proofs beyond intro-ductory courses, makes reference to a different version of natural deduction. This version was invented by Gerhard Gentzen in 1935 as refined by Dag Prawitz in 1965. Our set of basic rules coincides with theirs. In other words, the rules we use are those that are standard in philosophical and mathematical discussion of natural deduction proofs outside of introductory courses. 2.3 Alternative quantification rules An alternative approach to the quantifiers is to take as basic the rules for ∀I and ∀E from §34, and also two CQ rules which allow us to move from ∀x¬A to ¬∃xA and vice versa.4 Taking only these rules as basic, we could have derived the ∃I and ∃E rules provided in §34. To derive the ∃I rule is fairly sim-ple. Suppose A contains the name c, and contains no instances of the variable x, and that we want to do the following: 4Warren Goldfarb follows this line in Deductive Logic, 2003, Hackett Pub-lishing Co. APPENDIX B. ALTERNATIVE PROOF SYSTEMS 399 m A(. . . c. . . c. . .) k ∃xA(. . . x. . . c. . .) This is not yet permitted, since in this new system, we do not have the ∃I rule. We can, however, offer the following: m A(. . . c. . . c. . .) m + 1 ¬∃xA(. . . x. . . c. . .) m + 2 ∀x¬A(. . . x. . . c. . .) CQ m + 1 m + 3 ¬A(. . . c. . . c. . .) ∀E m + 2 m + 4 ⊥ ¬E m + 3, m m + 5 ∃xA(. . . x. . . c. . .) IP m + 1–m + 4 To derive the ∃E rule is rather more subtle. This is because the ∃E rule has an important constraint (as, indeed, does the ∀I rule), and we need to make sure that we are respecting it. So, suppose we are in a situation where we want to do the following: m ∃xA(. . . x. . . x. . .) i A(. . . c. . . c. . .) j B k B where c does not occur in any undischarged assumptions, or in B, or in ∃xA(. . . x. . . x. . .). Ordinarily, we would be allowed to use the ∃E rule; but we are not here assuming that we have access to this rule as a basic rule. Nevertheless, we could offer the following, more complicated derivation: APPENDIX B. ALTERNATIVE PROOF SYSTEMS 400 m ∃xA(. . . x. . . x. . .) i A(. . . c. . . c. . .) j B k A(. . . c. . . c. . .) →B →I i–j k + 1 ¬B k + 2 ¬A(. . . c. . . c. . .) MT k, k + 1 k + 3 ∀x¬A(. . . x. . . x. . .) ∀I k + 2 k + 4 ¬∃xA(. . . x. . . x. . .) CQ k + 3 k + 5 ⊥ ¬E k + 4, m k + 6 B IP k + 1–k + 5 We are permitted to use ∀I on line k +3 because c does not occur in any undischarged assumptions or in B. The entries on lines k + 4 and k + 1 contradict each other, because c does not occur in ∃xA(. . . x. . . x. . .). Armed with these derived rules, we could now go on to derive the two remaining CQ rules, exactly as in §38. So, why did we start with all of the quantifier rules as basic, and then derive the CQ rules? Our first reason is that it seems more intuitive to treat the quantifiers as on a par with one another, giving them their own basic rules for introduction and elimination. Our second reason relates to the discussion of alternative negation rules. In the derivations of the rules of ∃I and ∃E that we have offered in this section, we invoked IP. But, as we men-tioned earlier, IP is a contentious rule. So, if we want to move to a system which abandons IP, but which still allows us to use existential quantifiers, we will want to take the introduction and elimination rules for the quantifiers as basic, and take the CQ rules as derived. (Indeed, in a system without IP, LEM, and APPENDIX B. ALTERNATIVE PROOF SYSTEMS 401 DNE, we will be unable to derive the CQ rule which moves from ¬∀xA to ∃x¬A.) APPENDIX C Quick reference 3.1 Characteristic truth tables A ¬A T F F T A B A∧B A∨B A →B A ↔B T T T T T T T F F T F F F T F T T F F F F F T T 402 APPENDIX C. QUICK REFERENCE 403 3.2 Symbolization Sentential Connectives It is not the case that P ¬P Either P or Q (P ∨Q) Neither P nor Q ¬(P ∨Q) or (¬P ∧¬Q) Both P and Q (P ∧Q) If P then Q (P →Q) P only if Q (P →Q) P if and only if Q (P ↔Q) P unless Q (P ∨Q) Predicates All F s are Gs ∀x(F (x) →G (x)) Some F s are Gs ∃x(F (x) ∧G (x)) Not all F s are Gs ¬∀x(F (x) →G (x)) or ∃x(F (x) ∧¬G (x)) No F s are Gs ∀x(F (x) →¬G (x)) or ¬∃x(F (x) ∧G (x)) Only F s are Gs ∀x(G (x) →F (x)) ¬∃x(¬F (x) ∧G (x)) Identity Only c is G ∀x(G (x) ↔x = c) Everything other than c is G ∀x(¬x = c →G (x)) Everything except c is G ∀x(¬x = c ↔G (x)) The F is G ∃x(F (x) ∧∀y(F (y) →x = y) ∧G (x)) It is not the case that the F is G ¬∃x(F (x) ∧∀y(F (y) →x = y) ∧G (x)) The F is non-G ∃x(F (x) ∧∀y(F (y) →x = y) ∧¬G (x)) APPENDIX C. QUICK REFERENCE 404 3.3 Using identity to symbolize quantities There are at least F s. one ∃x F (x) two ∃x1∃x2(F (x1) ∧F (x2) ∧¬x1 = x2) three ∃x1∃x2∃x3(F (x1) ∧F (x2) ∧F (x3) ∧ ¬x1 = x2 ∧¬x1 = x3 ∧¬x2 = x3) four ∃x1∃x2∃x3∃x4(F (x1) ∧F (x2) ∧F (x3) ∧F (x4) ∧ ¬x1 = x2 ∧¬x1 = x3 ∧¬x1 = x4 ∧ ¬x2 = x3 ∧¬x2 = x4 ∧¬x3 = x4) n ∃x1 . . . ∃xn(F (x1) ∧. . . ∧F (xn) ∧ ¬x1 = x2 ∧. . . ∧¬xn−1 = xn) There are at most F s. One way to say ‘there are at most n F s’ is to put a negation sign in front of the symbolization for ‘there are at least n + 1 F s’. Equivalently, we can offer: one ∀x1∀x2 [︁ (F (x1) ∧F (x2)) →x1 = x2 ]︁ two ∀x1∀x2∀x3 [︁ (F (x1) ∧F (x2) ∧F (x3)) → (x1 = x2 ∨x1 = x3 ∨x2 = x3) ]︁ three ∀x1∀x2∀x3∀x4 [︁ (F (x1) ∧F (x2) ∧F (x3) ∧F (x4)) → (x1 = x2 ∨x1 = x3 ∨x1 = x4 ∨ x2 = x3 ∨x2 = x4 ∨x3 = x4) ]︁ n ∀x1 . . . ∀xn+1 [︁ (F (x1) ∧. . . ∧F (xn+1)) → (x1 = x2 ∨. . . ∨xn = xn+1) ]︁ There are exactly F s. One way to say ‘there are exactly n F s’ is to conjoin two of the symbolizations above and say ‘there are at least n F s and there are at most n F s.’ The following equivalent formulas are shorter: APPENDIX C. QUICK REFERENCE 405 zero ∀x ¬F (x) one ∃x [︁ F (x) ∧∀y(F (y) →x = y) ]︁ two ∃x1∃x2 [︁ F (x1) ∧F (x2) ∧ ¬x1 = x2 ∧∀y (︁F (y) →(y = x1 ∨y = x2))︁]︁ three ∃x1∃x2∃x3 [︁ F (x1) ∧F (x2) ∧F (x3) ∧ ¬x1 = x2 ∧¬x1 = x3 ∧¬x2 = x3 ∧ ∀y (︁F (y) →(y = x1 ∨y = x2 ∨y = x3))︁]︁ n ∃x1 . . . ∃xn [︁ F (x1) ∧. . . ∧F (xn) ∧ ¬x1 = x2 ∧. . . ∧¬xn−1 = xn ∧ ∀y (︁F (y) →(y = x1 ∨. . . ∨y = xn))︁]︁ APPENDIX C. QUICK REFERENCE 406 3.4 Basic deduction rules for TFL Reiteration m A A R m Conjunction m A n B A∧B ∧I m, n m A∧B A ∧E m m A∧B B ∧E m Conditional i A j B A →B →I i–j m A →B n A B →E m, n Negation i A j ⊥ ¬A ¬I i–j m ¬A n A ⊥ ¬E m, n Indirect proof i ¬A j ⊥ A IP i–j Explosion m ⊥ A X m APPENDIX C. QUICK REFERENCE 407 Disjunction m A A∨B ∨I m m A B∨A ∨I m m A∨B i A j C k B l C C ∨E m, i–j, k–l Biconditional i A j B k B l A A ↔B ↔I i–j, k–l m A ↔B n A B ↔E m, n m A ↔B n B A ↔E m, n APPENDIX C. QUICK REFERENCE 408 3.5 Derived rules for TFL Disjunctive syllogism m A∨B n ¬A B DS m, n m A∨B n ¬B A DS m, n Modus T ollens m A →B n ¬B ¬A MT m, n Double-negation elimination m ¬¬A A DNE m Excluded middle i A j B k ¬A l B B LEM i–j, k–l De Morgan Rules m ¬(A∨B) ¬A∧¬B DeM m m ¬A∧¬B ¬(A∨B) DeM m m ¬(A∧B) ¬A∨¬B DeM m m ¬A∨¬B ¬(A∧B) DeM m APPENDIX C. QUICK REFERENCE 409 3.6 Basic deduction rules for FOL Universal elimination m ∀xA(. . . x. . . x. . .) A(. . . c. . . c. . .) ∀E m Universal introduction m A(. . . c. . . c. . .) ∀xA(. . . x. . . x. . .) ∀I m c must not occur in any undischarged assumption Existential introduction m A(. . . c. . . c. . .) ∃xA(. . . x. . . c. . .) ∃I m Existential elimination m ∃xA(. . . x. . . x. . .) i A(. . . c. . . c. . .) j B B ∃E m, i–j c must not occur in any undischarged assumption, in ∃xA(. . . x. . . x. . .), or in B Identity introduction c = c =I Identity elimination m a = b n A(. . . a. . . a. . .) A(. . . b. . . a. . .) =E m, n m a = b n A(. . . b. . . b. . .) A(. . . a. . . b. . .) =E m, n APPENDIX C. QUICK REFERENCE 410 3.7 Derived rules for FOL m ∀x¬A ¬∃xA CQ m m ¬∃xA ∀x¬A CQ m m ∃x¬A ¬∀xA CQ m m ¬∀xA ∃x¬A CQ m APPENDIX C. QUICK REFERENCE 411 3.8 Rules for chains of equivalences ¬¬P ⇔P (DN) (P →Q) ⇔(¬P ∨Q) (Cond) ¬(P →Q) ⇔(P ∧¬Q) (P ↔Q) ⇔((P →Q) ∧(Q →P)) (Bicond) ¬(P ∧Q) ⇔(¬P ∨¬Q) (DeM) ¬(P ∨Q) ⇔(¬P ∧¬Q) (P ∨Q) ⇔(Q ∨P) (Comm) (P ∧Q) ⇔(Q ∧P) (P ∧(Q ∨R)) ⇔((P ∧Q) ∨(P ∧R)) (Dist) (P ∨(Q ∧R)) ⇔((P ∨Q) ∧(P ∨R)) (P ∨(Q ∨R)) ⇔((P ∨Q) ∨R) (Assoc) (P ∧(Q ∧R)) ⇔((P ∧Q) ∧R) (P ∨P) ⇔P (Id) (P ∧P) ⇔P (P ∧(P ∨Q)) ⇔P (Abs) (P ∨(P ∧Q)) ⇔P (P ∧(Q ∨¬Q)) ⇔P (Simp) (P ∨(Q ∧¬Q)) ⇔P (P ∨(Q ∨¬Q)) ⇔(Q ∨¬Q) (P ∧(Q ∧¬Q)) ⇔(Q ∧¬Q) Glossary antecedent The sentence on the left side of a conditional. argument A connected series of sentences, divided into premises and conclusion. atomic sentence An expression used to represent a basic sen-tence; a sentence letter in TFL, or a predicate symbol followed by names in FOL. biconditional The symbol ↔, used to represent words and phrases that function like the English phrase “if and only if”; or a sentence formed using this connective. bound variable An occurrence of a variable in a formula which is in the scope of a quantifier followed by the same variable. complete truth table A table that gives all the possible truth values for a sentence (of TFL) or sentences in TFL, with a line for every possible valuation of all sentence letters. completeness A property held by logical systems if and only if ⊨implies ⊢. conclusion The last sentence in an argument. conclusion indicator A word or phrase such as “therefore” used to indicate that what follows is the conclusion of an ar-gument. conditional The symbol →, used to represent words and phrases that function like the English phrase “if ... then ...”; a sentence formed by using this symbol. 412 GLOSSARY 413 conjunct A sentence joined to another by a conjunction. conjunction The symbol ∧, used to represent words and phrases that function like the English word “and”; or a sentence formed using that symbol. conjunctive normal form (DNF) A sentence which is a con-junction of disjunctions of atomic sentences or negated atomic sentences. connective A logical operator in TFL used to combine sentence letters into larger sentences. consequent The sentence on the right side of a conditional. consistency Sentences are jointly consistent iff the contradic-tion ⊥cannot be proved from them. contingent sentence A sentence that is neither a necessary truth nor a necessary falsehood; a sentence that in some case is true and in some other case, false. contradiction (of FOL) A sentence of FOL that is false in every interpretation. contradiction (of TFL) A sentence that has only Fs in the col-umn under the main logical operator of its complete truth table; a sentence that is false on every valuation. disjunct A sentence joined to another by a disjunction. disjunction The connective ∨, used to represent words and phrases that function like the English word “or” in its inclusive sense; or a sentence formed by using this con-nective. disjunctive normal form (DNF) A sentence which is a dis-junction of conjunctions of atomic sentences or negated atomic sentences. domain The collection of objects assumed for a symbolization in FOL, or that gives the range of the quantifiers in an interpretation. empty predicate A predicate that applies to no object in the domain. GLOSSARY 414 equivalence (in FOL) A property held by pairs of sentence of FOLs if and only if the sentences have the same truth value in every interpretation. equivalence (in TFL) A property held by pairs of sentences if and only if the complete truth table for those sentences has identical columns under the two main logical oper-ators, i.e., if the sentences have the same truth value on every valuation. existential quantifier The symbol ∃of FOL used to symbolize existence; ∃x F (x) is true iff at least one member of the domain is F . formula An expression of FOL built according to the inductive rules in §26.2. free variable An occurrence of a variable in a formula which is not a bound variable. functional completeness Property of a collection of connectives which holds iff every possible truth table is the truth ta-ble of a sentence involving only those connectives. inconsistency Sentences are inconsistent iff the contradiction ⊥ can be proved from them. interpretation A specification of a domain together with the ob-jects the names pick out and which objects the predicates are true of. invalid A property of arguments that holds when the conclusion is not a consequence of the premises; the opposite of valid. joint possibility A property possessed by some sentences when they are all true in a single case. main logical operator The operator used last in the construc-tion of a sentence (of TFL) or a formula of FOL. metalanguage The language logicians use to talk about the ob-ject language. In this textbook, the metalanguage is En-GLOSSARY 415 glish, supplemented by certain symbols like metavari-ables and technical terms like “valid”. metavariables A variable in the metalanguage that can represent any sentence in the object language. name A symbol of FOL used to pick out an object of the domain. necessary equivalence A property held by a pair of sentences that, in every case, are either both true or both false. necessary falsehood A sentence that is false in every case. necessary truth A sentence that is true in every case. negation The symbol ¬, used to represent words and phrases that function like the English word “not”. object language A language that is constructed and studied by logicians. In this textbook, the object languages are TFL and FOL. predicate A symbol of FOL used to symbolize a property or relation. premise A sentence in an argument other than the conclusion. premise indicator A word or phrase such as “because” used to indicate that what follows is the premise of an argument. provable equivalence A property held by pairs of statements if and only if there is a derivation which takes you from each one to the other one. satisfiability (in FOL) A property held by sentence of FOLs if and only if some interpretation makes all the sentences true. satisfiability (in TFL) A property held by sentences if and only if the complete truth table for those sentences contains one line on which all the sentences are true, i.e., if some valuation makes all the sentences true. scope The subformula of a sentence (of TFL) or a formula of FOL for which the main logical operator is the operator. sentence (of FOL) A formula of FOL which has no free vari-ables. GLOSSARY 416 sentence (of TFL) A string of symbols in TFL that can be built up according to the inductive rules given on p. 51. sentence letter An letter used to represent a basic sentence in TFL. sound A property of arguments that holds if the argument is valid and has all true premises. soundness A property held by logical systems if and only if ⊢ implies ⊨. substitution instance The result of replacing every free occur-rence of a variable in a formula with a name. symbolization key A list that shows which English sentences are represented by which sentence letters in TFL. tautology A sentence that has only Ts in the column under the main logical operator of its complete truth table; a sen-tence that is true on every valuation. term Either a name or a variable. theorem A sentence that can be proved without any premises. truth value One of the two logical values sentences can have: True and False. truth-functional connective An operator that builds larger sen-tences out of smaller ones and fixes the truth value of the resulting sentence based only on the truth value of the component sentences. universal quantifier The symbol ∀of FOL used to symbolize generality; ∀x F (x) is true iff every member of the do-main is F . valid A property of arguments where there conclusion is a con-sequence of the premises. validity A sentence of FOL that is true in every interpretation. validity of arguments (in FOL) A property held by arguments; an argument is valid if and only if no interpretation makes all premises true and the conclusion false. validity of arguments (in TFL) A property held by arguments if and only if the complete truth table for the argument GLOSSARY 417 contains no rows where the premises are all true and the conclusion false, i.e., if no valuation makes all premises true and the conclusion false. valuation An assignment of truth values to particular sentence letters. variable A symbol of FOL used following quantifiers and as placeholders in atomic formulas; lowercase letters be-tween s and z. In the Introduction to his volume Symbolic Logic, Charles Lutwidge Dodgson advised: “When you come to any passage you don’t un-derstand, read it again: if you still don’t under-stand it, read it again: if you fail, even after three readings, very likely your brain is getting a lit-tle tired. In that case, put the book away, and take to other occupations, and next day, when you come to it fresh, you will very likely find that it is quite easy.” The same might be said for this volume, al-though readers are forgiven if they take a break for snacks after two readings. 418
5870	https://artofproblemsolving.com/wiki/index.php/Exponentiation?srsltid=AfmBOorjRooWayZOIQ79T5rggZ_Qamc2zEeG7570-i4MbWyRX5sp7Acz	Art of Problem Solving Exponentiation - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Exponentiation Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Exponentiation Exponentiation is an arithmetic operation, just like addition, multiplication, etc. It is often written in the form , where is the exponent (or power) and is the base. In the order of operations, it is the second operation performed if a equation has parentheses or the first one performed when there is no parentheses. Contents [hide] 1 Introduction 2 Basic Properties 3 Fractional Exponents 4 Problems 5 See also Introduction To understand how exponents arise, let's first review how we can build multiplication from addition. Let's say we wanted to capture the notion of "the amount equal to 3, ten times." We could write this out as , but this gets burdensome quickly: if we wanted to capture the idea of "the amount equal to two hundred 3s." Thus, we define the multiplication function, usually denoted or , such that where there are 200 threes in the sum. This process (actually an inductive definition) defines the operation of "multiplication by positive integers." We can then extend the notion of multiplication to non-integers. Similarly, the exponentiation is defined as the repetition of multiplication. For example, writing out can get boring fast, so we define the exponential function to express this in a much more compact form so that the preceeding example can be written as (read 3 to the 5th or 3 to the 5 power). What this means is that we are multiplying 3 by itself 5 times. The base is 3 (what is repeatedly multiplied) and the exponent (or power) is 5 (the number of times to repeat multiplication). Our definition of exponentiation makes sense if the exponent is a positive integer. How about negative integers such as ? How do we multiply 2 by itself -4 times!? Let's think about what a negative sign means a little more. When we append a negative sign to a number (say 4, for example), we are basically saying go four units in the opposite direction. We want to do the opposite of multiplication four times. In other words, we want to divide by 2 four times. Therefore, Basic Properties Listed below are some important properties of exponents (with explanations and notes): We are multiplying together times and then times, so in total, we are multiplying together times. if From the property of same base multiplication, , so dividing both sides by results in . is undefined. if From the property of same base multiplication and zero exponent, , so dividing both sides by results in . This results from using the property of same base multiplication and negative exponent property. We are multiplying by itself times, and because of the first property, we are multiplying by itself times. We are multiplying together times, which is the same as multiplying together times and together times. Fractional Exponents So far we discussed expressions with integer exponents. However, it is also possible to extend the exponential function to all non-integers. How could we make sense of an expression like ? If you don't already know the answer, this is a good exercise; I recommend puzzling over it for awhile. Answer: Hoping that property 1 will remain true when or is a fraction, we see that should (hopefully) be equal to . Thus, we define to be , in order to make this be true. For the time being, how to deal with other fractions in the exponent can be an exercise for the reader. Hint: What would be? What about ? Do you notice anything? Try to figure out -- how does it relate to ? Problems Practice Problems on Alcumus General Positive Exponents (Prealgebra) Negative and Zero Exponents (Prealgebra) See also Logarithms Algebra Hyperexponentiation Radical Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
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5872	https://www.wordsclarity.com/dictionary/incongruous	incongruous \| Meaning, Synonyms, Keydifference & Examples... ☰ Home About Contact Words Have Power, Let’s Unlock It! Every word tells a story, sparks an idea, and shapes the way we communicate. Dive into a world where meanings come alive, synonyms open new possibilities, and subtle distinctions make all the difference! Supercharge your vocabulary, sharpen your writing, and fall in love with the art of words. Ready to explore? Let’s begin! Word of the Day Fight A violent confrontation or struggle between people or groups, which can be physical or verbal, and ranges from small-scale personal disputes to large-scale conflicts like wars. Know its synonyms with key difference Find the Right Word with Clarity Search🎤 Voice Search incongruous🔊 Meaning: Not in harmony or keeping with the surroundings or other aspects of something; out of place. Key Difference: While 'incongruous' refers to something that is strikingly out of place or inconsistent, its synonyms may vary in intensity, context, or connotation. Examples: The modern glass skyscraper looked incongruous amidst the ancient stone temples. His cheerful demeanor was incongruous with the somber mood of the funeral. Synonyms of incongruous: Similar words with related meanings. inconsistent🔊 Meaning: Not staying the same throughout; containing parts that do not agree with one another. Key Difference: 'Inconsistent' refers to a lack of uniformity, while 'incongruous' implies a stark mismatch with surroundings. Examples: Her alibi was inconsistent with the witness statements. The team's performance was inconsistent throughout the season. discordant🔊 Meaning: Lacking harmony or agreement; clashing. Key Difference: 'Discordant' often refers to sounds or opinions, while 'incongruous' is broader, applying to visual or situational mismatches. Examples: The discordant notes from the untuned piano made the audience wince. Their discordant views on politics led to frequent arguments. jarring🔊 Meaning: Incongruous in a striking or shocking way. Key Difference: 'Jarring' emphasizes a sudden, unpleasant contrast, whereas 'incongruous' can be more subtle. Examples: The bright pink door was jarring against the otherwise neutral-colored house. His loud laughter was jarring in the silent library. anomalous🔊 Meaning: Deviating from what is standard or expected. Key Difference: 'Anomalous' suggests something is irregular or unusual, while 'incongruous' focuses on being out of place. Examples: The anomalous test results puzzled the scientists. An anomalous snowfall in May disrupted the city's routine. clashing🔊 Meaning: Incompatible or contrasting sharply. Key Difference: 'Clashing' implies active conflict, while 'incongruous' may simply denote a lack of harmony. Examples: The clashing colors of his outfit made it hard to look at. Their clashing personalities made teamwork difficult. mismatched🔊 Meaning: Unsuitably paired or combined. Key Difference: 'Mismatched' refers to poor pairing, while 'incongruous' suggests a broader lack of fit with surroundings. Examples: The mismatched furniture made the room look chaotic. Their mismatched skills made the project challenging. unfitting🔊 Meaning: Not appropriate or suitable. Key Difference: 'Unfitting' is more about suitability, while 'incongruous' emphasizes visual or situational contrast. Examples: His casual attire was unfitting for the formal event. The joke was unfitting for such a serious occasion. discrepant🔊 Meaning: Showing a difference or inconsistency. Key Difference: 'Discrepant' is often used in technical contexts, while 'incongruous' is more general. Examples: The discrepant data points raised questions about the experiment's validity. Their accounts of the event were discrepant. outlandish🔊 Meaning: Looking or sounding bizarre or unfamiliar. Key Difference: 'Outlandish' suggests something is strange or exotic, while 'incongruous' means it doesn't fit the context. Examples: His outlandish costume turned heads at the party. The outlandish proposal was met with skepticism. Conclusion: ✅ 'Incongruous' is best used when describing something that stands out due to a lack of harmony with its surroundings. ✅ 'Inconsistent' can be used when referring to logical or factual discrepancies without hesitation. ✅ For a more professional tone when describing conflicting elements, 'discordant' is a strong choice. ✅ 'Jarring' is best when emphasizing a sudden, unpleasant contrast. ✅ Use 'anomalous' in scientific or technical contexts where deviation from the norm is key. ✅ 'Clashing' works well when describing active conflict, such as in design or personalities. ✅ 'Mismatched' is ideal for describing poorly paired items or traits. ✅ When something is simply unsuitable, 'unfitting' is the right word. ✅ For technical inconsistencies, 'discrepant' is precise and effective. ✅ If something is bizarre or exotic, 'outlandish' captures that meaning perfectly. 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5873	https://www.grc.nasa.gov/www/k-12/VirtualAero/BottleRocket/airplane/termv.html	An object which is falling through the atmosphere is subjected to two external forces. One force is the gravitational force, expressed as the weight of the object. The other force is the air resistance, or drag of the object. The motion of a falling object can be described by Newton's second law of motion (Force equals mass times acceleration -- F = m a) which can be solved for the acceleration of the object in terms of the net external force and the mass of the object. a = F / m The net external force (F) is equal to the difference between the weight and the drag forces (W - D). When drag is equal to weight, there is no net external force on the object and the object will fall at a constant velocity as described by Newton's first law of motion. The constant velocity is called the terminal velocity . We can determine the value of the terminal velocity by doing a little algebra and using the drag equation. Drag (D) depends on a drag coefficient, (Cd) the air density, (r) the square of the air velocity (V) and some reference area (A) of the object. D = Cd r V ^2 A / 2 At terminal velocity, D = W. Solving for the velocity, we obtain the equation V = sqrt ( (2 W) / (Cd r A) ) Typical values of the drag coefficient are given on a separate slide. The terminal velocity equation tells us that an object with a large cross-sectional area or a high drag coefficient will fall slower than an object with a small area or low drag coefficient. (A large flat plate will fall slower than an a small ball with the same weight.) And if we had two objects with the same area and drag coefficient (two identically sized spheres), the lighter object would fall slower. This seems to contradict the findings of Galileo that all free falling objects would fall at the same rate with equal air resistance. But Galileo's principal only applies in a vacuum, where there is NO air resistance and drag is equal to zero. Guided Tours Falling Objects: Back to top Go to... Beginner's Guide Home Page byTom Benson Please send suggestions/corrections to: benson@grc.nasa.gov
5874	https://tasks.illustrativemathematics.org/content-standards/6/EE/A/tasks/931	Illustrative Mathematics Processing math: 0% Engage your students with effective distance learning resources. ACCESS RESOURCES>> Grade 6 Domain Expressions and Equations Cluster Apply and extend previous understandings of arithmetic to algebraic expressions. Task Rectangle Perimeter 3 Rectangle Perimeter 3 No Tags Alignments to Content Standards:6.EE.A Student View Task Sadie computes the perimeter of a rectangle by adding the length, l, and width, w, and doubling this sum. Eric computes the perimeter of a rectangle by doubling the length, l, doubling the width, w, and adding the doubled amounts. Write an expression for Sadie’s way of calculating the perimeter. Write an expression for Eric’s way as well. Use both of the expressions to find the perimeter of a rectangle with length 30 and width 75. Explain why Sadie and Eric always get the same answer, no matter what the length and width of the rectangle are. IM Commentary The purpose of this task is to ask students to write expressions and to consider what it means for two expressions to be equivalent. This is another variation of the Rectangle Perimeter problem shown in "6EE Rectangle Perimeter 1" and "6EE Rectangle Perimeter 2." It goes one step further than "6EE Rectangle Perimeter 1" by asking students to explore the idea of equivalent expressions in addition to writing and using an expression. Solution Sadie first adds l and w to obtain l + w and then doubles the sum. So her expression is 2(l + w). Eric first doubles the length and the width, getting 2l and 2w, respectively. He then adds these amounts to obtain the expression 2l + 2w for the perimeter of the rectangle. With l=30 and w=75 using Sadie’s expression we have 2(l+w)=2(30+75)=210. Using Eric’s expression we have 2l + 2w = 2(30) + 2(75) = 210. Sadie and Eric will always get the same answer, no matter the length and width of the rectangle, because they are both using correct methods for computing the perimeter of a rectangle. Sadie’s expression is equivalent to Eric’s expression by the distributive property: 2(l + w) = 2l + 2w. Rectangle Perimeter 3 Sadie computes the perimeter of a rectangle by adding the length, l, and width, w, and doubling this sum. Eric computes the perimeter of a rectangle by doubling the length, l, doubling the width, w, and adding the doubled amounts. Write an expression for Sadie’s way of calculating the perimeter. Write an expression for Eric’s way as well. Use both of the expressions to find the perimeter of a rectangle with length 30 and width 75. Explain why Sadie and Eric always get the same answer, no matter what the length and width of the rectangle are. Print Task Typeset May 4, 2016 at 18:58:52. Licensed by Illustrative Mathematics under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
5875	https://pubmed.ncbi.nlm.nih.gov/19233353/	American College of Surgeons trauma centre designation and mechanical ventilation outcomes - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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American College of Surgeons trauma centre designation and mechanical ventilation outcomes Joseph J DuBose1,Pedro G R Teixeira,Anthony Shiflett,Christine Trankiem,Bradley Putty,Gustavo Recinos,Kenji Inaba,Howard Belzberg Affiliations Expand Affiliation 1 Los Angeles County Hospital/University of Southern California School of Medicine, Los Angeles, CA 9003-4525, United States. PMID: 19233353 DOI: 10.1016/j.injury.2008.09.015 Item in Clipboard American College of Surgeons trauma centre designation and mechanical ventilation outcomes Joseph J DuBose et al. Injury.2009 Jul. Show details Display options Display options Format Injury Actions Search in PubMed Search in NLM Catalog Add to Search . 2009 Jul;40(7):708-12. doi: 10.1016/j.injury.2008.09.015. Epub 2009 Feb 23. Authors Joseph J DuBose1,Pedro G R Teixeira,Anthony Shiflett,Christine Trankiem,Bradley Putty,Gustavo Recinos,Kenji Inaba,Howard Belzberg Affiliation 1 Los Angeles County Hospital/University of Southern California School of Medicine, Los Angeles, CA 9003-4525, United States. PMID: 19233353 DOI: 10.1016/j.injury.2008.09.015 Item in Clipboard Cite Display options Display options Format Abstract Objective: The association between hospital volume and outcomes following mechanical ventilation has been previously examined in diverse patient populations. The American College of Surgeons (ACS) Committee on Trauma has outlined criteria for trauma centre level designations with specific requirements for both specialty capabilities and hospital volume. Our objective is to determine the relationship between ACS centre designation and outcomes for trauma patients undergoing mechanical ventilation. Methods: We conducted a retrospective cohort study using the National Trauma Databank (NTDB), identifying 13,933 adult (age>or=18) trauma patients receiving mechanical ventilation for greater than 48 h from 2000 to 2004 who were admitted to either an ACS Level I or Level II trauma centre. The primary endpoints examined were mortality, pneumonia and Acute Respiratory Distress Syndrome (ARDS). Univariate analysis defined differences between those patients admitted to ACS Level I and Level II facilities. Logistic regression analysis was used to identify if ACS level designation was an independent risk factor for the goal outcomes. Results: Patients admitted to a Level I facility and mechanically ventilated for greater than 48 h were more commonly greater than age 55 (71.3% vs. 67.9%, p<0.01), hypotensive (SBP<90) (16.1% vs. 12.8%, p<0.01), and likely to have sustained injury due to penetrating mechanism (11.1% vs. 5.1%, p<0.01). On univariate analysis, mortality and the incidence of pneumonia did not differ between the two groups. Level I admission was, however, less commonly associated with the development of ARDS (5.8% vs. 7.7%, p<0.01) and patients admitted to Level I facilities were significantly more likely to be discharged to home than Level II counterparts (29.7% vs. 22.9%, p<0.01). Logistic regression revealed that, while ACS Level designation was not a predictive factor for mortality or the development of pneumonia, admission to an ACS Level II facility was an independent predictor for the development of ARDS [p<0.01, odds ratio, 95% CI: 1.35 (1.18-1.59)]. Conclusion: For trauma patients requiring mechanical ventilation for >48 h, ACS trauma centre designation had no effect on overall mortality or the incidence of pneumonia. Compared to Level I counterparts, however, patients admitted to an ACS Level II facility were significantly more likely to develop ARDS following trauma. This finding needs further investigation in a large, prospective analysis. PubMed Disclaimer Similar articles Relationship between American College of Surgeons trauma center designation and mortality in patients with severe trauma (injury severity score > 15).Demetriades D, Martin M, Salim A, Rhee P, Brown C, Doucet J, Chan L.Demetriades D, et al.J Am Coll Surg. 2006 Feb;202(2):212-5; quiz A45. doi: 10.1016/j.jamcollsurg.2005.09.027. Epub 2005 Dec 19.J Am Coll Surg. 2006.PMID: 16427544 The impact of American College of Surgeons trauma center designation and outcomes after early thoracotomy: a National Trauma Databank analysis.Bukur M, Castelo Branco B, Inaba K, Cestero R, Kobayashi L, Tang A, Demetriades D.Bukur M, et al.Am Surg. 2012 Jan;78(1):36-41.Am Surg. 2012.PMID: 22273303 Effect of trauma center designation on outcome in patients with severe traumatic brain injury.DuBose JJ, Browder T, Inaba K, Teixeira PG, Chan LS, Demetriades D.DuBose JJ, et al.Arch Surg. 2008 Dec;143(12):1213-7; discussion 1217. doi: 10.1001/archsurg.143.12.1213.Arch Surg. 2008.PMID: 19075174 Impact of trauma center designation in pediatric renal trauma: National Trauma Data Bank analysis.Mahran A, Fernstrum A, Swindle M, Mishra K, Bukavina L, Raina R, Narayanamurthy V, Ross J, Woo L.Mahran A, et al.J Pediatr Urol. 2020 Oct;16(5):658.e1-658.e9. doi: 10.1016/j.jpurol.2020.07.019. Epub 2020 Jul 24.J Pediatr Urol. 2020.PMID: 32773248 Review. Care of the older trauma patient following low-energy transfer trauma-highlighting a research void.Tonkins M, Bradbury D, Bramley P, Sabir L, Wilkinson A, Lecky F.Tonkins M, et al.Age Ageing. 2022 Apr 1;51(4):afac074. doi: 10.1093/ageing/afac074.Age Ageing. 2022.PMID: 35380606 See all similar articles Cited by Nationwide cohort study of independent risk factors for acute respiratory distress syndrome after trauma.Tignanelli CJ, Hemmila MR, Rogers MAM, Raghavendran K.Tignanelli CJ, et al.Trauma Surg Acute Care Open. 2019 Feb 15;4(1):e000249. doi: 10.1136/tsaco-2018-000249. eCollection 2019.Trauma Surg Acute Care Open. 2019.PMID: 30899792 Free PMC article. The impact of level of the American College of Surgeons Committee on Trauma verification and state designation status on trauma center outcomes.Elkbuli A, Dowd B, Flores R, Boneva D, McKenney M.Elkbuli A, et al.Medicine (Baltimore). 2019 Jun;98(25):e16133. doi: 10.1097/MD.0000000000016133.Medicine (Baltimore). 2019.PMID: 31232965 Free PMC article. 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5876	https://www.quora.com/Does-evolution-favor-quality-over-quantity-If-not-then-why-do-we-trade-offs	Does evolution favor quality over quantity? If not, then why do we trade-offs? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Does evolution favor quality over quantity? If not, then why do we trade-offs? All related (41) Sort Recommended Floyd A. Associate Professor of Anthropology (2001–present) · Author has 7K answers and 8.1M answer views ·4y There are two “strategies” for reproductive success, known technically as “K-selection” and “r-selection.” (I am referring to these mechanisms as “strategies,” although obviously, that is metaphorical. Most organisms are not thinking about the process at all. They just do what they do, and if it works, great, they keep doing it. If it doesn’t work, the lineage becomes extinct. I am also describing the strategies as two different types, but like most processes in nature, they can be modeled as a continuum, from “extreme” r-selection to “extreme” K-selection, with most species falling somewhere i Continue Reading There are two “strategies” for reproductive success, known technically as “K-selection” and “r-selection.” (I am referring to these mechanisms as “strategies,” although obviously, that is metaphorical. Most organisms are not thinking about the process at all. They just do what they do, and if it works, great, they keep doing it. If it doesn’t work, the lineage becomes extinct. I am also describing the strategies as two different types, but like most processes in nature, they can be modeled as a continuum, from “extreme” r-selection to “extreme” K-selection, with most species falling somewhere in between the extremes.) All organisms have an “energy budget,” by which I mean each organism gains energy from the nutrients that it consumes, and spends energy on actions, including reproduction. No organism has an infinite amount of energy, so each organism can “spend” only as much energy as it consumes. If an organism expends a lot of energy on one activity, it has less energy left over to spend on other activities. The strategy known as “r-selection” is an emphasis on quantity of offspring. A spider, a salmon, or a patch of grass will produce many offspring. Spiders and salmon may lay hundreds of fertilized eggs, and a patch of grass might produce thousands of seeds. However, since each organism has a finite energy budget, the amount of energy spent in producing so many offspring leaves little or nothing left to invest in caring for the offspring and ensuring their survival. Most newly hatched spiders and salmon will get eaten by predators and most grass seeds will fall on unsuitable ground and won’t sprout. In a relatively stable environment, the few offspring that survive will be at just about “replacement” level, so the population will tend to be stable. K-selected organisms emphasize quality over quantity, and produce far fewer offspring, but invest a lot of energy into making sure those few offspring survive. Primates (especially humans) and most other large mammals tend to have few offspring over the course of their lives, but most of those offspring will survive to adulthood, so they will reach or (as in the case of humans over the past few centuries) exceed “replacement” levels and the population size will remain stable (or even grow). From the perspective of the species, neither of these strategies is favored, both work equally well at maintaining population sizes. If two parent organisms have two offspring that both survive, replacement is maintained and the species persists for another generation. If two parental organisms have 100 offspring, 98 of them die and only two survive, replacement levels are maintained, and the species persists for another generation. Although neither strategy is inherently “better” than the other, there are some environmental variables that will tend to favor one strategy over the other. In “wide open,” empty niches, where there are a lot of resources and few competitors, “r-strategists” will tend to do better over the course of several generations than their relatively K-selected conspecifics. For example, if two pregnant spiders are blown to an island with no spiders and lots of flies, and one spider produces 500 offspring while the other produces only 20, that environment will obviously favor the spider that reproduces the most. If there is a 90% “infant mortality” rate, the spider that produced 500 babies will have 50 survivors, while the spider that produced 20 will have only two surviving baby spiders. By contrast, in an environment where the population is already close to carrying capacity (there’s not enough food or other resources for the population to grow any bigger), a pair of humans who have just two kids might actually do better than a pair of humans who have 10 kids. If the amount of food available is enough to feed just two children, the relatively “K-selected” couple have enough food to feed both of their kids, but the relatively “r-selected” couple have to divide the food among all ten of their kids, and each child can get only 1/5 the amount of food it needs to survive, so they end up with zero surviving offspring, despite starting out with more kids. In highly variable environments, when the amount of food (or other vital resources) varies from one breeding season to the next (let’s assume, for simplicity, that a breeding season is one year), the optimal strategy may be much more complicated. An environment that has abundant resources in “good” years, but also has frequent catastrophic shortfalls in “bad” years (for example a drought every two or three years), maintaining a K-selection strategy might be favored even in the “rich” years. A mating pair that has two offspring and passes on the genes for lower fecundity and more parental investment will have two surviving children, followed by two surviving grandchildren, two surviving great-grandchildren, and so on. A mated pair that has ten surviving children in a good year might have 100 surviving grandchildren the following year, but the drought hits the third year and instead of 1000 great-grandchildren, they end up with few or none due to competition for food. Alternatively, in an environment that only rarely (“once in a century drought” or “thousand year storm”) experiences catastrophic shortfalls, the “r-selection” strategy might still out compete the K-strategists, simply by out-breeding them. By the time the one rare catastrophe comes around, the r-strategist descendants are so overwhelming a majority that even a 99% fatality rate will still leave them with more offspring than their K-strategist neighbors. A “simplistic” (“Evolution 101″) understanding of reproduction might at first seem to suggest that the “r-selection” strategy is always going to have more surviving descendants than their “K-selected” neighbors. But in reality, things in nature are always going to be more complicated than our simple models. This is actually one of my favorite topics in evolutionary theory, so if you are interested in discussing it, I would love to talk more about it, but I don’t want to bore you. Hope that helps. Upvote · 9 3 9 1 Sponsored by Complex Law Attention UK Drivers: Check PCP refund eligibility now. If you took out car finance between 2007–2021, use Complex Law to check your refund eligibility. Learn More 999 134 Related questions More answers below Does evolution support the claim that quantity is better than quality? In which cases is quality not preferred to quantity? While many people say quality is better than quantity, is there any examples where quantity is better? On what occasions does quantity over quality become the best outcome? Is it true that quantity is quality in its own way? Paul Lucas Read on evolution, including reading Origin of Species, Evolutionary Biology, et · Author has 11.4K answers and 10.9M answer views ·4y Like everything else concerning natural selection: it depends on the environment. Natural selection (what the question mistakenly called “evolution”, although evolution is more than natural selection) depends on leaving more surviving offspring to the next generation than other individuals of the population. Speaking in very broad generality, there are 2 reproduction strategies: r- and k- K and r reproductive strategies Which one works best for a population depends on the environment the species is in. And remember, “environment” is everything that affects the organism, not just the climate. Most Continue Reading Like everything else concerning natural selection: it depends on the environment. Natural selection (what the question mistakenly called “evolution”, although evolution is more than natural selection) depends on leaving more surviving offspring to the next generation than other individuals of the population. Speaking in very broad generality, there are 2 reproduction strategies: r- and k- K and r reproductive strategies Which one works best for a population depends on the environment the species is in. And remember, “environment” is everything that affects the organism, not just the climate. Most environments do not totally fit either list. Therefore there are trade-offs. Upvote · 9 2 9 1 Ernest W. Adams It's real. Don't try to deny it; you'll only look foolish. · Author has 64.3K answers and 308.1M answer views ·4y Evolution doesn't really favor anything. It's a sieve, not a force. It eliminates unfavorable qualities from the gene pool and leaves the rest. This means anything that causes you to die before you can reproduce (natural selection) and anything that causes the opposite sex to refuse you as a mate (sexual selection). But if you can survive to puberty and find a mate, your genes carry on, no matter what else is going on with you. Upvote · 99 45 Steph G Former Structural Engineer and Project Manager (1980–2010) · Author has 25.6K answers and 14.6M answer views ·4y “A key trade-off that organisms face concerning quantity versus quality in offspring: organisms that tend to invest heavily in offspring also tend to have lower numbers of offspring whereas organisms that invest little in offspring also tend to have higher numbers of lower quality offspring. Organisms continually face trade-offs for how to allocate limited energy and resources. One of the key trade-offs involves the quantity versus the quality of offspring. On the one hand, if organisms invest heavily in their offspring to better their developmental and survival outcomes, they tend to only hav Continue Reading “A key trade-off that organisms face concerning quantity versus quality in offspring: organisms that tend to invest heavily in offspring also tend to have lower numbers of offspring whereas organisms that invest little in offspring also tend to have higher numbers of lower quality offspring. Organisms continually face trade-offs for how to allocate limited energy and resources. One of the key trade-offs involves the quantity versus the quality of offspring. On the one hand, if organisms invest heavily in their offspring to better their developmental and survival outcomes, they tend to only have enough resources to produce a small number of “high-quality” offspring. On the other hand, if organisms make little parental investment per child, they can produce a large number of “low-quality” offspring – although each child has a lower chance of survival, there is a higher probability that at least some offspring will survive long enough to reproduce.” Continue reading at the link below. Quantity Versus Quality of Offspring Upvote · 9 4 Sponsored by Sync.com Sync.com secure cloud storage. Introducing Sync.com. Get secure cloud storage and collaboration for your business. Free signup. Learn More 999 175 Related questions More answers below What are some exceptions to quality over quantity? When is quantity over quality? When is quality over quantity not true? Will we ever see a return to quality over quantity? What is something that isn't better when it's quality over quantity? Nev Anderson Lives in Wellington, New Zealand · Author has 16.5K answers and 2.9M answer views ·4y “Does evolution favor quality over quantity? If not, then why do we trade-offs?” What is favoured is genetic diversity. Evolution is a consequence. The consequence arising from Natural Selection consistently and persistently removing the least adept from a species population, thus shaping future generations, by improving the gene pool. thus improving the gene pool and shaping future generations. Natural Selection favours advantage. Any adaption, or ability that increases survivability improves chances of breeding, and raising offspring. However, being perfectly adapted to One Set of Environmenta Continue Reading “Does evolution favor quality over quantity? If not, then why do we trade-offs?” What is favoured is genetic diversity. Evolution is a consequence. The consequence arising from Natural Selection consistently and persistently removing the least adept from a species population, thus shaping future generations, by improving the gene pool. thus improving the gene pool and shaping future generations. Natural Selection favours advantage. Any adaption, or ability that increases survivability improves chances of breeding, and raising offspring. However, being perfectly adapted to One Set of Environmental conditions, is a recipe for extinction. For conditions change over time, meaning what is perfect today, could become a liability. What is favoured is genetic diversity. Diversity provides a depth within a species population to enable a greater range of challenges to be met. One could compare it to a team, a variety of skills combine to contribute to overall success. That is why there are trade offs. Upvote · Anthony Faiell Studied 6 years in college as well as personal research · Author has 195 answers and 175.1K answer views ·3y A lot of people here gave great biology based answers. And they are not wrong, technically evolution is more a filter for failure than one for success. It merely favors reproduction. Critical thought on the other hand can add a lot to this. Sparta is a great example of a human society that pushed for quality over quantity, and most of us know what happened to them when quantity showed up and ran them over. Going even further back, it's interesting how religion is so rooted in so many cultures. It's because religion was the first force that brought large groups under the same flag. And it clearl Continue Reading A lot of people here gave great biology based answers. And they are not wrong, technically evolution is more a filter for failure than one for success. It merely favors reproduction. Critical thought on the other hand can add a lot to this. Sparta is a great example of a human society that pushed for quality over quantity, and most of us know what happened to them when quantity showed up and ran them over. Going even further back, it's interesting how religion is so rooted in so many cultures. It's because religion was the first force that brought large groups under the same flag. And it clearly shows that these quantities that religion built ensured their survival. In modern times we can look to biology to explain that humans have exponential reproduction. This means that, with medical advances and technology preventing natural mortality, higher quantities will create even higher quantities (hopefully we all understand how exponents work). So yea, people here gave you the cookie cutter biology answer that “evolution does not work towards any direct purpose or goal.” And they aren't completely wrong. Except that quantity ensured survival throughout history, and now we have no quality check to speak of, providing medicine to basically everyone. Meaning quantity is what it's still going to push towards. I could go deeper and start talking about psychological reward systems like welfare that encourage underperforming humans to reproduce, which also enforces quantity over quality. Our democracy is pretty much built as a numbers vs numbers game. So yea idk, I guess you are free to decide for yourself. Upvote · Sponsored by SIOOLTD Casual Men's Solid Color Button-Up Polo Shirt. Enhance your summer wardrobe with this casually stylish, USA-made home shirt. Shop Now 99 70 Zen Faulkes Ph.D. · Author has 11.7K answers and 10.9M answer views ·8y Related Does evolution support the claim that quantity is better than quality? Yes! Also: No! In ecology, a basic concept are that different organisms have different reproductive strategies. They’re called r and K strategies (r/K selection theory - Wikipedia). Some organisms live fast and produce many offspring (quantity), which is the r strategy. So, this strategy is better for those species, as your question hypothesized. Other organisms are long lived and invest a lot in a small number of offspring (quality), which is the K strategy. So, this strategy is better for those species, unlike as your question hypothesized. You see both in the natural world, leading to the concl Continue Reading Yes! Also: No! In ecology, a basic concept are that different organisms have different reproductive strategies. They’re called r and K strategies (r/K selection theory - Wikipedia). Some organisms live fast and produce many offspring (quantity), which is the r strategy. So, this strategy is better for those species, as your question hypothesized. Other organisms are long lived and invest a lot in a small number of offspring (quality), which is the K strategy. So, this strategy is better for those species, unlike as your question hypothesized. You see both in the natural world, leading to the conclusion that both can be successful. Upvote · 9 7 Ken Weiss Many decades of research, writing, and teaching in genetics and evolution · Author has 549 answers and 352.2K answer views ·4y Quality and quantity are subjective terms, so your question can’t really be answered in such terms. Evolution is an impersonal screen of current variation, removing variation that, for whatever reason, doesn’t do well in its current circumstances. The screen is strictly local and present-time, with no way to relate to the future. If in some circumstances or some way quality, or quantity, do better, they will (if also lucky) persist preferentially. If I understand your use of the term, ‘trade-offs’ would only refer to complex aspects of success, and what passes the screening is what we see, not Continue Reading Quality and quantity are subjective terms, so your question can’t really be answered in such terms. Evolution is an impersonal screen of current variation, removing variation that, for whatever reason, doesn’t do well in its current circumstances. The screen is strictly local and present-time, with no way to relate to the future. If in some circumstances or some way quality, or quantity, do better, they will (if also lucky) persist preferentially. If I understand your use of the term, ‘trade-offs’ would only refer to complex aspects of success, and what passes the screening is what we see, not because of trade-offs per se, though there can be circumstances when a species’ survival depends on various things and the most adaptive (and lucky) variation survives and proliferates. Upvote · Parag Jharia M.Sc in Master of Science in Chemistry&Science and Technology in Ancient India, Govt Model Science College Jabalpur (Graduated 2005) ·5y Originally Answered: Why does nature prefer quantity over quality? · Actually nature prefers quality rather than quantity. As per Darwin's survival of the fittest the qualified product is selected to proliferate and increase its population and the low quality products of nature like weak animals low quality seeds die. The quantity you see in the jungles Or anywhere are actually the quality products. They are healthy, they are competitive, they can fight or Run for their lives. The weak ones left Behind to get hunted or to die. Upvote · Democracy-Healthcare-Defence Evolution isn't for good, it's for passing on genes · Author has 1K answers and 773.5K answer views ·8y Related Does evolution support the claim that quantity is better than quality? No. The only thing which evolution does is make those more able to pass on their genes, pass on their genes in greater proportion to those less able. I’m not even sure how to define quality in this context as I know there are many things passing on their genes like crazy which I have no affection for at all. Such as cockroaches and Great White sharks. However, I don’t know how to judge quality betwe Continue Reading No. The only thing which evolution does is make those more able to pass on their genes, pass on their genes in greater proportion to those less able. I’m not even sure how to define quality in this context as I know there are many things passing on their genes like crazy which I have no affection for at all. Such as cockroaches and Great White sharks. However, I don’t know how to judge quality between them. Right now if you mean apex predator then that is not the most numer... Upvote · Geoffrey Walton Single father of Nelson (age 18 and Christian (age 15). · Author has 570 answers and 4.1M answer views ·Updated 10y Related Which is better in life: quality or quantity? My father was 74 when they told him he had liver cancer. Two weeks later they told him it was inoperable; that he had, perhaps, 90 days to live. A few days later he had his first chemotherapy. He said it was terrible, and decided it would also be his last. The doctors told him he was foolish, that chemo would prolong his life. His reply was consistent, "I will live longer, but I will feel terrible. Better is better." Their arguments were persuasive: you'll be in terrible pain, you'll be in and out of the hospital constantly, you'll need extensive medication and may be out of your mind. His repl Continue Reading My father was 74 when they told him he had liver cancer. Two weeks later they told him it was inoperable; that he had, perhaps, 90 days to live. A few days later he had his first chemotherapy. He said it was terrible, and decided it would also be his last. The doctors told him he was foolish, that chemo would prolong his life. His reply was consistent, "I will live longer, but I will feel terrible. Better is better." Their arguments were persuasive: you'll be in terrible pain, you'll be in and out of the hospital constantly, you'll need extensive medication and may be out of your mind. His reply was succinct, "I don't think so." Dad lived 16 months; he made it one month past his 45th wedding anniversary. He went to the hospital for regular checkups, but nothing more. He never took pain meds, nor appeared to need them. He was lucid - and pleasant - until one day he fell asleep in his own bed, holding my mother's hand, and didn't wake up. Better is better. Upvote · 2.2K 2.2K 99 54 9 1 Pierre Vigoureux Author has 35.1K answers and 13.2M answer views ·4y I sort of understand your question. Evolution is natural selection and random mutation. So the question simplifies to a question of quantity and quality in natural selection, which is a very good question. It is both. In some species we have hundreds of turtle eggs, most of which get eaten. In chimpanzee species (and therefore us) it is extended childhoods, 9 month long pregnancies, difficult births with a high proportion of something going wrong, and only a few children, where the tribe is lucky to just reproduce enough to replace dead children not old enough to reproduce. In terms of human beings Continue Reading I sort of understand your question. Evolution is natural selection and random mutation. So the question simplifies to a question of quantity and quality in natural selection, which is a very good question. It is both. In some species we have hundreds of turtle eggs, most of which get eaten. In chimpanzee species (and therefore us) it is extended childhoods, 9 month long pregnancies, difficult births with a high proportion of something going wrong, and only a few children, where the tribe is lucky to just reproduce enough to replace dead children not old enough to reproduce. In terms of human beings and chimpanzees, we are optimised for quality. In terms of ants and sea turtles they are optimised for quantity. Very good first question. Incomprehensible second question. So I won't bother answering that. Upvote · Ann Carlyle Author has 5.4K answers and 6.4M answer views ·8y Related Does evolution support the claim that quantity is better than quality? The word “better” has no definition. Is it “better” to be a nematode or to be a blue whale? Which is “better” — a cholera germ or a college graduate? Do you mean: Which produces the most individuals? Which results in the greater biomass? Which will survive the longest before becoming extinct? Are you comparing two specific species? Or are you comparing one species such as Homo sapiens with an entire Kingdom such as Monera? Let us say that E coli vastly outnumbers H sapiens, and even has the larger biomass. Does that make E coli “better”? Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 9 2 9 2 Related questions Does evolution support the claim that quantity is better than quality? In which cases is quality not preferred to quantity? While many people say quality is better than quantity, is there any examples where quantity is better? On what occasions does quantity over quality become the best outcome? Is it true that quantity is quality in its own way? What are some exceptions to quality over quantity? When is quantity over quality? When is quality over quantity not true? Will we ever see a return to quality over quantity? What is something that isn't better when it's quality over quantity? What is the difference between trade-off and pay-off? Can you explain what it means to say that “Quality has no quantity”? Which are the things where quality matters more than quantity? Why does nature prefer quantity over quality? What are some good examples where quantity matters more than quality? Related questions Does evolution support the claim that quantity is better than quality? In which cases is quality not preferred to quantity? While many people say quality is better than quantity, is there any examples where quantity is better? On what occasions does quantity over quality become the best outcome? Is it true that quantity is quality in its own way? What are some exceptions to quality over quantity? 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5877	https://math.libretexts.org/Courses/Fresno_City_College/Precalculus%3A__Algebra_and_Trigonometry_(Math_4_-_FCC)/07%3A_Systems_of_Equations/7.01%3A_Systems_of_Linear_Equations_-_Two_Variables	7.1: Systems of Linear Equations - Two Variables - Mathematics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 7: Systems of Equations Precalculus: Algebra and Trigonometry (Math 4 - FCC) { } { "7.01:Systems_of_Linear_Equations-Two_Variables" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.02:_Systems_of_Linear_Equations-_Three_Variables" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.03:_Partial_Fractions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.04:_Solve_Systems_of_Equations_Using_Matrices" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.05:_Graphing_Systems_of_Linear_Inequalities" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Solving_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Linear_and_Absolute_Value_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Polynomial_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Rational_and_Radical_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Exponential_and_Logarithmic_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Systems_of_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Sequences_Series_and_the_Binomial_Theorem" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_The_Unit_Circle-_Sine_and_Cosine_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Periodic_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Trigonometric_Identities_and_Equations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Further_Applications_of_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Thu, 10 Apr 2025 22:30:13 GMT 7.1: Systems of Linear Equations - Two Variables 186158 186158 David Yang { } Anonymous Anonymous 2 false false [ "article:topic", "inconsistent equation", "authorname:openstax", "system of linear equations", "independent system", "dependent system", "license:ccby", "showtoc:no", "transcluded:yes", "source-math-1383", "program:openstax", "licenseversion:40", "source@ "source-math-89809" ] [ "article:topic", "inconsistent equation", "authorname:openstax", "system of linear equations", "independent system", "dependent system", "license:ccby", "showtoc:no", "transcluded:yes", "source-math-1383", "program:openstax", "licenseversion:40", "source@ "source-math-89809" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. Home 2. Campus Bookshelves 3. Fresno City College 4. Precalculus: Algebra and Trigonometry (Math 4 - FCC) 5. 7: Systems of Equations 6. 7.1: Systems of Linear Equations - Two Variables Expand/collapse global location 7.1: Systems of Linear Equations - Two Variables Last updated Apr 10, 2025 Save as PDF 7: Systems of Equations 7.2: Systems of Linear Equations - Three Variables Page ID 186158 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. Introduction to Systems of Equations 1. TYPES OF LINEAR SYSTEMS 2. Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution 3. Example 7.1.1: Determining Whether an Ordered Pair Is a Solution to a System of Equations 1. Solution 4. Exercise 7.1.1/07:_Systems_of_Equations/7.01:_Systems_of_Linear_Equations_-_Two_Variables#Exercise_.5C(.5CPageIndex.7B1.7D.5C)) Solving Systems of Equations by Graphing Example 7.1.2: Solving a System of Equations in Two Variables by Graphing Solution Exercise 7.1.2 Q&A Solving Systems of Equations by Substitution How to: Given a system of two equations in two variables, solve using the substitution method. Example 7.1.3: Solving a System of Equations in Two Variables by Substitution Solution Exercise 7.1.3 Q&A Solving Systems of Equations in Two Variables by the Addition Method How to: Given a system of equations, solve using the addition method. Example 7.1.4: Solving a System by the Addition Method Solution Example 7.1.5: Using the Addition Method When Multiplication of One Equation Is Required Solution Exercise 7.1.4 Example 7.1.6: Using the Addition Method When Multiplication of Both Equations Is Required Solution Example 7.1.7: Using the Addition Method in Systems of Equations Containing Fractions Solution Exercise 7.1.5 Identifying Inconsistent Systems of Equations Containing Two Variables Example 7.1.8: Solving an Inconsistent System of Equations Solution Exercise 7.1.6 Expressing the Solution of a System of Dependent Equations Containing Two Variables Example 7.1.9: Finding a Solution to a Dependent System of Linear Equations Solution Exercise 7.1.7 Using Systems of Equations to Investigate Profits Example 7.1.10: Finding the Break-Even Point and the Profit Function Using Substitution 1. Solution Example 7.1.11: Writing and Solving a System of Equations in Two Variables Solution Exercise 7.1.8 Media Key Concepts Learning Objectives Solve systems of equations by graphing. Solve systems of equations by substitution. Solve systems of equations by addition. Identify inconsistent systems of equations containing two variables. Express the solution of a system of dependent equations containing two variables. A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section, we will consider linear equations with two variables to answer these and similar questions. Figure 7.1.1 Introduction to Systems of Equations In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution. In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables. 2⁢x+y=15 3⁢x–y=5 The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4,7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists. 2⁢(4)+(7)=15 True 3⁢(4)−(7)=5 True In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions. Another type of system of linear equations is an inconsistent system, which is one in which the equations represent two parallel lines. The lines have the same slope and different y- intercepts. There are no points common to both lines; hence, there is no solution to the system. TYPES OF LINEAR SYSTEMS There are three types of systems of linear equations in two variables, and three types of solutions. An independent system has exactly one solution pair (x,y). The point where the two lines intersect is the only solution. An inconsistent system has no solution. Notice that the two lines are parallel and will never intersect. A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations. Figure 7.1.2 compares graphical representations of each type of system. Figure 7.1.2 Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution Substitute the ordered pair into each equation in the system. Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution. Example 7.1.1: Determining Whether an Ordered Pair Is a Solution to a System of Equations Determine whether the ordered pair (5,1) is a solution to the given system of equations. x+3⁢y=8 2⁢x−9=y Solution Substitute the ordered pair (5,1) into both equations. (5)+3⁢(1)=8 8=8 True 2⁢(5)−9=(1)1=1 True The ordered pair (5,1) satisfies both equations, so it is the solution to the system. Analysis We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines. See Figure 7.1.3. Figure 7.1.3 Exercise 7.1.1 Determine whether the ordered pair (8,5) is a solution to the following system. 5⁢x−4⁢y=20 2⁢x+1=3⁢y Answer Not a solution. Solving Systems of Equations by Graphing There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes. Example 7.1.2: Solving a System of Equations in Two Variables by Graphing Solve the following system of equations by graphing. Identify the type of system. 2⁢x+y=−8 x−y=−1 Solution Solve the first equation for y. 2⁢x+y=−8 y=−2⁢x−8 Solve the second equation for y. x−y=−1 y=x+1 Graph both equations on the same set of axes as in Figure 7.1.4. Figure 7.1.4 The lines appear to intersect at the point (−3,−2). We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations. 2⁢(−3)+(−2)=−8−8=−8 True(−3)−(−2)=−1−1=−1 True The solution to the system is the ordered pair (−3,−2),so the system is independent. Exercise 7.1.2 Solve the following system of equations by graphing. 2⁢x−5⁢y=−25−4⁢x+5⁢y=35 Answer The solution to the system is the ordered pair (−5,3). Figure 7.1.5 Q&A Can graphing be used if the system is inconsistent or dependent? Yes, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system. Solving Systems of Equations by Substitution Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical. How to: Given a system of two equations in two variables, solve using the substitution method. Solve one of the two equations for one of the variables in terms of the other. Substitute the expression for this variable into the second equation, then solve for the remaining variable. Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair. Check the solution in both equations. Example 7.1.3: Solving a System of Equations in Two Variables by Substitution Solve the following system of equations by substitution. −x+y=−5 2⁢x−5⁢y=1 Solution First, we will solve the first equation for y. −x+y=−5 y=x−5 Now we can substitute the expression x−5 for y in the second equation. 2⁢x−5⁢y=1 2⁢x−5⁢(x−5)=1 2⁢x−5⁢x+25=1−3⁢x=−24 x=8 Now, we substitute x=8 into the first equation and solve for y. −(8)+y=−5 y=3 Our solution is (8,3). Check the solution by substituting (8,3) into both equations. −x+y=−5−(8)+(3)=−5 True 2⁢x−5⁢y=1 2⁢(8)−5⁢(3)=1 True Exercise 7.1.3 Solve the following system of equations by substitution. x=y+3 4=3⁢x−2⁢y Answer (−2,−5) Q&A Can the substitution method be used to solve any linear system in two variables? Yes, but the method works best if one of the equations contains a coefficient of 1 or –1 so that we do not have to deal with fractions. Solving Systems of Equations in Two Variables by the Addition Method A third method of solving systems of linear equations is the addition method. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition. How to: Given a system of equations, solve using the addition method. Write both equations with x- and y-variables on the left side of the equal sign and constants on the right. Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable. Solve the resulting equation for the remaining variable. Substitute that value into one of the original equations and solve for the second variable. Check the solution by substituting the values into the other equation. Example 7.1.4: Solving a System by the Addition Method Solve the given system of equations by addition. x+2⁢y=−1−x+y=3 Solution Both equations are already set equal to a constant. Notice that the coefficient of x in the second equation, –1, is the opposite of the coefficient of x in the first equation, 1. We can add the two equations to eliminate x without needing to multiply by a constant. x+2⁢y=−1−x+y―=3―3⁢y=2 Now that we have eliminated x, we can solve the resulting equation for y. 3⁢y=2 y=2 3 Then, we substitute this value for y into one of the original equations and solve for x. −x+y=3−x+2 3=3−x=3−2 3−x=7 3 x=−7 3 The solution to this system is (−7 3,2 3). Check the solution in the first equation. x+2⁢y=−1(−7 3)+2⁢(2 3)=−7 3+4 3=−3 3−1=−1 True Analysis We gain an important perspective on systems of equations by looking at the graphical representation. See Figure 7.1.6 to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution. Figure 7.1.6 Example 7.1.5: Using the Addition Method When Multiplication of One Equation Is Required Solve the given system of equations by the addition method. 3⁢x+5⁢y=−11 x−2⁢y=11 Solution Adding these equations as presented will not eliminate a variable. However, we see that the first equation has 3⁢x in it and the second equation has x. So if we multiply the second equation by −3,the x-terms will add to zero. x−2⁢y=11−3⁢(x−2⁢y)=−3⁢(11)Multiply both sides by−3.−3⁢x+6⁢y=−33 Use the distributive property. Now, let’s add them. 3⁢x+5⁢y=−11−3⁢x+6⁢y―=−33―11⁢y=−44 y=−4 For the last step, we substitute y=−4 into one of the original equations and solve for x. 3⁢x+5⁢y=−11 3⁢x+5⁢(−4)=−11 3⁢x−20=−11 3⁢x=9 x=3 Our solution is the ordered pair (3,−4). See Figure 7.1.7. Check the solution in the original second equation. x−2⁢y=11(3)−2⁢(−4)=3+8=11 True Figure 7.1.7 Exercise 7.1.4 Solve the system of equations by addition. 2⁢x−7⁢y=2 3⁢x+y=−20 Answer (−6,−2) Example 7.1.6: Using the Addition Method When Multiplication of Both Equations Is Required Solve the given system of equations in two variables by addition. 2⁢x+3⁢y=−16 5⁢x−10⁢y=30 Solution One equation has 2⁢x and the other has 5⁢x. The least common multiple is 10⁢x so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate x by multiplying the first equation by −5 and the second equation by 2. −5⁢(2⁢x+3⁢y)=−5⁢(−16)−10⁢x−15⁢y=80 2⁢(5⁢x−10⁢y)=2⁢(30)10⁢x−20⁢y=60 Then, we add the two equations together. −10⁢x−15⁢y=80 10⁢x−20⁢y―=60―−35⁢y=140 y=−4 Substitute y=−4 into the original first equation. 2⁢x+3⁢(−4)=−16 2⁢x−12=−16 2⁢x=−4 x=−2 The solution is (−2,−4). Check it in the other equation. 5⁢x−10⁢y=30 5⁢(−2)−10⁢(−4)=30−10+40=30 30=30 See Figure 7.1.8. Figure 7.1.8 Example 7.1.7: Using the Addition Method in Systems of Equations Containing Fractions Solve the given system of equations in two variables by addition. x 3+y 6=3 x 2−y 4=1 Solution First clear each equation of fractions by multiplying both sides of the equation by the least common denominator. 6⁢(x 3+y 6)=6⁢(3)2⁢x+y=18 4⁢(x 2−y 4)=4⁢(1)2⁢x−y=4 Now multiply the second equation by −1 so that we can eliminate the x-variable. −1⁢(2⁢x−y)=−1⁢(4)−2⁢x+y=−4 Add the two equations to eliminate the x-variable and solve the resulting equation. 2⁢x+y=18−2⁢x+y=−4 2⁢y=14 y=7 Substitute y=7 into the first equation. 2⁢x+(7)=18 2⁢x=11 x=11 2=7.5 The solution is (11 2,7). Check it in the other equation. x 2−y 4=1 11 2 2−7 4=1 11 4−7 4=1 4 4=1 Exercise 7.1.5 Solve the system of equations by addition. 2⁢x+3⁢y=8 3⁢x+5⁢y=10 Answer (10,−4) Identifying Inconsistent Systems of Equations Containing Two Variables Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different y-intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as 12=0. Example 7.1.8: Solving an Inconsistent System of Equations Solve the following system of equations. x=9−2⁢y x+2⁢y=13 Solution We can approach this problem in two ways. Because one equation is already solved for x, the most obvious step is to use substitution. x+2⁢y=13(9−2⁢y)+2⁢y=13 9+0⁢y=13 9=13 Clearly, this statement is a contradiction because 9≠13. Therefore, the system has no solution. The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows. x=9−2⁢y 2⁢y=−x+9 y=−1 2⁢x+9 2 We then convert the second equation expressed to slope-intercept form. x+2⁢y=13 2⁢y=−x+13 y=−1 2⁢x+13 2 Comparing the equations, we see that they have the same slope but different y-intercepts. Therefore, the lines are parallel and do not intersect. y=−1 2⁢x+9 2 y=−1 2⁢x+13 2 Analysis Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown in Figure 7.1.9. Figure 7.1.9 Exercise 7.1.6 Solve the following system of equations in two variables. 2⁢y−2⁢x=2 2⁢y−2⁢x=6 Answer No solution. It is an inconsistent system. Expressing the Solution of a System of Dependent Equations Containing Two Variables Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as 0=0. Example 7.1.9: Finding a Solution to a Dependent System of Linear Equations Find a solution to the system of equations using the addition method. x+3⁢y=2 3⁢x+9⁢y=6 Solution With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating x. If we multiply both sides of the first equation by −3, then we will be able to eliminate the x-variable. x+3⁢y=2(−3)⁢(x+3⁢y)=(−3)⁢(2)−3⁢x−9⁢y=−6 Now add the equations. −3⁢x−9⁢y=−6+3⁢x+9⁢y―=6―0=0 We can see that there will be an infinite number of solutions that satisfy both equations. Analysis If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let’s look at what happens when we convert the system to slope-intercept form. x+3⁢y=2 3⁢y=−x+2 y=−1 3⁢x+2 3 3⁢x+9⁢y=6 9⁢y=−3⁢x+6 y=−3 9⁢x+6 9 y=−1 3⁢x+2 3 See Figure 7.1.10. Notice the results are the same. The general solution to the system is (x,−1 3⁢x+2 3). Figure 7.1.10 Exercise 7.1.7 Solve the following system of equations in two variables. y−2⁢x=5−3⁢y+6⁢x=−15 Answer The system is dependent so there are infinite solutions of the form (x,2⁢x+5). Using Systems of Equations to Investigate Profits Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation R=x⁢p, where x=quantity and p=price. The revenue function is shown in orange in Figure 7.1.11. The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in Figure 7.1.11. The x-axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars. Figure 7.1.11 The point at which the two lines intersect is called the break-even point. We can see from the graph that if 700 units are produced, the cost is $⁢3,300 and the revenue is also $⁢3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money. The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as P⁡(x)=R⁡(x)−C⁡(x). Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses. Example 7.1.10: Finding the Break-Even Point and the Profit Function Using Substitution Given the cost function C⁡(x)=0.85⁢x+35,000 and the revenue function R⁡(x)=1.55⁢x, find the break-even point and the profit function. Solution Write the system of equations using y to replace function notation. y=0.85⁢x+35,000 y=1.55⁢x Substitute the expression 0.85⁢x+35,000 from the first equation into the second equation and solve for x. 0.85⁢x+35,000=1.55⁢x 35,000=0.7⁢x 50,000=x Then, we substitute x=50,000 into either the cost function or the revenue function. 1.55⁢(50,000)=77,500 The break-even point is (50,000,77,500). The profit function is found using the formula P⁡(x)=R⁡(x)−C⁡(x). P⁡(x)=1.55⁢x−(0.85⁢x+35,000)=0.7⁢x−35,000 The profit function is P⁡(x)=0.7⁢x−35,000. Analysis The cost to produce 50,000 units is $⁢77,500, and the revenue from the sales of 50,000 units is also $⁢77,500. To make a profit, the business must produce and sell more than 50,000 units. See Figure 7.1.12. Figure 7.1.12 We see from the graph in Figure 7.1.13 that the profit function has a negative value until x=50,000, when the graph crosses the x-axis. Then, the graph emerges into positive y-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss. Figure 7.1.13 Example 7.1.11: Writing and Solving a System of Equations in Two Variables The cost of a ticket to the circus is $⁢25.00 for children and $⁢50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $⁢70,000. How many children and how many adults bought tickets? Solution Let c = the number of children and a = the number of adults in attendance. The total number of people is 2,000. We can use this to write an equation for the number of people at the circus that day. c+a=2,000 The revenue from all children can be found by multiplying $⁢25.00 by the number of children, 25⁢c. The revenue from all adults can be found by multiplying $⁢50.00 by the number of adults, 50⁢a. The total revenue is $⁢70,000. We can use this to write an equation for the revenue. 25⁢c+50⁢a=70,000 We now have a system of linear equations in two variables. c+a=2,000 25⁢c+50⁢a=70,000 In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either c or a. We will solve for a. c+a=2,000 a=2,000−c Substitute the expression 2,000−c in the second equation fora a and solve for c. 25⁢c+50⁢(2,000−c)=70,000 25⁢c+100,000−50⁢c=70,000−25⁢c=−30,000 c=1,200 Substitute c=1,200 into the first equation to solve for a. 1,200+a=2,000 a=800 We find that 1,200 children and 800 adults bought tickets to the circus that day. Exercise 7.1.8 Meal tickets at the circus cost $⁢4.00 for children and $⁢12.00 for adults. If 1,650 meal tickets were bought for a total of, $⁢14,200, how many children and how many adults bought meal tickets? Answer 700 children, 950 adults Media Access these online resources for additional instruction and practice with systems of linear equations. Solving Systems of Equations Using Substitution Solving Systems of Equations Using Elimination Applications of Systems of Equations Key Concepts A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously. The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. See Example 7.1.1. Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution. One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes. See Example 7.1.2. Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation. See Example 7.1.3. A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See Example 7.1.4. It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together. See Example 7.1.5, Example 7.1.6, and Example 7.1.7. Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect. See Example 7.1.8. The solution to a system of dependent equations will always be true because both equations describe the same line. See Example 7.1.9. Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit. See Example 7.1.10 and Example 7.1.11. This page titled 7.1: Systems of Linear Equations - Two Variables is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. Back to top 7: Systems of Equations 7.2: Systems of Linear Equations - Three Variables Was this article helpful? 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5878	https://brilliant.org/wiki/complex-conjugates-problem-solving-easy/	Complex Conjugates Sign up with Facebook or Sign up manually Already have an account? Log in here. Beakal Tiliksew, Gene Keun Chung, Mira B, and Mahindra Jain Arron Kau Beakal Tiliksew Bradan Litzinger Jimin Khim contributed Contents Complex Conjugates Complex Conjugates - Arithmetic Complex Conjugates - Polynomials Complex Conjugates - Quadratics Complex Conjugates Problem Solving - Intermediate Complex Conjugates Problem Solving - Advanced Complex Conjugates Find the sum of real values of x and y for which the following equation is satisfied: 3+i(1+i)x−2i+3−i(2−3i)y+i=i. This problem is part of a set. The correct answer is: 2.00 Complex Conjugates - Arithmetic Given a complex number z=a+bi(a,b∈R), the complex conjugate of z, denoted z, is the complex number z=a−bi. The complex conjugate has the same real component a, but has opposite sign for the imaginary component b. Properties of complex conjugates: z+w=z+w z−w=z−w z⋅w=z⋅w (wz)=wz z=z, i.e. the conjugate of the conjugate of z is z. z=z, if and only if z is real. ∣z∣2=z⋅z=z⋅z, where ∣z∣ stands for the modulus of z. zn=(z)n Operations on z and z: a) z+z=(a+ib)+(a−ib)=2a (∈R) b) z−z=(a+ib)−(a−ib)=2bi (pure imaginary number) c) z⋅z=(a+ib)(a−ib)=a2−abi+abi−b2i2=a2+b2 (i2=(−1)2=−1) Based on these operations, we can add some more properties of conjugate: z+z=2Re(z), twice the real element of z. z−z=2Im(z), twice the imaginary element of z. Why conjugate? The need of conjugation comes from the fact that i2=−1. This means that the equation has two roots, namely i and −i. What this tells us is that from the standpoint of real numbers, both are indistinguishable. For example, for a polynomial f(x) with real coefficient, f(z=a+bi)=0 could be a solution if and only if its conjugate is also a solution f(z=a−bi)=0. This means they are basically the same in the real numbers frame. One importance of conjugation comes from the fact the product of a complex number with its conjugate, is a real number!! (See the operation c) above.) This can come in handy when simplifying complex expressions. It is like rationalizing a rational expression. Let's look at an example to see what we mean. Perform the necessary operations to put 5+2i4+3i to a+bi(a,b∈R) form. Multiply both the numerator and denominator with the conjugate of the denominator, in a way similar to when rationalizing an expression: 5+2i4+3i⇒a=5+2i4+3i⋅5−2i5−2i=52+22(4+3i)(5−2i)=2920−8i+15i−6i2=2926+297i=2926,b=297. □ Let's look at more examples to strengthen our understanding. Perform the necessary operation to put(4+5i2−3i)(1−3i4−i) to a+bi(a,b∈R) form. Performing the necessary operations, and using the properties of complex numbers and their conjugates, we have (4+5i2−3i)(1−3i4−i)=(4+5i2−3i)⋅(1−3i4−i)=4+5i2−3i⋅1−3i4−i=4−5i2+3i.1+3i4+i=19+7i5+14i. Use the rationalizing factor 19−7i to simplify: 19+7i5+14i⋅19−7i19−7i=410193−410231i. □ Perform the necessary operation to put1−5xi−3x+3+i3i to a+bi(a,b∈R) form. Rationalizing each term and summing up common terms, we have 1−5xi−3x+3+i3i=1−5xi−3x⋅1+5xi1+5xi+3+i3i⋅3−i3−i=(1+25x2−3x−15x2i)+(109i+3)=(1+25x2−3x−1+25x215x2i)+(109i+103)=(1+25x2−3x+103)+(1+25x2−15x2i+109i)=10+250x2−30x+3+75x2+(10+250x2−150x2+9+225x2)i. □ Complex Conjugates - Polynomials Given that x=5−i is a root of f(x)=x3−8x2+6x+52, factor f(x) completely. By the complex conjugate root theorem, we know that x=5+i is also a root of f(x). Hence, (x−(5−i))(x−(5+i))=((x−5)+i)((x−5)−i)=x2−10x+26 is a real factor of f(x). We can divide f(x) by this factor to obtain x2−10x+26x3−8x2+6x+52=x+2. Thus, f(x) can be factored as f(x)=(x−5+i)(x−5−i)(x+2). □ Find the cubic polynomial that has roots 5 and 3+i. According to the complex conjugate root theorem, 3−i which is the conjugate of 3+i is also a root of the polynomial. Hence, let f(x) be the cubic polynomial with roots 3+i, 3−i, and 5, then f(x)=(x−5)(x−(3+i))(x−(3−i))=(x−5)((x−3)−i)((x−3)+i)=(x−5)(x2−6x+9−i2)=(x−5)(x2−6x+10)=x3−11x2+40x−50. □ Complex Conjugates - Quadratics If p and q are real numbers and 2+3i is a root of x2+px+q=0, what are the values of p and q? Since the coefficients of the quadratic equation are all real numbers, 2−3i which is the conjugate of 2+3i is also a root of the quadratic equation. Thus, by Vieta's formular −p=(2+3i)+(2−3i),q=(2+3i)(2−3i). Therefore, p=−4 and q=7. □ Prove that if a+bi (b=0) is a root of x2+px+q=0 and a,b,p,q∈R, then a−bi is also a root of the quadratic equation. Since a+bi is a root of the quadratic equation, it must be true that (a+bi)2+p(a+bi)+q=0. If we rewrite above equation, we obtain (a2−b2+pa+q)+(2ab+pb)i=0. Since a,b,p,q∈R, we have a2−b2+pa+q=0,2ab+pb=0.(1) Now, if we substitute a−bi into x2+px+q, then we obtain (a−bi)2+p(a−bi)+q=(a2−b2+pa+q)−(2ab+pb)i, which is zero by (1). Therefore, it must be true that a−bi is also a root of the quadratic equation. □ Complex Conjugates Problem Solving - Intermediate Can the two complex numbers sinx+icos2x and cosx−isin2x be the conjugates of each other? If so, what is the possible real value for x? Let cosx−isin2x be the conjugate of sinx+icos2x, then we have sinx+icos2x⇒sinx−icos2x=cosx−isin2x=cosx−isin2x, which means sinx=cosx and cos2x=sin2x since the values of sine or cosine functions are real numbers. So we can rewrite above equations as follows: tanx=1 and tan2x=1. Observe that these two equations cannot hold simultaneously, then the two complex numbers in the problem cannot be the conjugates of each other for any real value x. □ If α2=3−4i, what is the value of αα? Since α2=3−4i, we have (α)2=α2=3+4i. Hence, (αα)2⇒αα=α2(α)2=(3−4i)(3+4i)=25=±5. Observe that if α=p+qi (p,q∈R) and α=p−qi, then αα=p2+q2≥0. Then αα=5. □ Find the sum of real values of x and y for which the following equation is satisfied: 3+i(1+i)x−2i+3−i(2−3i)y+i=i. This problem is part of a set. The correct answer is: 2.00 Complex Conjugates Problem Solving - Advanced For a non-real complex number α, if α+α1 is a real number, what is the value of αα? Since α+α1 is a real number, we have α+α1=(α+α1)=α+α1. Therefore, we obtain (α−α)+(α1−α1)(α−α)(1−αα1)=0=0. Since α is a non-real number, α=α. Thus, 1−αα1=0, which implies αα=1. □ If a complex number z=a+bi (a>0,b>0) satisfies z2+z=0, how many positive integers n less than 100 are there that make zn an integer? Since z2+z=0, we have z2+z=(a+bi)2+(a−bi)=(a2−b2+a)+(2ab−b)i=0. Hence, a2−b2+a2ab−b⇒b(2a−1)=0(1)=0=0.(2) Since b>0, we obtain a=21 from (2), and by substituting this into (1) we have b2=43 or b=23 since b>0. Therefore, z=21+3i. Now, observe that z2z3z4z5z6=2−1+3i=zz2=21+3i⋅2−1+3i=−1=zz3=21+3i⋅(−1)=2−1−3i=z2z3=2−1+3i⋅(−1)=21−3i=(z3)2=1⋮, then n must be a multiple of 3 to make zn an integer. Thus, there are 33 positive integers less then 100 that make zn an integer. □ Cite as: Complex Conjugates. Brilliant.org. Retrieved 02:57, September 28, 2025, from
5879	https://dp.la/item/b1d4d8a524a24e8892ba041ff84f6e8c	Guiding principles for stabilization and reconstruction \| DPLA Skip to Main Content Digital Public Library of AmericaShow Menu Home Browse by Topic Browse by Partner Exhibitions Primary Source Sets My Lists About DPLA News DPLA Pro Home Browse by Topic Browse by Partner Exhibitions Primary Source Sets My Lists About DPLA News DPLA Pro Search United States Government Publishing Office (GPO) United States Government Publishing Office (GPO) Guiding principles for stabilization and reconstruction Please read DPLA's Statement on Potentially Harmful Content. Guiding principles for stabilization and reconstruction Image View Full Item Created Date 2009 Description Includes bibliographical references (pages 205-231). Description based on online resource, PDF version; title from title page (USIP, viewed November 20, 2017). PartnerUnited States Government Publishing Office (GPO) Contributing InstitutionUnited States Government Publishing Office (GPO) Publisher Washington, D.C. : United States Institute of Peace ; Carlisle, PA : U.S. Army Peacekeeping and Stability Operations Institute SubjectsPeace-building, American--Handbooks, manuals, etc Postwar reconstruction--Handbooks, manuals, etc Nation-building--Handbooks, manuals, etc Nation-building Peace-building, American Postwar reconstruction Typetext Format Language material Electronic resource Computer Online resource URL Rights Pursuant to Title 17 Section 105 of the United States Code, this file is not subject to copyright protection and is in the public domain. For more information please see Cite this item Chicago citation style Guiding principles for stabilization and reconstruction. 2009. Retrieved from the Digital Public Library of America, (Accessed September 28, 2025.)APA citation style(2009) Guiding principles for stabilization and reconstruction. Retrieved from the Digital Public Library of America, MLA citation style Retrieved from the Digital Public Library of America Note: These citations are programmatically generated and may be incomplete. Add to a new list DPLA Frequently Asked Questions How Can I Use DPLA? Terms & Conditions Harmful Content About DPLA Contact Us Feedback News Tools Primary Source Sets Browse by Partner Browse by Topic Exhibitions My Lists Search DPLA Pro DPLA Pro Home Prospective Hubs Hub Network Developers Education Projects Ebooks Events Donate
5880	https://dictionary.law.com/Default.aspx?selected=2189	Menu Copyright Â© 2019 ALM Media Properties, LLC. All Rights Reserved. Search Legal Terms and Definitions \| \| \| \| --- \| Browse: A B C D E F G H I J K L M N O P Q R S T U V W Y Z \| \| \| \| \| \| \| \| Enter a Legal Term all wordsany wordsphrase \| \| Search the Definitions all wordsany wordsphrase \| all wordsany wordsphrase all wordsany wordsphrase \| \| \| \| \| \| --- --- \| \| \| \| \| \| --- \| \| \| \| --- \| \| undivided interest n. title to real property held by two or more persons without specifying the interests of each party by percentage or description of a portion of the real estate. Such interests are typical between joint tenants, tenants in common and tenants by the entirety. \| \| See also: joint tenancy real property tenancy by the entirety tenancy in common \| \| \| \| \| \| --- \| \| \| \| --- \| \| undivided interest n. title to real property held by two or more persons without specifying the interests of each party by percentage or description of a portion of the real estate. Such interests are typical between joint tenants, tenants in common and tenants by the entirety. \| \| See also: joint tenancy real property tenancy by the entirety tenancy in common \| \| \| \| --- \| \| undivided interest n. title to real property held by two or more persons without specifying the interests of each party by percentage or description of a portion of the real estate. Such interests are typical between joint tenants, tenants in common and tenants by the entirety. \| \| n. title to real property held by two or more persons without specifying the interests of each party by percentage or description of a portion of the real estate. Such interests are typical between joint tenants, tenants in common and tenants by the entirety. \| \| \| The People's Law Dictionary by Gerald and Kathleen Hill Publisher Fine Communications \| Publications Law Topics Rankings More Law.com Copyright © 2025 ALM Media Properties, LLC. All Rights Reserved.
5881	https://knowt.com/note/bd81c300-545d-4bac-b9be-ba3d7b367835/aphg-unit-5	aphg unit 5 Notes \| AP Human Geography \| Knowt Note 0.0(0)Rate it Practice Test Flashcards Knowt Play Call with Kai AP Human Geography Study GuidesAP Human Geography Ultimate GuideUnit 1: Thinking GeographicallyUnit 2: Population and Migration Patterns and ProcessesUnit 3: Cultural Patterns and ProcessesUnit 4: Political Patterns and ProcessesUnit 5: Agriculture and Rural Land-Use Patterns and ProcessesUnit 6: Cities and Urban Land-Use Patterns and ProcessesUnit 7: Industrial and Economic Development Patterns and Processes Top Exams AP English Language and CompositionAP BiologyAP United States History Studying for another AP Exam? Check out our other AP study guides HomeAP Human GeographyUnit 5: Agriculture and Rural Land-Use Patterns and Processes 240d ago aphg unit 5 Unit 5Introduction-Chapter 11- origins, patterns, and settlements of agriculture5.1- intro to agriculture5.2- settlement patterns and survey methods5.3- agricultural origins and diffusionsChapter 12- the second and third agricultural revolutions5.4- the second agricultural revolution5.5- the green revolutionChapter 13- spatial arrangement of agriculture5.6- agricultural production regions5.7- spatial organization of agriculture5.8- von thunen model5.9- the global system of agricultureChapter 14- challenges and consequences of agriculture5.10- consequences of agricultural practices5.11- challenges of contemporary agriculture5.12- women in agriculture everything is straight from the aphg textbook Unit 5 Agricultural and Rural Land-Use Patterns and Processes Introduction- Change in agriculture over decades: Smaller family farms More large corporate farms More machinery More chemicals Expanded farmland from things that used to be forests, wetlands, ect. GMOs Global supply chains (can go farther distances, faster, ect.) Improvements in transportation + infrastructure Growth of agribusiness (farms as corporations) Globalization of agriculture Complex system that connects producers and consumers across the world (globally) due to advances in biotech mechanization transportation food preservation Distance between producers and consumers can be farther bc of Improvements in agricultural technology advances in transportation Increased globalized economy Chapter 11- origins, patterns, and settlements of agriculture 5.1- intro to agriculture See flashcards 5.2- settlement patterns and survey methods Rural Survey Methods: (defining boundaries of land ownership) Metes and bounds (in the past) Utilization of landmarks and physical features to establish boundary lines Short distances Results in irregular shaped plots of land (people would just go mark their own land “from that mountain to the river is mine”) Not practical anymore, used to be in Great Britain and North America- now, rarely used (sometimes in the Middle East) (French) Long Lot (still used along rivers) Long strips of land that start at a river or lake with the intention of providing all landowners with equal access to the resources (soil and water) and transportation. (France, Spain, North America, ect.) In North America, Quebec and Louisiana Township and range (aka Public Land Survey System) (currently used) Made by thomas Jefferson Started in 1785 after U.S. switched to surveying rather than landscape features Rectangles and grid system townships- area 6mi x 6mi Keeps track of land sales and purchases Utilizes a uniform survey method Section- each square mile- consisted of 640 acres Rural Settlement Patterns:(shapes) Clustered- (aka nucleated) Farmland and pastures surround settlements and towns Share resources and community Have to walk to farmland survey method: Metes & Bounds (have to agree between themselves, could lead to fights) Throughout European history, rural residents lived in groups of homes in close proximity to one another. Dispersed- Used right now in North America Settlements are isolated and dispersed over the land area U.S. Government promoted Westward Expansion by giving farmers land (usually 160 acres) if they agreed to live and farm on it. Survey method: township & range ~Encourages self-sufficiency but makes shared things (like schools) hard~ Linear- Settlement organized along a line- transportation system, or physical feature like a river/coast (today, more demand for along a transportation route) Survey method: long lots 5.3- agricultural origins and diffusions First (neolithic/stone age) Agricultural Revolution- origin of farming (marked by domestication ꜜ of plants and animals- mostly was subsistence farming at the time) Domestication- the deliberate effort to grow plants and raise animals, making plants and animals adapt to human demands Animal domestication First was probably hunters in Central Asia- dogs and horses Then SW Asia- goats, pigs, sheep, cattle Then cats, horses, camels, donkeys, llamas, ect. Plant domestication Probably after animal domestication First was vegetative planting (using parts of existing plants to grow more) Then seeds Hearth- the geographic origin of a trait, characteristic, innovation, or other concept. It is where something is “born” Agricultural hearth- the separate location in which groups of people began to domesticate plants and animals agricultural hearths Southwest asia (fertile crescent in middle east) - first hearth (Middle east) east coast of mediterranean sea, along tigris and euphrates, to persian gulf. Southeast asia (includes islands like philippines, malaysia, thailand, indonesia) South asia (indus valley (north of india)- also called fertile crescent) East asia (incl. China, like wei-huang river valley) Sub-saharan africa (west africa + nile river valley (E Africa)) Mesoamerica [central america (guatemala, el salvador, nicaragua) + andean highlands (in ecuador, peru, chile)] more hearths around rivers: Chang Jiang (Yangtze) and (Wei) Huang He (yellow) river valleys in East Asia/China Nile valley in NE Africa Ganges valley in South Asia first communities made were usually along rivers- source of water + drinking, transportation, defense, source of food (fish) (flooding of rivers) also contributed to soil fertility, making agriculture more productive- people could focus on other tasks like building, military, ect. advancing all areas of human life. Commonalities of agricultural hearths: Fertile soil in river valleys availability of water Moderate climate Skilled residence Independent inventions- when a trait has many cultural hearths; crops and animals are domesticated in multiple regions but there was no interaction between the regions. (ex: wheat was domesticated in SW, E, and S Asia, but all did it independently without interacting with each other, they each had the idea themselves.) (civilization started around the agricultural hearts- they became the first urban centers, turned into civilizations, societies and powerful states. increased trade, large empires, and increased other specialized jobs. Diffusion of agriculture Contagious diffusion Agriculture first diffused to the immediate surrounding areas of the hearths through close contact and proximity between farmers Ex: agriculture and animal domestication originated in Fertile Crescent then diffused through Turkey into Eastern Europe Immigration and migration (relocation diffusion) As people migrate, they bring food, ingredients, seeds, and animals with them. Stimulus diffusion: as people moved into new locations, seeds may not be as prosperous due to different physical conditions → resulting in the adaptation of farming methods/crops, using the agricultural knowledge the farmer had previously acquired (the idea had diffused and stayed but adapted) Trade routes Ex: the silk road: over 4,000 miles of trade routes that connected China with Europe with a variety of different people interacting. Traded silk, rice, and more. → resulted in the spread of plants and animals throughout East Asia, South Asia, Middle East, Africa, and Europe Columbian exchange- (after columbus in 1492) Global movements of plants and animals between Afro-Eurasia and the Americas. (mainly caribbean) Examples: Coffee from east africa and bananas and sugar cane from New Guinea thrive in America today. Potatoes from NW South America and maize from Mexico are grown in Europe, Asia, and Africa. Tobacco and sugar are now new in Europe Also with animals and diseases Turkey- found in Americas, taken to Europe Smallpox, influenza, malaria, and measles came from the eastern hemisphere to western hemisphere. (after that, contagious diffusion makes things further spread in the new places) Chapter 12- the second and third agricultural revolutions 5.4- the second agricultural revolution Result: increased food supply (especially to countries part of global trade networks) and more people got access to a greater variety of food → higher life expectancy Fewer but larger, more productive farms, and less farmers/farm laborers. → More people in urban areas than rural (for first time in U.S. history) Began in 1700s “Used advances of the Industrial Revolution to increase food supplies and support population growth” Involved mechanization of agricultural production, advances in transportation (ex: could transport food over long distances, so the farming could be far away from markets), and more. New innovations increased yields, especially in Europe and the U.S. Examples: iron/steel plow Mechanized seed drilling mccormick reaper/harvester Grain elevator barbed wire Mixed nitrogen and nitric acid fertilizer New techniques also increased yields Crop rotation- planning different crops (in a specific sequence/order/timeline) on the same plot of land, to restore nutrients back into the soil. (ex: grains extract nitrogen from the soil, then alfalfa adds nitrogen back to the soil.) Irrigation- applying controlled amounts of water to crops using human-made devices (canals, pipes, sprinkler systems) instead of just rainfall. Enclosure acts- Laws made by the British government allowing people to purchase and enclose land for their own use. -Divided up common land that was shared by farmers into individual plots. result: allowed for larger farms, more efficient production, and commercial farming. But, many farmers were forced off their land and lost their traditional way of life- also less need for farm laborers, so people looked for jobs in the industrial/manufacturing sector, like factories, instead. Most of these factory jobs were in urban areas → created big migration of people from rural to urban areas 5.5- the green revolution Mid 20th century (main part of third agricultural revolution) Advances in plant biology Dr. Norman Borlaugh- father of green revolution Developed higher yielding, more disease resistant, and faster growing crops- specifically worked on wheat in Mexico. Concerns- increased use of chemical fertilizer & pesticides. Hybridization- breeding two plants to create one seed with both plant’s characteristics. (ex: long grain rice from Indonesia + dense grain short rice from Taiwan = longer and denser rice grain introduced in rest of Asia.) Green revolution also introduced more machinery into developing countries (tractors, tillers, broadcast seeders, grain carts, ect.) GMOs (genetically modified organisms)- humans changing DNA of a seed through engineering -to increase yields, resist diseases, and withstand chemicals used to kill pests/weeds. Positive impacts- Higher yields(on the same amount of land) Countries like India and Mexico moved from importers to exporters (for wheat) Reduced hunger, lowered death rates, and grew populations Money for research and business Lower food prices Negative impacts- Environmental damages pollution due to increased use of pesticides & fertilizers → runoff in water led to polluted water, hurt ecosystems, health issues soil erosion (b/c of intensive use of land) fossil fuels from machinery increased pollution Impact on gender roles Traditionally, much of the farming labor was done by women, but men dominated socially and when the green revolution technologies came, the men were educated on these new methods and benefited, while the women were excluded. (continued on 5.12) Economic changes Machinery, seeds, fertilizers and pesticides cost increased Farming tech heavily relied on investments (mostly from private corporations), but the corporations began to stop investing. Because of the migration from rural to urban, and these core wealthy areas got advantages over the poor outer areas. Struggle in Africa Africa did not benefit from the Green Revolution for many reasons: Diverse climate and soil Harsh environmental conditions Lack of well developed infrastructure → cost of investment in research, development, and transportation was too high. Africa's staple crops (sorghum, millet, cassava, yams, cowpeas, peanuts) were not target research audiences for hybridization. Africa grew the most during the Green Revolution, but hunger remained a problem since the Green Revolution was not effective there. Chapter 13- spatial arrangement of agriculture 5.6- agricultural production regions examples of intensive farming Paddy rice farming in S, SE, and E Asia is labor intensive.(terraced fields makes using machinery difficult) Truck farming (Cali, Teas, Florida, and near large cities) is capital intensive b/c uses expensive machinery and more. Also labor intensive because use low-pay migrant workers to harvest (large scale farms) Factory farming- capital intensive raising of livestock where animals are in close quarters and bred and fed in a controlled environment. (animals move like a conveyor belt) Aquaculture (aquafarming)- intensive farming. Fish,shellfish and water plants are raised in netted areas in the sea, tanks, or more. extensive examples include shifting cultivation, pastoral nomadism/nomadic herding, ranching Lots of extensive farming has turned into intensive bc of increased demand for food (higher population) and global competition. (now, people use land continuously without fallow period, and use irrigation, chemical fertilizers, GMOs, ect.) Double cropping- planting + harvesting a crop two (or three) times a year on the same field. intercropping/multicropping- two or more crops grow simultaneously on the same field. monocropping/monoculture- (the opposite) - only growing one type of crop at a time on a piece of land. (like plantations ) Examples: World dependent on India for Rice Russia- wheat Sri- lanka- tea (In the past) Iraq- dates U.S.- corn Iran- saffron Pros: Maximized efficiency Lower per-unit cost of production Higher yield More profit Cons: Soil depletion Increased reliance on chemicals Increased risk since all of investments are in and depending on this one crop. Increased demand for meat (50% higher between 1998 and 2018) results in feedlots- confined spaces where cattle have limited movement. (aka CAFOs- concentrated animal feeding operations) -maximizes space, makes cattle bigger bc of limited movement → maximizes profit. Some things combine intensive + extensive, ex: raising cattle in wyoming- roam and feed on grass for months which is extensive, then intensive phase where they are transported to fatten quickly before processed for meat. 5.7- spatial organization of agriculture Agribusiness- farms run as corporations - Large- Scale Production. Various steps: Research and development Processing and production Transportation Marketing Retail (sale) of agricultural goods (replacing small family farms) More cost effective than family owned, cheaper for us. Higher yields than family-owned, so still make more money. Owned by: Transnational corporations- operate in many countries Commercial (intention is making a profit) Use lots of machinery Use lots of biotech (GMOs) and chemicals Home scale vs agribusiness scale examples: Growing crops in a backyard garden vsowning thousands of acres worked by many employees selling at local market vs selling to wholesale distributors who ship it regional + globally Growing diff types to see which grows best vs investing thousands in research and development of new seeds, fertilizers, and pesticides Vertical integration- Large corporate farms own other businesses that are involved in the steps of producing a particular product. This gives the large farm more control + greater profits. Ex: they might own a research and development company that improves seeds, a transporting company that transports farm products, a processing factory, and a wholesaler that distributes the food to stores. Ex: Tyson Foods owns the farms where they raise the animals, then own the processing plants where they slaughter, process, and package the meat products. Economies of scale- increase in efficiency to lower per-unit production cost. (usually occurs with larger farming operations, but could happen at any scale) Ex: someone buys more land, but not more machinery, just uses the already existing machinery more efficiently- this will increase the earning without increasing the expenses too much, and the cost per unit will decrease, and profit increase. Another example, at a smaller farm: increasing level of fertilization and irrigation- increases expenses by 10%, but yields increase by 25%, so cost per unit decreases, and profits increase. Disappearance of small family farms because they can’t compete with the cheaper cots of the larger farms. They also cannot afford the newest tech but need to get it to compete in the global market. Commodity chain- process by corporations to gather resources, turn them into goods, and transport to consumers. template commodity chain: (simplified) planting (using seeds, fertilizers, water) → growing (fertile land and moisture) → harvesting (dried and stored until processing) → processing (packaged or used to create other products) → marketing (sold for food, animals, ect.) Additionally includes financial institutions (banks), transportation companies, distributors, and governments. Carrying capacity- total number of people farmers can support given the available resources. - risen a lot over the years. ex: in 1962, avg. U.S. farmers fed ~26 people. Today, it’s 166 people! Cool chains- transportation networks that keep food cool throughout the trip. (now perishables can travel long distances for a low cost) 5.8- von thunen model Location theory- why people choose certain locations for economic activity (factories, stores, restaurants, or agriculture) Johann Von thunen- North German farmer and economist Published “an isolated state” in 1828 Von thunen model- economic model that gives a pattern for the types of products farmers would produce at different positions relative to the market. farmers make this decision based on: transportation cost land cost Intensity of land use Perishability of the product His assumptions: farming = economic activity Farmers in business to make profit 1 market where farmers sold products 1 transportation system farmers paid for transportation, which was different with distance market was in isotropic plain- flat with similar fertility + climate throughout area beyond market and farmland was wilderness (center = market) First circle= horticulture (dairy farming + market gardening) Needs to be close to marked because perishability- gets spoiled easily Perishability makes transport difficult Intensive farming because of the high value + price of land near the market. (not really anymore because of improvements in transportation, refrigeration, trucks, ect.) Second= forests Timber Not perishable, but close to the market because transportation is difficult , heavy and expensive. Firewood was an essential good in the 1800s (building fires for cooking, heating homes, building materials) Third= grain + cereal crops perishability= less perishable transportation= not fragile, bulky, or heavy Easier transportation to market, doesn’t need to be close to market Extensive farming becomes profitable due to cost of land (low, because far from market) Fourth= livestock ranching perishability= not perishable while alive :) transportation= cheap because would just walk to market Land is less desirable because far from market + cheaper → ranchers can rent/buy large quantities of animals. Bid- Rent Theory: The value of the land is influenced by its relationship to the market. Graph: bid-price curve/bid-rent curve- determines starting position for each land use relative to the market, and where each land use would end Each line on the graph shows farmers’ willingness to pay for land at various distances from the market. (willing to pay more for close land and less for far land) Free market-economy: where supply and demand decide who gets what land, instead of government policy.The farmer with greatest profit will pay most at each location Where the two uppermost lines intersect show the end of one zone and start of the other The closer land is to the market/transportation route, the more expensive -so you can only buy a small amount of land, farming needs to be intensive to make a profit. The farther land is to the market/transportation route, the cheaper -so you can buy more land, farming is extensive. shortcomings/limitations of his theory Mainly, it was limited to his time and location Assumed isotropic plain- but: Ex: a river runs through it, making transportation easier → zones would then be along the river Some places may also have better climates or soil. comparative advantage: naturally occurring beneficial conditions (advantages that exist naturally in a environment, which can contribute to its economic success (ex: fertile land, climate) Now: there isn’t just one central market, there are multiple markets Now: Food preservation techniques (tin cans, refrigeration) refrigeration in trucks- goods can go longer distances Improved transportation infrastructure! - complex transportation like ships, air routes, ect. (ex: NYC flowers (perishable) are actually grown in south america, and flown there) wood is no longer used as a major source of heating, cooking and building (U.S. still uses wood but not most of the world.) 5.9- the global system of agriculture Supply chains- all the steps required to get a product/service to customers. (there can be thousands of miles between producers and consumers- ex: seeds could be developed in the UK, sent to Ghana where it’s grown and harvested, then sent to China where its turned into the final product and sent to the U.S. to sell.) regional interdependence Low- latitude countries with tropical climates (many of which are periphery/semi-periphery) produce crops like coffee, tea, bananas, and pineapples, and sell them to core countries. These are luxury crops- not necessary to human survival but have a high profit margin. (ex: cocoa beans processed into chocolate, grown on large plantations (that practice monoculture) that are controlled by transnational corporations - these corporations provide capital to run the plantations. Problems for periphery and semi periphery countries (producers): Farmers may not be able to afford their own luxury crops As the supply of locally grown food decreases, prices for local consumers increase. For countries that are dependent on one thing they export to core countries, if that thing is not in demand anymore then they become vulnerable. Competition may cause farmers to do environmentally damaging things (chemicals, soil erosion, ect.) The exchange of food around the world is dependent on -political systems, -strong infrastructure, and -trade policies Connections between Europe and the rest of the world are because of colonization. Relationships between core countries + semi periphery and periphery countries still resemble colonialism. Neocolonialism- the use of economic, political, and social pressures to control former colonies (describes current state of global food distribution) (ex: while growing/processing coffee beans is expensive, the profit margin in selling coffee drinks is very high. But most of the money stays with the transnational corporations in wealthy countries and little goes back to the producers.) Fair trade movement- promotes higher incomes for producers +more sustainable farming practices. Principles: direct trade between producers and consumers so that producers get more money (no intermediate step) fair price paid to farmers by importers, and workers paid a fair price. good conditions for laborers (safe working environment, no child labor) environmental sustainability (no GMOs, safe practices for the environment) respect for local culture Subsides- financial support that governments provide to farmers Ex: they subsidize rice in Japan, wheat and corn in the U.S. and soybean in China. Do this to: ensure dependable food supply Help farmers by increasing agricultural exports help consumers by reducing food costs (by highering supply) Infrastructure- roads, bridges, ect, that help transportation/exports of agricultural products. Governments subsidize infrastructure as well [ex: U.S gov indirectly subsidizes things from the midwest by spending money on the Mississippi river to be used for exporting/transportation products like corn, soybean, ect.] This makes the products go into the market at a lower price (because water transportation is cheap,) so ex: people from mexico can purchase it at a lower price from the U.S. than from mexico- this benefits the people of Mexico ,but hurts the farmers of Mexico Chapter 14- challenges and consequences of agriculture 5.10- consequences of agricultural practices Environmental Issues: Environmental issues rose when commercial overtook subsistence. Land cover change: study of how land is used and impact of changing land use: Pollution- increased use of chemical fertilizers and pesticides polluted air, water, and land. + health issues. Desertification- formerly fertile lands become infertile (ex: deforestation or overgrazing can increase wind erosion, losing the topsoil.) Soil salinization- salt from water remains in the soil. (decreases plants ability to absorb water and nutrients- lowers yields + makes soil useless.) Efforts- governments have made limits on chemicals, and protected zones are created where development is not allowed, to preserve the land. (ex: “save the rainforest” in 1980s) Humans Altering landscape: Terrace farming- Typically practiced by subsistence farmers in mountainous areas (common in China and India, with rice.) Steps are built into the sides of hills/mountain sides in order to cultivate water-intensive crops like rice. During rainfall, the water floods through the steps without pulling out the plants or causing soil erosion. Allows land to be cultivated that would otherwise be non-arable. Negative when not carefully maintained → leads to mudslides Irrigation when rainfall is not enough, applying controlled amounts of water from other sources[sprinkler systems, dams, canals, pipes, aquifers,(underground reserves of water that are piped to the fields - could be overused) ect.] Used a lot in California- turns dry central + southern regions into farmland Dams can destroy river ecosystems and displace people from their land. center-pivot irrigation- pivot is rotated around the center, giving specific amounts of water (or pesticides/fertilizer) to the field. Creates circular pattern. (used with large-scale commercial farming.) Negative effects/problems: Disrupt natural drainage of water Reduce normal regeneration of soil caused by natural flooding Reduce amount of surface water in rivers/lakes Overwatering can lead to chemical-contaminated water seeping into rivers + underground systems. history/example problem: 1960s, soviets diverted water from the aral sea to incr. cotton production. aral sea (once 4th largest lake global) reduced to 10% by 1997. (remaining water was saline) Draining wetlands Wetland: (low-lying) area of land that is covered by water (swamps, marshes, and bogs) Drained and converted to arable land. Results in loss of biodiversity and other environmental issues Loss of habitat for fish, mammals, ect Increases storm and flood damage (b/c during floods, wetlands hold water then slowly release) Deforestation Occurs today mostly in SE Asia, Africa, and Amazon rainforest Leads to soil erosion, decreased rainfall, lower temps, desertification, and increase in CO2 Shifting cultivation Practiced in periphery and semi- periphery countries in South America, C and W Africa, and SE Asia. Farmers cultivate the land until the soil becomes infertile, wasted. It is overused constantly without rest and without switching crops + no fallow period → until loses hydration and land is depleted, then move somewhere else. Ex: on the great plains/prairies of the U.S, during a period known as the dust bowl, 35 million acres of arable land became useless. Slash and burn agriculture (type of shifting cultivation that permanently alters the landscape.) Happens a lot in the Amazon. Cutting and burning forests in order to create fields for cultivation- ash from burning gives nutrients + fertilizes crops Results in deforestation and soil erosion. Half of the world’s topsoil has been lost in the past 150 years. Pastoral nomadism Practiced in semi-arid (dry, hot) and arid climates, subsistence farming. Everywhere, but most common in Asia and Africa. Herders move animals seasonally to find grass to graze. Results in soil erosion and desertification (formerly fertile lands become infertile) Typically as a result of droughts + overgrazing Societal effects of agriculture Demand for meat- About 35% of crops get fed to livestock, so there are les crops available for human consumption. Environmental effects of increased meat: large amounts of fresh water they need disposal of their waste. Increased animal waste that could have bacteria that contaminates water, soil, and causes illness. Increased consumption for processed food is affecting people's health. (High BP, Cholesterol, obesity, ect.) Role of women- In semi-periphery and periphery countries, women play important role in agriculture. (helped into the fields.) Now, the men in these areas have accepted more jobs in urban centers and left the farm, and women are left to manage the farm. → Feminization of agriculture International aid agencies have given more education, training, and financial assistance in agriculture to women. Economics Agriculture is part of Gross Domestic Product (GDP) of countries- it is a small percent of U.S. and Canada’s GDPs, but since their GDPs are very big (over $22 trillion), the value of agriculture is over $200 billion Examples: country Percentage of GDP from agriculture Percentage of labor force in agriculture United States 0.9%1.3 Canada 1.6%2 Afghanistan 23%44.3 Ghana 18.3%44.7 because of improvements in tech, the percentage of labor in the U.S. and Canada is much lower than Afghanistan and Ghana, which doesn’t have as much tech. 5.11- challenges of contemporary agriculture answer to this question is in this section- how do consumers influence agricultural practices? GMOs (genetically modified organisms)- part of biotech- plants scientists have modified by putting genes of one species into the DNA of another one (to make food more nutritious, resistant to weather or pests, and less spoiled.) - results in farmers reducing chemical use to control disease,weeds, and pests. history: produced in 1970s, grown popular in 1990s. The U.S, Brazil and Argentina use lots of GMOs. but many countries (in Europe) restricted them because: too expensive for poor farmers and new seeds often need to be bought every yr. GMO seeds resistant to pests and herbicides may result in more superpests or superweeds. Potential (not proven) health risks to consumers aquaculture/aquafarming- (talked about in 5.6- in general, the practice of raising fish + sea things) China and SE Asia have been using it for a while, but it's rising in other places. Blue Revolution- (what practicing aquaculture is referred to) - now the fastest growing form of food production in the world. (responsible for 50% of the world’s seafood.) Problems with aquaculture: high fish density increases spread of diseases + gives diseases to nearby wild (in open-pen system, the cages/net is on the seafloor so fishes can somewhat interact with the wild surroundings) chemicals/antibiotics to counter the diseases can hurt the ecosystem around it. fish can escape pens and breed/compete with fish nearby excess feed + fish waste can make dangerous levels of organic matter in nearby bodies of water Social concerns: can challenge traditional fishers and lead to conflict between groups of fishers owners of the aquaculture operations may pay workers low wages/bad working conditions, and may hurt the environment fish from these farms may have high levels of pesticides and harm humans. Environmental issues of agriculture: (elaborated, talked about in other sections as well.) chemicals from pesticides and fertilizers → contaminates water and growth of toxic algae that uses the oxygen in the water, therefore killing fish and other sea life. antibiotics given to livestock (to prevent disease and incr growth) may create superbugs that could be transferred to and hurt us. fossil fuels from machinery water may be wasted by farmers (overwatering, leaks in irrigation systems, growing crops in hot, dry places that require lots of water, ect.) almost 70% of freshwater is used for agriculture. Loss of biodiversity (bc of monoculture and farmers abandoning old varieties of crops for new improved ones) When theres too many animals or too little land, animals overgraze → makes the vegetation disappear (won't come back even after the animals leave) → soil erosion Happening a lot in Africa, and in pastoral nomadism practices b/c these people have access to less land (b/c now not that much land is just public, or now used for housing, farming, mining, petrol, ect.) organic foods- crops are non-GMO and produced without chemical pesticides + fertilizers. animals have to be fed 100% organic feed and no antibiotics/hormones. rising as consumers believe it to be more healthy pros: More labor intensive so more jobs Cons: More expensive Need more land to get same quantity Produce more greenhouse gasses Can still use naturally-occuring chemicals value-added crops- crops consumers are willing to pay more for b/c of special qualities or they are difficult to acquire: organic foods rare subtropical plants (passion fruit, cherimoya, longan, star fruit) grass-fed beef and free-range chickens/eggs Farmers try this when their profits decline value-added farming- turning their crops into a higher value product instead of just selling as is. Ex: grapes into wine, strawberries and other fruit into jam, tomato/pepper into salsa. Dairy farmings do this by making + selling their own ice cream or cheese. local-food movement- consumers seek out food produced nearby (advocates for this are called locavores.) supports local farmers Reduces fossil fuels from transportation (high price) Starting ~1990s, farmers markets became more popular. urban farming- production of farm goods within an urban area so you can provide locally grown food. reduced distance between producer and consumer in urban areas (consumers enjoy fresher products) A variation of urban farming is community gardens -share agricultural products with family/friends, and maybe the needy, instead of a business. People grow and harvest their own food, and allows poorer people to get more high-quality food. vertical farms- new type of urban farming that grows crops indoors in stackable trays, using greenhouses, artificial light (to mimic sunlight,) and hydroponics- growing plants in mineral-enriched solutions instead of soil. Uses less land and water, but uses lots of energy for the lights used to grow the crops. community-supported agriculture (CSA)- connects producers and consumers when consumers subscribe to a certain quantity of crop for a season, so the consumer gets a local supply of fresh product and the farmer earns throughout the season instead of just at the end. food insecurity- lack of access to enough food (b/c of economic issues.) Reasons: Poverty Unemployment under-employment In other countries: (non core) High population growth Political instability Environmental challenges (obviously, most prevalent in poor neighborhoods and among homeless) places: food desert- neighborhood that has access to little to no healthy + affordable foods Low income/ high poverty areas urban area - 33% of the population located more than 1 mile from a grocery store. rural area- more than 10 miles away usually home to racial/ethnic minorities or large % of elderly/disabled bc of economic + physical restraints (rare car ownership, city buses unavailable/unaffordable) residents cannot travel to grocery stores outside of their area and have to shop in their own neighborhood. (grocery chains removed stores in these areas bc of little profit) What's left are small convenience stores which don't have healthy food. → residents get health issues. food distribution system- network of trade and transportation to get food from farm to consumer Local level: if farmers can not reach the market, product will go to waste, and if consumer cannot reach the market, they will face hunger. Regional/global level: (modern transportation systems have improved distribution, but:) Ships, trucks, ect. can only work well with good infrastructure (ports, roads, rail lines.)- usually “the people in great need of food live in remote areas without good transportation links.” governments can affect food distribution- they might use food as a weapon during political unrest. (to suppress opposition) Weather problems: In core countries, farmers have crop insurance for weather disasters, but that’s not in non-core countries- weather disasters can impact the population a lot and another region needs to diliver food to the population. Climate change will affect farmers a lot. (increase droughts and storms → hurt crop yields and livestock) For hot areas: could cause drought, for livestock: heat-wave death, reduced milk, and reduced wait gain for beef cattle. For cold areas: could be good, increase growing season with shorter & milder winters. (+ longer life for young animals and less energy cost.) Farmland is threatened by expansion of cities and suburbs (suburbanization). Being converted for urban development (housing, shopping centers, business buildings, ect.) Since cities began as agricultural centers, the cities are where the fertile land IS. food processing- transformation of agricultural products into food or transforming food items into a different food [ex: milling wheat into flour (which is then sold to consumers or to factories to make bread)] Core countries (because of advanced infrastructure), can move goods rapidly, without waste. But in periphery countries, because of poor infrastructure, there is product loss. (can’t go far distances fast) Used to be: Food processing facilities near the farms, in rural areas (so they could work with fresh products, and have lower labor and land cost in rural areas) Now: because of improvements in transportation and infrastructure, the small, old facilities have closed for larger and more efficient ones, but are farther from the farms- they allow companies to process products at a lower cost per unit. (economies of scale, elaborated on 5.7) In commercial farming, products have to go from farm to food processor, wholesaler, and market, in a timely manner. With perishable crops, this means the pickup and delivery has to be fast- If that's not possible, then the farmers and shippers have to have access to a storage system that can preserve the quality of the crops for a couple days. Agricultural industries started Just-in-time delivery to reduce the need for storage space. (during covid this was affected and product shortages happened) Since food is vital to national security, governments help protect agriculture: Regulations on migrant workers who work in this industry. Financial assistance like low-interest loans and subsidies.(5.9) investment in transportation + storage infrastructure(5.9) (to increase efficiency + storage capacity) Regulations on the amount a product is grown/sold, to prevent surplus/shortages. taxes to discourage production/use of unhealthy things (like tobacco) International politics can lead to changes in production: Tariff- tax on imports Quota- There is a limit to the quantity that a good can be exported/imported A country might have to reduce exports of a certain crop, which could lead to less production. both are there to protect domestically produced goods- these restrictions raise the price of imports and of that product. - if this continues, countries change their production of crops. 5.12- women in agriculture Today, women make up about 40% of the agricultural labor force. In subsistence farming areas, it's about 70% (in developing world) men migrate to urban areas looking for work, and women stay at home and work the farms w/ the children. (if they sell their goods at the local market, women are the sellers) when machinery was introduced, women became less involved with field work. So in large-scale agribusiness, along with raising crops, tending animals and processing products, they work on the business side of it- management, sales, distribution and research. where people live/work changed how they prepare food. In urban areas, there is less growing and more buying. Today, women work more outside the house than olden days, so they are also doing less preparing food. people purchase way more food for convenience now -cake mix, frozen microwave meals, ect. (so food companies are investing in this to make prettier, tastier, healthier foods) In 2015 (for the first time in history) the restaurant/eating out bill was more than the grocery bill. Gender inequality- unequal opportunities, treatment, or rights of a person based on gender. Examples in agriculture: (as written before) men move to urban areas for work and women stay to tend to the farm and take care of the children- but, they still have little authority in the eyes of government officials, businesses, society, or male farmers. Gender-specific obstacles -”discriminatory practices that prevent (female) farmers from reaching their potential productivity” Women are: denied access to finance and capital → can’t afford machinery or chemicals, can’t modernize and increase productivity. denied training and education → cant lrean abou the new improve practices to improve productivity/safety on the farms denied property ownership limited mobility → in some countries, the male wont let them go out, so its hard for them to take products to market to sell/trade, or go to market to get food. lack political power → rarely allowed leadership positions or are even approached when development projects are planned- the plans dont have the knowledge or ideas of the women who play a big role in the plan’s success or failure. As a result of gender equality leading women to be less productive, there's a crop gap of 20-30% between male and female-run farms. The worst food insecurities are periphery countries, which is where gender-specific obstacles are the most- if these went away, f emale-run farms would have more productivity, and result in economic growth and enough food available in the periphery (no more food insecurity!) Note 0.0(0)Rate it Practice Test Flashcards Knowt Play Call with Kai AP Human Geography Study GuidesAP Human Geography Ultimate GuideUnit 1: Thinking GeographicallyUnit 2: Population and Migration Patterns and ProcessesUnit 3: Cultural Patterns and ProcessesUnit 4: Political Patterns and ProcessesUnit 5: Agriculture and Rural Land-Use Patterns and ProcessesUnit 6: Cities and Urban Land-Use Patterns and ProcessesUnit 7: Industrial and Economic Development Patterns and Processes Top Exams AP English Language and CompositionAP BiologyAP United States History Studying for another AP Exam? Check out our other AP study guides aphg unit 5 everything is straight from the aphg textbook Unit 5 Agricultural and Rural Land-Use Patterns and Processes Introduction- Change in agriculture over decades: Smaller family farms More large corporate farms More machinery More chemicals Expanded farmland from things that used to be forests, wetlands, ect. GMOs Global supply chains (can go farther distances, faster, ect.) Improvements in transportation + infrastructure Growth of agribusiness (farms as corporations) Globalization of agriculture Complex system that connects producers and consumers across the world (globally) due to advances in biotech mechanization transportation food preservation Distance between producers and consumers can be farther bc of Improvements in agricultural technology advances in transportation Increased globalized economy Chapter 11- origins, patterns, and settlements of agriculture 5.1- intro to agriculture See flashcards 5.2- settlement patterns and survey methods Rural Survey Methods: (defining boundaries of land ownership) Metes and bounds (in the past) Utilization of landmarks and physical features to establish boundary lines Short distances Results in irregular shaped plots of land (people would just go mark their own land “from that mountain to the river is mine”) Not practical anymore, used to be in Great Britain and North America- now, rarely used (sometimes in the Middle East) (French) Long Lot (still used along rivers) Long strips of land that start at a river or lake with the intention of providing all landowners with equal access to the resources (soil and water) and transportation. (France, Spain, North America, ect.) In North America, Quebec and Louisiana 1. Township and range (aka Public Land Survey System) (currently used) Made by thomas Jefferson Started in 1785 after U.S. switched to surveying rather than landscape features Rectangles and grid system townships- area 6mi x 6mi Keeps track of land sales and purchases Utilizes a uniform survey method Section- each square mile- consisted of 640 acres Rural Settlement Patterns:(shapes) 1. Clustered- (aka nucleated) Farmland and pastures surround settlements and towns Share resources and community Have to walk to farmland survey method: Metes & Bounds (have to agree between themselves, could lead to fights) Throughout European history, rural residents lived in groups of homes in close proximity to one another. Dispersed- Used right now in North America Settlements are isolated and dispersed over the land area U.S. Government promoted Westward Expansion by giving farmers land (usually 160 acres) if they agreed to live and farm on it. Survey method: township & range ~Encourages self-sufficiency but makes shared things (like schools) hard~ Linear- Settlement organized along a line- transportation system, or physical feature like a river/coast (today, more demand for along a transportation route) Survey method: long lots 5.3- agricultural origins and diffusions First (neolithic/stone age) Agricultural Revolution- origin of farming (marked by domestication ꜜ of plants and animals- mostly was subsistence farming at the time) Domestication- the deliberate effort to grow plants and raise animals, making plants and animals adapt to human demands Animal domestication First was probably hunters in Central Asia- dogs and horses Then SW Asia- goats, pigs, sheep, cattle Then cats, horses, camels, donkeys, llamas, ect. Plant domestication Probably after animal domestication First was vegetative planting (using parts of existing plants to grow more) Then seeds Hearth- the geographic origin of a trait, characteristic, innovation, or other concept. It is where something is “born” Agricultural hearth- the separate location in which groups of people began to domesticate plants and animals agricultural hearths Southwest asia (fertile crescent in middle east) - first hearth (Middle east) east coast of mediterranean sea, along tigris and euphrates, to persian gulf. Southeast asia (includes islands like philippines, malaysia, thailand, indonesia) South asia (indus valley (north of india)- also called fertile crescent) East asia (incl. China, like wei-huang river valley) Sub-saharan africa (west africa + nile river valley (E Africa)) Mesoamerica [central america (guatemala, el salvador, nicaragua) + andean highlands (in ecuador, peru, chile)] more hearths around rivers: Chang Jiang (Yangtze) and (Wei) Huang He (yellow) river valleys in East Asia/China Nile valley in NE Africa Ganges valley in South Asia first communities made were usually along rivers- source of water + drinking, transportation, defense, source of food (fish) (flooding of rivers) also contributed to soil fertility, making agriculture more productive- people could focus on other tasks like building, military, ect. advancing all areas of human life. Commonalities of agricultural hearths: Fertile soil in river valleys availability of water Moderate climate Skilled residence Independent inventions- when a trait has many cultural hearths; crops and animals are domesticated in multiple regions but there was no interaction between the regions. (ex: wheat was domesticated in SW, E, and S Asia, but all did it independently without interacting with each other, they each had the idea themselves.) (civilization started around the agricultural hearts- they became the first urban centers, turned into civilizations, societies and powerful states. increased trade, large empires, and increased other specialized jobs. Diffusion of agriculture Contagious diffusion Agriculture first diffused to the immediate surrounding areas of the hearths through close contact and proximity between farmers Ex: agriculture and animal domestication originated in Fertile Crescent then diffused through Turkey into Eastern Europe Immigration and migration (relocation diffusion) As people migrate, they bring food, ingredients, seeds, and animals with them. Stimulus diffusion: as people moved into new locations, seeds may not be as prosperous due to different physical conditions → resulting in the adaptation of farming methods/crops, using the agricultural knowledge the farmer had previously acquired (the idea had diffused and stayed but adapted) Trade routes Ex: the silk road: over 4,000 miles of trade routes that connected China with Europe with a variety of different people interacting. Traded silk, rice, and more. → resulted in the spread of plants and animals throughout East Asia, South Asia, Middle East, Africa, and Europe Columbian exchange- (after columbus in 1492) Global movements of plants and animals between Afro-Eurasia and the Americas. (mainly caribbean) Examples: Coffee from east africa and bananas and sugar cane from New Guinea thrive in America today. Potatoes from NW South America and maize from Mexico are grown in Europe, Asia, and Africa. Tobacco and sugar are now new in Europe Also with animals and diseases Turkey- found in Americas, taken to Europe Smallpox, influenza, malaria, and measles came from the eastern hemisphere to western hemisphere. (after that, contagious diffusion makes things further spread in the new places) Chapter 12- the second and third agricultural revolutions 5.4- the second agricultural revolution Result: increased food supply (especially to countries part of global trade networks) and more people got access to a greater variety of food → higher life expectancy Fewer but larger, more productive farms, and less farmers/farm laborers. → More people in urban areas than rural (for first time in U.S. history) Began in 1700s “Used advances of the Industrial Revolution to increase food supplies and support population growth” Involved mechanization of agricultural production, advances in transportation (ex: could transport food over long distances, so the farming could be far away from markets), and more. New innovations increased yields, especially in Europe and the U.S. Examples: iron/steel plow Mechanized seed drilling mccormick reaper/harvester Grain elevator barbed wire Mixed nitrogen and nitric acid fertilizer New techniques also increased yields Crop rotation- planning different crops (in a specific sequence/order/timeline) on the same plot of land, to restore nutrients back into the soil. (ex: grains extract nitrogen from the soil, then alfalfa adds nitrogen back to the soil.) Irrigation- applying controlled amounts of water to crops using human-made devices (canals, pipes, sprinkler systems) instead of just rainfall. Enclosure acts- Laws made by the British government allowing people to purchase and enclose land for their own use. -Divided up common land that was shared by farmers into individual plots. result: allowed for larger farms, more efficient production, and commercial farming. But, many farmers were forced off their land and lost their traditional way of life- also less need for farm laborers, so people looked for jobs in the industrial/manufacturing sector, like factories, instead. Most of these factory jobs were in urban areas → created big migration of people from rural to urban areas 5.5- the green revolution Mid 20th century (main part of third agricultural revolution) Advances in plant biology Dr. Norman Borlaugh- father of green revolution Developed higher yielding, more disease resistant, and faster growing crops- specifically worked on wheat in Mexico. Concerns- increased use of chemical fertilizer & pesticides. Hybridization- breeding two plants to create one seed with both plant’s characteristics. (ex: long grain rice from Indonesia + dense grain short rice from Taiwan = longer and denser rice grain introduced in rest of Asia.) Green revolution also introduced more machinery into developing countries (tractors, tillers, broadcast seeders, grain carts, ect.) GMOs (genetically modified organisms)- humans changing DNA of a seed through engineering -to increase yields, resist diseases, and withstand chemicals used to kill pests/weeds. Positive impacts- Higher yields(on the same amount of land) Countries like India and Mexico moved from importers to exporters (for wheat) Reduced hunger, lowered death rates, and grew populations Money for research and business Lower food prices Negative impacts- Environmental damages pollution due to increased use of pesticides & fertilizers → runoff in water led to polluted water, hurt ecosystems, health issues soil erosion (b/c of intensive use of land) fossil fuels from machinery increased pollution Impact on gender roles Traditionally, much of the farming labor was done by women, but men dominated socially and when the green revolution technologies came, the men were educated on these new methods and benefited, while the women were excluded. (continued on 5.12) Economic changes Machinery, seeds, fertilizers and pesticides cost increased Farming tech heavily relied on investments (mostly from private corporations), but the corporations began to stop investing. Because of the migration from rural to urban, and these core wealthy areas got advantages over the poor outer areas. Struggle in Africa Africa did not benefit from the Green Revolution for many reasons: Diverse climate and soil Harsh environmental conditions Lack of well developed infrastructure → cost of investment in research, development, and transportation was too high. Africa's staple crops (sorghum, millet, cassava, yams, cowpeas, peanuts) were not target research audiences for hybridization. Africa grew the most during the Green Revolution, but hunger remained a problem since the Green Revolution was not effective there. Chapter 13- spatial arrangement of agriculture 5.6- agricultural production regions examples of intensive farming Paddy rice farming in S, SE, and E Asia is labor intensive.(terraced fields makes using machinery difficult) Truck farming (Cali, Teas, Florida, and near large cities) is capital intensive b/c uses expensive machinery and more. Also labor intensive because use low-pay migrant workers to harvest (large scale farms) Factory farming- capital intensive raising of livestock where animals are in close quarters and bred and fed in a controlled environment. (animals move like a conveyor belt) Aquaculture (aquafarming)- intensive farming. Fish,shellfish and water plants are raised in netted areas in the sea, tanks, or more. extensive examples include shifting cultivation, pastoral nomadism/nomadic herding, ranching Lots of extensive farming has turned into intensive bc of increased demand for food (higher population) and global competition. (now, people use land continuously without fallow period, and use irrigation, chemical fertilizers, GMOs, ect.) Double cropping- planting + harvesting a crop two (or three) times a year on the same field. intercropping/multicropping- two or more crops grow simultaneously on the same field. monocropping/monoculture- (the opposite) - only growing one type of crop at a time on a piece of land. (like plantations ) Examples: World dependent on India for Rice Russia- wheat Sri- lanka- tea (In the past) Iraq- dates U.S.- corn Iran- saffron Pros: Maximized efficiency Lower per-unit cost of production Higher yield More profit Cons: Soil depletion Increased reliance on chemicals Increased risk since all of investments are in and depending on this one crop. Increased demand for meat (50% higher between 1998 and 2018) results in feedlots- confined spaces where cattle have limited movement. (aka CAFOs- concentrated animal feeding operations) -maximizes space, makes cattle bigger bc of limited movement → maximizes profit. Some things combine intensive + extensive, ex: raising cattle in wyoming- roam and feed on grass for months which is extensive, then intensive phase where they are transported to fatten quickly before processed for meat. 5.7- spatial organization of agriculture Agribusiness- farms run as corporations - Large- Scale Production. Various steps: Research and development Processing and production Transportation Marketing Retail (sale) of agricultural goods (replacing small family farms) More cost effective than family owned, cheaper for us. Higher yields than family-owned, so still make more money. Owned by: Transnational corporations- operate in many countries Commercial (intention is making a profit) Use lots of machinery Use lots of biotech (GMOs) and chemicals Home scale vs agribusiness scale examples: Growing crops in a backyard garden vsowning thousands of acres worked by many employees selling at local market vs selling to wholesale distributors who ship it regional + globally Growing diff types to see which grows best vs investing thousands in research and development of new seeds, fertilizers, and pesticides Vertical integration- Large corporate farms own other businesses that are involved in the steps of producing a particular product. This gives the large farm more control + greater profits. Ex: they might own a research and development company that improves seeds, a transporting company that transports farm products, a processing factory, and a wholesaler that distributes the food to stores. Ex: Tyson Foods owns the farms where they raise the animals, then own the processing plants where they slaughter, process, and package the meat products. Economies of scale- increase in efficiency to lower per-unit production cost. (usually occurs with larger farming operations, but could happen at any scale) Ex: someone buys more land, but not more machinery, just uses the already existing machinery more efficiently- this will increase the earning without increasing the expenses too much, and the cost per unit will decrease, and profit increase. Another example, at a smaller farm: increasing level of fertilization and irrigation- increases expenses by 10%, but yields increase by 25%, so cost per unit decreases, and profits increase. Disappearance of small family farms because they can’t compete with the cheaper cots of the larger farms. They also cannot afford the newest tech but need to get it to compete in the global market. Commodity chain- process by corporations to gather resources, turn them into goods, and transport to consumers. template commodity chain: (simplified) planting (using seeds, fertilizers, water) → growing (fertile land and moisture) → harvesting (dried and stored until processing) → processing (packaged or used to create other products) → marketing (sold for food, animals, ect.) Additionally includes financial institutions (banks), transportation companies, distributors, and governments. Carrying capacity- total number of people farmers can support given the available resources. - risen a lot over the years. ex: in 1962, avg. U.S. farmers fed ~26 people. Today, it’s 166 people! Cool chains- transportation networks that keep food cool throughout the trip. (now perishables can travel long distances for a low cost) 5.8- von thunen model Location theory- why people choose certain locations for economic activity (factories, stores, restaurants, or agriculture) Johann Von thunen- North German farmer and economist Published “an isolated state” in 1828 Von thunen model- economic model that gives a pattern for the types of products farmers would produce at different positions relative to the market. farmers make this decision based on: transportation cost land cost Intensity of land use Perishability of the product His assumptions: farming = economic activity Farmers in business to make profit 1 market where farmers sold products 1 transportation system farmers paid for transportation, which was different with distance market was in isotropic plain- flat with similar fertility + climate throughout area beyond market and farmland was wilderness (center = market) First circle= horticulture (dairy farming + market gardening) Needs to be close to marked because perishability- gets spoiled easily Perishability makes transport difficult Intensive farming because of the high value + price of land near the market. (not really anymore because of improvements in transportation, refrigeration, trucks, ect.) Second= forests Timber Not perishable, but close to the market because transportation is difficult , heavy and expensive. Firewood was an essential good in the 1800s (building fires for cooking, heating homes, building materials) Third= grain + cereal crops perishability= less perishable transportation= not fragile, bulky, or heavy Easier transportation to market, doesn’t need to be close to market Extensive farming becomes profitable due to cost of land (low, because far from market) Fourth= livestock ranching perishability= not perishable while alive :) transportation= cheap because would just walk to market Land is less desirable because far from market + cheaper → ranchers can rent/buy large quantities of animals. Bid- Rent Theory: The value of the land is influenced by its relationship to the market. Graph: bid-price curve/bid-rent curve- determines starting position for each land use relative to the market, and where each land use would end Each line on the graph shows farmers’ willingness to pay for land at various distances from the market. (willing to pay more for close land and less for far land) Free market-economy: where supply and demand decide who gets what land, instead of government policy.The farmer with greatest profit will pay most at each location Where the two uppermost lines intersect show the end of one zone and start of the other The closer land is to the market/transportation route, the more expensive -so you can only buy a small amount of land, farming needs to be intensive to make a profit. The farther land is to the market/transportation route, the cheaper -so you can buy more land, farming is extensive. shortcomings/limitations of his theory Mainly, it was limited to his time and location Assumed isotropic plain- but: Ex: a river runs through it, making transportation easier → zones would then be along the river Some places may also have better climates or soil. comparative advantage: naturally occurring beneficial conditions (advantages that exist naturally in a environment, which can contribute to its economic success (ex: fertile land, climate) Now: there isn’t just one central market, there are multiple markets Now: Food preservation techniques (tin cans, refrigeration) refrigeration in trucks- goods can go longer distances Improved transportation infrastructure! - complex transportation like ships, air routes, ect. (ex: NYC flowers (perishable) are actually grown in south america, and flown there) wood is no longer used as a major source of heating, cooking and building (U.S. still uses wood but not most of the world.) 5.9- the global system of agriculture Supply chains- all the steps required to get a product/service to customers. (there can be thousands of miles between producers and consumers- ex: seeds could be developed in the UK, sent to Ghana where it’s grown and harvested, then sent to China where its turned into the final product and sent to the U.S. to sell.) regional interdependence Low- latitude countries with tropical climates (many of which are periphery/semi-periphery) produce crops like coffee, tea, bananas, and pineapples, and sell them to core countries. These are luxury crops- not necessary to human survival but have a high profit margin. (ex: cocoa beans processed into chocolate, grown on large plantations (that practice monoculture) that are controlled by transnational corporations - these corporations provide capital to run the plantations. Problems for periphery and semi periphery countries (producers): Farmers may not be able to afford their own luxury crops As the supply of locally grown food decreases, prices for local consumers increase. For countries that are dependent on one thing they export to core countries, if that thing is not in demand anymore then they become vulnerable. Competition may cause farmers to do environmentally damaging things (chemicals, soil erosion, ect.) The exchange of food around the world is dependent on -political systems, -strong infrastructure, and -trade policies Connections between Europe and the rest of the world are because of colonization. Relationships between core countries + semi periphery and periphery countries still resemble colonialism. Neocolonialism- the use of economic, political, and social pressures to control former colonies (describes current state of global food distribution) (ex: while growing/processing coffee beans is expensive, the profit margin in selling coffee drinks is very high. But most of the money stays with the transnational corporations in wealthy countries and little goes back to the producers.) Fair trade movement- promotes higher incomes for producers +more sustainable farming practices. Principles: direct trade between producers and consumers so that producers get more money (no intermediate step) fair price paid to farmers by importers, and workers paid a fair price. good conditions for laborers (safe working environment, no child labor) environmental sustainability (no GMOs, safe practices for the environment) respect for local culture Subsides- financial support that governments provide to farmers Ex: they subsidize rice in Japan, wheat and corn in the U.S. and soybean in China. Do this to: ensure dependable food supply Help farmers by increasing agricultural exports help consumers by reducing food costs (by highering supply) Infrastructure- roads, bridges, ect, that help transportation/exports of agricultural products. Governments subsidize infrastructure as well [ex: U.S gov indirectly subsidizes things from the midwest by spending money on the Mississippi river to be used for exporting/transportation products like corn, soybean, ect.] This makes the products go into the market at a lower price (because water transportation is cheap,) so ex: people from mexico can purchase it at a lower price from the U.S. than from mexico- this benefits the people of Mexico ,but hurts the farmers of Mexico Chapter 14- challenges and consequences of agriculture 5.10- consequences of agricultural practices Environmental Issues: Environmental issues rose when commercial overtook subsistence. Land cover change: study of how land is used and impact of changing land use: Pollution- increased use of chemical fertilizers and pesticides polluted air, water, and land. + health issues. Desertification- formerly fertile lands become infertile (ex: deforestation or overgrazing can increase wind erosion, losing the topsoil.) Soil salinization- salt from water remains in the soil. (decreases plants ability to absorb water and nutrients- lowers yields + makes soil useless.) Efforts- governments have made limits on chemicals, and protected zones are created where development is not allowed, to preserve the land. (ex: “save the rainforest” in 1980s) Humans Altering landscape: Terrace farming- Typically practiced by subsistence farmers in mountainous areas (common in China and India, with rice.) Steps are built into the sides of hills/mountain sides in order to cultivate water-intensive crops like rice. During rainfall, the water floods through the steps without pulling out the plants or causing soil erosion. Allows land to be cultivated that would otherwise be non-arable. Negative when not carefully maintained → leads to mudslides Irrigation when rainfall is not enough, applying controlled amounts of water from other sources[sprinkler systems, dams, canals, pipes, aquifers,(underground reserves of water that are piped to the fields - could be overused) ect.] Used a lot in California- turns dry central + southern regions into farmland Dams can destroy river ecosystems and displace people from their land. center-pivot irrigation- pivot is rotated around the center, giving specific amounts of water (or pesticides/fertilizer) to the field. Creates circular pattern. (used with large-scale commercial farming.) Negative effects/problems: Disrupt natural drainage of water Reduce normal regeneration of soil caused by natural flooding Reduce amount of surface water in rivers/lakes Overwatering can lead to chemical-contaminated water seeping into rivers + underground systems. history/example problem: 1960s, soviets diverted water from the aral sea to incr. cotton production. aral sea (once 4th largest lake global) reduced to 10% by 1997. (remaining water was saline) Draining wetlands Wetland: (low-lying) area of land that is covered by water (swamps, marshes, and bogs) Drained and converted to arable land. Results in loss of biodiversity and other environmental issues Loss of habitat for fish, mammals, ect Increases storm and flood damage (b/c during floods, wetlands hold water then slowly release) Deforestation Occurs today mostly in SE Asia, Africa, and Amazon rainforest Leads to soil erosion, decreased rainfall, lower temps, desertification, and increase in CO2 Shifting cultivation Practiced in periphery and semi- periphery countries in South America, C and W Africa, and SE Asia. Farmers cultivate the land until the soil becomes infertile, wasted. It is overused constantly without rest and without switching crops + no fallow period → until loses hydration and land is depleted, then move somewhere else. Ex: on the great plains/prairies of the U.S, during a period known as the dust bowl, 35 million acres of arable land became useless. Slash and burn agriculture (type of shifting cultivation that permanently alters the landscape.) Happens a lot in the Amazon. Cutting and burning forests in order to create fields for cultivation- ash from burning gives nutrients + fertilizes crops Results in deforestation and soil erosion. Half of the world’s topsoil has been lost in the past 150 years. Pastoral nomadism Practiced in semi-arid (dry, hot) and arid climates, subsistence farming. Everywhere, but most common in Asia and Africa. Herders move animals seasonally to find grass to graze. Results in soil erosion and desertification (formerly fertile lands become infertile) Typically as a result of droughts + overgrazing Societal effects of agriculture Demand for meat- About 35% of crops get fed to livestock, so there are les crops available for human consumption. Environmental effects of increased meat: large amounts of fresh water they need disposal of their waste. Increased animal waste that could have bacteria that contaminates water, soil, and causes illness. Increased consumption for processed food is affecting people's health. (High BP, Cholesterol, obesity, ect.) Role of women- In semi-periphery and periphery countries, women play important role in agriculture. (helped into the fields.) Now, the men in these areas have accepted more jobs in urban centers and left the farm, and women are left to manage the farm. → Feminization of agriculture International aid agencies have given more education, training, and financial assistance in agriculture to women. Economics Agriculture is part of Gross Domestic Product (GDP) of countries- it is a small percent of U.S. and Canada’s GDPs, but since their GDPs are very big (over $22 trillion), the value of agriculture is over $200 billion Examples: country Percentage of GDP from agriculture Percentage of labor force in agriculture United States 0.9%1.3 Canada 1.6%2 Afghanistan 23%44.3 Ghana 18.3%44.7 because of improvements in tech, the percentage of labor in the U.S. and Canada is much lower than Afghanistan and Ghana, which doesn’t have as much tech. 5.11- challenges of contemporary agriculture answer to this question is in this section- how do consumers influence agricultural practices? GMOs (genetically modified organisms)- part of biotech- plants scientists have modified by putting genes of one species into the DNA of another one (to make food more nutritious, resistant to weather or pests, and less spoiled.) - results in farmers reducing chemical use to control disease,weeds, and pests. history: produced in 1970s, grown popular in 1990s. The U.S, Brazil and Argentina use lots of GMOs. but many countries (in Europe) restricted them because: too expensive for poor farmers and new seeds often need to be bought every yr. GMO seeds resistant to pests and herbicides may result in more superpests or superweeds. Potential (not proven) health risks to consumers aquaculture/aquafarming- (talked about in 5.6- in general, the practice of raising fish + sea things) China and SE Asia have been using it for a while, but it's rising in other places. Blue Revolution- (what practicing aquaculture is referred to) - now the fastest growing form of food production in the world. (responsible for 50% of the world’s seafood.) Problems with aquaculture: high fish density increases spread of diseases + gives diseases to nearby wild (in open-pen system, the cages/net is on the seafloor so fishes can somewhat interact with the wild surroundings) chemicals/antibiotics to counter the diseases can hurt the ecosystem around it. fish can escape pens and breed/compete with fish nearby excess feed + fish waste can make dangerous levels of organic matter in nearby bodies of water Social concerns: can challenge traditional fishers and lead to conflict between groups of fishers owners of the aquaculture operations may pay workers low wages/bad working conditions, and may hurt the environment fish from these farms may have high levels of pesticides and harm humans. Environmental issues of agriculture: (elaborated, talked about in other sections as well.) chemicals from pesticides and fertilizers → contaminates water and growth of toxic algae that uses the oxygen in the water, therefore killing fish and other sea life. antibiotics given to livestock (to prevent disease and incr growth) may create superbugs that could be transferred to and hurt us. fossil fuels from machinery water may be wasted by farmers (overwatering, leaks in irrigation systems, growing crops in hot, dry places that require lots of water, ect.) almost 70% of freshwater is used for agriculture. Loss of biodiversity (bc of monoculture and farmers abandoning old varieties of crops for new improved ones) When theres too many animals or too little land, animals overgraze → makes the vegetation disappear (won't come back even after the animals leave) → soil erosion Happening a lot in Africa, and in pastoral nomadism practices b/c these people have access to less land (b/c now not that much land is just public, or now used for housing, farming, mining, petrol, ect.) organic foods- crops are non-GMO and produced without chemical pesticides + fertilizers. animals have to be fed 100% organic feed and no antibiotics/hormones. rising as consumers believe it to be more healthy pros: More labor intensive so more jobs Cons: More expensive Need more land to get same quantity Produce more greenhouse gasses Can still use naturally-occuring chemicals value-added crops- crops consumers are willing to pay more for b/c of special qualities or they are difficult to acquire: organic foods rare subtropical plants (passion fruit, cherimoya, longan, star fruit) grass-fed beef and free-range chickens/eggs Farmers try this when their profits decline value-added farming- turning their crops into a higher value product instead of just selling as is. Ex: grapes into wine, strawberries and other fruit into jam, tomato/pepper into salsa. Dairy farmings do this by making + selling their own ice cream or cheese. local-food movement- consumers seek out food produced nearby (advocates for this are called locavores.) supports local farmers Reduces fossil fuels from transportation (high price) Starting ~1990s, farmers markets became more popular. urban farming- production of farm goods within an urban area so you can provide locally grown food. reduced distance between producer and consumer in urban areas (consumers enjoy fresher products) A variation of urban farming is community gardens -share agricultural products with family/friends, and maybe the needy, instead of a business. People grow and harvest their own food, and allows poorer people to get more high-quality food. vertical farms- new type of urban farming that grows crops indoors in stackable trays, using greenhouses, artificial light (to mimic sunlight,) and hydroponics- growing plants in mineral-enriched solutions instead of soil. Uses less land and water, but uses lots of energy for the lights used to grow the crops. community-supported agriculture (CSA)- connects producers and consumers when consumers subscribe to a certain quantity of crop for a season, so the consumer gets a local supply of fresh product and the farmer earns throughout the season instead of just at the end. food insecurity- lack of access to enough food (b/c of economic issues.) Reasons: Poverty Unemployment under-employment In other countries: (non core) High population growth Political instability Environmental challenges (obviously, most prevalent in poor neighborhoods and among homeless) places: food desert- neighborhood that has access to little to no healthy + affordable foods Low income/ high poverty areas urban area - 33% of the population located more than 1 mile from a grocery store. rural area- more than 10 miles away usually home to racial/ethnic minorities or large % of elderly/disabled bc of economic + physical restraints (rare car ownership, city buses unavailable/unaffordable) residents cannot travel to grocery stores outside of their area and have to shop in their own neighborhood. (grocery chains removed stores in these areas bc of little profit) What's left are small convenience stores which don't have healthy food. → residents get health issues. food distribution system- network of trade and transportation to get food from farm to consumer Local level: if farmers can not reach the market, product will go to waste, and if consumer cannot reach the market, they will face hunger. Regional/global level: (modern transportation systems have improved distribution, but:) Ships, trucks, ect. can only work well with good infrastructure (ports, roads, rail lines.)- usually “the people in great need of food live in remote areas without good transportation links.” governments can affect food distribution- they might use food as a weapon during political unrest. (to suppress opposition) Weather problems: In core countries, farmers have crop insurance for weather disasters, but that’s not in non-core countries- weather disasters can impact the population a lot and another region needs to diliver food to the population. Climate change will affect farmers a lot. (increase droughts and storms → hurt crop yields and livestock) For hot areas: could cause drought, for livestock: heat-wave death, reduced milk, and reduced wait gain for beef cattle. For cold areas: could be good, increase growing season with shorter & milder winters. (+ longer life for young animals and less energy cost.) Farmland is threatened by expansion of cities and suburbs (suburbanization). Being converted for urban development (housing, shopping centers, business buildings, ect.) Since cities began as agricultural centers, the cities are where the fertile land IS. food processing- transformation of agricultural products into food or transforming food items into a different food [ex: milling wheat into flour (which is then sold to consumers or to factories to make bread)] Core countries (because of advanced infrastructure), can move goods rapidly, without waste. But in periphery countries, because of poor infrastructure, there is product loss. (can’t go far distances fast) Used to be: Food processing facilities near the farms, in rural areas (so they could work with fresh products, and have lower labor and land cost in rural areas) Now: because of improvements in transportation and infrastructure, the small, old facilities have closed for larger and more efficient ones, but are farther from the farms- they allow companies to process products at a lower cost per unit. (economies of scale, elaborated on 5.7) In commercial farming, products have to go from farm to food processor, wholesaler, and market, in a timely manner. With perishable crops, this means the pickup and delivery has to be fast- If that's not possible, then the farmers and shippers have to have access to a storage system that can preserve the quality of the crops for a couple days. Agricultural industries started Just-in-time delivery to reduce the need for storage space. (during covid this was affected and product shortages happened) Since food is vital to national security, governments help protect agriculture: Regulations on migrant workers who work in this industry. Financial assistance like low-interest loans and subsidies.(5.9) investment in transportation + storage infrastructure(5.9) (to increase efficiency + storage capacity) Regulations on the amount a product is grown/sold, to prevent surplus/shortages. taxes to discourage production/use of unhealthy things (like tobacco) International politics can lead to changes in production: Tariff- tax on imports Quota- There is a limit to the quantity that a good can be exported/imported A country might have to reduce exports of a certain crop, which could lead to less production. both are there to protect domestically produced goods- these restrictions raise the price of imports and of that product. - if this continues, countries change their production of crops. 5.12- women in agriculture Today, women make up about 40% of the agricultural labor force. In subsistence farming areas, it's about 70% (in developing world) men migrate to urban areas looking for work, and women stay at home and work the farms w/ the children. (if they sell their goods at the local market, women are the sellers) when machinery was introduced, women became less involved with field work. So in large-scale agribusiness, along with raising crops, tending animals and processing products, they work on the business side of it- management, sales, distribution and research. where people live/work changed how they prepare food. In urban areas, there is less growing and more buying. Today, women work more outside the house than olden days, so they are also doing less preparing food. people purchase way more food for convenience now -cake mix, frozen microwave meals, ect. (so food companies are investing in this to make prettier, tastier, healthier foods) In 2015 (for the first time in history) the restaurant/eating out bill was more than the grocery bill. Gender inequality- unequal opportunities, treatment, or rights of a person based on gender. Examples in agriculture: (as written before) men move to urban areas for work and women stay to tend to the farm and take care of the children- but, they still have little authority in the eyes of government officials, businesses, society, or male farmers. Gender-specific obstacles -”discriminatory practices that prevent (female) farmers from reaching their potential productivity” Women are: denied access to finance and capital → can’t afford machinery or chemicals, can’t modernize and increase productivity. denied training and education → cant lrean abou the new improve practices to improve productivity/safety on the farms denied property ownership limited mobility → in some countries, the male wont let them go out, so its hard for them to take products to market to sell/trade, or go to market to get food. lack political power → rarely allowed leadership positions or are even approached when development projects are planned- the plans dont have the knowledge or ideas of the women who play a big role in the plan’s success or failure. As a result of gender equality leading women to be less productive, there's a crop gap of 20-30% between male and female-run farms. The worst food insecurities are periphery countries, which is where gender-specific obstacles are the most- if these went away, f emale-run farms would have more productivity, and result in economic growth and enough food available in the periphery (no more food insecurity!) Insert Your Equation Cancel Insert
5882	https://cs.uwaterloo.ca/~r5olivei/courses/2024-spring-cs466/lecture-notes/lecture13/	Contents Linear Programming Relaxations and Rounding Many important optimization problems are NP-hard. A prominent example is the Integer Linear Programming (ILP) problem, which is defined as follows: where , , and are given. The goal is to find an integer vector that minimizes the objective function while satisfying the constraints. One advantage of ILPs is that they can model a wide range of combinatorial optimization problems. However, this very same property makes them computationally intractable in general. But we do know how to solve linear programming (LP) problems efficiently. So, a natural question is: can we get good solutions when we relax the integrality constraints from an ILP and solve the corresponding LP instead? Can we round the solution we obtained from the LP to get a good solution for the ILP? If we manage to do the above, we would be getting “pretty good” solutions for a wide range of combinatorial optimization problems. This is the idea behind the method of linear programming relaxations and rounding: we are content with approximately optimal solutions that we can find efficiently. Here is a high-level overview of the method: Formulate the combinatorial optimization problem as an ILP. (say by minimizing some objective function) Relax the integrality constraints to get an LP. (This is called the linear programming relaxation of the ILP.) 2.1. We are still minimizing the same objective function, but over a (potentially) larger feasible region. Hence 3. Solve the LP to get a fractional solution. This can be done efficiently. 3.1. If the fractional solution is already integral, we are done, as it will be a solution to the ILP. 3.2. Otherwise, we need to round the fractional solution to get an integral solution. This is the tricky part. In this case, we need to design a rounding algorithm that transforms the fractional solution into an integral solution while preserving the objective function value as much as possible. Thus, we would like to find a rounding algorithm and a value such that Example: Vertex Cover Given a weighted graph , where , we would like to find a minimum weight vertex cover. That is, we want to find a set such that for every edge , at least one of or is in , and the total weight of the vertices in is minimized. Following our strategy, we can formulate the problem as an ILP: We can interpret the above ILP as follows: if and only if vertex is in the vertex cover . The inequality constraints ensure that for every edge , at least one of or is in the vertex cover. The objective function is the total weight of the vertices in the vertex cover, which we want to minimize. The linear programming relaxation of the above ILP is obtained by relaxing the integrality constraints on . This gives us the following LP: Now we use an efficient LP solver to solve the above LP. Let the the optimal solution we obtain. Now we need to devise a rounding algorithm that transforms the fractional solution into an integral solution . In this case, we can use the following simple rounding algorithm: round each to the nearest integer. Let be the rounded value of . Then, if , and otherwise. Note that , for every vertex . Moreover, we can see that encodes a vertex cover: for every edge , since , we have , which means that at least one of or is in the vertex cover encoded by . Finally, we can analyze the approximation guarantee of the above rounding algorithm. The cost of the solution given by is: Thus, we have obtained a 2-approximation algorithm for the vertex cover problem using linear programming relaxations and rounding. Main Example: Set Cover The set cover problem is a classic combinatorial optimization problem, generalizing the vertex cover problem. Input: A finite set (the universe) and a collection of subsets of . Output: A minimum-size subcollection such that We could also have a weighted version of the set cover problem, where each set has a non-negative weight associated with it, and the goal is to minimize the total weight of the sets in the subcollection . Let us now formulate the weighted set cover problem as an ILP: The above ILP can be interpreted as follows: if and only if set is in the subcollection . The inequality constraints ensure that every element is covered by at least one set in . The objective function is the total weight of the sets in the subcollection , which we want to minimize. Now we can proceed with our method of linear programming relaxations and rounding. The linear programming relaxation of the above ILP is obtained by relaxing the integrality constraints on . With this, we get the following LP: Suppose we solve the above LP and obtain an optimal solution . We now need to come up with a rounding algorithm that transforms the fractional solution into an integral solution . Can we just round each to the nearest integer as we did in the vertex cover example? Not really. Say a given vertex is an element of 20 of the sets in , and the optimal solution assigns a value of to each of these sets. Then the above rounding algorithm would round each of these values to 0, which is not good, as it would not cover . Instead, we can think of as the “probability” that we would pick set in an optimal solution. This way, is describing a set of “optimal probability distributions” over each of the sets in . Given this interpretation, we can use the following randomized rounding algorithm: Algorithm 1: Random Pick Input: A fractional solution to the LP relaxation of the weighted set cover problem. Output: A subcollection of sets in . (which hopefully covers ) Set . For : Pick set with probability . If is picked, add it to . Return . Note that the expected weight of the subcollection output by the above algorithm is exactly the value of the fractional solution . But will be be a set cover? Let us consider the Random Pick algorithm from the perspective of an element . Let (for simplicity) and . As long as we select at least one of , we are good (with respect to ). Note that we select set with probability , and we know that as is a feasible solution to the LP. What is the probability that is covered by Random Pick? It is definitely not 1, as you can see from the case where and , for instance. The following lemma gives us a bound on the probability that an element is covered by the Random Pick algorithm: Lemma 1 (Probability of covering an element): in a sequence of independent experiments, in which the experiment succeeds with probability , and the probability that at least one of the experiments succeeds is at least . Proof: The probability that no experiment succeeds is Thus, the probability that at least one experiment succeeds is at least . By the above lemma, we see that the probability that an element is covered by the Random Pick algorithm is at least . But we could still have many elements that are not covered by . How do we deal with this? By perseverance! Algorithm 2: Randomized Rounding Input: A fractional solution to the LP relaxation of the weighted set cover problem. Output: A subcollection of sets in that covers . Set . While there is an uncovered element : . Return . The Randomized Rounding algorithm repeatedly applies the Random Pick algorithm until all elements in are covered. To show that the above is a good algorithm, we need to show that we will not execute the Random Pick algorithm too many times (with high probability). This is captured by the following lemma: Lemma 2 (Probability decay): Let . The probability that the Randomized Rounding algorithm executes the Random Pick algorithm more than times is at most . Proof: The probability that the Randomized Rounding algorithm executes the Random Pick algorithm more than times is the probability that there is an uncovered element after iterations. Let . For each iteration, by Lemma 1, the probability that is covered is at least . Hence, the probability that is not covered after iterations is at most By the union bound, the probability that there is an uncovered element after iterations is at most . Now that we know that we wil cover with high probability, we need to bound the cost of the solution we came up with. Suppose that the Randomized Rounding algorithm executes the Random Pick algorithm times. Let be the total weight of the subcollection output by the Randomized Rounding algorithm. At each implementation of the Random Pick process, the expected weight of the subcollection output is exactly the value of the fractional solution . That is . After calls to Random Pick, expected total weight of the subcollection is at most By Markov: We are now ready to get a bound on the cost of the solution output by the Randomized Rounding algorithm: Lemma 3 (Cost of Rounding): Given a fractional solution to the LP relaxation of the weighted set cover problem, the Randomized Rounding algorithm outputs, with probability at least , a set cover of weight at most . Proof: Let . By Lemma 2, there is a probability of at most that the Randomized Rounding algorithm executes the Random Pick algorithm more than times. After calls to Random Pick, the expected total weight of the subcollection is . By Markov, the probability that the total weight of the subcollection is at least is at most . By the union bound, with probability at most , the algorithm either makes more than calls to Random Pick or our solution has weight . Thus, with probability we stop at iterations and construct a solution with weight at most .
5883	https://e2e.ti.com/support/power-management-group/power-management/f/power-management-forum/1430810/lm4051-n-precision-current-source	LM4051-N: Precision current source for Hall sensor - Power management forum - Power management - TI E2E support forums TI E2E support forums Search User Site Search User E2E™ design support > Forums Amplifiers API solutions Audio Clock & timing Data converters DLP® products Interface Isolation Logic Microcontrollers Motor drivers Power management Processors RF & microwave Sensors Site support Switches & multiplexers Tools Wireless connectivity Archived forums Archived groups Technical articles TI training Tech days Online training Live events Power Supply Design Seminar Getting started 简体中文 Power management Power management forum Mentions Tags More Ask a related question What is a related question? A related question is a question created from another question. When the related question is created, it will be automatically linked to the original question. Ask a new question State Locked Locked Replies 4 replies Subscribers 64 subscribers Views 443 views Users 0 members are here REF50E REF35 LM4041-N LM4051 LM4051-N Support feedback Options Tags Options Share More Related LM4051-N: LM4051AIM3-1.2/NOPB Part Number: LM4051-N Tool/software: What is the failure rate in FITs from life test data for part LM4051AIM3-1.2/NOPB TPS63900: CFGx Resistor Precision (Output Voltage Configuration) Part Number: TPS63900 Tool/software: Hello, I was just wondering if CFG1-CFG3 allows any tolerances for the resistor values. For example, to configure V OUT we should select R CFG1 = 36.5 kΩ and R... UCC5350: TYPICAL SOURCE AND SINK CURRENT Part Number: UCC5350 Tool/software: 想确认下，那个是正确的参数，thanks TPS7A57: How to connect a TPS7A57 as a constant current source Part Number: TPS7A57 Tool/software: Hello, good afternoon. How to connect a TPS7A57 as a current source for a very specific application. My load has low resistance (between 0.5 ohms and 1.5 ohms) and I... TPS26750: Controlling BQ charger output voltage/current via TPS25750 for USB-C Source in energy management project Part Number: TPS26750 Other Parts Discussed in Thread: BQ25756 , TPS25750 Tool/software: Hello, I am currently working on an energy management project and would like to clarify a few points regarding... This thread has been locked. If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question. LM4051-N: Precision current source for Hall sensor Chris van der Aar Intellectual570 points Part Number:LM4051-N Other Parts Discussed in Thread:LM4041-N, REF50E, REF35 Tool/software: For flux sensor mesuerements we want to use a Hall sensor where we need a constant current source of 2mA. The manufacturer of the Hall sensor describes the circuit below for offset compensation. The sensor needs a constant current of 2mA. Maybe you can give us some suggestions about the current source that can be used (e.g. LM4051?) and the instrumentation opamp and the output opamp shown in the circuit below? Or maybe a reference design with a hall sensor? We prefer a differential analog output for minimizing noise because it has to be connected with a measurement circuit (probably with a differential analog input but we have to select something for that). The Hall sensor output is 200mV max (positive or negative) for the Hall sensor we are using. The output should be around 5V max I expect (depending on the controller/measuring system we are going to use but this is still under investigation) 11 months ago Edit Insert Format Table Tools Add a reply... Edit Insert Format Table Tools - [x] Suggest as answer Login 0Jackson Wightman11 months ago TI__Expert8940 points Hello, There are a few options to create a current source. LM4051AIM3X-ADJ/NOPB (LM4051-N) or LM4041CEM3X-ADJ/NOPB (LM4041-N) are good options to do so. An example using one of these devices can be seen in section 3.2 of this application note: For increased accuracy, you could consider a configuration as described in section 3.3 of the application note, using a series voltage reference, such as REF35 or REF50E: I will let the op-amp experts comment on which amplifier is best to use here. Thanks, Jackson 0Ron Michallick11 months ago in reply to Jackson Wightman TI__Guru165146 points Chris, What are the supply voltages available? 0Chris van der Aar11 months ago in reply to Ron Michallick Intellectual570 points I gueess +24V from the system but in fact I can make any voltage with a low noide LDO when the current consumption remains low. Or otherwise a low noise converter. I have some examples already from T.I. 0Ron Michallick11 months ago in reply to Chris van der Aar TI__Guru165146 points Chris, Does the processed linear output voltage go an ADC or another linear circuit? I'll assume ADC below, for now using 5V supply and ADC reference. This uses OP1 and NMOS to make a typical current (based on VCC). Use VCC as the ADC reference to cancel error in VCC voltage. After the hall, INA can amplify then signal goes to a single ended (hall up to +/-200mV = [0.5V to 4.5V]) or difference ADC input [-2V to 2V]. Another option is to mirror the VCC to hall current. This changes the VCM for OPA and INA. Otherwise it works the same Of course the INA needs to work with the VCM and output range. On the other hand; If the next stage is also linear, then a precision reference should be used for 2mA generation and any other linear translation or comparison. Edit Insert Format Table Tools Edit Insert Format Table Tools [x] Suggest as answer Reply About TI Company Careers Newsroom Events Investor relations Manufacturing Corporate citizenship Quick links Contact us TI E2E™ design support forums Cross-reference search Customer support center Packaging Quality & reliability myTI account FAQs Buying TI API suites myTI company accounts Shipping, payment & taxes Ordering FAQs Authorized distributors Connect with us Texas Instruments has been making progress possible for decades. We are a global semiconductor company that designs, manufactures, tests and sells analog and embedded processing chips. Our products help our customers efficiently manage power, accurately sense and transmit data and provide the core control or processing in their designs. Accessibility Cookie policy Privacy policy Terms of sale Terms of use Trademarks Website feedback © Copyright 1995-2025 Texas Instruments Incorporated. All rights reserved. Quote
5884	https://www.youtube.com/watch?v=rkNGAQK8jL8	Evaluating Algebraic Expressions in Fractional Form \| Math with Mr. J Math with Mr. J 1730000 subscribers 778 likes Description 56496 views Posted: 24 Oct 2022 Welcome to Evaluating Algebraic Expressions in Fractional Form with Mr. J! Need help with evaluating algebraic expressions? You're in the right place! Whether you're just starting out, or need a quick refresher, this is the video for you if you're looking for help with how to evaluate algebraic expressions. Mr. J will go through examples of evaluating algebraic expressions in fractional form (algebraic expressions with fraction bars) and explain the steps of how to evaluate an algebraic expression using substitution. MORE VIDEOS ON ALGEBRAIC EXPRESSIONS: ✅ Intro to Evaluating Algebraic Expressions = ✅ How to Evaluate Algebraic Expressions with One Variable = ✅ How to Evaluate Algebraic Expressions with Two Variables = ✅ Evaluating Algebraic Expressions with Exponents = ✅ Evaluating Algebraic Expressions with Variables Next to Each Other = ✅ Evaluating Algebraic Expressions in Fractional Form = ✅ What Does it Mean When a Number is Next to a Variable? = About Math with Mr. J: This channel offers instructional videos that are directly aligned with math standards. Teachers, parents/guardians, and students from around the world have used this channel to help with math content in many different ways. All material is absolutely free. #MathWithMrJ Click Here to Subscribe to the Greatest Math Channel On Earth: Follow Mr. J on Twitter: @MrJMath5 Email: math5.mrj@gmail.com Music: Hopefully this video is what you're looking for when it comes to evaluating algebraic expressions in fractional form. Have a great rest of your day and thanks again for watching! ✌️✌️✌️ ✅ Thanks to Aloud, this video has been dubbed into Spanish and Portuguese. #DubbedWithAloud English This video has been dubbed into Spanish (United States) and Portuguese (Brazil) using an artificial voice via to increase accessibility. You can change the audio track language in the Settings menu. Spanish Este video ha sido doblado al español con voz artificial con para aumentar la accesibilidad. Puede cambiar el idioma de la pista de audio en el menú Configuración. Portuguese Este vídeo foi dublado para o português usando uma voz artificial via para melhorar sua acessibilidade. Você pode alterar o idioma do áudio no menu Configurações. 52 comments Transcript: welcome to math with Mr J [Music] in this video I'm going to cover how to evaluate expressions in fractional form so expressions with a fraction bar now when we evaluate an expression we find the value of the expression so what it equals once we work through it for our examples we are going to evaluate the Expressions when x equals 6 and Y equals 10. let's jump into number one where we have 6X over x minus 3. now that fraction bar just means division we have the numerator the top divided by the denominator the bottom so we work through the top work through the bottom then divide you can think of this as the fraction bar splitting the expression into two groups the top and the bottom the numerator and the denominator work through each using the order of operations and then divide now that we know that we will be working through the top and the bottom then dividing the first thing that we need to do is replace or plug in the given values for the variables this is called substitution for number one we have one variable X but it's in two different places let's plug in 6 for x and rewrite this expression we will start up top where we have 6X which is 6 times x that's going to be 6. times 6 over x minus 3 so 6 minus 3. now we can evaluate starting with the top so we have six times six let's work to the side here 6 times 6 equals 36. over 6 minus three now we can work through the bottom so this is going to equal 36 over well 6 minus three equals three so we have 36 over 3 36 divided by 3. that equals 12. so the value of this expression is 12 when x equals 6. to recap we work through the top then we worked through the bottom and we ended by dividing let's move on to number two where we have y squared minus 20 over X Plus y let's plug in and evaluate starting with y squared let's plug in 10 for y so we have 10 squared minus 20 over X Plus y so plug in 6 for x plus 10 for y now we can evaluate let's start with the top so we have 10 squared minus 20. we have an exponent and subtraction exponents come before subtraction as far as the order of operations go so we need to start with 10 squared which means 10 times 10 so 10 squared equals 100. so let's work to the side here starting with the top so 10 squared equals 100 minus 20. over 6 Plus 10. now we can subtract so we have 100 minus 20. let's continue down here so subtract 100 minus 20. that equals 80. over 6 Plus 10. now we can work through the bottom so we have addition this is going to equal 80 over 6 plus 10 equals 16. so now we have 80 over 16 or 80 divided by 16. that equals 5. the value of this expression is 5 when x equals 6 and Y equals 10. so there you have it there's how you evaluate expressions in fractional form I hope that helped thanks so much for watching until next time peace foreign
5885	https://www.emathhelp.net/calculators/algebra-1/slope-intercept-form-calculator-with-two-points/	\| \| \| Home Calculators Calculators: Algebra I Algebra Calculator Slope-Intercept Form Calculator with Two Points Find the slope-intercept form of a line step by step The slope-intercept form calculator will find the slope of the line passing through the two given points, its y-intercept, and the slope-intercept form of the line, with steps shown. Related calculators: Line Calculator, Slope Calculator, Parallel and Perpendicular Line Calculator Your Input Find the equation of a line given two points $$$P = \left(-1, 5\right)$$$ and $$$Q = \left(3, 7\right)$$$. Solution The slope of a line passing through two points $$$P = \left(x_{1}, y_{1}\right)$$$ and $$$Q = \left(x_{2}, y_{2}\right)$$$ is given by $$$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$$$. We have that $$$x_{1} = -1$$$, $$$y_{1} = 5$$$, $$$x_{2} = 3$$$, and $$$y_{2} = 7$$$. Plug the given values into the formula for a slope: $$$m = \frac{7 - 5}{3 - \left(-1\right)} = \frac{1}{2}$$$. Now, the y-intercept is $$$b = y_{1} - m x_{1}$$$ (or $$$b = y_{2} - m x_{2}$$$, the result is the same): $$$b = 5 - \left(\frac{1}{2}\right)\cdot \left(-1\right) = \frac{11}{2}$$$ Finally, the equation of the line can be written in the form $$$y = b + m x$$$: $$$y = \frac{x}{2} + \frac{11}{2}$$$ Answer The slope of the line is $$$m = \frac{1}{2} = 0.5$$$A. The y-intercept is $$$\left(0, \frac{11}{2}\right) = \left(0, 5.5\right)$$$A. The x-intercept is $$$\left(-11, 0\right)$$$A. The equation of the line in the slope-intercept form is $$$y = \frac{x}{2} + \frac{11}{2} = 0.5 x + 5.5$$$A.
5886	https://chem.libretexts.org/Courses/Purdue/Purdue%3A_Chem_26605%3A_Organic_Chemistry_II_(Lipton)/Chapter_13._Addition-Elimination_Sequences/13.1%3A_Nucleophilic_Addition-Elimination/13.1.4_Aldol_Condensation	Skip to main content 13.1.4 Aldol Condensation Last updated : Jun 5, 2019 Save as PDF 13.1.3 Reaction of Aldehydes and Ketones with Nitrogen Nucleophiles 13.1.5 The Wittig Reaction Page ID : 23018 ( \newcommand{\kernel}{\mathrm{null}\,}) Enolizable aldehydes and enolizable ketones, in the presence of an acid or base catalyst in aqueous medium at high temperature, undergo a reaction, giving an α, β-unsaturated aldehyde or an α, β-unsaturated ketone, respectively, as the product. This reaction is known as aldol condensation. The base-catalyzed aldol condensation, in which the catalyst is usually the hydroxide ion, is more common. eg: Mechanism Step 1: The hydroxide ion deprotanates the aldehyde reversibly. Step 2: Enolate ion 1 adds to the unreacted aldehyde. Step 3: Alkoxide ion 2 is protonated by water. Step 4: Aldol 3 is an enolizable aldehyde. A small amount of it is converted to the corresponding enolate ion (4) by the hydroxide ion. Step 5: Enolate ion 4 loses a hydroxide ion. Step 1 through 3 are an aldol reaction, steps 4 and 5 a 1,2-elimination via E1cB mechanism. Thus, aldol condensation is aldol reaction Gamini Gunawardena from the OChemPal site (Utah Valley University) Further Reading Wikipedia Aldol Condensation Khan Academy Aldol Condensation Mixed Aldol Condensation Intramolecular Aldol Condensation Carey 5th Ed Online Aldol Condensation of RCHO Aldol Condensation of Ketones 13.1.3 Reaction of Aldehydes and Ketones with Nitrogen Nucleophiles 13.1.5 The Wittig Reaction
5887	https://www.einstein-online.info/en/spotlight/atombombe/	From E=mc² to the atomic bomb The subtle connections between Einstein’s formula, nuclear fission and nuclear fusion An article by Markus Pössel When Einstein’s most famous formula E=mc² is mentioned, the atomic bomb is usually not far behind. Indeed there is a connection between the two, but it is subtle, and sadly, some popular science texts get it wrong: they will tell you that a nuclear explosion is “caused by the transformation of matter and energy” according to Einstein’s formula, and that the gigantic conversion factor c2 is responsible for the immense power of such weapons. Ten seconds after the ignition of the first atomic bomb, New Mexico, July 16, 1945 [Image: Los Alamos National Laboratory] But first things first. Let’s have a look at what Einstein really did say about the relation between mass and energy. Equivalence or transformation? For Einstein, mass (more precisely: relativistic mass; the property that determines how difficult it is to change a body’s speed or its direction of motion) and energy are simply two different names for one and the same physical quantity. Whenever a system has an energy E, it automatically has the relativistic mass m=E/c2; whenever a system has the mass m, you need to assign it an energy E=mc2. Once the mass is known, so is the energy, and vice versa. In that context, it makes no sense to talk about the “transformation of mass into energy” – where there’s one, there’s the other. The context in which “transformation of mass into energy” does make sense is a bit different. It is intimately connected with the fact that there are different kinds of energy. Already in classical, pre-Einstein physics, the concept of “energy” comprises a plethora of sub-definitions for different sorts of energy, sub-definitions like those for the kinetic energy associated with any moving body, the energy of electromagnetic radiation, thermal energy or the binding energy that needs to be taken into account whenever there is a force holding together two objects to form a composite object. Yet all these different definitions can be viewed as facets of a single physical quantity, energy. The reason is the possibility of transformations between the different energy forms. For instance, you can increase a body’s temperature (and thus its thermal energy) by letting it absorb electromagnetic radiation energy. In these transformations, the total sum of all the different kinds of energy – the total energy – is constant over time. Energy can be transformed from one variety into another, but it can neither vanish nor be created from nothing. A new kind of energy This conservation of energy holds not only in classical physics, but also in special relativity. However, in relativity, the definitions of the different species of energy are a bit different and, most importantly, there is a completely new type of energy: even if a particle is neither moving nor part of a bound system, it has an associated energy, simply because of its mass. This is called the particle’s rest energy, and it is related to the particle’s rest mass as rest energy = (rest mass)· c2. Compared with other types of energy, rest energy is very much concentrated. For example: If you use a television tube to accelerate an electron to 20,000 kilometres per second, the kinetic energy gained is still only about five hundred times smaller than the electron’s rest energy. Also, this rest energy is about a hundred times larger than the radiation energy of a high-energy X-ray photon. This high concentration is important for processes where rest energy (or, equivalently, rest mass) is converted to more common forms of energy. For instance, when a particle and its anti-particle annihilate and vanish in a puff of electromagnetic radiation, comparatively little matter is transformed into rather a lot radiation. Studying the masses of different types of atomic nuclei, you will find that in nuclear fission – the process that powers an ordinary atomic bomb -, some “nuclear rest energy” or “nuclear rest mass” is transformed into other forms of energy. For example, the rest mass of a nucleus of uranium-235 is slightly larger than the combined rest masses of the nuclear fragments into which it splits during nuclear fission. Here’s where E=mc2 comes into play: This mass difference corresponds to the energy set free during nuclear fission. So is it, after all, true that Einstein’s formula explains the power of the nuclear bomb – and that the large conversion factor c2 is responsible for the immense amounts of energy released? Binding energies: nuclei vs. molecules Not at all. Different process, same calculation: For chemical reactions, there are tiny mass differences as well. To pick an example: When hydrogen and oxygen explosively combine to make water, the sum of the rest masses of the initial hydrogen and oxygen atoms is just a little bit more than the sum of the rest masses of the resulting water molecules. The same is true for the chemical reactions involving spontaneous oxydation – in other words: burning. The same formula applies: The mass difference, multiplied by c2, gives the energy set free during the chemical reaction. Same formula, same conversion factor – yet chemical reactions are much less violent than nuclear explosions. This clearly shows that the difference between nuclear and chemical reactions must be due to something other than E=mc2. To see where the difference lies, one must take a closer look. Atomic nuclei aren’t elementary and indivisible. They have component parts, namely protons and neutrons. In order to understand nuclear fission (or fusion), it is necessary to examine the bonds between these components. First of all, there are the nuclear forces binding protons and neutrons together. Then, there are further forces, for instance the electric force with which all the protons repel each other due to the fact they all carry the same electric charge. Associated with all of these forces are what is called binding energies – the energies you need to supply to pry apart an assemblage of protons and neutrons, or to overcome the electric repulsion between two protons. (More information about these binding energies and their role in nuclear fission and fusion can be found in the spotlight topic Is the whole the sum of its parts?) Only with the systematics of these forces and binding energies well understood were physicists able to uncover the laws behind nuclear fission and fusion: The strength of the nuclear bond depends on the number of neutrons and protons involved. It varies in such a way that binding energy is released both in splitting up a heavy nucleus into smaller parts and in fusing light nuclei into heavier ones. This, as well as the chain reaction phenomenon, explains the immense power of nuclear bombs. Einstein’s formula plays second fiddle in that derivation – it’s all about different kinds of energy. Sure, there are some radioactive decay processes following nuclear fission, and, if so inclined, one can view the decay of a neutron decaying into a slightly lighter proton as a transformation of rest energy into other energy forms. But these additional processes contribute a mere 10 per cent of the total energy set free in nuclear fission. The main contribution is due to binding energy being converted to other forms of energy – a consequence not of Einstein’s formula, but of the fact that nuclear forces are comparatively strong, and that certain lighter nuclei are much more strongly bound than certain more massive nuclei. Still, E=mc2 had a supporting role in the story of nuclear fission research. Not as the mechanism behind nuclear power, but as a tool: Because energy and mass are equivalent, highly sensitive measurements of the masses of different atomic nuclei gave the researchers important clues about the strength of the nuclear bond. Einstein’s formula does not tell us why the nuclear binding energies are as large as they are, but it opens up one way (among several) to measure these binding energies. (More about this application of Einstein’s formula can be found in the spotlight topic Is the whole the sum of its parts?) In fact, Einstein’s politics played a more decisive role in the story of the atomic bomb than his physics. Following a request by the physicist Leo Szilard, Einstein wrote a letter to president Roosevelt, explaining about the potential power of nuclear weapons and the possibility of Nazi Germany developing such weapons, and urging the president to take action. Einstein’s letter played its part in setting into motion the political process that culminated in the Manhattan project – the development, construction and testing of the first nuclear bombs. Further Information E=mc2 is a consequence of special relativity. More on that theory can be found in the chapter Special relativity of Elementary Einstein. Related Spotlights on Relativity can be found in the section Special relativity. Information about the historical background can be found on the page The Nuclear Age on “Einstein – Image and Impact”, a website of the American Institute of Physics. Colophon Markus Pössel is the managing scientist at Haus der Astronomie, the Center for Astronomy Education and Outreach in Heidelberg, and senior outreach scientist at the Max Planck Institute for Astronomy. He initiated Einstein Online. Citation Cite this article as: Markus Pössel, “From E=mc² to the atomic bomb” in: Einstein Online Band 04 (2010), 01-1004 The case of the travelling twins The dark heart of the Milky Way Weitere Artikel Special relativity / Elementary Tour part 6: E=mc² Is the whole the sum of its parts? Gravitationswellen / Einsteiger-Tour Teil 3: Gravitationswellen-Astronomie Kosmologie / Einsteiger-Tour Teil 1: Kosmos auf Expansionskurs Kosmologie / Einsteiger-Tour Teil Fazit Special relativity / Elementary Tour part 2: The principle of relativity Tag Cloud astronomy Basic properties Basics Basics black holes Black Holes & Co. Close encounters with black holes Cosmology Cosmology and quantum gravity Detecting gravitational waves Energy and mass From Newton to Einstein - and beyond General relativity General theory of relativity Gravitational wave gravitational wave detector Gravitational wave sources Gravitational waves How to track compact objects interferometrischer Detektor Light in general relativity Loop quantum gravity Mass and other sources of gravity Miscellaneous Miscellaneous neutron stars Neutronenstern Numerical relativity Other approaches to the problem of quantum gravity Physics in the background Relativity and the Quantum Singularities Special relativity Special relativity and time String theory The basics of general relativity The first chemical elements The mathematics behind general relativity The physics of black holes What is Einstein Online? Einstein Online is a web portal with comprehensible information on Einstein's theories of relativity and their most exciting applications from the smallest particles to cosmology. Einstein Online is provided by the Max Planck Institute for Gravitational Physics (Albert-Einstein-Institut) About Einstein Online Includes information on our authors and contributing Institutions, and a brief history of the website. Glossary More than 400 entries from "absolute zero" to "XMM Newton" - whenever you see this type of link on an Einstein Online page, it'll take you to an entry in our relativistic dictionary. a b c d e f g h i j k l m n o p q r s t u v w x y z Links & Literatur Our recommendations for books and websites on relativity and its history.
5888	https://www.youtube.com/watch?v=Mq8-1F8nZgM	USING SYMMETRY IN PROBABILITY COOL MATH 190 subscribers 17 likes Description 760 views Posted: 11 Jun 2021 We will use symmetry to solve a probability problem. But there is a slight problem. The problem is not symmetric! 1 comments Transcript: Intro hello welcome to another cool math video my name is ali today we are going to look at a probability problem and use symmetry in an unusual way all right let's let me pull out the problem Read the problem it's a past amc 10 problem let's read it first tina randomly selects two distinct numbers from the set one up to five and sergio randomly selects a number from the set volume to 10. the probability that sergio's number is larger than the sum of the two numbers chosen by tina is what so let's first underline the Finding symmetry keywords here two distinct numbers from this set and schedule selects a random number from one up to 10 and we want sergio's number to be larger than the sum of the two numbers by tina okay you may ask where's the symmetry here they don't select the numbers from the same interval and we have a sum of two numbers here we have just one number here where is the symmetry well the answer is there is no symmetry at least not yet so we'll develop we'll change it a little bit to get a symmetry so first let's set up our unknowns let's say tina's numbers are t1 and t2 and their sum is t and sergio's number is s so the range for t is the smallest is three and the biggest we can get is nine four plus five and s is from one up to ten now they are not the same interval right and we don't i mean there's symmetry in s but this do we have symmetry and t let's see i had seen this before maybe not this tank numbers but if you roll it die twice and get the sum right then the distribution is is going to be between 2 and 12 and it's going to be symmetric around 7. right the number of cases for 2 more than 7 nine is going to be the same as the number of cases for two less than seven five so they're probably the the product of five and nine are the same they're symmetry for the probability distribution around seven and here we have the same case there's only one case to get three one plus two there's also one case to get nine four plus five right so three and nine have the same number of cases or the same probabilities and the same way you can check that four and eight five and seven have the same they appear wise they get the same probability so it's the distribution for t is symmetric around six how do you use that we still have s that's not in that range we can make it in that range if we deal with the other cases separately so let's divide into three cases s could be one or two this could be 10 or it could be from 3 up to 9 in the symmetric range we want which is most of the cases okay so s when s or two and since sergio is picking up his number randomly the chance of this happening is two-tenths then his number the probability that s is going to be bigger than t is none right because t is always at least 3. and if s is 10 with probability 1 10 then the chance for x bigger than t is one it's always going to be bigger than t and for this interval and there are seven numbers left now so 7 out of 10 cases the chance that s is going to be bigger than t now this probability now we have the symmetry so we expect half the time s bigger than t half the times to be which will be smaller than t except they could be the same right so other than the cases where they are the same it should be half the time s bigger than t now how do you find the problem that they are the same so t is a little bit more complicated but sergio is simple he's a simple guy he picks a number randomly from one up to ten in this case randomly from three up to nine right so whatever tina's sum is whatever t is what is the chance that s is that particular t no matter what t is if t is three the chance that sergio picks three is one out of seven if t is six the chance that sergio picks six is one out of seven in this particular case i'm talking about when we restrict the sergious numbers to three to nine so always it's gonna be one seventh chance whatever t is one stable chance sergio's number is going to be the same as t and then in the remaining cases half the time sergio's number is going to be strictly bigger half the time it's going to be strictly smaller so this is our probability and then it remains to just multiply these with their initial chances and add them so 2 out of 10 times the product is 0 1 out of 10 times the probability is 1 and 7 out of 10 times the probability is what is this number six sevenths over two so three sevenths that probably is three sevenths then we get one-tenths plus another three-tenths so that's four-tenths which is same as two-fifths yeah it's in the choices all right so if we didn't see the symmetric solution how could we approach it so let's say this is the symmetric nice cool solution the standard solution does casework or based on we can we can check there are two ways other either you can pick sergio's number and see what's the problem that tina's sum is going to be bigger than or smaller than that and do casework based on that or you can pick tina's some and see what's the chance that sergio's number is bigger than that so you have to do it's like double case work for both t and s so you have to fix one of them first and then deal with the other and and change those so uh you you could do s equals uh let's test private t so t equals three four five six seven eight nine if t is three the chance of t being three is one out of ten and then in these chances what's the chance that sergio's number is bigger than three that's seven out of ten if tina's sum is four with probably two out of ten feet it's still one out of ten because they are distinct right one and three so one out of ten times then sergio's number bigger than four is six out of ten five one four two three so there are two possibilities to get five out of five choose to ten cases and then you see sergio's probabilities decrease as tina's sum increase right that makes sense and then for t equals six one five two four still two possibilities and this is decreased to four out of 10 now 7 2 5 3 4 2 possibilities and then 8 and you get eight three and five that's it one tenth two tenths and then nine only one way and then serious number has to be ten so you would have to express this sum and then you know multiply and add them and it you should get 40 out of i'm cheating here cutting from the answer but you should get 40 out of 100 all the denominators are 100 the numerators should add up to 40. it's pretty long but you can see the symmetry for tina's sum here the distribution is you know around the ends 1 10 to 1 10 and then another 1 10 to 1 10 and then 2 10 2 10 then at the middle it's another two tenths all right so that's the that's how you Conclusion can tweak a problem even if it's not symmetric at the beginning and deal with the unsymmetric cases separately and then use the symmetry exploit the symmetry in a probability problem hope you enjoyed it for me more videos like this please subscribe thank you yeah that was much better than before good good
5889	https://www.youtube.com/watch?v=jsXETt1mf8A	Scale Factors - Word Problems (7th Grade) Math Swap Tutoring with Shawn, Tutor and Engineer 151 subscribers 2 likes Description 187 views Posted: 4 Oct 2024 Scale factors are a concept related to the constant of proportionality. They're used to show how a shape is mathematically scaled up or down from its original size. Let's apply these to some word problems. Free Resources Get a free copy of my studying protocol by becoming a subscriber Register for my next free SAT Math workshop Need math tutoring? Schedule your free consultation at or call (478)254-0515 Need SAT math help? One-on-one tutoring - Schedule your free consultation at or call (478)254-0515 SAT Math Mastery Online Course Follow us on Social Media Facebook Instagram Timestamps Join the Math Swap Tutoring e-mail list: Don't forget to LIKE this video and to SUBSCRIBE to this channel! mathswap Transcript: what's going on y'all Sean of massat tutoring here with a video on scale factor specifically word problems um having to do with scale factors now before you uh continue to watch make sure you like the video make sure you subscribe as well um and let's get into it so from a previous video um I talked about scale factors um and you can find the scale factor by using the following equation K is equal to y/x y is a measurement um from your scaled shape X is a measurement from your original shape I just write o r i g all right now so if you want more explanation on that go to my previous video so let's apply it to a word problem here um that reads as follows Sam drew a scale of a box the scale he used was 3 in is equal to 1 foot um what scale factor does his drawing use so going by the formula for scale factor K the scale factor is going to be equal to a measurement from the scaled version the scale shape is going to be 3 Ines because the scale is actually the drawing if you think about it okay and it's going to be measured in inches so 3 in and and then one foot is the Bottom now the key with scale F the SK the sorry the key with converting this scale to a scale factor is you have to have the same units um I say in the numerator and denominator and the reason why and I I wrote that kind of fast so hopefully you follow what I wrote but the reason why that's important is because think about what a scale factor is it's just a number you want the same units because when you have the same unit in the top the numerator and the denominator the units canceled out the same way if you were finding like um um greatest common factor between a numerator and the denominator and they divide out to one the the the units divide out to one as well so you want to be able to have the same units in the top and the bottom so that that will be so that the the scale factor will only be a number and let me see if I can help that make help you make sense of that by doing it here so we need to ask ourselves so 3 in um for every one foot now let's conver convert this unit down here to inches how many inches are in a foot okay 12 in so we're going to say what we're going to do here is we're going to say this is um also equal to 3 Ines for every 12 Ines all right so what did we do all we did was all we did was we um so we got the scale we got the scale here and then because we want the the units to match we went ahead and ask ourselves separately how many inches are in the foot so we can have inches at the bottom too and so we know that 12 inches are in the foot and so we now rewrote this scale as 3 in over 12 in the units cancel so now we have 3 over 12 3 over 12 both of those share a factor of three that divides out so K is equal to 1 over4 that is our scale factor for this example okay so we have another example here and let's go over it now I do want you to pause the video on this one to try it yourself to see if you can kind of replicate what I did the previous example your on your own okay so try it yourself okay so this example says Janet made a scale a scale drawing of a house the scale she used was 3 cm equal 6 M what is the scale factor of the drawing so remember K the scale factor is going to be equal to Y over X so remember y corresponds to the scale X corresponds to the original now the scale you know that this measurement right here the centimeters the smaller the smaller measurement that's going to be the scale cuz that's where the the drawing isn't be is going to be drawn in centimeters right so that goes up top and then the 6 M goes on the bottom because that's going to be from the original the house right okay now what do we need to do we need to convert so that we have the same units up top and on the bottom now you got centimeters up top you got meters on the bottom let's convert meters to cenim now we know how many centimeters how many centimeters are in a meter one meter is 100 cm therefore we need to multiply 6 times 100 to get how many centimet are in 6 M so that's 600 cm in in the um on the bottom still got 3 cmet on the top centimeters up top centimeters on the bottom cancel so we have three over 600 both of these have a common fact of three so the scale factor is going to be one over and then 600 by three is 200 okay all right let me know if you got that um in the comments make sure you like make sure you subscribe make sure you share with a friend otherwise have a great day hey thank you for watching this video if you need math tutoring or you need SAT Math help look in the description to find out how you can sign up today also don't forget to like share and subscribe so you stay in touch with mass swap tutoring on YouTube have a good one
5890	https://www.quora.com/What-are-the-roots-or-quadratic-of-ax2-bx-c-0	Something went wrong. Wait a moment and try again. Roots of Quadratic Equati... Basic Algebra Solving Quadratic Equatio... Quadratic Formula Quadratic Polynomials Mathematics and Algebra 5 What are the roots or quadratic of ax2+bx+c=0? Ak Murthy Former Retired as Prof. of mathematics 30 yrs of service · Author has 1.2K answers and 800.7K answer views · 4y The specific word 'quadratic' w.r.t the equation ax²+bx+c = 0 tacitly means that a≠0. In that case this equation has exactly two roots imaginary or real ,distict are equal. These two roots are given by the formula x = (—b±√( b²—4ac)/2a. The nature of the two roots is decided by the expression “ b²—4ac” For this reason " b²—4ac” is called the discriminant of the quadratic equation denoted by D( or by ∆) . When a,b c are real If D > 0, the two roots are real and distinct roots. If D = 0, then the two roots real and equal. If D< 0 then the roots are complex conjugates . If a,b,c are integers and D is a The specific word 'quadratic' w.r.t the equation ax²+bx+c = 0 tacitly means that a≠0. In that case this equation has exactly two roots imaginary or real ,distict are equal. These two roots are given by the formula x = (—b±√( b²—4ac)/2a. The nature of the two roots is decided by the expression “ b²—4ac” For this reason " b²—4ac” is called the discriminant of the quadratic equation denoted by D( or by ∆) . When a,b c are real If D > 0, the two roots are real and distinct roots. If D = 0, then the two roots real and equal. If D< 0 then the roots are complex conjugates . If a,b,c are integers and D is a nonzero perfect square ( that is D is the square of an integer), then the the two roots are rational and distinct . If D= 0 then the roots are rational and equal. Condition for the' equation ‘ ax² +bx +c = 0 ( carefully note the avoidance of the adjective ' quadratic’ for the equation)to have exactly one root ‘∞’ is a= 0, b ≠ 0. ( and a finite root —c/b ) The condition for the equation to have two infinite roots is a=b= 0 and c ≠ 0 This case will look very funny in that the equality 7 = 0 as a statement is a contradiction. But this equality stated as an equation( from ax²+bx + 7 = 0 where a=b= 0) refers to the fact that the equation 7 = 0 has two infinite roots! These ideas of infinite roots are used in Analytic Geometry when we find the equations of the two asymptoes of the hyperbola (x²/a² )—(y²/b²)= 1 Finally when a= b=c= ,0, the equation ax²+bx+c = 0:reduces to an identity and the solution set is the eniire R( or C according as the universal set is R or C ) Related questions What is the relation of a quadratic equation's sum of roots and product of roots, i.e. ax2+bx+c=0 and let its roots are α and β? What is the solution of the quadratic equation ax2 + bx + c = 0 given by? What are roots of a quadratic equation? What would they be for a x 2 + b x + c What is the roots of quadratic equation x²+6x-14=0? How do you solve the quadratic equation: ax^2 + bx + c = 0? Mohammad Afzaal Butt B.Sc in Mathematics & Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views · 6y Originally Answered: What is the root of equation ax^2 +bx+c = 0? · ax2+bx+c=0⟹x2+bax+ca=0 ⟹x2+bax+b24a2−b24a2+ca=0 ⟹(x+b2a)2−b2−4ac4a2=0 ⟹(x+b2a)2−(√b2−4ac2a)2=0 ⟹(x+b2a−√b2−4ac2a)(x+b2a+√b2−4ac2a)=0 ⟹(x+b−√b2−4ac2a)(x+b+√b2−4ac2a)=0 ⟹x ax2+bx+c=0⟹x2+bax+ca=0 ⟹x2+bax+b24a2−b24a2+ca=0 ⟹(x+b2a)2−b2−4ac4a2=0 ⟹(x+b2a)2−(√b2−4ac2a)2=0 ⟹(x+b2a−√b2−4ac2a)(x+b2a+√b2−4ac2a)=0 ⟹(x+b−√b2−4ac2a)(x+b+√b2−4ac2a)=0 ⟹x=−b±√b2−4ac2a Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views · 4y Related Let x1 and x2 be the roots of the quadratic equation ax2+bx+c=0 By using quadratic formula, find the roots of ax2+bx+c=0 Find x1+ x2 and x1 x2. What relationships do you observe? The use of x1 and x2 will make this question must too messy so I am going to let the solutions (or roots) be α and β The above is a very awkward w... The use of x1 and x2 will make this question must too messy so I am going to let the solutions (or roots) be α and β The above is a very awkward w... Assistant Bot · Sep 6 For the quadratic equation ax^2 + bx + c = 0 (with a ≠ 0), the two roots are x = [ -b ± sqrt(b^2 − 4ac) ] / (2a). Details and cases: Discriminant Δ = b^2 − 4ac Δ > 0: two distinct real roots. Δ = 0: one real repeated root x = −b / (2a). Δ < 0: two complex conjugate roots x = [−b ± i·sqrt(4ac − b^2)] / (2a). Alternative forms and notes: If b^2 ≫ 4ac, direct evaluation is fine. If subtraction causes loss of significance (when b and sqrt(Δ) are nearly equal), use numerically stable formulas: Let q = −(b + sign(b)·sqrt(Δ)) / 2, then roots are x1 = q / a and x2 = c / q. For a = 0 the equation reduces to line For the quadratic equation ax^2 + bx + c = 0 (with a ≠ 0), the two roots are x = [ -b ± sqrt(b^2 − 4ac) ] / (2a). Details and cases: Discriminant Δ = b^2 − 4ac Δ > 0: two distinct real roots. Δ = 0: one real repeated root x = −b / (2a). Δ < 0: two complex conjugate roots x = [−b ± i·sqrt(4ac − b^2)] / (2a). Alternative forms and notes: If b^2 ≫ 4ac, direct evaluation is fine. If subtraction causes loss of significance (when b and sqrt(Δ) are nearly equal), use numerically stable formulas: Let q = −(b + sign(b)·sqrt(Δ)) / 2, then roots are x1 = q / a and x2 = c / q. For a = 0 the equation reduces to linear bx + c = 0 (root x = −c / b when b ≠ 0). Related questions What are the roots of the quadratic equation ax^2+bx+c=0, if c=a-b? What condition will make the roots of the quadratic equation be a x 2 + b x + c = 0 ( a ≠ 0 ) ? How do you solve the quadratic equation (ax^2 + bx + c = 0) for (x)? What are the roots of the equation ax^2+bx+c=0? 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I wrote out a whole answer, which I’ve left below for reference, before seeing what’s going on here. So, what is our equation? ax2+bx+c=0 And what is the condition that says the sum of the coefficients is zero? a+b+c=0 Do you see it yet? a.12+b.1+c=0 How about now? The condition that the sum of the coefficients be zero is precisely the condition that x=1 be a solution of the equation. And this will work for a polynomial of any degree, not just a Sometimes in mathematics, you work through a problem and get to a solution, and only at the end does it become clear that there is a much simpler approach. I wrote out a whole answer, which I’ve left below for reference, before seeing what’s going on here. So, what is our equation? ax2+bx+c=0 And what is the condition that says the sum of the coefficients is zero? a+b+c=0 Do you see it yet? a.12+b.1+c=0 How about now? The condition that the sum of the coefficients be zero is precisely the condition that x=1 be a solution of the equation. And this will work for a polynomial of any degree, not just a quadratic. The sum of the coefficients of a polynomial is zero if and only if it has 1 as a root. Here, then, is my first and much more complicated answer: Let’s ignore the case when a=0 for now, because it’s not the interesting part of the answer. Since we can multiply all the coefficients by a non-zero constant without changing the roots or the condition that the coefficients sum to zero, we can assume a=1. Then we have x2−(c+1)x+c=0 Conveniently, this can be factorised! (x−c)(x−1)=0 So the roots are 1 and c. Are there any conditions on c? None at all! So, if the coefficients of a quadratic equation sum to zero, it has 1 as a root. And the converse holds too, as it’s easy to check by multiplying out a(x−c)(x−1)=0. Now would be a good time to return to the linear case a=0. Then we have bx+c=0, so b+c=0 if and only if x=1 or we have the further degenerate case a=b=c=0. In summary, a quadratic or linear equation has coefficients summing to zero if and only if 1 is a root. I wonder if this works for polynomials of higher degree. Wait a minute… Robert Nichols Author has 5K answers and 15.6M answer views · 5y Originally Answered: What are the roots of the quadratic equation ax²+bx+c=0 by square method? · What are the roots of the quadratic equation ax²+bx+c=0 by square method? Move the constant to the other side, by subtracting c from both sides ax²+bx = -c Divide all terms by a, x²+(b/a)x = -c/a Take half of the coefficient of the x-term, in this case, b/2a Square the term you got from step 2, (b/2a)² Add this term to both sides, x²+(b/a)x + (b/2a)²= -c/a + (b/2a)² Rewrite the left half as a perfect square, (x + b/2a)² = -c/a + (b/2a)² Write -c/a as -4ac/4a ² , so that you have a common denominator for the right side, (x + b/2a)² = (-4ac)/(4a² )+ (b²)/4a² Write the right side as a single fraction, (x + What are the roots of the quadratic equation ax²+bx+c=0 by square method? Move the constant to the other side, by subtracting c from both sides ax²+bx = -c Divide all terms by a, x²+(b/a)x = -c/a Take half of the coefficient of the x-term, in this case, b/2a Square the term you got from step 2, (b/2a)² Add this term to both sides, x²+(b/a)x + (b/2a)²= -c/a + (b/2a)² Rewrite the left half as a perfect square, (x + b/2a)² = -c/a + (b/2a)² Write -c/a as -4ac/4a ² , so that you have a common denominator for the right side, (x + b/2a)² = (-4ac)/(4a² )+ (b²)/4a² Write the right side as a single fraction, (x + b/2a)² = (-4ac + b²)/4a² Take the square root of both sides, x + b/2a = ±√{(-4ac + b²)/4a²} Simplify the right side ± √{(-4ac + b²)/4a²} => ±{√(-4ac + b²)}/√(4a²) => ±√(-4ac + b²)/2a To isolate x subtract b/2a from both sides, x = -b/(2a) ±√(-4ac + b²)/2a Combine the like terms on the left since they have the same denominator , x = {-b±√(-4ac + b²)}/2a Rewrite it by switching the order of (-4ac + b²) into (b² -4ac), so it now is the quadratic formula: x = {-b±√(b²- 4ac)}/2a Promoted by US Auto Insurance Now US Auto Insurance Now Helping Drivers Find Great Car Insurance Deals · Tue What are some of the most effective ways to save money? 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Whether it's driving for a rideshare service, freelancing your skills (writing, graphic design, social media), or selling crafts online, a side hustle can be a powerful way to accelerate debt payoff or build up your savings. The extra income can be a great boost to your financial goals. Dean Rubine Author of: It's not just π; all of trig is wrong! · Author has 10.6K answers and 23.7M answer views · 3y Originally Answered: What is a pseudocode that will calculate the roots of quadratic equation ax²+bx+c=0? · ``` include #include #include voidQuadraticRoots(double a, double b, double c){ double d = bb - 4ac; double r, s, t; if(a == 0) { printf("Not a quadratic equation\n"); return; } if(d == 0) { // might want fabs(d) 0) { t = sqrt(d); r = (-b + t) / (2a); s = (-b - t) / (2a); printf("Two real roots: %g and %g\n", r, s); } else { r = - b / (2a); t = sqrt(-d)/ fabs(2a); printf("Two complex roots: %g + %g i and %g - %g i\n", r, t, r, t); }} ``` That’s some actual C code. I haven’t attemp ``` include #include #include voidQuadraticRoots(double a, double b, double c){ double d = bb - 4ac; double r, s, t; if(a == 0) { printf("Not a quadratic equation\n"); return; } if(d == 0) { // might want fabs(d) 0) { t = sqrt(d); r = (-b + t) / (2a); s = (-b - t) / (2a); printf("Two real roots: %g and %g\n", r, s); } else { r = - b / (2a); t = sqrt(-d)/ fabs(2a); printf("Two complex roots: %g + %g i and %g - %g i\n", r, t, r, t); }} ``` That’s some actual C code. I haven’t attempted to compile it so there are probably some issues lurking. Theoretically it has the problem that it’s calculating an approximation to the roots, because the square roots can’t generally be done exactly. Also the code assumes real coefficients a , b , c ; in general they might be complex. But approximations to the roots of an equation with real coefficients is probably what’s intended, so I’ll leave it here. Jishnu Mukherjee B.Tech from Indian Institute of Technology, Kharagpur (IIT KGP) (Expected 2026) · Author has 156 answers and 851.9K answer views · 6y Originally Answered: What are the roots of the quadratic equation ax²+bx+c=0, if the sum and products of roots are 4 and roots are also positive? · Let the roots be:- α,β Given:- α+β=4 ⟹α=4−β……………….(1) α∗β=4 ⟹(4−β)∗β=4[See(1)] ⟹4β−β2=4 ⟹β2–4β+4=0[Multiplying both sides by -1] ⟹β2–2β−2β+4=0[Middle term break] ⟹β(β−2)−2(β−2)=0[Taking common] ⟹(β−2)(β−2)=0 ⟹ Either:-β−2=0 β=2 or, β−2=0 β=2. Both values show:-β=2 Now from (1) we know that :- α=4−β ⟹α=4–2 ⟹α=2 So both the roots of the given quadratic equation are equal and real. The roots are:- α=2 β Let the roots be:- α,β Given:- α+β=4 ⟹α=4−β……………….(1) α∗β=4 ⟹(4−β)∗β=4[See(1)] ⟹4β−β2=4 ⟹β2–4β+4=0[Multiplying both sides by -1] ⟹β2–2β−2β+4=0[Middle term break] ⟹β(β−2)−2(β−2)=0[Taking common] ⟹(β−2)(β−2)=0 ⟹ Either:-β−2=0 β=2 or, β−2=0 β=2. Both values show:-β=2 Now from (1) we know that :- α=4−β ⟹α=4–2 ⟹α=2 So both the roots of the given quadratic equation are equal and real. The roots are:- α=2 β=2 Hope it helps you. Thanks for the question. Promoted by Webflow Metis Chan Works at Webflow · Sep 10 What’s the best CMS for teams to build and publish websites quickly? Most CMS platforms are slow and require dev help for even small changes. Webflow combines website building and a visual CMS in one platform, so lean teams can move much faster. Build custom pages visually, no code required Update and publish content instantly with a visual CMS Keep everything secure with hosting, SEO, and workflows built in Fivetran publishes 4x more content with Webflow’s visual CMS. Try Webflow for free today and launch websites and content faster. … (more) Donald Hartig PhD in Mathematics, University of California, Santa Barbara (Graduated 1970) · Author has 7.4K answers and 2.8M answer views · Sep 3 Originally Answered: What are the roots of the quadratic equation ax²+bx+c=0 by square method? · 0=a(x2+bax+ca) ⟹0=x2+bax+ca=(x+b2a)2+ca−b24a2 Consequently,x=−b2a±√b2−4ac2a. Ambar Chatterjee Former Head Nuclear Physics Division at Bhabha Atomic Research Centre (BARC) (1976–2012) · Author has 495 answers and 902.3K answer views · 1y Originally Answered: What is the solution of the quadratic equation ax2 + bx + c = 0 given by? · The solution x=(-b+-sqrt(b^2–4ac)/2a can derived by a procedure called “completing the square”. Divide the equation by a: x^2 + (b/a)x + (c/a) = 0. Transpose the last term: x^2 + (b/a)x = -c/a. Add (b/2a)^2 on both sides: x^2 + (b/a)x + (b/2a)^2 = (-c/a) + (b/2a)^2. Using the identity a^2 + 2ab + b^2 = (a + b)^2, we can simplify the LHS to [x + (b/2a)]^2 = (b^2 – 4ac)/4a^2. Taking the square root on both sides, we get x + (b/2a) = ±√(b^2-4ac)/2a. Therefore, the quadratic formula is x = [-b ± √(b^2-4ac)]/2a. If you know high school algebra basics you could derive the result in numerous oth The solution x=(-b+-sqrt(b^2–4ac)/2a can derived by a procedure called “completing the square”. Divide the equation by a: x^2 + (b/a)x + (c/a) = 0. Transpose the last term: x^2 + (b/a)x = -c/a. Add (b/2a)^2 on both sides: x^2 + (b/a)x + (b/2a)^2 = (-c/a) + (b/2a)^2. Using the identity a^2 + 2ab + b^2 = (a + b)^2, we can simplify the LHS to [x + (b/2a)]^2 = (b^2 – 4ac)/4a^2. Taking the square root on both sides, we get x + (b/2a) = ±√(b^2-4ac)/2a. Therefore, the quadratic formula is x = [-b ± √(b^2-4ac)]/2a. If you know high school algebra basics you could derive the result in numerous other ways. The Indian mathematician Brahmagupta (597–668 AD) explicitly described the quadratic formula in his treatise Brāhmasphuṭasiddhānta published in 628 AD. Probably Diaphontus knew this in 250 AD. Michael Tam Technical Services/HACCP Coordinator · Author has 453 answers and 463.3K answer views · 6y ax^2 + bx + c = 0 Divide the whole equation by a, where a does not equal 0. x^2 + (b/a)x + (c/a) = 0 Complete the square on x^2 + (b/a)x You would halve b/a to get b/2a, then square it to get b^2/(4a^2) (x^2 + (b/a)x + (b^2/(4a^2)) - (b^2/(4a^2)) + c/a = 0 (x + b/2a)^2 -b^2/4a^2 + 4c/(4a) - 0 (x + b/2a)^2 +(-b^2 + 4ac)/(4a^2) = 0 (x + b/2a)^2 = (b^2 - 4ac)/(4a^2) x + b/2a = +/- sqrt (b^2 - 4ac)/(2a) x = - b/2a +/- sqrt (b^2 - 4ac)/(2a) Express this with a common denominator of (2a) x = (-b +/- sqrt (b^2 - 4ac))/(2a) This is how the quadratic equation was derived. Dodie Cowan Former Professor of Mathematics at Polk State College (2004–2015) · Author has 2.2K answers and 1.3M answer views · 5y Originally Answered: What are the roots of the quadratic equation ax²+bx+c=0 by square method? · The square method means that you want to express the sides in a form where you can take the square root of each side. It is also known as “Completing the Square”. To use this method, you want the right side to be in the form (x-k)^2. So, first, subtract the constant term: ax2+bx=−c. Now, divide through by the leading coefficient (a): x2+bax=−ca. Now, take half of the coefficient of the x term, square it, and add to both sides: x2+bax+b24a2=−ca+b24a2 The left side can now be factored as a perfect square. On the right side, get a common denom The square method means that you want to express the sides in a form where you can take the square root of each side. It is also known as “Completing the Square”. To use this method, you want the right side to be in the form (x-k)^2. So, first, subtract the constant term: ax2+bx=−c. Now, divide through by the leading coefficient (a): x2+bax=−ca. Now, take half of the coefficient of the x term, square it, and add to both sides: x2+bax+b24a2=−ca+b24a2 The left side can now be factored as a perfect square. On the right side, get a common denominator. (x+b2a)2=−4ac4a2+b24a2 (x+b2a)2=b2−4ac4a2 Next, we take the square root of each side: x+b2a=±√b2−4ac4a2 x+b2a=±√b2−4ac2a Finally, solve for x: x=±√b2−4ac2a−b2a x=−b±√b2−4ac2a We usually memorize this as the Quadratic Formula so that we don’t have to go through all these steps every time we do it. Related questions What is the relation of a quadratic equation's sum of roots and product of roots, i.e. ax2+bx+c=0 and let its roots are α and β? What is the solution of the quadratic equation ax2 + bx + c = 0 given by? What are roots of a quadratic equation? What would they be for a x 2 + b x + c What is the roots of quadratic equation x²+6x-14=0? How do you solve the quadratic equation: ax^2 + bx + c = 0? What are the roots of the quadratic equation ax^2+bx+c=0, if c=a-b? What condition will make the roots of the quadratic equation be a x 2 + b x + c = 0 ( a ≠ 0 ) ? How do you solve the quadratic equation (ax^2 + bx + c = 0) for (x)? What are the roots of the equation ax^2+bx+c=0? What are the roots of quadratic equations AX 2 +BX+C=0 6. GCD and LCM of two numbers? Does the quadratic formula hold is the equation is ax^2+bx- c=0? The equation a x 2 + b x + c = 0 will not represent a quadratic equation, if? If the roots of the quadratic equation x²-2kx+2k²-4=0 are real, then what is the range of values of k? What are the roots of the quadratic 21x^2-37x-28=0? What are the roots/solutions of quadratic equation x² + 3x -10 =0? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
5891	https://www.securitycameraking.com/securitynews/badge-door-entry-system/?srsltid=AfmBOooHfipvv1tGLS_Lm5quvGjYWhdWix03AO5A3x_vr-rFzEHAjW1d	Badge Door Entry System: Top 5 Secure Solutions 2025 Skip to content 561.288.5258 Support Close Support Open Support Downloads Support Ticket Remote Support Support Fourm Quick Start Guides Resellers Close Resellers Open Resellers Become a Reseller 561.288.5258 Contact Us About Us My Account 561.288.5258 Support Close Support Open Support Downloads Support Ticket Remote Support Support Fourm Quick Start Guides Resellers Close Resellers Open Resellers Become a Reseller 561.288.5258 Contact Us About Us My Account Surveillance Close Surveillance Open Surveillance Security Cameras Coaxial HD Cameras Network IP Cameras Cameras By Style Video Recorders Coaxial Recorders Network Recorders Full Systems Accessories CCTV Video Storage Security Camera Cables Security Camera Connectors Camera Tools & Accessories Camera Mounting Brackets Power Supplies DVR Lockboxes Surveillance Camera Monitors Networking Special Solutions Elite Software Licenses Access ControlClose Access Control Open Access Control Access Control Packages Access Control Panels and Boards Access Control Readers Standalone Access Control Readers Access Control Cards Electronic Door Strikes Electromagnetic Locks Request to Exit Buttons Access Control Power Supplies Access Control Cable Access Control Accessories Overstock CCTV Learning Center Close CCTV Learning Center Open CCTV Learning Center CCTV Knowledge Base Calculators Downloads Security Camera Blog Video Vault Quick Start Guides What is HD-CVI Smart Phone Access Videos Security Camera Videos Security NVR/DVR Videos Lendschart and Information Resolution Comparisons Cannabis Compliance Archived Blog Articles Surveillance Security Cameras Coaxial HD Cameras Network IP Cameras Video Recorders Coaxial Recorders Network Recorders Software Licences Cameras By Style Video Doorbells Video Intercoms Bullet Cameras Indoor Dome Cameras Vandal Dome Cameras Pan Tilt Cameras License Plate Cameras Box Security Cameras Hidden Cameras Full Systems Special Solutions Accessories CCTV Video Storage Security Camera Cables Security Camera Connectors Camera Tools & Accessories Camera Mounting Brackets Power Supplies DVR Lockboxes Surveillance Camera Monitors Networking Access Control Access Control Packages Access Control Panels and Boards Access Control Readers Access Control Standalone Readers Access Control Cards Electronic Door Strikes Electromagnetic Locks Request To Exit Buttons Access Control Power Supplies Access Control Cable Access Control Accessories Access Control Software Overstock CCTV Learning Center CCTV Forum Calculators Downloads Security Camera Blog Video Vault Quick Start Guides What is HD-CVI? 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Smartphone Access Videos Security DVR/NVR Videos Lens Chart Information Resolution Comparisons Cannabis Compliance No products in the cart. $0.00 0 Cart 10% off + Free Shipping over $100 Big Sale Now! 00 Days 00 Hours 00 Minutes 00 Seconds Sorry :) The Sale Has Ended! Use Coupon: LaborDay25 Products search Search Access Control Entry Made Easy: Best Badge Door Systems for Your Business February 19, 2025 Exploring the convenience and security of a badge door entry system is essential for modern businesses looking to improve access control without compromising daily operations. A badge door entry system offers a secure and efficient way to control access to your facility. With advancements in technology, these systems use methods such as RFID, magnetic strips, and smart technologies to manage entry and exit, helping to protect assets and ensure the safety of your staff and customers. Key benefits of a badge door entry system: Improved Security: Controls unauthorized access and keeps records of entry actions. Ease of Use: Simple for employees to access areas using badges instead of keys. Scalability: Fits businesses of any size, from small retail shops to large corporate buildings. My name is Brad Besner SCK, founder and president of Security Camera King, Inc., based in Boca Raton, FL. With years of experience in the security industry, I specialize in solutions like badge door entry system to help businesses safeguard their premises effectively. Understanding Badge Door Entry Systems Components of Badge Door Entry Systems A badge door entry system is an effective way to manage who can enter specific areas of your business. It typically consists of three main components: ID badges, card readers, and door locks. Each part plays a crucial role in ensuring seamless and secure access. ID Badges: These are the credentials your employees or authorized personnel carry. They can be swipe badges with magnetic stripes, RFID badges, or scannable badges with QR codes. Each type has its own technology and level of security. Card Readers: These devices read the information stored in the ID badges. Depending on the technology, card readers can detect information through swiping, proximity, or scanning. The reader then communicates with the system to verify the credentials. Door Locks: Once the card reader verifies the credentials, it sends a signal to the door lock to open up the door. Magnetic locks are often used in these systems because they are robust and reliable. They engage and disengage quickly, allowing for smooth and secure access. How Badge Door Entry Systems Work The operation of a badge door entry system can be broken down into a few simple steps: Credential Verification: When a user presents their ID badge to a card reader, the reader checks the badge’s credentials against a database. This process is quick, often taking just a few seconds. Access Authorization: If the credentials are valid, the system authorizes access. This step ensures that only those with the correct permissions can enter the area. It helps maintain the security of sensitive areas within your business. Electronic Locks: Once access is authorized, the card reader sends a signal to the door’s electronic lock to open. This lock can be a magnetic lock or another type of electronic lock, depending on the system’s configuration. This combination of components offers a streamlined and secure way to manage entry and exit within your facility. By leveraging such technology, businesses can improve security while providing convenient access to authorized personnel. Let’s explore the different types of badge credentials that can be integrated into these systems. Types of Badge Credentials When it comes to badge door entry systems, choosing the right type of badge credential is key to balancing security, convenience, and cost. Let’s explore the three main types of badge credentials: RFID badges, swipe badges, and scannable badges. RFID Badges RFID (Radio Frequency Identification) badges are a popular choice for businesses that need a contactless entry solution. These badges use radio waves to communicate with card readers. The reader emits an electromagnetic field that powers the RFID chip in the badge, allowing it to transmit data back to the reader. Range: Depending on the system, RFID readers can detect badges from 2.5 inches to 20 feet away. Convenience: Users simply hold their badge near the reader for access, making it quick and easy. Security: RFID badges are more secure than some other types because they’re harder to replicate. Swipe Badges Swipe badges are a more traditional option. They use a magnetic stripe to store data, which is read when the badge is swiped through a card reader. Reusability: The magnetic stripe can be rewritten, so if an employee’s access needs to change, you can update their existing badge without issuing a new one. Durability: These badges can wear out over time due to repeated swiping, which might lead to more frequent replacements. Security: While effective, swipe badges can be less secure than RFID since the magnetic stripe can be damaged or demagnetized easily. Scannable Badges Scannable badges are the most cost-effective option, often featuring a QR code printed on the badge. Users simply hold their badge up to a scanner for access. Affordability: These badges are inexpensive to produce, making them ideal for large organizations. Ease of Use: Scanning a QR code is straightforward and quick. Security Risks: QR codes are easier to replicate than magnetic stripes or RFID chips, potentially posing a security risk if not managed properly. Each type of badge credential has its own strengths and weaknesses. When selecting the right one for your business, consider factors like security needs, budget, and user convenience. By choosing the right credential, you can ensure that your badge door entry system effectively balances security and ease of access. Next, let’s look at the advantages and disadvantages of implementing these systems in your business. Advantages and Disadvantages of Badge Door Entry Systems When considering a badge door entry system for your business, it’s crucial to weigh the pros and cons. Let’s break down the key advantages and disadvantages. Advantages Increased Security One of the biggest perks of a badge entry system is the boost in security. With these systems, you know exactly who is entering your building and when. This is because the system logs every entry, helping you monitor access in real-time. If a badge is lost or stolen, you can easily deactivate it, minimizing security risks. Moreover, badge systems work well with multiple entry points. This is perfect for buildings with restricted areas, like server rooms or executive offices. Only authorized personnel can enter these zones, keeping sensitive areas secure. Simplified Access Gone are the days of fumbling with keys. With a badge system, entry is as easy as a swipe or tap. Employees simply hold their badge up to a reader, and voilà, they’re in. This not only speeds up entry but also reduces the need for security personnel to manually check IDs. Disadvantages Unreliable Cards Despite their benefits, badge systems aren’t without flaws. Badges can be unreliable due to their slim size and sensitivity to damage. They might stop working if exposed to magnets or cell phones. This can be frustrating for users and disrupt smooth entry. Expensive Replacements Replacing badges can quickly become costly. While a single badge might cost around $10, the expenses add up if many badges are lost or damaged. Plus, staff time is needed to deactivate old badges and set up new ones, which can be inefficient. In summary, badge door entry systems offer increased security and simplified access, but they come with potential downsides like unreliable cards and expensive replacements. When choosing a system, consider these factors to ensure it aligns with your business needs. Next, we’ll explore real-world use cases for these systems to see how they perform in different environments. Use Cases for Badge Door Entry Systems Badge door entry systems have become a cornerstone for security in various environments. Let’s explore how different sectors use these systems to improve security and streamline access. Office Suites Office suites benefit significantly from badge door entry systems. In shared office spaces, these systems help segment access between different businesses. This ensures that only authorized personnel can enter specific office suites. Tenants can customize access levels, granting full building access to some employees while restricting others to certain floors or areas. Additionally, temporary digital badges can be issued to guests, streamlining visitor management and maintaining security. Medical Offices In medical offices, safeguarding patient records and treatment areas is paramount. Badge entry systems restrict access to healthcare personnel and authorized staff, ensuring compliance with HIPAA security standards. These systems also control access to medical equipment and labs, allowing only qualified personnel into sensitive areas. In emergencies, real-time staff location information provided by badge systems improves evacuation procedures and patient safety. Financial Institutions For financial institutions like banks, security is non-negotiable. Badge systems restrict access to sensitive areas such as vaults and teller stations. This ensures that only authorized personnel can enter high-security zones, protecting valuable assets. Compliance with strict regulatory standards is also crucial, and badge entry logs provide a valuable audit trail to demonstrate adherence to access control policies. Moreover, temporary badges with limited permissions can be issued to manage visitor entry for clients and vendors. Schools In schools, ensuring the safety of students and faculty is a top priority. Badge systems secure classrooms, labs, and faculty offices, preventing unauthorized access. Schools can issue temporary access badges for parents, guest speakers, or maintenance staff, ensuring the building remains secure while allowing necessary access. Additionally, these systems track movement across campus buildings, improving overall security and enabling quick responses to emergencies or unauthorized entries. In conclusion, badge door entry systems offer custom solutions for various sectors, enhancing security and simplifying access. Next, we’ll address some frequently asked questions about these systems to provide further clarity. Frequently Asked Questions about Badge Door Entry Systems What technology do badge readers use? Badge readers use several technologies to identify and authenticate users. RFID (Radio Frequency Identification) is one of the most popular methods. RFID badges, also known as proximity cards, use radio waves to communicate with the reader. This allows for contactless entry, which is both convenient and efficient. Another common technology is the magnetic stripe, found in swipe badges. These badges have a magnetized stripe that stores data. When swiped through a reader, the system checks the data and grants access if it’s valid. QR codes are also used in some systems. These are simple to produce and inexpensive, making them a popular choice for temporary or low-security access. However, they can be less secure than RFID or magnetic stripe options. How much do badge entry systems cost? The cost of a badge entry system can vary widely depending on several factors. On average, the cost per door can range from a few hundred to several thousand dollars. This depends on the complexity of the system and the technology used. Installation fees are another consideration. It’s essential to hire a certified installer to ensure the system is set up correctly and that warranties remain valid. Installation costs can vary based on the complexity of the setup and the location of the facility. Do badges use RFID? Yes, many badges use RFID technology. RFID badges are popular because they offer a contactless way to grant access, making them convenient for users. These badges are often referred to as proximity cards due to their ability to work from a short distance. While not all badges use RFID, it is a common choice for businesses looking for a reliable and efficient access control solution. In summary, badge door entry systems incorporate various technologies to meet different security needs. Understanding these options can help businesses choose the right system for their requirements. Next, we’ll dig into the benefits and potential drawbacks of these systems. Conclusion In today’s world, ensuring the security of your business premises is more critical than ever. Badge door entry systems offer a robust solution for controlling access and enhancing security. By using technologies like RFID, magnetic stripes, and QR codes, these systems provide a seamless way to manage who enters your building and when. At Security Camera King, Inc., we specialize in delivering top-notch access control solutions that cater to businesses of all sizes. Our products are designed to be reliable, efficient, and easy to use, ensuring that your security needs are met without hassle. One of the significant advantages of implementing a badge door entry system is improved security. By keeping unauthorized individuals out and maintaining detailed records of access events, these systems provide peace of mind and protection for your assets. Moreover, with the flexibility to integrate with other security measures like video surveillance, you can create a comprehensive security strategy custom to your unique needs. Another benefit is the convenience offered by these systems. Gone are the days of managing countless keys or worrying about lost access cards. With features like remote access management and automatic credential updates, you can easily control access permissions and respond swiftly to any security concerns. In conclusion, badge door entry systems are a smart investment for any business aiming to improve its security framework. With the expertise and support from Security Camera King, Inc., you can implement a system that not only meets your current needs but also adapts to future challenges. Contact us today to explore how our solutions can safeguard your business and provide you with the security you deserve. 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5892	https://www.pharmacytimes.com/view/balancing-cancer-risks-and-benefits-of-oral-contraceptives	Advertisement News Article Balancing Cancer Risks and Benefits of Oral Contraceptives Author(s): Danielle Valletti, Assistant Editor Fact checked by: Laura Joszt, MA Listen Key Takeaways Hormonal contraceptives slightly increase breast cancer risk, normalizing within 10 to 15 years post discontinuation, with similar risk across delivery methods. Long-term use of hormonal contraceptives significantly reduces ovarian and endometrial cancer risk, with protection lasting over 20 years post cessation. Pharmacists should clearly communicate relative and absolute cancer risks, using visual tools and age-based risk charts to aid patient understanding. Screening for contraindications and co-risk factors is essential, with method selection reflecting individual health backgrounds and reproductive goals. SHOW MORE Hormonal contraceptives present a nuanced risk-benefit profile, slightly increasing breast and cervical cancer risk while significantly reducing ovarian and endometrial cancer rates. Hormonal contraceptives are widely used for reproductive health and family planning but carry a complex risk-benefit profile in relation to cancer. Evidence from large studies, including Oxford and US cohorts, consistently shows a small, short-term increase in breast and cervical cancer risk with current or recent use, but substantial, long-term reductions in ovarian and endometrial cancers.1,2 A landmark UK nested case-control study published in PLOS Medicine analyzed nearly 9500 women younger than 50 years diagnosed with invasive breast cancer and over 18,000 matched controls.2 Findings show that current or recent use of any hormonal method, including combined oral contraceptives, oral progestagen-only pills, injectable progestagens, implants, and intrauterine devices (IUDs), was associated with a 20% to 30% relative increase in breast cancer risk. Absolute excess risks were age dependent, estimated at 8 cases per 100,000 users aged 16 to 20 years, and up to 265 per 100,000 at age 35 to 39 years, after 5 years of use.Elevated risk declined and returned to baseline within 10 to 15 years post discontinuation.Risk magnitude was similar across delivery methods from pills to IUDs, highlighting that progestagen-only and combined formulations confer comparable breast cancer risk.2,3 Image Credit: New Africa \| stock.adobe.com Ovarian and Endometrial Cancer Evidence from large meta-analyses confirms that hormonal contraceptive use yields approximately 40% to 50% reduced risk of both ovarian and endometrial cancers, with protection lasting at least 20 years post-cessation.3,4 The Collaborative Group’s meta-analysis encompassing 36 epidemiologic studies involving over 27,000 women with endometrial cancer found that each 5-year use was associated with a 24% lower risk, and that reduction persisted for more than 30 years after stopping.4 Nurses’ Health Study II also showed nearly 57% lower risk for endometrial cancer among women using contraceptives for over 10 years.3 These long-term benefits must be weighed against short-term risks when counseling patients. Strategies for Pharmacists When educating patients, pharmacists should communicate both relative and absolute risks clearly. Although hormonal contraceptive use is linked to a relative 20% to 30% increase in breast cancer risk, absolute excess risk remains low, especially in younger women.2 Visual tools and age-based risk charts, such as illustrating that at ages 35 to 39 years, 5 years of use raises breast cancer rates from 2.0% to 2.2%, can help clarify risk.Meanwhile, protective effects such as reducing the risk of ovarian and endometrial cancer remain durable and clinically significant.3,4 Screening for contraindications and co–risk factors like smoking, hypertension, migraines with aura, or coagulation disorders is essential. While estrogen-containing contraceptives carry a small venous thrombosis risk, progestagen-only methods may reduce vascular risk, though they maintain an equivalent breast cancer risk compared to combined contraceptives.2 Method selection should therefore reflect each patient’s health background and reproductive goals. Pharmacists can develop patient education materials that present short-term trade-offs and long-term benefits of contraceptive use. Incorporate electronic medical record alerts for cervical and breast cancer screening, risk assessment prompts, and follow-up planning. Collaboration with ob-gyns and primary care providers ensures contraceptive planning aligns with broader health considerations and cancer risk reduction strategies. Conclusion Both combined and progestagen-only hormonal contraceptives are associated with a slight, time-limited increase in breast cancer risk, particularly during current or recent use, with risk normalizing within a decade after stopping.Conversely, these methods provide powerful and enduring protection against ovarian and endometrial cancers.3,4 Pharmacists equipped with accurate, evidence-based information can guide women toward informed contraceptive choices that prioritize both reproductive autonomy and long-term health. REFERENCES National Cancer Institute. Oral contraceptives and cancer risk. National Cancer Institute. Published February 22, 2018. Accessed July 30, 2025 Fitzpatrick D, Pirie K, Reeves G, Green J, Beral V. Combined and progestagen-only hormonal contraceptives and breast cancer risk: a UK nested case-control study and meta-analysis. PLOS Med. 2023;20(3):e1004188. doi:10.1371/journal.pmed.1004188 Burchardt NA, Shafrir AL, Kaaks R, Tworoger SS, Fortner RT. Oral contraceptive use by formulation and endometrial cancer risk among women born in 1947-1964: the Nurses' Health Study II, a prospective cohort study. Eur J Epidemiol. 2021;36(8):827-839. doi:10.1007/s10654-020-00705-5 Collaborative Group on Epidemiological Studies on Endometrial Cancer. Endometrial cancer and oral contraceptives: an individual participant meta-analysis of 27 276 women with endometrial cancer from 36 epidemiological studies.Lancet Oncol. 2015;16(9):1061-1070. doi:10.1016/S1470-2045(15)00212-0 Newsletter Stay informed on drug updates, treatment guidelines, and pharmacy practice trends—subscribe to Pharmacy Times for weekly clinical insights. Subscribe Now! 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5893	https://www.chemguide.co.uk/basicorg/isomerism/ez.html	\| \| \| E-Z NOTATION FOR GEOMETRIC ISOMERISM This page explains the E-Z system for naming geometric isomers. \| \| Important!If you have come straight here via a search engine, you should be aware that this page follows on from an introductory page about geometric isomerism. Unless you are already confident about how geometric isomers arise, and the cis-trans system for naming them, you should follow this link first. You will find links back to this current page at suitable points on that page. \| \| The E-Z system The problem with the cis-trans system for naming geometric isomers Consider a simple case of geometric isomerism which we've already discussed on the previous page. You can tell which is the cis and which the trans form just by looking at them. All you really have to remember is that trans means "across" (as in transatlantic or transcontinental) and that cis is the opposite. It is a simple and visual way of telling the two isomers apart. So why do we need another system? There are problems as compounds get more complicated. For example, could you name these two isomers using cis and trans? Because everything attached to the carbon-carbon double bond is different, there aren't any obvious things which you can think of as being "cis" or "trans" to each other. The E-Z system gets around this problem completely - but unfortunately makes things slightly more difficult for the simple examples you usually meet in introductory courses. How the E-Z system works We'll use the last two compounds as an example to explain how the system works. You look at what is attached to each end of the double bond in turn, and give the two groups a "priority" according to a set of rules which we'll explore in a minute. In the example above, at the left-hand end of the bond, it turns out that bromine has a higher priority than fluorine. And on the right-hand end, it turns out that chlorine has a higher priority than hydrogen. If the two groups with the higher priorities are on the same side of the double bond, that is described as the (Z)- isomer. So you would write it as (Z)-name of compound. The symbol Z comes from a German word (zusammen) which means together. \| \| Note:I'm not getting bogged down in the names of these more complex compounds. As soon as I put the proper full names in, the whole thing suddenly looks much more complicated than it really is, and you will start to focus on where the whole name comes from rather than on if it is a (Z)- or (E)- isomer. \| \| If the two groups with the higher priorities are on opposite sides of the double bond, then this is the (E)- isomer. E comes from the German entgegen which means opposite. So the two isomers are: Summary (E)- : the higher priority groups are on opposite sides of the double bond. (Z)- : the higher priority groups are on the same side of the double bond. \| \| Note:Three possible suggestions for remembering this: E is for "Enemies", which are on opposite sides. You don't, of course, need a way of remembering the Z as well - it's just the other way around from E. In Z isomers, the higher priority groups are on zee zame zide. That works best if you imagine you are an American speaking with a stage German accent! This is the way I remembered it when I first came across E-Z notation. It is more visual than the other methods. It relies on the fact that the shapes of E and Z isomers are the opposite of the shapes of the letters E and Z. In the letter E, the horizontal strokes are all on the same side; in the E isomer, the higher priority groups are on opposite sides. In the letter Z, the horizontal strokes are on opposite sides; in the Z isomer, the groups are on the same side. Choose whichever of these methods make most sense to you. \| \| Rules for determining priorities These are known as Cahn-Ingold-Prelog (CIP) rules after the people who developed the system. The first rule for very simple cases You look first at the atoms attached directly to the carbon atoms at each end of the double bond - thinking about the two ends separately. The atom which has the higher atomic number is given the higher priority. Let's look at the example we've been talking about. Just consider the first isomer - and look separately at the left-hand and then the right-hand carbon atom. Compare the atomic numbers of the attached atoms to work out the various priorities. Notice that the atoms with the higher priorities are both on the same side of the double bond. That counts as the (Z)- isomer. The second isomer obviously still has the same atoms at each end, but this time the higher priority atoms are on opposite sides of the double bond. That's the (E)- isomer. What about the more familiar examples like 1,2-dichloroethene or but-2-ene? Here's 1,2-dichloroethene. Think about the priority of the two groups on the first carbon of the left-hand isomer. Chlorine has a higher atomic number than hydrogen, and so has the higher priority. That, of course, is equally true of all the other carbon atoms in these two isomers. In the first isomer, the higher priority groups are on opposite sides of the bond. That must be the (E)- isomer. The other one, with the higher priority groups on the same side, is the (Z)- isomer. And now but-2-ene . . . This adds the slight complication that you haven't got a single atom attached to the double bond, but a group of atoms. That isn't a problem. Concentrate on the atom directly attached to the double bond - in this case the carbon in the CH3 group. For this simple case, you can ignore the hydrogen atoms in the CH3 group entirely. However, with more complicated groups you may have to worry about atoms not directly attached to the double bond. We'll look at that problem in a moment. Here is one of the isomers of but-2-ene: The CH3 group has the higher priority because its carbon atom has an atomic number of 6 compared with an atomic number of 1 for the hydrogen also attached to the carbon-carbon double bond. The isomer drawn above has the two higher priority groups on opposite sides of the double bond. The compound is (E)-but-2-ene. A minor addition to the rule to allow for isotopes of, for example, hydrogen Deuterium is an isotope of hydrogen having a relative atomic mass of 2. It still has only 1 proton, and so still has an atomic number of 1. However, it isn't the same as an atom of "ordinary" hydrogen, and so these two compounds are geometric isomers: The hydrogen and deuterium have the same atomic number - so on that basis, they would have the same priority. In a case like that, the one with the higher relative atomic mass has the higher priority. So in these isomers, the deuterium and chlorine are the higher priority groups on each end of the double bond. That means that the left-hand isomer in the last diagram is the (E)- form, and the right-hand one the (Z)-. Extending the rules to more complicated molecules If you are reading this because you are doing a course for 16 - 18 year olds such as UK A level, you may well not need to know much about this section, but it really isn't very difficult! Let's illustrate this by taking a fairly scary-looking molecule, and seeing how easy it is to find out whether it is a (Z)- or (E)- isomer by applying an extra rule. Focus on the left-hand end of the molecule. What is attached directly to the carbon-carbon double bond? In both of the attached groups, a carbon atom is attached directly to the bond. Those two atoms obviously have the same atomic number and therefore the same priority. So that doesn't help. In this sort of case, you now look at what is attached directly to those two carbons (but without counting the carbon of the double bond) and compare the priorities of these next lot of atoms. You can do this in your head in simple cases, but it is sometimes useful to write the attached atoms down, listing them with the highest priority atom first. It makes them easier to compare. Like this . . . In the CH3 group: The atoms attached to the carbon are H H H. In the CH3CH2 group: The atoms attached directly to the carbon of the CH2 group are C H H. In the second list, the C is written first because it has the highest atomic number. Now compare the two lists atom by atom. The first atom in each list is an H in the CH3 group and a C in the CH3CH2 group. The carbon has the higher priority because it has the higher atomic number. So that gives the CH3CH2 group a higher priority than the CH3 group. Now look at the other end of the double bond. The extra thing that this illustrates is that if you have a double bond, you count the attached atom twice. Here is the structure again. So, again, the atoms attached directly to the carbon-carbon double bond are both carbons. We therefore need to look at what is attached to those carbons. In the CH2OH group: The atoms attached directly to the carbon are O H H. In the CHO group: The atoms attached directly to the carbon are O O H. Remember that the oxygen is counted twice because of the carbon-oxygen double bond. In both lists, the oxygens are written first because they have a higher atomic number than hydrogen. So, what is the priority of the two groups? The first atom in both lists is an oxygen - that doesn't help. Look at the next atom in both lists. In the CH2OH group, that's a hydrogen; in the CHO list, it's an oxygen. The oxygen has the higher priority - and that gives the CHO group a higher priority than the CH2OH group. The isomer is therefore a (Z)- form, because the two higher priority groups (the CH3CH2 group and the CHO group) are both on the same side of the bond. That's been a fairly long-winded explanation just to make clear how it works. With a bit of practice, it takes a few seconds to work out in any but the most complex cases. One more example to make a couple of additional minor points . . . Here's an even more complicated molecule! Before you read on, have a go at working out the relative priorities of the two groups on the left-hand end of the double bond, and the two on the right-hand end. There's another bit of rule that I haven't specifically told you yet, but it isn't hard to guess what it might be when you start to look at the problem. If you can work this out, then you won't have any difficulty with any problem you are likely to come across at this level. Look first at the left-hand groups. In both the top and bottom groups, you have a CH2 group attached directly to the carbon-carbon double bond, and the carbon in that CH2 group is also attached to another carbon atom. In each case, the list will read C H H. There is no difference between the priorities of those groups, so what are you going to do about it? The answer is to move out along the chain to the next group. And if necessary, continue to do this until you have found a difference. Next along the chain at the top left of the molecule is another CH2 group attached to a further carbon atom. The list for this group is again C H H. But the next group along the chain at the bottom left is a CH group attached to two more carbon atoms. Its list is therefore C C H. Comparing these lists atom by atom, leads you to the fact that the bottom group has the higher priority. Now look at the right-hand groups. Here is the molecule again: The top right group has C H H attached to the first carbon in the chain. The bottom right one has Cl H H. The chlorine has a higher atomic number than carbon, and so the bottom right group has the higher priority of these two groups. The extra point I am trying to make with this bit of the example is that you must just focus on one bit of a chain at a time. We never get around to considering the bromine at the extreme top right of the molecule. We don't need to go out that far along the chain - you work out one link at a time until you find a difference. Anything beyond that is irrelevant. For the record, this molecule is a (Z)- isomer because the higher priority groups at each end are on the same side of the double bond. Can you easily translate cis- and trans- into (Z)- and (E)-? You might think that for simple cases, cis- will just convert into (Z)- and trans- into (E)-. Look for example at the 1,2-dichloroethene and but-2-ene cases. But it doesn't always work! Think about this relatively uncomplicated molecule . . . This is clearly a cis- isomer. It has two CH3 groups on the same side of the double bond. But work out the priorities on the right-hand end of the double bond. The two directly attached atoms are carbon and bromine. Bromine has the higher atomic number and so has the higher priority on that end. At the other end, the CH3 group has the higher priority. That means that the two higher priority groups are on opposite sides of the double bond, and so this is an (E)- isomer - NOT a (Z)-. Never assume that you can convert directly from one of these systems into the other. The only safe thing to do is to start from scratch in each case. Does it matter that the two systems will sometimes give different results? No! The purpose of both systems is to enable you to decode a name and write a correct formula. Properly used, both systems will do this for you - although the cis-trans system will only work for very straightforward molecules. \| \| \| Questions to test your understanding If this is the first set of questions you have done, please read the introductory page before you start. You will need to use the BACK BUTTON on your browser to come back here afterwards. questions on E/Z notation \| Where would you like to go now? To the isomerism menu. . . To menu of basic organic chemistry. . . To Main Menu . . . © Jim Clark 2007 (last modified November 2012) \|
5894	https://www.youtube.com/watch?v=SeIGDiERb0k	Ex: Find the Parametric Equations of a Line Perpendicular to a Plane Through a Point Mathispower4u 330000 subscribers 597 likes Description 135320 views Posted: 8 Feb 2014 This video explains how to determine the parametric equations of a line that is perpendicular to a plane through a given point. Site: 23 comments Transcript: we want to complete the parametric equations of the line through the point 1A 6 comma -5 and perpendicular to the plane given by the equation 3x + 8 y - Z = -2 we're looking at the point normal form of the equation of a plane that contains the point x sub 1 comma y sub 1 comma Z sub1 and has a normal Vector n with components a b and c which has an equation as we see here notice if we distribute the coefficients of X Y and Z would give us the components to a normal Vector to the plane and since our line is perpendicular to the plane this normal Vector would be a direction Vector of the line or the line would be parallel to this normal Vector which is all we need in order to find the equation of the line a line parallel to a vector v with components a B and C passing through the point X sub1 Y sub 1 comma Z sub1 has the following parametric equations before we find our equations though let's take a look at this graphically we're given the equation of this yellow plane and we're asked to find the parametric equations of the line that's perpendicular to the plane passing through this blue point well looking at the normal Vector here to the plane notice how the line is parallel to this normal vector and therefore we can use this normal vector and this point to find the equation of the line perpendicular to the plane so going back to our problem looking at the equation of the plane we need to be able to recognize that this tells us a normal Vector to the plane Vector n would have an X component of three a y component of 8 and a z component of1 and this is also the direction Vector to the line perpendicular to the plane or the line will be parallel to this Vector which means looking at the rric equations of our line this tells us that a is equal to 3 B = 8 and cal-1 and because we're given our line passes through the point 1 comma 6 comma -5 this tells th that x sub 1 = 1 y sub 1 = 6 and Z sub 1al 5 notice for this problem they gave us the first equation for x of T they gave us this so we knew exactly which normal vector to use as the vector that the line would be parallel to any scalar multiple of this Vector n would also be a vector that the line would be parallel to but because the coefficient of T is positive3 we know this is the exact Vector we want to use to find our equations so again we know that X of T even though it's given would be equal to x sub 1 + a T which would be 1 + 3 t or 3 t y of T would be equal to Y sub 1 + B T which would be 6 + 8 T and Z of T would be equal to Z sub 1 + C T which would be -5 minus 1 t or just minus t these parametric equations would give us the line perpendicular to the plane passing through the given point just remember these parameters equations are not unique any scalar multiple of vector n would also give us a vector that's parallel to the line perpendicular to the plane passing through the given point again we just found the parametric equations of the line in our graph here that's perpendicular to the plane passing through this blue point I hope you found this helpful
5895	https://onlinelibrary.wiley.com/doi/10.1155/2013/828214	A Comparison of Hemoglobin A2 Levels in Untreated and Treated Groups of HIV Patients on ART Including Zidovudine - Nigam - 2013 - Pathology Research International - Wiley Online Library Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Skip to Article Content Skip to Article Information Search within Search term Advanced SearchCitation Search Search term Advanced SearchCitation Search Login / Register Individual login Institutional login REGISTER Pathology Research International Volume 2013, Issue 1 828214 Research Article Open Access A Comparison of Hemoglobin A 2 Levels in Untreated and Treated Groups of HIV Patients on ART Including Zidovudine Jitendra Singh Nigam, Jitendra Singh Nigam Department of Pathology, D.D.U. Hospital, Hari Nagar, New Delhi 110066, India Search for more papers by this author Jyotsna Naresh Bharti, Jyotsna Naresh Bharti Department of Pathology, Maulana Azad Medical College, New Delhi 110002, India mamc.ac.in Search for more papers by this author Dinker Kumar, Corresponding Author Dinker Kumar dinker.kumar25@gmail.com Department of Pathology, D.D.U. Hospital, Hari Nagar, New Delhi 110066, India Search for more papers by this author Ankit Sharma, Ankit Sharma Department of Pathology, D.D.U. Hospital, Hari Nagar, New Delhi 110066, India Search for more papers by this author Jitendra Singh Nigam, Jitendra Singh Nigam Department of Pathology, D.D.U. Hospital, Hari Nagar, New Delhi 110066, India Search for more papers by this author Jyotsna Naresh Bharti, Jyotsna Naresh Bharti Department of Pathology, Maulana Azad Medical College, New Delhi 110002, India mamc.ac.in Search for more papers by this author Dinker Kumar, Corresponding Author Dinker Kumar dinker.kumar25@gmail.com Department of Pathology, D.D.U. Hospital, Hari Nagar, New Delhi 110066, India Search for more papers by this author Ankit Sharma, Ankit Sharma Department of Pathology, D.D.U. Hospital, Hari Nagar, New Delhi 110066, India Search for more papers by this author First published: 26 December 2013 Citations: 2 Academic Editor: Piero Tosi This article is part of Special Issue: About References ---------- Related ------- Information ----------- PDF Sections Abstract 1. Introduction 2. Materials and Methods 3. Results 4. Discussion 5. Conclusion References Citing Literature PDF Tools Export citation Add to favorites Track citation ShareShare Give access Share full text access Close modal Share full-text access Please review our Terms and Conditions of Use and check box below to share full-text version of article. [x] I have read and accept the Wiley Online Library Terms and Conditions of Use Shareable Link Use the link below to share a full-text version of this article with your friends and colleagues. Learn more. Copy URL Share a link Share on Email Facebook x LinkedIn Reddit Wechat Bluesky Abstract Objective. To assess the level of haemoglobin A 2 in HIV patients on antiretroviral therapy (ART) including zidovudine with untreated HIV patients. Material and Methods. The study was a case control study. A total of 185 patients were included in the study; the case group included 125 HIV patients who were on antiretroviral therapy (ART) including zidovudine and 60 were in the control group who were not receiving ART. The high-performance liquid chromatography (HPLC) was done and hemoglobin A 2 level was observed; value more than 3.5% was considered significant. The Hb A 2 percentages of HIV patients were compared with those of control using an unpaired t-test. Results. The mean of Hb A 2 in case group was 3.462% (SD 0.675) and in control group it was 2.815% (SD0.246). The higher Hb A 2 value was seen in significant number of treated patients than control group (P< 0.0001). Conclusion. The clinicians, pathologists, haematologists, and genetic counsellors should be aware of effects of nutritional anaemia and ART on Hb A 2 to reduce the chances of misdiagnosis of β-thalassaemia especially in developing countries and for centres for antenatal screening. 1. Introduction Cooley and Lee were the first to describe the diminished rate of synthesis of one or more globin chains or its part, which consequently results in reduced rate of synthesis of the haemoglobin. Whipple and Bradford in 1936 gave the term “Thalassaemia” . β-thalassaemia is commonly seen in Mediterranean region, Indian subcontinent, Southeast Asia, and African ancestry . For β-thalassaemia, screening of populations was done either by measuring the haemoglobin A 2 percentage or by red cell indices in patients with MCV or MCH below a certain cut-off point . The haemoglobin A 2 is 4%-5% in most cases of heterozygous β 0 or severe β+ thalassaemia and 3.6–4.2% in heterozygous mild β+ thalassaemia . The raised Hb A 2 values are seen in unstable haemoglobins, hyperthyroidism, and megaloblastic anaemia, and human-immunodeficiency-virus- (HIV-) infected patients on antiretroviral therapy, while hypochromic microcytic red cell picture is seen only in heterozygous β-thalassaemia . History, clinical details, and coexisting condition help to rule out the causes of increase in Hb A 2, which can avoid unnecessary investigation and reduce the financial burden of institutional setting in developing countries. Awareness of the fact that there are increases in Hb A 2 due to antiretroviral drugs among clinicians and pathologists should be must. Spiga et al. reported that zidovudine inhibits β-globin gene expression in human erythroid progenitor cells . HIV-infected patients receiving highly active ARV therapy (HAART) or ZDV are associated with elevation of Hb A 2 value [2–5]. Galacteros et al. observed that HIV-infected patients at stage IV of disease have moderate elevation of Hb A 2 level and beta-thalassemia-like unbalanced biosynthetic globin ratio . 2. Materials and Methods This study was a case control study. A total of 185 patients were included in the study, who were recruited from the ART center of the DDU Hospital. The case group included 125 HIV patients who were on antiretroviral therapy (ART) including zidovudine, nevirapine, and lamivudine for the last three years and 60 untreated newly diagnosed patients were in the control group who were not receiving ART. All these patients were tested with rapid ELISA method for HIV and V.D.R.L for syphilis. In case group, 75 patients were male and 50 were female. The complete clinical data, haemogram, osmotic fragility (OF) test, and biochemical investigations were done. All the patients were tested for HIV. The red cell indices, red cell distribution width (RDW), and OF test were done for the evaluation of thalassaemia. High-performance liquid chromatography (HPLC) was done and hemoglobin A 2 level was observed; value more than 3.5% was considered significant. The Hb A 2 percentages of HIV patients were compared with those of control using an unpaired t-test. HbE trait, history of alcohol, and abnormal liver function test were the exclusion criteria for the study. 3. Results The red cells indices show mean MCV 96.88 fl, and 54.4% of (68/125) patients had MCV > 100 fl. The rest of the patients had MCV within normal range. In control group, all patients have MCV within normal range with mean value of MCV 82 fl. RDW is increased in 60.8% (76/125) of patients with mean RDW 20.9. Serum ferritin, folic acid levels, and vitamin B12 level were within normal limits. The mean of Hb A 2 in HIV patients on treatment was 3.462% (SD 0.675) and control has a mean Hb A 2 of 2.815% (SD 0.246). The high Hb A 2 was seen in significantly higher number of patients than controls (P< 0.0001). Osmotic fragility (OF) test of all patients and control was normal (Table 1). Table 1. Results. \| \| Case \| Control \| --- \| Number of patients \| 125 (75 M/50 F) \| 60 \| \| Mean MCV \| 96.88 fl \| 82 fl \| \| MCV > 100 fl \| 54.4% (68/125) \| Nil \| \| Mean RDW \| 20.9 \| 14.6 \| \| RDW ↑ \| 60.8% (76/125) \| 16.67% (10/60) \| \| Serum ferritin \| Normal \| Normal \| \| Folic acid \| Normal \| Normal \| \| Vitamin B 12 \| Normal \| Normal \| \| Mean Hb A 2 \| 3.462% (SD 0.675) \| 2.815% (SD 0.246) \| \| <3 Hb A 2 \| 11.2% (14/125) \| 70% (42/60) \| \| 3–3.5 Hb A 2 \| 56.8% (71/125) \| 30% (18/60) \| \| >3.5 Hb A 2 \| 32% (40/125) 37.33% (28/75) male; 24% (12/50) female \| None \| \| Osmotic fragility (OF) \| Normal \| Normal \| 4. Discussion The β-thalassemia is one of the most common single gene disorders in India with an overall prevalence of 3-4%, and β-thalassemia carriers vary between 8 and 10% in certain communities like Sindhis, Muslims, Cutchi Bhanushalis, and some tribal groups . The diagnosis of heterozygous β-thalassaemia is of key importance in the antenatal setting because it is helpful in concern to haemoglobinopathy counseling, partner screening, and possibly extensive haematological and molecular analyses of the patient samples . Rao et al. observed that patients with megaloblastic anaemia have raised Hb A 2 (P< 0.001) as compared to normal cases . In the present study, vitamin B 12 and folic acid levels were within normal limits in both study and control groups, but mean MCV in study group was 96.88 fl and 54.4% showed macrocytosis (MCV > 100 fl). The impair DNA synthesis in megaloblastic anaemia occurs due to delayed nuclear maturation, while zidovudine causes inhibition of nucleoside reverse transcriptase ; therefore, in both conditions, the increased synthesis of δ chains leads to higher Hb A 2 values because more Hb synthesis occurs in less mature erythroid precursors, and the synthesis of δ chains is relatively greater in less mature cells . Thyroid hormone affects the δ gene transcription and causes increase in both the percentage and absolute amount of Hb A 2 . In the present study, thyroid functions test was within normal limits in both study and control groups. Wilkinson et al. observed that incidence of increased Hb A 2 value was significantly high in patients treated with zidovudine containing HAART than in those treated with non-zidovudine HAART and normal volunteers . In the present study all patients on zidovudine containing ART had raised Hb A 2 value than control group. 37.33% (28/75) of the males and 24% (12/50) of the females showed raised Hb A 2. Pornprasert et al. also observed that ART increases the Hb A 2 value and alters some haematological parameters that might affect the diagnosis of thalassaemia carriers, especially in the investigation of β-thalassaemia trait . Kosalaraksa et al. observed that abnormal high % Hb A 2 was found in more than a half of ZDV-exposed HIV-infected children; therefore, low MCV and MCH were important coparameters to reduce the misinterpretation of β-thalassemia trait; however, the DNA analysis should be performed to confirm the diagnosis in such situation . Howard et al. concluded that Hb A 2 should be estimated before the initiation antiretroviral drugs in all HIV-infected women of child-bearing age to allow a diagnosis of heterozygous β-thalassaemia to be made and help to prevent the inaccurate diagnosis of heterozygous β-thalassaemia, and unnecessary genetic counselling and genetic analysis . Pornprasert et al. also concluded that Hb A 2 values should be measured in all HIV-1-infected couples before the initiation of antiretroviral drugs to rule out misdiagnosis of β-thalassaemia . In the present study, all cases showed normal Hb A 2 before starting ART. Few drugs have effect on various haemoglobins like butyrate, which induces fetal hemoglobin (HbF) synthesis in cultures of erythroid progenitors in man by beta oxidation by mitochondrial enzymes resulting in the formation of two acetate molecules from each molecule of butyrate . The fatty acid analogue valproic acid (n-dipropylacetic acid) may increase the synthesis of fetal hemoglobin . Macrocytosis was a striking result of hydroxy urea therapy and paralleled the increase in HbF synthesis . In the present study, no history of intake of other drugs was present. 5. Conclusion In India, HIV patients have wider access to ART to reduce the annual AIDS-related deaths, and all patients with increased Hb A 2 should be investigated for detailed history, identified high risk factor for HIV, and all haematological parameters including OF test, MCV, MCH, MCHC, and RDW. The clinicians, pathologists, haematologists, and genetic counsellors should be aware of effects of nutritional anaemia and ART on Hb A 2 to reduce the chances of misdiagnosis of β-thalassaemia. References 1 Bain B. J., Haemoglobinopathy Diagnosis, 2006, 2nd edition, Blackwell Publishing, Malden, Mass, USA. Google Scholar 2 Howard J., Henthorn J. S., Murphy S., and Davies S. 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J., Increased fetal hemoglobin production in patients receiving valproic acid for epilepsy, Blood. (1994) 84, no. 5, 1690–1691, 2-s2.0-0028064768. 10.1182/blood.V84.5.1690.1690 Google Scholar 16 Alter B. P. and Gilbert H. S., The effect of hydroxyurea on hemoglobin F in patients with myeloproliferative syndromes, Blood. (1985) 66, no. 2, 373–379, 2-s2.0-0021853293. Google Scholar Citing Literature All articles> References ---------- Related ------- Information ----------- Recommended Increase of hemoglobin A 2 in human immunodeficiency virus-1-infected patients treated with zidovudine Jean-Pierre Routy MD,Marc Monte,Raymond Beaulieu,Emil Toma,Line St-Pierre,Marc Dumont, American Journal of Hematology Haemoglobin A 2′ and its significance in β thalassaemia diagnosis D. ABDEL-GADIR,L. PHELAN,B. J. BAIN, International Journal of Laboratory Hematology Zidovudine therapy in HIV infection: which patients should be treated and when. MG Barry,DJ Back,AM Breckenridge, British Journal of Clinical Pharmacology The prevalence of reduced zidovudine susceptibility in zidovudine‐naive, antiretroviral‐experienced HIV‐1‐infected patients Y Gilleece,C Torti,S Mandalia,BG Gazzard,D Pillay,AL Pozniak, HIV Medicine Genotype of subjects with borderline hemoglobin A2 levels: Implication for, β‐thalassemia carrier screening R. Galanello,S. Barella,A. Ideo,D. Gasperini,C. Rosatelli,L. Paderi,E. Paglietti,C. Sollaino,L. Perseu,D. Loi,A. Cao, American Journal of Hematology Metrics Citations: 2 Details Copyright © 2013 Jitendra Singh Nigam et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 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5896	https://stackoverflow.com/questions/32337659/calculate-dot-product-using-map-and-lambda-in-scheme	Calculate dot-product using map and lambda in Scheme - Stack Overflow Join Stack Overflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google Sign up with GitHub OR Email Password Sign up Already have an account? Log in Skip to main content Stack Overflow 1. About 2. Products 3. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Calculate dot-product using map and lambda in Scheme Ask Question Asked 10 years, 1 month ago Modified10 years ago Viewed 1k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Calculate dot product of two vectors v and w. My thinking is get the product of first items in vectors, and use accumulate function to add them all together. lisp (define (dot-product v w) (accumulate + 0 (map (lambda(v,w)( (car v) (car w))) (list v w)))) I don't understand why this does not work: lisp (dot-product (list 1 2) (list 3 4)) //output is 12, not the 11 that I anticipated I think I have big trouble understanding how map function works in this case, thank you for any explanation. lambda scheme map-function Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Sep 2, 2015 at 22:04 Will Ness 71.5k 10 10 gold badges 105 105 silver badges 192 192 bronze badges asked Sep 1, 2015 at 17:31 user3366240user3366240 85 5 5 bronze badges Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. The way you're invoking map won't have the desired effect. Giving the list of (list v w) means that your function is called twice, once with v, once with w (no matter how many items each of those lists have). Instead, do this: lisp (define (dot-product v w) (apply + (map v w))) Here, you are passing two lists to map, which means it will call your function ( in this case) with two arguments: one for each list given. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Sep 1, 2015 at 17:40 C. K. YoungC. K. Young 224k 47 47 gold badges 393 393 silver badges 445 445 bronze badges 6 Comments Add a comment user3366240 user3366240Over a year ago Thank you! sorry that I still confused about one thing: isn't map going to go through each element in the (list v w)? why this function only be called twice? 2015-09-01T18:02:22.717Z+00:00 0 Reply Copy link C. K. Young C. K. YoungOver a year ago @user3366240 (list v w) is a list of two elements (the first element is v, and the second element is w). 2015-09-01T18:04:07.573Z+00:00 0 Reply Copy link user3366240 user3366240Over a year ago Thanks, that makes sense. But do you know how that 12 was computed? when the fun called twice the result should be 3. 2015-09-01T18:24:01.587Z+00:00 0 Reply Copy link soegaard soegaardOver a year ago (dot-product '(1 2) '(3 4)) = (apply + map '(1 2) '(3 4)) = (apply+ (list ( 1 3) ( 2 4)) = (apply + (list 3 8)) = (+ 3 8) = 11 2015-09-01T18:25:44.34Z+00:00 0 Reply Copy link user3366240 user3366240Over a year ago Thanks @soegaard, but I am still trying to figure out how my code calculate that wrong answer, so I can learn not to make that mistake again. I assume in (lambda(v,w)( (car v) (car w)), the inputs are 1 and 3, and output is 3, but somehow I get 12. 2015-09-01T18:47:52.447Z+00:00 0 Reply Copy link Add a comment\|Show 1 more comment This answer is useful 1 Save this answer. Show activity on this post. This not an answer, but I needed to show some code. This shows why your original didn't work. Let's define v and w so we can use them in examples. ```lisp (define v '(1 2)) (define w '(3 4)) ``` Now let's see where your computation begins: ```lisp (map (lambda (v,w) ( (car v) (car w))) (list v w)) '(3 9) ``` This is probably not the result you expected. The problems is that (v,w) is turned into (v (unquote w)) by the reader. What you meant was: ```lisp (map (lambda (v w) ( v w)) v w) '(3 8) ``` And if you put that into your original example, you get 12. Note that (lambda (v w) ( v w)) is the same as . ```lisp (map v w) '(3 8) ``` Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Sep 1, 2015 at 19:13 soegaardsoegaard 31.1k 4 4 gold badges 62 62 silver badges 109 109 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. 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5897	https://web.ma.utexas.edu/users/m408m/Display12-5-4.shtml	\| \| \| \| \| \| \| \| --- --- --- \| M408M Learning Module Pages Main page Chapter 10: Parametric Equations and Polar Coordinates Chapter 12: Vectors and the Geometry of Space Learning module LM 12.1: 3-dimensional rectangular coordinates: Learning module LM 12.2: Vectors: Learning module LM 12.3: Dot products: Learning module LM 12.4: Cross products: Learning module LM 12.5: Equations of Lines and Planes: Equations of a line Equations of planes Finding the normal to a plane Distances to lines and planes Learning module LM 12.6: Surfaces: Chapter 13: Vector Functions Chapter 14: Partial Derivatives Chapter 15: Multiple Integrals \| \| Finding the normal to a planeFinding the normal to a plane As promised, we return the the question of finding the equation for a plane from the location of three points, say Q(x1,y1,z1),R(x2,y2,z2),S(x3,y3,z3) The fact that the cross-product a×b is perpendicular to both a and b makes it very useful when dealing with normals to planes. \| \| \| --- \| \| Let b=⟨x1,y1,z1⟩, r=⟨x2,y2,z2⟩, s=⟨x3,y3,z3⟩. The vectors QR−→− = r−b,QS−→ = s−b, then lie in the plane. The normal to the plane is given by the cross product n=(r−b)×(s−b). Once this normal has been calculated, we can then use the point-normal form to get the equation of the plane passing through Q,R, and S. \| \| In practice, it's usually easier to work out n in a given example rather than try to set up some general equation for the plane. \| \| \| --- \| \| Example: Find an equation for the plane passing through the points Q(−1,1,2), R(−4,2,2), S(−2,1,5). Solution: when the plane passes through Q,R, and S, then the vectors QR−→−=⟨−3,1,0⟩,QS−→=⟨−1,0,3⟩, lie in the plane. Thus the cross-product n = ∣∣∣∣i−3−1j10k03∣∣∣∣ = = 3i+9j+k is normal to the plane. \| If r=⟨x,y,z⟩ determines an arbitrary point P in the plane, the vector v = QP−→− = ⟨x+1,y−1,z−2⟩ lies in the plane and so is perpendicular to n. In this case, n⋅v = 3(x+1)+9(y−1)+(z−2) = 0, which after simplification becomes 3x+9y+z−8 = 0. Consequently, the plane 3x+9y+z=8 passes through Q,R, and S. \| \|
5898	https://api.pageplace.de/preview/DT0400.9781292451992_A46750396/preview-9781292451992_A46750396.pdf	Introductory Circuit Analysis Fourteenth Edition Robert L. Boylestad Brian A. Olivari GLOBAL EDITION Summary of Equations to Accompany INTRODUCTORY CIRCUIT ANALYSIS, Fourteenth Edition, Global Edition by Robert L. Boylestad / Brian A. Olivari © Copyright 2024 by Pearson Education, Ltd. All Rights Reserved. ac Sinusoidal Alternating Waveforms Sine wave υ α α ω π = = = = = V t ft f T sin , 2 , 1/ , 1 radian 57.3°, m π ( ) ( ) = × radians /180° degrees , π ( ) ( ) = × degrees 180°/ radians Identities t t t t sin 90° cos , sin cos /2 , ω ω ω ω π ( ) ( ) ( ) + = = − sin sin , cos cos α α α α ( ) ( ) − = − − = Average value = G algebraic sum of areas/ length of curve Effective (rms) value I I I I 0.707 , 2 , m m rms rms = = I i t T area / rms 2 ( ) [ ] = V V V V V v t T 0.707 2 area / m m rms rms rms 2 ( ) [ ] = = = The Basic Elements and Phasors R L I V R X L i : : / , in phase , leads by 90° m m L L L ω υ = = C X C i : 1/ , leads by 90° C C C ω υ = Power P V I /2 cos m m θ ( ) = = R V I P V I I R V R : cos / rms rms rms rms 2 rms 2 rms θ = = = Power factor θ = = F P V I cos / p rms rms Rectangular form = ± A jB C Polar form θ = ∠ C C Conversions C A B , 2 2 θ = + = B A A C B C tan / , cos , sin 1 θ θ ( ) = = − Operations j 1, = − j j j A A j B B C C 1, 1/ , , 2 1 2 1 2 1 2 ( ) ( ) = − = − ± = ± ± + ± ± C C C C C C C C , / / 1 2 1 2 1 2 1 2 1 2 1 2 ⋅ θ θ θ θ ( ) ( ) ( ) = ∠ + = ∠ − Series and Parallel ac Circuits Elements ∠ ∠ ∠− R X X 0°, 90°, 90° L C Series = + + + + = = F R Z Z Z Z Z Z I E Z . . . , / , / T N s T p T 1 2 3 Voltage divider rule x x V Z E Z / T = Parallel Y Y Y T 1 2 = + + Y Y Z Z Z Z Z ... , / , N T 3 1 2 1 2 ( ) + + = + G B 0°, 90°, L ∠ ∠− B F G Y 90°, cos / C P T T θ ∠ = = Current divider rule I1 = Z I Z Z I Z I Z Z /( ), / T T 2 1 1 2 2 1 2 ( ) + = + Equivalent circuits R R X X R X R X X R / , / , s p p p p s p p p p 2 2 2 2 2 2 ( ) ( ) = + = + R R X R / , p s s s 2 2 ( ) = + ( ) = + X R X X / p s s s 2 2 Series-Parallel ac Networks Employ block impedances and obtain general solution for reduced network. Then substitute numerical values. General approach similar to that for dc networks. Methods of Analysis and Selected Topics (ac) Source conversion = = = p s p s E IZ Z Z I E Z , , / Bridge networks = Z Z Z Z / / 1 3 2 4 ∆ ∆ Y Y - , - conversions See dc coverage, replacing R by Z. Network Theorems (ac) Review dc content on other side. Thévenin’s theorems (dependent sources) E E Z E I , / , OC Th Th OC SC = = Z E I / Th g g = Norton’s theorems (dependent sources) I I , SC N = Z E I Z E I / , / N OC SC N g g = = Maximum power transfer theorem θ θ = = − = Z Z P E R , , /4 L Th L Th Th Th max 2 Z Power (ac) R P VI V I I R V R : /2 / m m 2 2 = = = = Apparent power S VI, = θ θ = = = P S F P S cos , cos / p Reactive power Q VI sin θ = L Q VI I X V X C Q VI I X V X : / , : / , L L L C C C 2 2 2 2 = = = = = = S P Q F P S , / T T T p T T 2 2 = + = Resonances Series X X f LC Z R Q X R Q X R , 1/(2 ), , / , / L C S TS l L l S L π = = = = = = R L C (1/ ) / , V Q E V Q E P P , , 1/2 , L S Cs S HPF max ( ) = = = π ( ] ( ) ( ) = − + + − f R L R L LC f R L 1/2 [ /2 (1/2) / ) 4/ , use /2 , 1 2 2 π = − = = BW f f R L f Q /2 / S S 2 1 Parallel X X X , Lp C Lp = = R X X f LC R C L Z R R / , 1/ 2 1 / , , l L L p l TP s p 2 2 2 π ( ) [ ] ( ) ( ) + = − = R R X R / , p l L l 2 2 ( ) = + = = − = Q R R X BW f f f Q ( )/ , / p s p Lp p p 2 1 Q Z R Q R X X X X f LC 10: , , , 1/(2 ), Tp s l Lp L L C p 2 π ≥ ≅ ≅ = ≅ Q Q I I QI BW f Q R L , , / /2 p l L C T p p l π = = ≅ = = Decibels, Filters, and Bode Plots Logarithms N b x N x x ab , log , log 2.3 log , log x b e 10 10 = = = = a b log log , 10 10 + a b a b a n a log / log log , log log , n 10 10 10 10 10 = − = P P V V dB 10log / , dB 20log / 10 2 1 10 2 1 = = υ R-C filters f RC high-pass 1/ 2 , c π ( ) ( ) = π ( ) ( ) ( ) = + ∠ = − R R X X R f RC V V / tan / low-pass 1/ 2 , o i C C c 2 2 1 = + ∠− − X R X R X V V / / tan o i C C C 2 2 1 Octave 2 : 1, 6 dB/octave Decade 10 : 1, 20 dB/decade Transformers Mutual inductance = M k L L p s Iron-core = Φ E fN 4.44 , p p m = Φ = = = E fN E E N N a N N I I N N 4.44 , / / , / , / / , S S m p S p S p S p S S p a E I E I P P Z Z , , ideal p L p p S S i O 2 ( ) = = = Air-core ω = + + M Z Z Z Z ( ) /( ) i p S L 2 Polyphase Systems Y-Y system = = = = φ φ φ φ φ I I I V E E V , , 3 g L L L Y-Δ system = = φ φ V E I I , 3 L L Δ-Δ system V E E I I , 3 L L = = = φ φ φ Δ-Y system E V I I E E 3 , , L L L = = = φ φ φ Power P P 3 , T = φ Q Q S S E I F P S 3 , 3 3 , / T T L L p T T = = = = φ φ Pulse Waveforms and the R-C Response % tilt ( ) [ ] ( ) = − × = + V V V V V V / 100% with /2 1 2 1 2 Pulse repetition frequency ( ) = T prf 1/ Duty cycle ( ) = × t T / 100% p ( ) ( )( ) ( ) = + − × V V duty cycle peak value 1 duty cycle b av R-C circuits V V V e 1 C i f i t RC / v ( )( ) = + − − − Compensated attenuator = R C R C p p s s Nonsinusoidal Circuits Fourier series f A A t A t sin sin 2 0 1 2 α ω ω ( ) = + + + + A n t B t B t B n t sin cos cos 2 ... cos n n 1 2 ω ω ω ω + + + + Even function α α ( ) ( ) = − f f B , no terms n Odd function f α ( ) = f A , no terms, n α ( ) − − no odd harmonics if f t f T t ( /2) , [ ] ( ) = + no even harmonics if ( ) ( ) ( ) = − + f t f T t /2 Effective (rms) value V V V V V V ( )/2 m m m m rms 0 2 2 2 2 2 n n 1 1 = + + + ′ + ′ + + ′ ( ) Power θ θ = + + … + = = P V I V I V I I R V R cos cos / T n n n 0 0 1 1 rms 2 rms 2 Standard Resistor Values Ohms Ω ( ) Kilohms kΩ ( ) Megohms MΩ ( ) 0.10 1.0 10 100 1000 10 100 1.0 10.0 0.11 1.1 11 110 1100 11 110 1.1 11.0 0.12 1.2 12 120 1200 12 120 1.2 12.0 0.13 1.3 13 130 1300 13 130 1.3 13.0 0.15 1.5 15 150 1500 15 150 1.5 15.0 0.16 1.6 16 160 1600 16 160 1.6 16.0 0.18 1.8 18 180 1800 18 180 1.8 18.0 0.20 2.0 20 200 2000 20 200 2.0 20.0 0.22 2.2 22 220 2200 22 220 2.2 22.0 0.24 2.4 24 240 2400 24 240 2.4 0.27 2.7 27 270 2700 27 270 2.7 0.30 3.0 30 300 3000 30 300 3.0 0.33 3.3 33 330 3300 33 330 3.3 0.36 3.6 36 360 3600 36 360 3.6 0.39 3.9 39 390 3900 39 390 3.9 0.43 4.3 43 430 4300 43 430 4.3 0.47 4.7 47 470 4700 47 470 4.7 0.51 5.1 51 510 5100 51 510 5.1 0.56 5.6 56 560 5600 56 560 5.6 0.62 6.2 62 620 6200 62 620 6.2 0.68 6.8 68 680 6800 68 680 6.8 0.75 7.5 75 750 7500 75 750 7.5 0.82 8.2 82 820 8200 82 820 8.2 0.91 9.1 91 910 9100 91 910 9.1 Fourteenth Edition Global Edition Robert L. Boylestad Brian A. Olivari Introductory Circuit Analysis Product Management: K. K. Neelakantan Content Strategy: Moasenla Jamir and Shahana Bhattacharya Supplements: Nitin Shankar, Nikhil Rakshit, and Tooba Shafique Production and Digital Studio: Vikram Medepalli, Naina Singh, and Niharika Thapa Rights and Permissions: Anjali Singh and Ashish Vyas Cover image: Vasylchenko / Shutterstock Please contact with any queries on this content. Pearson Education Limited KAO Two KAO Park Hockham Way Harlow, Essex CM17 9SR United Kingdom and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsonglobaleditions.com © Pearson Education Limited 2024 The rights of Robert L. Boylestad and Brian A. Olivari to be identified as the authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988. 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Please contact us with concerns about any potential bias at For accessibility-related issues, such as using assistive technology with Pearson products, alternative text requests, or accessibility documentation, email the Pearson Disability Support team at disability.support@pearson.com Pearson is dedicated to creating bias-free content that reflects the diversity, depth, and breadth of all learners’ lived experiences. Pearson’s Commitment to Diversity, Equity, and Inclusion • Everyone has an equitable and lifelong opportunity to succeed through learning. • Our educational content accurately reflects the histories and lived experiences of the learners we serve. • Our educational products and services are inclusive and represent the rich diversity of learners. • Our educational content prompts deeper discussions with students and motivates them to expand their own learning (and worldview). This page is intentionally left blank Robert L. Boylestad Brian A. Olivari In loving memory of my wife Else To my wife Margie and our children Eric, Alison and Stacey and our children Jenny, Brittany, and our grandchildren Kelcy, Morgan, Meredith and Ben. Britt, Samantha, Codie, Skylar, Lindsey and Aspen. This page is intentionally left blank The author would like to thank Don Johnson, an entrepreneur, author, and educator for his contribution. 7 Acknowledgments Kathleen Annis—AEMC Instruments Geno Anderson—Taitung Onesiphorus Children’s Home Jen Brophy—Red River Camps, Portage, Maine Tom Brown—LRAD Corporation Professor Leon Chua—University of California, Berkeley Professor Timothy Cochran—Alfred State College Professor Kenneth Cook—Lawrence Technological University Professor Raymond Cwieka—Essex College Professor Raj Desai—Midwestern State University Iulian Dobre—IMSAT Maritime Jessica Fiui—Honda Corporation Ron Forbes—B+K Precision, Inc. Felician Frentiu—IMSAT Maritime F. Halkiadakis—California University of Pennsylvania Professor Gerald Herder—Cal Poly Pomona Don Johnson—Professional Photographer Professor Ron Kalin—Penn State University Cara Kugler—Texas Instruments, Inc. Professor Ohbong Kwon—New York City College of Technology Professor David Malooly—Indiana State University Professor Don Mccann—Maine Maritime Academy, Castine, Maine Professor Kristen Dagan McGee—Naugatuck Valley Community College Cheryl Mendenhall—Cadence Design Systems, Inc Professor Henry C. Miller—Bluefield State College Professor Mack Mofidi—DeVry University Professor Mostafa Mortezaie—DeVry University Rajashree Narendra—Dayananda Sagar University, Bangalore, India Professor Louis Nashelsky—CUNY Todd R. Nelson—Author and Educator Professor Brian Norton—Oklahoma State University Katie Parker—EarthRoamer Corp. Professor Alireza Rahrooh—Daytona State College Professor Gilberto Medeiras Ribeiro—Universidade Federal de Minar Gerais, Brazil Professor Patrick Riley—California University of Pennsylvania Peter Sanburn—Itron, Inc. Professor William Shepherd—Owens Community College Peggy Suggs—Edison Electric Institute Don Szymanski—Owens Community College Professor William Tefft—Maine Maritime Academy, Castine, Maine Mark Walters—National Instruments, Inc. Stanley Williams—Hewlett Packard, Inc. Professor Chen Xijou—Dalian University of Technology Professor Jianhua Joshua Yang—University of Massachusetts CURRENT AND PREVIOUS EDITION REVIEWERS ACKNOWLEDGMENTS FOR THE GLOBAL EDITION Pearson would like to acknowledge and thank the following for the Global Edition: CONTRIBUTORS D. C. Meena—Delhi Technological University Nalan Özkurt—Yaşar University REVIEWERS Gokhan Bora Esmer—Marmara University V. Jagadeesh Kumar—Indian Institute of Technology Madras Ha Pham—University of Technology Sydney This page is intentionally left blank This page is intentionally left blank Brief Contents Chapter 1 Introduction 21 Chapter 2 Voltage and Current 55 Chapter 3 Resistance 89 Chapter 4 Ohm’s Law, Power, and Energy 127 Chapter 5 Series dc Circuits 153 Chapter 6 Parallel dc Circuits 207 Chapter 7 Series-Parallel Circuits 261 Chapter 8 Methods of Analysis and Selected T opics (dc) 305 Chapter 9 Network Theorems 367 Chapter 10 Capacitors 415 Chapter 11 Inductors 475 Chapter 12 Magnetic Circuits 519 Chapter 13 Sinusoidal Alternating Waveforms 545 Chapter 14 The Basic Elements and Phasors 595 Chapter 15 Series ac Circuits 641 Chapter 16 Parallel ac Circuits 685 Chapter 17 Series-Parallel ac Networks 725 Chapter 18 Methods of Analysis and Selected Topics (ac) 747 Chapter 19 Network Theorems (ac) 787 Chapter 20 Power (ac) 831 Chapter 21 Resonance 865 Chapter 22 Decibels, Filters, and Bode Plots 909 Chapter 23 Transformers 973 Chapter 24 Polyphase Systems 1015 Chapter 25 Pulse Waveforms and the R-C Response 1055 Chapter 26 Nonsinusoidal Circuits 1081 Appendices 1105 Index 1130 9 This page is intentionally left blank 11 Contents Chapter 1 Introduction 21 1.1 The Electrical/Electronics Industry 21 1.2 A Brief History 23 1.3 Units of Measurement 28 1.4 Systems of Units 29 1.5 Significant Figures, Accuracy, and Rounding Off 32 1.6 Powers of Ten 34 1.7 Fixed-Point, Floating-Point, Scientific, and Engineering Notation 38 1.8 Conversion Between Levels of Powers of Ten 40 1.9 Conversion Within and Between Systems of Units 42 1.10 Symbols 44 1.11 Conversion Tables 44 1.12 Calculators 45 1.13 Computer Analysis 49 Chapter 2 Voltage and Current 55 2.1 Introduction 55 2.2 Atoms and Their Structure 55 2.3 Voltage 58 2.4 Current 61 2.5 Voltage Sources 64 2.6 Ampere-Hour Rating 74 2.7 Battery Life Factors 75 2.8 Conductors and Insulators 77 2.9 Semiconductors 78 2.10 Ammeters and Voltmeters 78 2.11 Applications 81 Chapter 3 Resistance 89 3.1 Introduction 89 3.2 Resistance: Circular Wires 90 3.3 Wire Tables 93 3.4 Temperature Effects 96 3.5 Conductors and the Inferred Absolute Temperature ( ) Ti 96 3.6 Temperature Coefficient of Resistance 98 3.7 PPM/°C 99 3.8 Types of Resistors 99 3.9 Color Coding and Standard Resistor Values 104 3.10 Surface Mount Resistors 108 3.11 Conductance 109 3.12 Ohmmeters 110 3.13 Resistance: Metric Units 111 3.14 The Fourth Element—The Memristor 113 3.15 Superconductors 114 3.16 Thermistors 116 3.17 Photoconductive Cell 117 3.18 Varistors 117 3.19 Applications 118 Chapter 4 Ohm’s Law, Power, and Energy 127 4.1 Introduction 127 4.2 Ohm’s Law 127 4.3 Plotting Ohm’s Law 130 4.4 Power 133 4.5 Energy 135 4.6 Efficiency 139 4.7 Circuit Breakers, GFCIs, and Fuses 142 4.8 Applications 143 12 \| \| \| CONTENTS Chapter 5 Series dc Circuits 153 5.1 Introduction 153 5.2 Series Resistors 154 5.3 Series Circuits 157 5.4 Power Distribution in a Series Circuit 163 5.5 Voltage Sources in Series 164 5.6 Kirchhoff’s Voltage Law 166 5.7 Voltage Division in a Series Circuit 170 5.8 Interchanging Series Elements 174 5.9 Notation 175 5.10 Ground Connection Awareness 180 5.11 Voltage Regulation and the Internal Resistance of Voltage Sources 181 5.12 Loading Effects of Instruments 186 5.13 Protoboards (Breadboards) 188 5.14 Applications 190 Chapter 6 Parallel dc Circuits 207 6.1 Introduction 207 6.2 Parallel Resistors 207 6.3 Parallel Circuits 217 6.4 Power Distribution in a Parallel Circuit 223 6.5 Kirchhoff’s Current Law 225 6.6 Current Divider Rule 229 6.7 Voltage Sources in Parallel 235 6.8 Open and Short Circuits 236 6.9 Voltmeter Loading Effects 239 6.10 Summary Table 241 6.11 Troubleshooting Techniques 242 6.12 Protoboards (Breadboards) 243 6.13 Applications 244 Chapter 7 Series-Parallel Circuits 261 7.1 Introduction 261 7.2 Series-Parallel Networks 262 7.3 Reduce and Return Approach 263 7.4 Block Diagram Approach 265 7.5 Descriptive Examples 269 7.6 Ladder Networks 277 7.7 Voltage Divider Supply (Unloaded and Loaded) 279 7.8 Potentiometer Loading 282 7.9 Impact of Shorts and Open Circuits 284 7.10 Ammeter, Voltmeter, and Ohmmeter Design 287 7.11 Applications 291 Chapter 8 Methods of Analysis and Selected Topics (dc) 305 8.1 Introduction 305 8.2 Current Sources 306 8.3 Branch-Current Analysis 312 8.4 Mesh Analysis (General Approach) 319 8.5 Mesh Analysis (Format Approach) 326 8.6 Nodal Analysis (General Approach) 330 8.7 Nodal Analysis (Format Approach) 338 8.8 Bridge Networks 342 8.9 Y-∆ (T-π) and ∆-Y (π-T) Conversions 346 8.10 Applications 351 Chapter 9 Network Theorems 367 9.1 Introduction 367 9.2 Superposition Theorem 367 9.3 Thévenin’s Theorem 375 9.4 Norton’s Theorem 387 9.5 Maximum Power Transfer Theorem 391 9.6 Millman’s Theorem 400 9.7 Substitution Theorem 403 9.8 Reciprocity Theorem 405 Chapter 10 Capacitors 415 10.1 Introduction 415 10.2 The Electric Field 415 10.3 Capacitance 417 CONTENTS \| \| \| 13 10.4 Capacitors 421 10.5 Transients in Capacitive Networks: The Charging Phase 434 10.6 Transients in Capacitive Networks: The Discharging Phase 442 10.7 Initial Conditions 448 10.8 Instantaneous Values 451 10.9 Thévenin Equivalent: TH R C τ = 452 10.10 The Current iC 455 10.11 Capacitors in Series and in Parallel 457 10.12 Energy Stored by a Capacitor 461 10.13 Stray Capacitances 461 10.14 Applications 462 Chapter 11 Inductors 475 11.1 Introduction 475 11.2 Magnetic Field 475 11.3 Inductance 480 11.4 Induced Voltage L υ 486 11.5 R-L Transients: The Storage Phase 488 11.6 Initial Conditions 491 11.7 R-L Transients: The Release Phase 493 11.8 Thévenin Equivalent: τ = L R / Th 498 11.9 Instantaneous Values 500 11.10 Average Induced Voltage: υLav 501 11.11 Inductors in Series and in Parallel 503 11.12 Steady-State Conditions 504 11.13 Energy Stored by an Inductor 506 11.14 Applications 507 Chapter 12 Magnetic Circuits 519 12.1 Introduction 519 12.2 Magnetic Field 519 12.3 Reluctance 520 12.4 Ohm’s Law for Magnetic Circuits 521 12.5 Magnetizing Force 521 12.6 Hysteresis 522 12.7 Ampère’s Circuital Law 526 12.8 Flux Φ 527 12.9 Series Magnetic Circuits: Determining NI 527 12.10 Air Gaps 531 12.11 Series-Parallel Magnetic Circuits 533 12.12 Determining Φ 535 12.13 Applications 537 Chapter 13 Sinusoidal Alternating Waveforms 545 13.1 Introduction 545 13.2 Sinusoidal ac Voltage Characteristics and Definitions 546 13.3 Frequency Spectrum 549 13.4 The Sinusoidal Waveform 553 13.5 General Format for the Sinusoidal Voltage or Current 557 13.6 Phase Relations 560 13.7 Average Value 566 13.8 Effective (rms) Values 572 13.9 Converters and Inverters 578 13.10 ac Meters and Instruments 581 13.11 Applications 584 Chapter 14 The Basic Elements and Phasors 595 14.1 Introduction 595 14.2 Response of Basic R, L, and C Elements to a Sinusoidal Voltage or Current 598 14.3 Frequency Response of the Basic Elements 605 14.4 Average Power and Power Factor 611 14.5 Complex Numbers 617 14.6 Rectangular Form 617 14.7 Polar Form 618 14.8 Conversion Between Forms 619 14.9 Mathematical Operations with Complex Numbers 621 14.10 Calculator Methods with Complex Numbers 627 14.11 Phasors 629 Chapter 15 Series ac Circuits 641 15.1 Introduction 641 15.2 Resistive Elements 642 15.3 Inductive Elements 643 15.4 Capacitive Elements 645 15.5 Impedance Diagram 647 15.6 Series Configuration 648 15.7 Voltage Divider Rule 655 15.8 Frequency Response for Series ac Circuits 658 15.9 Summary: Series ac Circuits 671 15.10 Phase Measurements 671 15.11 Applications 674 Chapter 16 Parallel ac Circuits 685 16.1 Introduction 685 16.2 Total Impedance 685 16.3 Total Admittance 687 16.4 Parallel ac Networks 691 16.5 Current Divider Rule (CDR) 698 16.6 Frequency Response of Parallel Elements 698 16.7 Summary: Parallel ac Networks 708 16.8 Equivalent Circuits 709 16.9 Applications 714 Chapter 17 Series-Parallel ac Networks 725 17.1 Introduction 725 17.2 Illustrative Examples 725 17.3 Ladder Networks 735 17.4 Grounding 736 17.5 Applications 739 Chapter 18 Methods of Analysis and Selected T opics (ac) 747 18.1 Introduction 747 18.2 Independent Versus Dependent (Controlled) Sources 747 18.3 Source Conversions 748 18.4 Mesh Analysis 751 18.5 Nodal Analysis 759 18.6 Bridge Networks (ac) 770 18.7 Δ-Y, Y-Δ Conversions 775 Chapter 19 Network Theorems (ac) 787 19.1 Introduction 787 19.2 Superposition Theorem 787 19.3 Thévenin’s Theorem 795 19.4 Norton’s Theorem 807 19.5 Maximum Power Transfer Theorem 813 19.6 Substitution, Reciprocity, and Millman’s Theorems 817 19.7 Application 818 Chapter 20 Power (ac) 831 20.1 Introduction 831 20.2 General Equation 831 20.3 Resistive Circuit 832 20.4 Apparent Power 834 20.5 Inductive Circuit and Reactive Power 836 20.6 Capacitive Circuit 839 20.7 The Power Triangle 841 20.8 The Total P, Q, and S 843 20.9 Power-Factor Correction 848 20.10 Power Meters 853 20.11 Effective Resistance 853 20.12 Applications 856 Chapter 21 Resonance 865 21.1 Introduction 865 21.2 Series Resonant Circuit 867 21.3 The Quality Factor (Q) 870 21.4 ZT Versus Frequency 872 21.5 Selectivity 873 21.6 V , V , and V R L C 876 21.7 Practical Considerations 878 21.8 Summary 878 21.9 Examples (Series Resonance) 879 14 \| \| \| CONTENTS CONTENTS \| \| \| 15 21.10 Parallel Resonant Circuit 881 21.11 Selectivity Curve for Parallel Resonant Circuits 885 21.12 Effect of ≥ Q 10 l 888 21.13 Summary Table 891 21.14 Examples (Parallel Resonance) 892 21.15 Applications 899 Chapter 22 Decibels, Filters, and Bode Plots 909 22.1 Introduction 909 22.2 Properties of Logarithms 914 22.3 Decibels 915 22.4 Filters 921 22.5 R-C Low-Pass Filter 922 22.6 R-C High-Pass Filter 927 22.7 Band-Pass Filters 930 22.8 Band-Stop Filters 934 22.9 Double-Tuned Filter 936 22.10 Other Filter Configurations 938 22.11 Bode Plots 941 22.12 Sketching the Bode Response 948 22.13 Additional Properties of Bode Plots 953 22.14 Crossover Networks 960 22.15 Applications 961 Chapter 23 T ransformers 973 23.1 Introduction 973 23.2 Basic Operation 973 23.3 Reflected Impedance and Power 979 23.4 Impedance Matching, Isolation, and Displacement 981 23.5 Equivalent Circuit (Iron-Core Transformer) 986 23.6 Frequency Considerations 989 23.7 Series Connection of Mutually Coupled Coils 990 23.8 Air-Core Transformer 993 23.9 Nameplate Data 995 23.10 Types of Transformers 997 23.11 Tapped and Multiple-Load Transformers 999 23.12 Networks with Magnetically Coupled Coils 1000 23.13 Current Transformers 1001 23.14 Applications 1002 Chapter 24 Polyphase Systems 1015 24.1 Introduction 1015 24.2 Three-Phase Generator 1016 24.3 Y-Connected Generator 1017 24.4 Phase Sequence (Y-Connected Generator) 1019 24.5 Y-Connected Generator with a Y-Connected Load 1021 24.6 Y-∆ System 1023 24.7 ∆-Connected Generator 1025 24.8 Phase Sequence (∆-Connected Generator) 1026 24.9 ∆-∆ ∆-Y Three-Phase Systems 1026 24.10 Power 1028 24.11 Three-Wattmeter Method 1034 24.12 Two-Wattmeter Method 1035 24.13 Unbalanced, Three-Phase, Four-Wire, Y-Connected Load 1038 24.14 Unbalanced, Three-Phase, Three-Wire, Y-Connected Load 1040 24.15 Residential and Industrial Service Distribution Systems 1043 Chapter 25 Pulse Waveforms and the R-C Response 1055 25.1 Introduction 1055 25.2 Ideal Versus Actual 1055 25.3 Pulse Repetition Rate and Duty Cycle 1059 25.4 Average Value 1062 25.5 Transient R-C Networks 1063 25.6 R-C Response to Square-Wave Inputs 1066 25.7 Oscilloscope Attenuator and Compensating Probe 1073 25.8 Application 1075 Chapter 26 Nonsinusoidal circuits 1081 26.1 Introduction 1081 26.2 Fourier Series 1082 26.3 Fourier Expansion of a Square Wave 1089 26.4 Fourier Expansion of a Half-Wave Rectified Waveform 1091 26.5 Fourier Spectrum 1092 26.6 Circuit Response to a Nonsinusoidal Input 1093 26.7 Addition and Subtraction of Nonsinusoidal Waveforms 1099 Appendices 1105 Appendix A Conversion Factors 1106 Appendix B Determinants 1109 Appendix C Greek Alphabet 1117 Appendix D Magnetic Parameter Conversions 1118 Appendix E Maximum Power Transfer Conditions 1119 Appendix F Answers to Selected Odd-Numbered Problems 1121 Index 1130 16 \| \| \| CONTENTS 17 without having to refer to the calculator manual provided. 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We realize there is always room for improvement, and while not as obvious as it was in earlier editions, there is always something you can do to take the text one step higher. Robert L. Boylestad Brian A. Olivari Welcome to the fourteenth edition. Over the fifty plus years since the first edition in 1968, the revisions have typically involved adding additional content based on the changes in curriculum. This new fourteenth edition is more about refining and updating content to provide a revision that ensures that the most current and relevant content is provided. We are also excited about the ways students can pur-chase this textbook to ensure that they have the most effec-tive and affordable options. Students can choose from the print book or eText options. With the Pearson eText, students can search the text, use the study tools such as flashcards, make notes online, print out reading assignments that incorporate the notes they take during lectures, and bookmark important passages for later review. The most notable changes include the following: • The over 2000 images are as current as possible with many new or modified for this edition. • The overall design of the text has been updated and enhanced for a more modern, accessible layout. • The Multisim and PSpice computer material has been removed based on the vast majority of reviewers stating it was not covered. This valuable feedback indicated the importance of covering fundamental theories clearly rather than focusing on the mechanics of applying com-puter methods. • Topics requiring a solid understanding of power factor, lead, and lag concepts have been significantly enhanced throughout the text. • Laboratory experiments and summary of equations to accompany the text have been carefully reviewed for accuracy and changes made where required. • Problems were carefully reviewed to ensure that they progressed from the simple to the more complex in each section. • All solutions were checked carefully to ensure accuracy. The TI 89 Titanium calculator continues to be the choice for this text, and we have provided sufficient detail in the text to perform all the mathematical techniques required Preface SUPPLEMENTS To enhance the learning process, a full supplements pack-age accompanies this text and is available to instructors using the text for a course. Instructor Resources Instructor Resources can be downloaded at www. pearsonglobaleditions.com. If you don’t already have a username and password for access, you can request access at www.pearsonglobaleditions.com. • Instructor’s Resource Manual, containing text solutions. • PowerPoint Lecture Notes. • TestGen, a computerized test bank. This page is intentionally left blank 19 About the Authors ROBERT L. BOYLESTAD The original contract for Introductory Circuit Analysis was signed in 1965 resulting in a publishing date of 1968—a current life history of some 60 years. The first 13 editions were written by me with an average time interval between editions of some 4.5 years. Upon completion of each edition, I felt that I had done all I can to make it the best available. However, as the years passed, there was always something that could withstand some improvement. For the 14th edition, I felt it wise to join hands with Professor Brian Olivari (Maine Maritime Academy) to be sure that I was fully aware of recent trends in the scope and depth of coverage in various subject areas. I feel it has been a wonderful union in developing the content of this edition. My history in the educational field includes some 45 years at various professorial ranks in the City University, New York. There was also a period of some 10 years as assistant dean at the Thayer School of Engineering at Dartmouth. During my teaching career, I was fortunate enough to find the time to write some seven texts related to the electrical/electronics field. The two most successful are Introductory Circuit Anal-ysis and Electronic Devices and Circuit Theory (co-authored with Professor Louis Nashelsky). The first one is in its 14th edition now, and the latter is in its 11th edition. Both texts have been translated into foreign languages with the ICA text translated into seven languages: Arabic, Chinese, French, Korean, Portuguese, Spanish, and Taiwanese. In addition to Introductory Circuit Analysis, a number of manuals and texts related to computer languages such as BASIC Applied to Circuit Analysis, Manual for the IBM PC jr., and BASIC for Electronics and Computer Technologies were also published. I look forward to any response from users of this 14th edition to ensure that the text remains current in the years to come. BRIAN A. OLIVARI I was honored to have been asked by Professor Boylestad to collaborate on the 14th edition of Introductory Circuit Analysis. I could never have imagined being co-author on the latest edition when I first used Professor Boylestad’s Electronic Devices and Circuit Theory textbook as a student back in 1977, and then the 11th edition of Introductory Cir-cuit Analysis as a professor in 2008. As a faculty member at Maine Maritime Academy for over 15 years, my teaching has focused on both electrical engineer-ing and mathematics. In engineering, my primary interest has been in circuit analysis and electronics. In mathematics, I have been committed to preparing my math students to be successful at solving engineering problems. Prior to teaching at Maine Maritime Academy, much of my professional experience came from working in the telecommu-nications industry at Western Electric, Rochester Telephone, and, most notably, Bell Labs where I was part of a team of engineers (systems, mechanical, electrical, and software) that developed the first-of-its-kind, stored-program-controlled D5 Digital Channel Bank System. Additionally, I served as an associate editor for the IEEE Solid-State Circuits Conference Digest of Technical Papers and taught briefly at Rochester Institute of Technology. This page is intentionally left blank S S I 1 Introduction 1.1 THE ELECTRICAL /ELECTRONICS INDUSTRY Over the past few decades, technology has been changing at an ever-increasing rate. The pressure to develop new products, improve the performance and reliability of existing sys-tems, and create new markets will only accelerate that rate. This pressure, however, is also what makes this field so exciting. New ways of storing and manipulating information, manu-facturing integrated circuits, and developing software-embedded hardware systems that can “think” on their own based on input data are only a few possibilities. Change has always been part of the human experience, but in the past, it was gradual. In recent years, however, the rate at which change in the electrical/electronics industry has been taking place is mind-boggling. Just think, for example, it was only a few years ago that “flatscreen” TVs, with wide, flat screens were introduced. Already, these have been eclipsed by much wider 4K ultra high definition (UHD) and 3D models. Miniaturization has resulted in huge advances in electronic systems. For example, advancement in this technology has allowed cell phones that originally were the size of note-books to now be smaller than a deck of playing cards. These so-called smartphones (devices), such as an iPhone 12 (2020), provide a multitude of extremely desirable features, in addition to being able to make simple telephone calls. Such features include a multi-touch glass capac-itive screen, mp3 music player (iPod capability), web browser, email, text messaging, calen-dar and contacts, mobile game player, facial recognition (for security purposes), 4K HD video recording, high-resolution camera, video conferencing (FaceTime), GPS, Bluetooth, Wi-Fi, and the ability to run over 2 million software applications “Apps,” to name a few. Remarkably, this iPhone can also simultaneously store in its 256 GB memory: 6000 songs, 30,000 photos, as well as 18 hours of 4K video recorded at 30 fps. The Apple Watch, and other “smartwatches,” provide smartphone users with the ability to extend some of their devices’ capabilities to their wrists. Of course, laptop computers, iPads, and the like, have become ubiquitous in our society, primarily due to rapid technological advances in electronic systems. Nearly invisible and more powerful hearing aids with background noise cancellation features are now available for the hearing impaired. The list of new or improved products • Become aware of the rapid growth of the electrical/ electronics industry over the past century. • Understand the importance of applying a unit of measurement to a result or measurement and to ensure that the numerical values substituted into an equation are consistent with the unit of measurement of the various quantities. • Become familiar with the International System of Units (SI) also referred to as the metric system. • Understand the importance of powers of ten and how to work with them in any numerical calculation. • Be able to convert any quantity, in any system of units, to another system with confidence. Objectives 22 \| \| \| Introductio S S I continues to expand because significantly smaller electronic systems have been developed. Most of us would agree that the computer has revolutionized our daily lives, in ways never imagined not that long ago—and it continues to do so in an accelerated fashion. When we compare the Guidance Computer aboard the Apollo 11 spacecraft, which landed on the moon back in 1969, with a commonly used cell phone in use today (the iPhone), we discover Apollo’s computer weighed 70 pounds and offered 2.3 MHz of processing power. Whereas, the performance data for even a much older generation iPhone 6 (2014) is staggering. At 3.36 billion instructions per second, the iPhone 6’s clock is 32,600 times faster than the best on board computers of the Apollo era and could perform instructions 120,000,000 times faster. You wouldn’t be wrong in saying an iPhone could be used to guide 120,000,000 Apollo-era spacecrafts to the moon, all at the same time. And . . . the iPhone weighs a mere 4.6 ounces! Spurred on by the continuing process of miniaturization is a serious and growing interest in artificial intelligence, a term first used in 1955, as a drive to replicate the brain’s function with a packaged electronic equivalent. Although only about 3 pounds in weight, a size equivalent to about 2.5 pints of liquid with a power drain of about 20 watts (half that of a 40-watt light bulb), the brain contains over 100 billion neurons that have the ability to “fire” 200 times a second. Imagine the number of decisions made per second if all are firing at the same time! This num-ber, however, is undaunting to researchers who feel that an electronic package capable of emulating the human brain is a genuine possibility in the next 10 to 15 years. Of course, including emotional qualities will be the biggest challenge, but otherwise researchers feel the advances of recent years are clear evidence that it is a real possibility. Consider how much of our daily lives is already decided for us with automatic brake control, programmed parallel parking, GPS, Web searching, and so on. The move is obviously strong and on its way. Also, when you consider how far we have come since the development of the first transistor some 67 years ago, who knows what might develop in the next decade or two? This reduction in size of electronic systems is due primarily to an important innovation introduced in 1958—the integrated circuit (IC). An integrated circuit can now contain features less than 50 nanometers across. The fact that measurements are now being made in nanometers has resulted in the terminology nanotechnology to refer to the production of integrated circuits called nanochips. To better appreciate the impact of nanometer measurements, consider drawing 100 lines within the boundar-ies of 1 inch. Then attempt drawing 1000 lines within the same length. Cutting 50-nanometer features would require drawing over 500,000 lines in 1 inch. The integrated circuit shown in Fig. 1.1 is an intel® CoreTM i7 6-core processor that has around 3 billion transistors—a number hard to comprehend. However, before a decision is made on such dramatic reductions in size, the system must be designed and tested to determine if it is worth constructing as an integrated circuit. That design process requires engi-neers who know the characteristics of each device used in the system, including undesirable characteristics that are part of any electronic element. In other words, there are no ideal (perfect) elements in an elec-tronic design. Considering the limitations of each component is necessary A BRIEF HISTORY \| \| \| 23 S S I to ensure a reliable response under all conditions of temperature, vibra-tion, and effects of the surrounding environment. To develop this aware-ness requires time and must begin with understanding the basic characteristics of the device, as covered in this text. One of the objectives of this text is to explain how ideal components work and their function in a network. Another is to explain conditions in which components may not be ideal. One of the very positive aspects of the learning process associated with electric and electronic circuits is that once a concept or procedure is clearly and correctly understood, it will be useful throughout the career of the individual at any level in the industry. Once a law or equa-tion is understood, it will not be replaced by another equation as the material becomes more advanced and complicated. For instance, one of the first laws to be introduced is Ohm’s law. This law provides a rela-tionship between forces and components that will always be true, no matter how complicated the system becomes. In fact, it is an equation that will be applied in various forms throughout the design of the entire system. The use of the basic laws may change, but the laws will not change and will always be applicable. It is vitally important to understand that the learning process for cir-cuit analysis is sequential. That is, the first few chapters establish the foundation for the remaining chapters. Failure to properly understand the opening chapters will only lead to difficulties understanding the material in the chapters to follow. This first chapter provides a brief his-tory of the field followed by a review of mathematical concepts neces-sary to understand the rest of the material. 1.2 A BRIEF HISTORY In the sciences, once a hypothesis is proven and accepted, it becomes one of the building blocks of that area of study, permitting additional investigation and development. Naturally, the more pieces of a puzzle available, the more obvious is the avenue toward a possible solution. In fact, history demonstrates that a single development may provide the key that will result in a mushrooming effect that brings the science to a new plateau of understanding and impact. If the opportunity presents itself, read one of the many publications reviewing the history of this field. Space requirements are such that only a brief review can be provided here. There are many more contributors FIG. 1.1 Intel® Core™ i7 6-core processer. Postage stamp 24 \| \| \| Introductio S S I than could be listed, and their efforts have often provided important keys to the solution of some very important concepts. Throughout history, some periods were characterized by what appeared to be an explosion of interest and development in particular areas. As you will see from the discussion of the late 1700s and the early 1800s, inven-tions, discoveries, and theories came fast and furiously. Each new concept broadens the possible areas of application until it becomes almost impos-sible to trace developments without picking a particular area of interest and following it through. In the review, as you read about the development of radio, television, and computers, keep in mind that similar progressive steps were occurring in the areas of the telegraph, the telephone, power generation, the phonograph, appliances, and so on. There is a tendency when reading about the great scientists, inventors, and innovators to believe that their contribution was a totally individual effort. In many instances, this was not the case. In fact, many of the great contributors had friends or associates who provided support and encour-agement in their efforts to investigate various theories. At the very least, they were aware of one another’s efforts to the degree possible in the days when a letter was often the best form of communication. In particular, note the closeness of the dates during periods of rapid development. One contributor seemed to spur on the efforts of the others or possibly pro-vided the key needed to continue with the area of interest. In the early stages, the contributors were not electrical, electronic, or computer engineers as we know them today. In most cases, they were physicists, chemists, mathematicians, or even philosophers. In addition, they were not from one or two communities of the Old World. The home country of many of the major contributors introduced in the paragraphs to follow is provided to show that almost every established community had some impact on the development of the fundamental laws of electri-cal circuits. As you proceed through the remaining chapters of the text, you will find that a number of the units of measurement bear the name of major contributors in those areas—volt after Count Alessandro Volta, ampere after André Ampère, ohm after Georg Ohm, and so forth—fitting recogni-tion for their important contributions to the birth of a major field of study. Time charts indicating a limited number of major developments are provided in Fig. 1.2, primarily to identify specific periods of rapid devel-opment and to reveal how far we have come in the last 40 years. In essence, the current state of the art is a result of efforts that began in earnest some 250 years ago, with progress in the last 100 years being almost exponential. As you read through the following brief review, try to sense the grow-ing interest in the field and the enthusiasm and excitement that must have accompanied each new revelation. Although you may find some of the terms used in the review new and essentially meaningless, the remaining chapters will explain them thoroughly. The Beginning The phenomenon of static electricity has intrigued scholars throughout history. The Greeks called the fossil resin substance so often used to demonstrate the effects of static electricity elektron, but no extensive study was made of the subject until William Gilbert researched the phe-nomenon in 1600. In the years to follow, there was a continuing investi-gation of electrostatic charge by many individuals, such as Otto von A BRIEF HISTORY \| \| \| 25 S S I Guericke, who developed the first machine to generate large amounts of charge, and Stephen Gray, who was able to transmit electrical charge over long distances on silk threads. Charles DuFay demonstrated that charges either attract or repel each other, leading him to believe that there were two types of charge—a theory we subscribe to today with our defined positive and negative charges. There are many who believe that the true beginnings of the electrical era lie with the efforts of Pieter van Musschenbroek and Benjamin Franklin. In 1745, van Musschenbroek introduced the Leyden jar for the storage of electrical charge (the first capacitor) and demonstrated electrical shock (and therefore the power of this new form of energy). Franklin used the Leyden jar some 7 years later to establish that light-ning is simply an electrical discharge, and he expanded on a number of other important theories, including the definition of the two types of charge as positive and negative. From this point on, new discoveries and theories seemed to occur at an increasing rate as the number of individu-als performing research in the area grew. In 1784, Charles Coulomb demonstrated in Paris that the force between charges is inversely related to the square of the distance between the charges. In 1791, Luigi Galvani, professor of anatomy at the University of Bologna, Italy, performed experiments on the effects of electricity on animal nerves and muscles. The first voltaic cell, with its ability to produce electricity through the chemical action of a metal dissolving in an acid, was developed by another Italian, Alessandro Volta, in 1799. The fever pitch continued into the early 1800s, with Hans Christian Oersted, a Danish professor of physics, announcing in 1820 a relationship between magnetism and electricity that serves as the foundation for the theory of electromagnetism as we know it today. In the same year, a French physicist, André Ampère, demonstrated that there are magnetic Pentium 4 chip Wi-Fi (1996) Intel® Core™ 2 1.5 GHz (2001) 1950 First laptop Vacuum tube ampliﬁers B&W TV (1932) Electronic Microwave oven computers (1945) Solid-state era (1947) FM Radar radio (1929) 1900 1877 Floppy disk (1970) Apple’s mouse (1983) 2000 Mobile telephone (1946) Color TV (1940) ICs (1958) First assembled PC (Apple II in 1977) Telephone (b) A.D. 0 2000 1750s Gilbert 1600 Development 1000 1900 Fundamentals (a) Amazing Development Last 40 years Nanotechnology iPhone (2007) iPod (2001) iPhone 6S (2014) Apple watch (2015) Intel i7 (2018) Fuel-cell cars (2014) iPad (2010) Electric car (the Volt) (2011) Memristor computer (1979) processor 3 GHz (2006) Cell phone (1991) GPS (1993) Electrical ⁄ Electronics Era FIG. 1.2 Time charts: (a) long-range, (b) expanded. 26 \| \| \| Introductio S S I effects around every current-carrying conductor and that current-carrying conductors can attract and repel each other just like magnets. In the period 1826 to 1827, a German physicist, Georg Ohm, introduced an important relationship between potential, current, and resistance that we now refer to as Ohm’s law. In 1831, an English physicist, Michael Faraday, demon-strated his theory of electromagnetic induction, whereby a changing cur-rent in one coil can induce a changing current in another coil, even though the two coils are not directly connected. Faraday also did extensive work on a storage device he called the condenser, which we refer to today as a capacitor. He introduced the idea of adding a dielectric between the plates of a capacitor to increase the storage capacity (Chapter 10). James Clerk Maxwell, a Scottish professor of natural philosophy, performed extensive mathematical analyses to develop what are currently called Maxwell’s equations, which support the efforts of Faraday linking electric and mag-netic effects. Maxwell also developed the electromagnetic theory of light in 1862, which, among other things, revealed that electromagnetic waves travel through air at the velocity of light (186,000 miles per second or × 3 108 meters per second). In 1888, a German physicist, Heinrich Rudolph Hertz, through experimentation with lower-frequency electro-magnetic waves (microwaves), substantiated Maxwell’s predictions and equations. In the mid-1800s, Gustav Robert Kirchhoff introduced a series of laws of voltages and currents that find application at every level and area of this field (Chapters 5 and 6). In 1895, another German physicist, Wilhelm Röntgen, discovered electromagnetic waves of high frequency, commonly called X-rays today. By the end of the 1800s, a significant number of the fundamental equations, laws, and relationships had been established, and various fields of study, including electricity, electronics, power generation and distribution, and communication systems, started to develop in earnest. The Age of Electronics Radio The true beginning of the electronics era is open to debate and is sometimes attributed to efforts by early scientists in applying poten-tials across evacuated glass envelopes. However, many trace the begin-ning to Thomas Edison, who added a metallic electrode to the vacuum of the tube and discovered that a current was established between the metal electrode and the filament when a positive voltage was applied to the metal electrode. The phenomenon, demonstrated in 1883, was referred to as the Edison effect. In the period to follow, the transmission of radio waves and the development of the radio received widespread attention. In 1887, Heinrich Hertz, in his efforts to verify Maxwell’s equations, transmitted radio waves for the first time in his laboratory. In 1896, an Italian scientist, Guglielmo Marconi (often called the father of the radio), demonstrated that telegraph signals could be sent through the air over long distances (2.5 kilometers) using a grounded antenna. In the same year, Aleksandr Popov sent what might have been the first radio message some 300 yards. The message was the name “Heinrich Hertz” in respect for Hertz’s earlier contributions. In 1901, Marconi established radio communication across the Atlantic. In 1904, John Ambrose Fleming expanded on the efforts of Edison to develop the first diode, commonly called Fleming’s valve—actually the first of the electronic devices. The device had a profound impact on the design of detectors in the receiving section of radios. In 1906, Lee De Forest added a third element to the vacuum structure and created the A BRIEF HISTORY \| \| \| 27 S S I first amplifier, the triode. Shortly thereafter, in 1912, Edwin Armstrong built the first regenerative circuit to improve receiver capabilities and then used the same contribution to develop the first nonmechanical oscillator. By 1915, radio signals were being transmitted across the United States, and in 1918 Armstrong applied for a patent for the super-heterodyne circuit employed in virtually every television and radio to permit amplification at one frequency rather than at the full range of incoming signals. The major components of the modern-day radio were now in place, and sales in radios grew from a few million dollars in the early 1920s to over $1 billion by the 1930s. The 1930s were truly the golden years of radio, with a wide range of productions for the listening audience. Television (TV) The 1930s were also the true beginnings of the television era, although development on the picture tube began in ear-lier years with Paul Nipkow and his electrical telescope in 1884 and John Baird and his long list of successes, including the transmission of television pictures over telephone lines in 1927 and over radio waves in 1928, and simultaneous transmission of pictures and sound in 1930. In 1932, NBC installed the first commercial television antenna on top of the Empire State Building in New York City, and RCA began regular broadcasting in 1939. World War II slowed development and sales, but in the mid-1940s the number of TV sets grew from a few thousand to a few million. Color television became popular in the early 1960s. Computers The earliest computer system can be traced back to Blaise Pascal in 1642 with his mechanical machine for adding and subtracting numbers. In 1673, Gottfried Wilhelm von Leibniz used the Leibniz wheel to add multiplication and division to the range of opera-tions, and in 1823 Charles Babbage developed the difference engine to add the mathematical operations of sine, cosine, logarithms, and several others. In the years to follow, improvements were made, but the system remained primarily mechanical until the 1930s when electromechanical systems using components such as relays were introduced. It was not until the 1940s that totally electronic systems became the new wave. It is interesting to note that, even though IBM was formed in 1924, it did not enter the computer industry until 1937. An entirely electronic system known as ENIAC was dedicated at the University of Pennsylvania in 1946. It contained 18,000 tubes and weighed 30 tons but was several times faster than most electromechanical systems. Although other vac-uum tube systems were built, it was not until the birth of the solid-state era that computer systems experienced a major change in size, speed, and capability. The Solid-State Era In 1947, physicists William Shockley, John Bardeen, and Walter H. Brattain of Bell Telephone Laboratories demonstrated the point-contact transistor (Fig. 1.3), an amplifier constructed entirely of solid-state materials with no requirement for a vacuum, glass envelope, or heater voltage for the filament. Although reluctant at first due to the vast amount of material available on the design, analysis, and synthesis of tube networks, the industry eventually accepted this new technology as the wave of the future. In 1958, the first integrated circuit (IC) chip was developed at Texas Instruments, and in 1961 the first commercial SSPL/Getty Images FIG. 1.3 The first transistor. 28 \| \| \| Introductio S S I integrated circuit was manufactured by the Fairchild Corporation. Today some one trillion integrated circuits are manufactured each year. It is impossible to review properly the entire history of the electrical/ electronics field in a few pages. The effort here, both through the discus-sion and the time graphs in Fig. 1.2, was to reveal the amazing progress of this field in the last 50 years. The growth appears to be truly exponen-tial since the early 1900s, raising the interesting question, Where do we go from here? The time chart suggests that the next few decades will probably contain many important innovative contributions that may cause an even faster growth curve than we are now experiencing. 1.3 UNITS OF MEASUREMENT One of the most important rules to remember and apply when working in any field of technology is to use the correct units when substituting numbers into an equation. Too often we are so intent on obtaining a numerical solution that we overlook checking the units associated with the numbers being substituted into an equation. Results obtained, there-fore, are often meaningless. Consider, for example, the following very fundamental physics equation: velocity = v d t = v d distance = (1.1) = t time Assume, for the moment, that the following data are obtained for a mov-ing object: = = d t 4000 ft 1 min and v is desired in miles per hour. Often, without a second thought or consideration, the numerical values are simply substituted into the equa-tion, with the result here that = = = d t 4000 ft 1 min 4000 mph v As indicated above, the solution is totally incorrect. If the result is desired in miles per hour, the unit of measurement for distance must be miles, and that for time, hours. In a moment, when the problem is ana-lyzed properly, the extent of the error will demonstrate the importance of ensuring that the numerical value substituted into an equation must have the unit of measurement specified by the equation. The next question is normally, How do I convert the distance and time to the proper unit of measurement? A method is presented in Section 1.9 of this chapter, but for now it is given that = 1 mi 5280 ft 4000 ft 0.76 mi = = = 1 min h 0.017 h 1 60 Substituting into Eq. (1.1), we have = = = v d t 44 71 mph 0.76 mi 0.017 h . which is significantly different from the result obtained before. SYSTEMS OF UNITS \| \| \| 29 S S I To complicate the matter further, suppose the distance is given in kilometers, as is now the case on many road signs. First, we must real-ize that the prefix kilo stands for a multiplier of 1000 (to be introduced in Section 1.5), and then we must find the conversion factor between kilometers and miles. If this conversion factor is not readily available, we must be able to make the conversion between units using the con-version factors between meters and feet or inches, as described in Section 1.9. Before substituting numerical values into an equation, try to mentally establish a reasonable range of solutions for comparison purposes. For instance, if a car travels 4000 ft in 1 min, does it seem reasonable that the speed would be 4000 mph? Obviously not! This self-checking pro-cedure is particularly important in this day of the handheld calculator, when ridiculous results may be accepted simply because they appear on the digital display of the instrument. Finally, if a unit of measurement is applicable to a result or piece of data, then it should be applied to the numerical value. To state that = 44.71 v without including the unit of measurement mph is meaningless. Eq. (1.1) is not a difficult one. A simple algebraic manipulation will result in the solution for any one of the three variables. However, in light of the number of questions arising from this equation, the reader may wonder if the difficulty associated with an equation will increase at the same rate as the number of terms in the equation. In the broad sense, this will not be the case. There is, of course, more room for a mathematical error with a more complex equation, but once the proper system of units is chosen and each term properly found in that system, there should be very little added difficulty associated with an equation requiring an increased number of mathematical calculations. In review, before substituting numerical values into an equation, be absolutely sure of the following: 1. Each quantity has the proper unit of measurement as defined by the equation. 2. The proper magnitude of each quantity as determined by the defining equation is substituted. 3. Each quantity is in the same system of units (or as defined by the equation). 4. The magnitude of the result is of a reasonable nature when compared to the level of the substituted quantities. 5. The proper unit of measurement is applied to the result. 1.4 SYSTEMS OF UNITS In the past, the systems of units most commonly used were the English and metric, as outlined in Table 1.1. Note that while the English sys-tem is based on a single standard, the metric is subdivided into two interrelated standards: the MKS and the CGS. Fundamental quantities of these systems are compared in Table 1.1 along with their abbrevia-tions. The MKS and CGS systems draw their names from the units of measurement used with each system; the MKS system uses Meters, Kilograms, and Seconds, while the CGS system uses Centimeters, Grams, and Seconds. 30 \| \| \| Introductio S S I Understandably, the use of more than one system of units in a world that finds itself continually shrinking in size, due to advanced technical developments in communications and transportation, would introduce unnecessary complications to the basic understanding of any technical data. The need for a standard set of units to be adopted by all nations has become increasingly obvious. The International Bureau of Weights and Measures located at Sèvres, France, has been the host for the General Conference of Weights and Measures, attended by representatives from all nations of the world. In 1960, the General Conference adopted a sys-tem called Le Système International d’Unités (International System of Units), which has the international abbreviation SI. It was adopted by the Institute of Electrical and Electronic Engineers (IEEE) in 1965 and by the United States of America Standards Institute (USASI) in 1967 as a standard for all scientific and engineering literature. For comparison, the SI units of measurement and their abbreviations appear in Table 1.1. These abbreviations are those usually applied to each unit of measurement, and they were carefully chosen to be the most effective. Therefore, it is important that they be used whenever applica-ble to ensure universal understanding. Note the similarities of the SI sys-tem to the MKS system. This text uses, whenever possible and practical, all of the major units and abbreviations of the SI system in an effort to support the need for a universal system. Those readers requiring addi-tional information on the SI system should contact the information office of the American Society for Engineering Education (ASEE). Fig. 1.4 should help you develop some feeling for the relative magnitudes of the units of measurement of each system of units. Note in the figure the rela-tively small magnitude of the units of measurement for the CGS system. TABLE 1.1 Comparison of the English and metric systems of units. ENGLISH METRIC SI MKS CGS Length: Yard (yd) (0.914 m) Meter (m) (39.37 in.) (100 cm) Centimeter (cm) (2.54 cm = 1 in.) Meter (m) Mass: Slug (14.6 kg) Kilogram (kg) (1000 g) Gram (g) Kilogram (kg) Force: Pound (lb) (4.45 N) Newton (N) (100,000 dynes) Dyne Newton (N) Temperature: Fahrenheit ° ( F) ( ) = ° + 9 5 C 32 Celsius or Centigrade ° ( C) ( ) ( ) = ° − 5 9 F 32 Centigrade ° ( C) Kelvin (K) = + ° K 273.15 C Energy: Foot-pound (ft-lb) (1.356 joules) Newton-meter ( ) N • m or joule (J) (0.7376 ft-lb) Dyne-centimeter or erg ( ) = 1 joule 10 ergs 7 Joule (J) Time: Second (s) Second (s) Second (s) Second (s) American Society for Engineering Education (ASEE), 1818 N Street N.W., Suite 600, Washington, D.C. 20036-2479; (202) 331-3500; SYSTEMS OF UNITS \| \| \| 31 S S I A standard exists for each unit of measurement of each system. The standards of some units are quite interesting. The meter was originally defined in 1790 to be 1/10,000,000 the dis-tance between the equator and either pole at sea level, a length preserved on a platinum–iridium bar at the International Bureau of Weights and Measures at Sèvres, France. The meter is now defined with reference to the speed of light in a vacuum, which is 299,792,458 m/s. The kilogram is defined as a mass equal to 1000 times the mass of 1 cubic centimeter of pure water at 4°C. This standard is preserved in the form of a platinum–iridium cylinder in Sèvres. The second was originally defined as 1/86,400 of the mean solar day. However, since Earth’s rotation is slowing down by almost 1 second every 10 years, 1 slug English 1 kg SI and MKS 1 g CGS 1 yd 1 m 1 ft English English SI and MKS 1 yard (yd) = 0.914 meter (m) = 3 feet (ft) Length: Mass: 1 slug = 14.6 kilograms Temperature: English (Boiling) (Freezing) (Absolute zero) Fahrenheit Celsius or Centigrade Kelvin – 459.7˚F –273.15˚C 0 K 0˚F 32˚F 212˚F 0˚C 100˚C 273.15 K 373.15 K SI MKS and CGS K = 273.15 + ˚C (˚F – 32˚) ˚C = 5 9 _ ˚F = 9 5 ˚C + 32˚ _ English 1 ft-lb SI and MKS 1 joule (J) 1 erg (CGS) 1 dyne (CGS) SI and MKS 1 newton (N) 1 ft-lb = 1.356 joules 1 joule = 107 ergs 1 pound (lb) = 4.45 newtons (N) 1 newton = 100,000 dynes (dyn) 1 m = 100 cm = 39.37 in. 2.54 cm = 1 in. English CGS 1 cm 1 in. Actual lengths English 1 pound (lb) Force: Energy: 1 kilogram = 1000 g FIG. 1.4 Comparison of units of the various systems of units. 32 \| \| \| Introductio S S I the second was redefined in 1967 as 9,192,631,770 periods of the electromagnetic radiation emitted by a particular transition of the cesium atom. 1.5 SIGNIFICANT FIGURES, ACCURACY , AND ROUNDING OFF This section emphasizes the importance of knowing the source of a piece of data, how a number appears, and how it should be treated. Too often we write numbers in various forms with little concern for the for-mat used, the number of digits that should be included, and the unit of measurement to be applied. For instance, measurements of 22.1 in. and 22.10 in. imply different levels of accuracy. The first suggests that the measurement was made by an instrument accurate only to the tenths place; the latter was obtained with instrumentation capable of reading to the hundredths place. The use of zeros in a number, therefore, must be treated with care, and the implications must be understood. In general, there are two types of numbers: exact and approximate. Exact numbers are precise to the exact number of digits presented, just as we know that there are 12 apples in a dozen and not 12.1. Through-out the text, the numbers that appear in the descriptions, diagrams, and examples are considered exact, so that a battery of 100 V can be writ-ten as 100.0 V, 100.00 V, and so on, since it is 100 V at any level of precision. The additional zeros were not included for purposes of clar-ity. However, in the laboratory environment, where measurements are continually being taken and the level of accuracy can vary from one instrument to another, it is important to understand how to work with the results. Any reading obtained in the laboratory should be consid-ered approximate. The analog scales with their pointers may be diffi-cult to read, and even though the digital meter provides only specific digits on its display, it is limited to the number of digits it can provide, leaving us to wonder about the less significant digits not appearing on the display. The precision of a reading can be determined by the number of signif-icant figures (digits) present. Significant digits are those integers (0 to 9) that can be assumed to be accurate for the measurement being made. The result is that all nonzero numbers are considered significant, with zeros being significant in only some cases. For instance, the zeros in 1005 are considered significant because they define the size of the num-ber and are surrounded by nonzero digits. For the number 0.4020, the zero to the left of the decimal point is not significant but clearly defines the location of the decimal point. The other two zeros define the magni-tude of the number and the fourth-place accuracy of the reading. When adding approximate numbers, it is important to be sure that the accuracy of the readings is consistent throughout. To add a quantity accurate only to the tenths place to a number accurate to the thousandths place will result in a total having accuracy only to the tenths place. One cannot expect the reading with the higher level of accuracy to improve the reading with only tenths-place accuracy. In the addition or subtraction of approximate numbers, the entry with the lowest level of accuracy determines the level of accuracy of the solution. SIGNIFICANT FIGURES, ACCURACY , AND ROUNDING OFF \| \| \| 33 S S I For the multiplication and division of approximate numbers, the result has the same number of significant figures as the number with the least number of significant figures. For approximate numbers (and exact numbers, for that matter), there is often a need to round off the result; that is, you must decide on the appropriate level of accuracy and alter the result accordingly. The accepted procedure is simply to note the digit following the last to appear in the rounded-off form, add a 1 to the last digit if it is greater than or equal to 5, and leave it alone if it is less than 5. For example, ≅ ≅ 3.186 3.19 3.2, depending on the level of precision desired. The symbol ≅ means approximately equal to. EXAMPLE 1.1 Perform the indicated operations with the following approximate numbers and round off to the appropriate level of accuracy. a. + + = ≅536 7 532.6 4.02 0.036 536.656 . (as determined by 532.6) b. + + = ≅0 05 0.04 0.003 0.0064 0.0494 . (as determined by 0.04) EXAMPLE 1.2 Round off the following numbers to the hundredths place. a. = 32 42 32.419 . b. = 0 05 0.05328 . EXAMPLE 1.3 Round off the result 5.8764 to a. tenths-place accuracy. b. hundredths-place accuracy. c. thousandths-place accuracy. Solution: a. 5.9 b. 5.88 c. 5.876 For this text the level of accuracy to be carried through a series of calculations will be hundredths place. That is, at each stage of a devel-opment, exercise, or problem, the level of accuracy will be set using hundredths-place accuracy. Over a series of calculations this will natu-rally affect the accuracy of the final result but a limit has to be set or solutions will be carried to unwieldy levels. For instance, let us examine the following product: 9.64 0.4896 4.68504 ( )( ) = Clearly, we don’t want to carry this level of accuracy through any fur-ther calculations in a particular example. Rather, using hundredths-place accuracy, we will write it as 4.69. The next calculation may be 4.69 1.096 5.14024 ( )( ) = 34 \| \| \| Introductio S S I which to hundredths-place accuracy is 5.14. However, if we had carried the original product to its full accuracy, we would have obtained 4.68504 1.096 5.1348 ( )( ) = or, to hundredths-place accuracy, 5.13. Obviously, 5.13 is the more accurate solution, so there is a loss of accuracy using rounded-off results. However, as indicated above, this text will round off the final and intermediate results to hundredths place for clarity and ease of comparison. 1.6 POWERS OF TEN It should be apparent from the relative magnitude of the various units of measurement that very large and very small numbers are frequently encountered in the sciences. To ease the difficulty of mathematical oper-ations with numbers of such varying size, powers of ten are usually employed. This notation takes full advantage of the mathematical prop-erties of powers of ten. The notation used to represent numbers that are integer powers of ten is as follows: = = = = = = = = = = = = − − − − 1 10 1 10 0.1 10 10 10 1 100 0.01 10 100 10 1 1000 0.001 10 1000 10 1 10,000 0.0001 10 0 1 1 2 2 3 3 4 In particular, note that = 10 1 0 , and, in fact, any quantity to the zero power is 1 ( = = x 1, 1000 1, 0 0 and so on). Numbers in the list greater than 1 are associated with positive powers of ten, and numbers in the list less than 1 are associated with negative powers of ten. A quick method of determining the proper power of ten is to place a caret mark to the right of the numeral 1 wherever it may occur; then count from this point to the number of places to the right or left before arriving at the decimal point. Moving to the right indicates a positive power of ten, whereas moving to the left indicates a negative power. For example, 10,000.0 5 1 0 , 0 0 0 . 5 1014 0.00001 5 0 . 0 0 0 0 1 5 1025 1 2 3 4 1 2 3 4 5 Some important mathematical equations and relationships pertaining to powers of ten are listed below, along with a few examples. In each case, n and m can be any positive or negative real number. = = − − 1 10 10 1 10 10 n n n n (1.2) Eq. (1.2) clearly reveals that shifting a power of ten from the denom-inator to the numerator, or the reverse, requires simply changing the sign of the power. POWERS OF TEN \| \| \| 35 S S I EXAMPLE 1.4 a. 10 1 1000 1 10 3 3 = = − + b. 10 1 0.00001 1 10 5 5 = = + − The product of powers of ten: 10 10 10 n m n m ( )( ) ( ) = ( ) + (1.3) EXAMPLE 1.5 a. 10 1000 10,000 10 10 10 7 3 4 3 4 ( )( ) ( )( ) = = = ( ) + b. 10 0.00001 100 10 10 10 3 5 2 5 2 ( )( ) ( )( ) = = = − ( ) − −+ The division of powers of ten: 10 10 10 n m n m = ( ) − (1.4) EXAMPLE 1.6 a. 10 100,000 100 10 10 10 3 5 2 5 2 = = = ( ) − b. 10 1000 0.0001 10 10 10 10 7 3 4 3 4 3 4 = = = = ( ) ( ) ( ) − −− + Note the use of parentheses in part (b) to ensure that the proper sign is established between operators. The power of powers of ten: 10 10 n m nm ( ) = (1.5) EXAMPLE 1.7 a. 10 100 10 10 8 4 2 4 2 4 ( ) ( ) = = = ( )( ) b. 10 1000 10 10 6 2 3 2 3 2 ( ) ( ) = = = ( )( ) − − − − c. 10 0.01 10 10 6 3 2 3 2 3 ( ) ( ) = = = ( )( ) − − − − − For those of you with the interest, there is a YouTube video, “Powers of Ten,” produced in 1977 that is quite well done. The video presents the subject through the lens of the world we live in. A recommended viewing. Basic Arithmetic Operations Let us now examine the use of powers of ten to perform some basic arithmetic operations using numbers that are not just powers of ten. The number 5000 can be written as × = × 5 1000 5 10 , 3 and the number 0.0004 can be written as × = × − 4 0.0001 4 10 4. Of course, 10 5 can also be written as × 1 10 5 if it clarifies the operation to be performed. 36 \| \| \| Introductio S S I Addition and Subtraction To perform addition or subtraction using powers of ten, the power of ten must be the same for each term; that is, A B A B 10 10 10 n n n ( ) × ± × = ± × (1.6) Eq. (1.6) covers all possibilities, but students often prefer to remember a verbal description of how to perform the operation. Eq. (1.6) states when adding or subtracting numbers in a power-of-ten format, be sure that the power of ten is the same for each number. Then separate the multipliers, perform the required operation, and apply the same power of ten to the result. EXAMPLE 1.8 a. ( ) ( ) ( ) ( ) + = + 6300 75,000 6.3 1000 75 1000 = × + × 6.3 10 75 10 3 3 ( ) = + × 6.3 75 10 3 81 3 10 . 3 = × b. ( ) ( ) ( ) ( ) − = − 0.00096 0.000086 96 0.00001 8.6 0.00001 = × − × − − 96 10 8.6 10 5 5 ( ) = − × − 96 8.6 10 5 87 4 10 . 5 = × − Multiplication In general, A B A B 10 10 10 n m n m ( )( ) ( )( ) × × = × + (1.7) revealing that the operations with the power of ten can be separated from the operation with the multipliers. Eq. (1.7) states when multiplying numbers in the power-of-ten format, first find the product of the multipliers and then determine the power of ten for the result by adding the power-of-ten exponents. EXAMPLE 1.9 a. ( )( ) ( )( ) [ ] ( )( ) [ ] = 0.0002 0.000007 2 0.0001 7 0.000001 ( ) ( ) = × × − − 2 10 7 10 4 6 ( ) ( ) ( )( ) = × − − 2 7 10 10 4 6 14 10 10 = × − b. ( )( ) ( ) ( ) = × × − 340,000 0.00061 3.4 10 61 10 5 5 ( )( ) ( ) ( ) = × − 3.4 61 10 10 5 5 = × 207.4 10 0 = 207 4 . POWERS OF TEN \| \| \| 37 S S I Division In general, × × = × − A B A B 10 10 10 n m n m (1.8) revealing again that the operations with the power of ten can be sepa-rated from the same operation with the multipliers. Eq. (1.8) states when dividing numbers in the power-of-ten format, first find the result of dividing the multipliers. Then determine the associated power for the result by subtracting the power of ten of the denominator from the power of ten of the numerator. EXAMPLE 1.10 a. ( ) ( ) = × × = × = × − − − − − 0.00047 0.002 47 10 2 10 47 2 10 10 23.5 10 5 3 5 3 2 b. ( ) ( ) = × × = × = × − − 690,000 0.00000013 69 10 13 10 69 13 10 10 5.31 10 4 8 4 8 12 Powers In general, A A 10 10 n m m nm ( ) × = × (1.9) which again permits the separation of the operation with the power of ten from the multiplier. Eq. (1.9) states when finding the power of a number in the power-of-ten format, first separate the multiplier from the power of ten and determine each separately. Determine the power-of-ten component by multiplying the power of ten by the power to be determined. EXAMPLE 1.11 a. ( ) ( ) ( ) ( ) = × = × − − 0.00003 3 10 3 10 3 5 3 3 5 3 27 10 15 = × − b. ( ) ( ) ( ) ( ) = × = × 90,800,000 9.08 10 9.08 10 2 7 2 2 7 2 82 45 10 . 14 = × In particular, remember that the following operations are not the same. One is the product of two numbers in the power-of-ten format, while the other is a number in the power-of-ten format taken to a power. As noted below, the results of each are quite different: ≠ ( )( ) ( ) 10 10 10 3 3 3 3 ( )( ) = = 10 10 10 1,000,000 3 3 6 ( ) ( )( )( ) = = = 10 10 10 10 10 1,000,000,000 3 3 3 3 3 9 38 \| \| \| Introductio S S I 1.7 FIXED-POINT , FLOATING-POINT , SCIENTIFIC, AND ENGINEERING NOT A TION When you are using a computer or a calculator, numbers generally appear in one of four ways. If powers of ten are not employed, numbers are written in the fixed-point or floating-point notation. The fixed-point format requires that the decimal point appear in the same place each time. In the floating-point format, the decimal point appears in a location defined by the number to be displayed. Most computers and calculators permit a choice of fixed- or floating-point notation. In the fixed format, the user can choose the level of accuracy for the output as tenths place, hundredths place, thousandths place, and so on. Every output will then fix the decimal point to one location, such as the following examples using thousandths-place accuracy: = = = 0 333 0 063 1150 000 1 3 . 1 16 . 2300 2 . If left in the floating-point format, the results will appear as follows for the above operations: = = = 0 333333333333 0 0625 1150 1 3 . 1 16 . 2300 2 Powers of ten will creep into the fixed- or floating-point notation if the number is too small or too large to be displayed properly. Scientific (also called standard) notation and engineering notation make use of powers of ten, with restrictions on the mantissa (multiplier) or scale factor (power of ten). Scientific notation requires that the decimal point appear directly after the first digit greater than or equal to 1 but less than 10. A power of ten will then appear with the number (usually following the power notation E), even if it has to be to the zero power. A few examples: 3 33333333333E 1 6 25E 2 1 15E3 1 3 . 1 16 . 2300 2 . = − = − = Within scientific notation, the fixed- or floating-point format can be chosen. In the above examples, floating was employed. If fixed is cho-sen and set at the hundredths-point accuracy, the following will result for the above operations: 3 33E 1 6 25E 2 1 15E3 1 3 . 1 16 . 2300 2 . = − = − = Engineering notation specifies that all powers of ten must be 0 or multiples of 3, and the mantissa must be greater than or equal to 1 but less than 1000. This restriction on the powers of ten is because specific powers of ten have been assigned prefixes that are introduced in the next few paragraphs. FIXED-POINT , FLOATING-POINT , SCIENTIFIC, AND ENGINEERING NOTATION \| \| \| 39 S S I Using scientific notation in the floating-point mode results in the following for the above operations: 333 333333333E 3 62 5E 3 1 15E3 1 3 . 1 16 . 2300 2 . = − = − = Using engineering notation with two-place accuracy will result in the following: 333 33E 3 62 50E 3 1 15E3 1 3 . 1 16 . 2300 2 . = − = − = Prefixes Specific powers of ten in engineering notation have been assigned pre-fixes and symbols, as appearing in Table 1.2. They permit easy recogni-tion of the power of ten and an improved channel of communication between technologists. TABLE 1.2 Multiplication Factors SI Prefix SI Symbol 1000 000 000 000 000 000 10 18 = exa E 1 000 000 000 000 000 10 15 = peta P 1 000 000 000 000 10 12 = tera T 1 000 000 000 10 9 = giga G 1 000 000 10 6 = mega M 1 000 10 3 = kilo k 0 001 10 . 3 = − milli m 0 000 001 10 . 6 = − micro µ 0 000 000 001 10 . 9 = − nano n 0 000 000 000 001 10 . 12 = − pico p 0 000 000 000 000 001 10 . 15 = − femto f 0 000 000 000 000 000 001 10 . 18 = − atto a EXAMPLE 1.12 a. = × = = 1 megohm 1 MΩ 1,000,000 ohms 1 10 ohms 6 b. 100 kilometers 100 km 100,000 meters 100 10 meters 3 = × = = c. 0 1 millisecond 0 1 ms 0.0001 second 0.1 10 second . . 3 = × = = − d. µ = × = = − 1 microfarad 1 F 0.000001 farad 1 10 farad 6 Here are a few examples with numbers that are not strictly powers of ten. 40 \| \| \| Introductio S S I EXAMPLE 1.13 a. 41,200 m is equivalent to 41.2 10 m 41.2 kilometers 3 × = = 41.2 km. b. 0.00956 J is equivalent to 9.56 10 J 9.56 millijoules 3 × = = − 9.56 mJ. c. 0.000768 s is equivalent to 768 10 s 768 microseconds 6 × = = − 768 µs. d. ( ) ( ) = × × = × − − 8400 m 0.06 8.4 10 m 6 10 8.4 6 10 10 m 3 2 3 2 140 km 1.4 10 m 140 10 m 140 kilometers 5 3 = × = × = = e. ( ) ( ) = × = × − − 0.0003 s 3 10 s 81 10 s 4 4 4 16 0 0081 ps 0.0081 10 s 0.0081 picosecond . 12 = × = = − 1.8 CONVERSION BETWEEN LEVELS OF POWERS OF TEN It is often necessary to convert from one power of ten to another. For instance, if a meter measures kilohertz (kHz—a unit of measurement for the frequency of an ac waveform), it may be necessary to find the correspond-ing level in megahertz (MHz). If time is measured in milliseconds (ms), it may be necessary to find the corresponding time in microseconds µ ( ) s for a graphical plot. The process is not difficult if we simply keep in mind that an increase or a decrease in the power of ten must be asso-ciated with the opposite effect on the multiplying factor. The procedure is best described by the following steps: 1. Replace the prefix by its corresponding power of ten. 2. Rewrite the expression, and set it equal to an unknown multiplier and the new power of ten. 3. Note the change in power of ten from the original to the new format. If it is an increase, move the decimal point of the original multiplier to the left (smaller value) by the same number. If it is a decrease, move the decimal point of the original multiplier to the right (larger value) by the same number. EXAMPLE 1.14 Convert 20 kHz to megahertz. Solution: In the power-of-ten format: = × 20 kHz 20 10 Hz 3 The conversion requires that we find the multiplying factor to appear in the space below: 20 3 103 Hz 3 106 Hz Increase by 3 Decrease by 3 CONVERSION BETWEEN LEVELS OF POWERS OF TEN \| \| \| 41 S S I Since the power of ten will be increased by a factor of three, the mul-tiplying factor must be decreased by moving the decimal point three places to the left, as shown below: 0.002 3 103 m 3 1023 m Decrease by 6 Increase by 6 0.002000 5 2000 6 020. 5 0.02 3 0.01 3 1023 s 3 1026 s Decrease by 3 Increase by 3 0.010 5 10 3 and × = × = 0 02 MHz 20 10 Hz 0.02 10 Hz . 3 6 EXAMPLE 1.15 Convert 0.01 ms to microseconds. Solution: In the power-of-ten format: = × − 0.01 ms 0.01 10 s 3 and Since the power of ten will be reduced by a factor of three, the multi-plying factor must be increased by moving the decimal point three places to the right, as follows: and 10 s 0.01 10 s 10 10 s 3 6 µ × = × = − − There is a tendency when comparing −3 to −6 to think that the power of ten has increased, but keep in mind when making your judgment about increasing or decreasing the magnitude of the multiplier that − 10 6 is a great deal smaller than − 10 3. EXAMPLE 1.16 Convert 0.002 km to millimeters. Solution: In this example we have to be very careful because the difference between +3 and −3 is a factor of 6, requiring that the multiplying factor be modified as follows: and × = × = − 2000 mm 0.002 10 m 2000 10 m 3 3 42 \| \| \| Introductio S S I 1.9 CONVERSION WITHIN AND BETWEEN SYSTEMS OF UNITS The conversion within and between systems of units is a process that cannot be avoided in the study of any technical field. It is an operation, however, that is performed incorrectly so often that this section was included to provide one approach that, if applied properly, will lead to the correct result. There is more than one method of performing the conversion pro-cess. In fact, some people prefer to determine mentally whether the con-version factor is multiplied or divided. This approach is acceptable for some elementary conversions, but it is risky with more complex operations. The procedure to be described here is best introduced by examining a relatively simple problem such as converting inches to meters. Specifi-cally, let us convert 48 in. (4 ft) to meters. If we multiply the 48 in. by a factor of 1, the magnitude of the quan-tity remains the same: 1 48 in. 48 in.( ) = (1.10) Let us now look at the conversion factor for this example: = 1 m 39.37 in. Dividing both sides of the conversion factor by 39.37 in. results in the following format: = 1 1 m 39.37 in. ( ) Note that the end result is that the ratio 1 m/39.37 in. equals 1, as it should since they are equal quantities. If we now substitute this factor (1) into Eq. (1.10), we obtain =           1 48 in.( ) 48 in . 1 m 39.37 in . which results in the cancellation of inches as a unit of measure and leaves meters as the unit of measure. In addition, since the 39.37 is in the denominator, it must be divided into the 48 to complete the operation: = 1 219 m 48 39.37 m . Let us now review the method: 1. Set up the conversion factor to form a numerical value of (1) with the unit of measurement to be removed from the original quantity in the denominator. 2. Perform the required mathematics to obtain the proper magnitude for the remaining unit of measurement. EXAMPLE 1.17 Convert 6.8 min to seconds. Solution: The conversion factor is = 1 min 60 s Since the minute is to be removed as the unit of measurement, it must appear in the denominator of the (1) factor, as follows: CONVERSION WITHIN AND BETWEEN SYSTEMS OF UNITS \| \| \| 43 S S I Step 1: ( ) = 1 60 s 1 min ( ) Step 2: 1 6.8 min( ) 6.8 min 60 s 1 min 6.8 60 s ( )( ) =          = = 408 s EXAMPLE 1.18 Convert 0.24 m to centimeters. Solution: The conversion factor is 1 m 100 cm = Since the meter is to be removed as the unit of measurement, it must appear in the denominator of the (1) factor as follows: Step 1: ( ) = 1 100 cm 1 m Step 2: ( ) ( )( ) =           = 1 0.24 m 0.24 m 100 cm 1 m 0.24 100 cm 24 cm = The products (1)(1) and (1)(1)(1) are still 1. Using this fact, we can perform a series of conversions in the same operation. EXAMPLE 1.19 Determine the number of minutes in half a day. Solution: Working our way through from days to hours to minutes, always ensuring that the unit of measurement to be removed is in the denominator, results in the following sequence: 0.5 day 24 h 1 day 60 min 1 h 0.5 24 60 min ( ) ( ) ( )                       = 720 min = EXAMPLE 1.20 Convert 2.2 yards to meters. Solution: Working our way through from yards to feet to inches to meters results in the following: 2 012 m 2.2 yards 3 ft 1 yard 12 in. 1 ft 1 m 39.37 in. 2.2 3 12 39.37 m . ( )( )( )                                = = The following examples are variations of the above in practical situations. EXAMPLE 1.21 In Europe, Canada, and many other countries, the speed limit is posted in kilometers per hour. How fast in miles per hour is 100 km/h? 44 \| \| \| Introductio S S I Solution: ( )( )( )( )( ) 1 1 1 1 100 km h 62.14 100 km h 1000 m 1 km 39.37 in. 1 m 1 ft 12 in. 1 mi 5280 ft 100 1000 39.37 12 5280 mi h mph ( ) ( )( )( ) ( ) =                                                        = = Many travelers use 0.6 as a conversion factor to simplify the math involved; that is, and ( )( ) ( )( ) ≅ ≅ 100 km/h 0.6 60 mph 60 km/h 0.6 36 mph EXAMPLE 1.22 Determine the speed in miles per hour of a compet-itor who can run an amazing 4-min mile. Solution: Inverting the factor 4 min/1 mi to 1 mi/4 min, we can pro-ceed as follows:                    = = 15 mph 1 mi 4 min 60 min 4 60 4 mi h 1.10 SYMBOLS Throughout the text, various symbols will be employed that you may not have had occasion to use. Some are defined in Table 1.3, and others will be defined in the text as the need arises. 1.11 CONVERSION TABLES Conversion tables such as those appearing in Appendix A can be very useful when time does not permit the application of methods described in this chapter. However, even though such tables appear easy to use, frequent errors occur because the operations appearing at the head of the table are not performed properly. In any case, when using such tables, try to establish mentally some order of magnitude for the quantity to be determined compared to the magnitude of the quantity in its original set of units. This simple operation should prevent several impossible results that may occur if the conversion operation is improperly applied. For example, consider the following from such a conversion table: × To convert from Miles To Meters Multiply by 1.609 10 3 A conversion of 2.5 mi to meters would require that we multiply 2.5 by the conversion factor; that is, 4 02 10 m 2.5 mi(1.609 10 ) . 3 3 × = × A conversion from 4000 m to miles would require a division process: × = × = − 2 49 mi 4000 m 1.609 10 2486.02 10 . 3 3 TABLE 1.3 Symbol Meaning ≠ Not equal to ≠ 6.12 6.13 > Greater than > 4.78 4.20 > > Much greater than 840 16 > > < Less than < 430 540 < < Much less than 0.002 46 < < ≥ Greater than or equal to ≥ x y is satisfied for = y 3 and > x 3 or = x 3 ≤ Less than or equal to ≤ x y is satisfied for = y 3 and < x 3 or = x 3 ≅ Appoximately equal to ≅ 3.14159 3.14 ∑ Sum of ∑( ) + + = 4 6 8 18 \|\| Absolute magnitude of = a 4, where a 4 or 4 = − + ∴ Therefore = ∴ = ± x x 4 2 ≡ By definition Establishes a relationship between two or more quantities a:b Ratio defined by a b a b c d : : = Proportion defined by = a b c d CALCULATORS \| \| \| 45 S S I In each of the above, there should have been little difficulty realizing that 2.5 mi would convert to a few thousand meters and 4000 m would be only a few miles. As indicated above, this kind of anticipatory think-ing will eliminate the possibility of ridiculous conversion results. 1.12 CALCULATORS In most texts, the calculator is not discussed in detail. Instead, students are left with the general exercise of choosing an appropriate calculator and learning to use it properly on their own. However, some discussion about the use of the calculator is needed to eliminate some of the impos-sible results obtained (and often strongly defended by the user—because the calculator says so) through a correct understanding of the process by which a calculator performs the various tasks. Time and space do not permit a detailed explanation of all the possible operations, but the fol-lowing discussion explains why it is important to understand how a cal-culator proceeds with a calculation and that the unit cannot accept data in any form and still generate the correct answer. TI-89 Calculator Although the calculator chosen for this text is one of the more expen-sive, a great deal of thought went into its choice. The TI-89 Titanium model calculator (Fig. 1.5) was used in the previous edition and, before preparing the manuscript for this 13th edition, a study was made of the calculators available today. In all honesty, some of the cheapest calculators on the market can perform the necessary func-tions required in this text. However, the time it will take to perform some of the basic operations required in the ac section of this text may result in a high level of frustration because it takes so long to do a sim-ple analysis. The TI-89 has the ability to significantly reduce the time required and number of operations needed to complete the same analy-sis and, therefore, was chosen for this edition also. However, it is cer-tainly possible that your instructor is recommending a different calculator for the course or your chosen field. In such situations there is no doubt your professor has balanced the needs of the course with the financial obligations you face and has suggested a calculator that will perform very well. For those using the TI-89 calculator, there will be times when it seems to require more steps than you expected to perform a simple task. However, be assured that as you work through the content of this text you will be very pleased with the performance of the calculator. Bear in mind that the TI-89 has capabilities that could be very helpful in other areas of study such as mathematics and physics. In addition, it is tool that will serve you well not only in your college years but in your future career as well. When using any calculator for the first time, the unit must be set up to provide the answers in a desired format. Following are the steps needed to set up the TI-89 calculator correctly. Initial Settings In the following sequences, the arrows within the square indicate the direction of the scrolling required to reach the desired location. The scrolling may continue for a number of levels but eventu-ally the desired heading will appear on the screen. (Don Johnson Photo) FIG. 1.5 Texas Instruments Titanium TI-89 calculator. 46 \| \| \| Introductio S S I Notation The first sequence sets the engineering notation (Section 1.7) as the choice for all answers. It is particularly important to take note that you must select the ENTER key twice to ensure the process is set in memory. Exponential Format ENGINEERING Accuracy Level Next, the accuracy level can be set to two places as follows: Display Digits 3:FIX 2 Approximate Mode For all solutions, the solution should be in decimal form to second-place accuracy. If this is not set, some answers will appear in fractional form to ensure the answer is EXACT (another option). This selection is made with the following sequence: Exact/Approx 3: APPROXIMATE Clear Screen To clear the screen of all entries and results, use the following sequence: 8: Clear Home Clear Current Entries To delete the sequence of current entries at the bottom of the screen, select the CLEAR key. Turn Off To turn off the calculator, apply the following sequence: Calculator Fundamentals Order of Operations Although setting the correct format and accu-rate input is important, improper results occur primarily because users fail to realize that no matter how simple or complex an equation, the calculator performs the required operations in a specific order. This is a fact that is true for any calculator you may use. The content below is for the majority of commercially available calculators. Consider the operation + 8 3 1 which is often entered as + = + = + = 3 67 8 3 1 2.67 1 . This is incorrect (2 is the answer). The calculator will not perform the addition first and then the divi-sion. In fact, addition and subtraction are the last operations to be per-formed in any equation. It is therefore very important that you carefully study and thoroughly understand the next few paragraphs in order to use the calculator properly. 1. The first operations to be performed by a calculator can be set using parentheses (). It does not matter which opera-tions are within the parentheses. The parentheses simply CALCULATORS \| \| \| 47 S S I dictate that this part of the equation is to be determined first. There is no limit to the number of parentheses in each equation—all operations within parentheses will be per-formed first. For instance, for the example above, if paren-theses are added as shown below, the addition will be performed first and the correct answer obtained: + = 8 3 1 ( ) + = 2 00 . 2. Next, powers and roots are performed, such as x2, x , and so on. 3. Negation (applying a negative sign to a quantity) and single-key operations such as sin, tan−1, and so on, are performed. 4. Multiplication and division are then performed. 5. Addition and subtraction are performed last. It may take a few moments and some repetition to remember the order, but at least you are now aware that there is an order to the opera-tions and that ignoring them can result in meaningless results. EXAMPLE 1.23 Determine 9 3 Solution: = 1 73 . In this case, the left bracket is automatically entered after the square root sign. The right bracket must then be entered before performing the calculation. For all calculator operations, the number of right brackets must always equal the number of left brackets. EXAMPLE 1.24 Find + 3 9 4 Solution: If the problem is entered as it appears, the incorrect answer of 5.25 will result. + = + = 3 9 4 5.25 Using brackets to ensure that the addition takes place before the divi-sion will result in the correct answer as shown below: + ( ) = + = = 3 00 3 9 4 12 4 . EXAMPLE 1.25 Determine + + 1 4 1 6 2 3 48 \| \| \| Introductio S S I Solution: Since the division will occur first, the correct result will be obtained by simply performing the operations as indicated. That is, + + = + + = 1 08 1 4 1 6 2 3 . Powers of Ten The key is used to set the power of ten of a number. Setting up the number = × 2200 2.2 10 3 requires the follow-ing keypad selections: = 2 20E3 . Setting up the number × − 8.2 10 6 requires the negative sign (–) from the numerical keypad. Do not use the negative sign from the mathemati-cal listing of ÷, ×, –, and +. That is, 8 20E 6 . = − EXAMPLE 1.26 Perform the addition × + × 6.3 10 75 10 3 3 and compare your answer with the longhand solution of Example 1.8(a). Solution: + = 81 30E3 . which confirms the results of Example 1.8(a). EXAMPLE 1.27 Perform the division ( ) ( ) × × − 69 10 13 10 4 8 and compare your answer with the longhand solution of Example 1.10(b). Solution: = 5 31E12 . which confirms the results of Example 1.10(b). EXAMPLE 1.28 Using the provided format of each number, per-form the following calculation in one series of keypad entries: ( ) ( ) ( ) × × = − − 0.004 6 10 2 10 ? 4 3 2 Solution: 600 00E 3 0 6 . . = − = Brackets were used to ensure that the calculations were performed in the correct order. Note also that the number of left brackets equals the num-ber of right brackets. COMPUTER ANALYSIS \| \| \| 49 S S I 1.13 COMPUTER ANAL YSIS There is no question that a basic knowledge of computer methods is something that the graduating student must learn in an educational pro-gram. Industry now requires students to be proficient in the use of a computer. Two general directions can be taken to develop the necessary com-puter skills: the study of computer languages or the use of software packages. Languages There are several languages that provide a direct line of communication with the computer and the operations it can perform. A language is a set of symbols, letters, words, or statements that the user can enter into the computer. The computer system will “understand” these entries and will perform them in the order established by a series of commands called a program. The program tells the computer what to do on a sequential, line-by-line basis in the same order a student would perform the calcula-tions in longhand. The computer can respond only to the commands entered by the user. This requires that the programmer understand fully the sequence of operations and calculations required to obtain a particu-lar solution. A lengthy analysis can result in a program having hundreds or thousands of lines. Once written, the program must be checked care-fully to ensure that the results have meaning and are valid for an expected range of input variables. Some of the popular languages applied in the electrical/electronics field today include C++, QBASIC, Java, and FORTRAN. Each has its own set of commands and statements to communicate with the computer, but each can be used to perform the same type of analysis. Software Packages The second approach to computer analysis—software packages— avoids the need to know a particular language; in fact, the user may not be aware of which language was used to write the programs within the package. All that is required is a knowledge of how to input the network parameters, define the operations to be performed, and extract the results; the package will do the rest. However, there is a problem with using packaged programs without understanding the basic steps the pro-gram uses. You can obtain a solution without the faintest idea of either how the solution was obtained or whether the results are valid or way off base. It is imperative that you realize that the computer should be used as a tool to assist the user—it must not be allowed to control the scope and potential of the user! Each software package has a menu, which defines the range of appli-cation of the package. Once the software is entered into the computer, the system will perform all the functions appearing in the menu, as it was preprogrammed to do. Be aware, however, that if a particular type of analysis is requested that is not on the menu, the software package cannot provide the desired results. The package is limited solely to those maneuvers developed by the team of programmers who developed the software package. In such situations the user must turn to another soft-ware package or write a program using one of the languages listed above. 50 \| \| \| Introductio S S I In broad terms, if a software package is available to perform a partic-ular analysis, then that package should be used rather than developing new routines. Most popular software packages are the result of many hours of effort by teams of programmers with years of experience. How-ever, if the results are not in the desired format, or if the software pack-age does not provide all the desired results, then the user’s innovative talents should be put to use to develop a software package. As noted above, any program the user writes that passes the tests of range and accuracy can be considered a software package of his or her authorship for future use. PROBLEMS Note: More difficult problems are denoted by an asterisk () throughout the text. SECTION 1.2 A Brief History 1. Visit your local library (at school or home) and describe the extent to which it provides literature and computer support for the technologies—in particular, electricity, electronics, electromagnetics, and computers. 2. Choose an area of particular interest in this field and write a very brief report on the history of the subject. 3. Choose an individual of particular importance in this field and write a very brief review of his or her life and important contributions. SECTION 1.3 Units of Measurement 4. a. In a recent Tour de France time trial, a participant trav-eled 60.5 mi in a time trial in 2 h, 15 min. What was his average speed in mph? b. What is the speed in km/h? 5. In most countries outside the United States speed is mea-sured in kilometers per hour (km/h) rather than miles per hour (mph). Although the exact conversion from km/h to mph is 0.57, a conversion factor of 0.6 is generally used because it is easier to remember. For the reverse conversion a factor of 1.7 is normally used. Perform the following con-versions using the approximate conversion factors. a. What is the speed in mph of an automobile on the Ger-man Autobahn traveling at 160 km/h? b. What is the speed in km/h of a motorcycle traveling at 70 mph in the United States? 6. How long in seconds will it take a car traveling at 80 mph to travel the length of a football field (100 yd)? 7. A pitcher has the ability to throw a baseball at 95 mph. a. How fast is the speed in ft/s? b. How long does the hitter have to make a decision about swinging at the ball if the plate and the mound are sepa-rated by 60 ft? c. If the batter wanted a full second to make a decision, what would the speed in mph have to be? SECTION 1.4 Systems of Units 8. Are there any relative advantages associated with the metric system compared to the English system with respect to length, mass, force, and temperature? If so, explain. 9. Which of the four systems of units appearing in Table 1.1 has the smallest units for length, mass, and force? When would this system be used most effectively? 10. Which system of Table 1.1 is closest in definition to the SI system? How are the two systems different? Why do you think the units of measurement for the SI system were cho-sen as listed in Table 1.1? Give the best reasons you can without referencing additional literature. 11. What is room temperature (23°C) in the English and SI systems? 12. How many foot-pounds of energy are associated with 4000 J? 13. In Europe the height of a man or woman is measured in centimeters and his or her weight in kilograms. a. What is the weight in pounds (lb) of a man who weighs 70.8 kg? b. What is the weight in kg of a woman who weighs 145 lb? c. What is the height of a man in cm who is 6 ft tall? d. In Norway, the average height of a man is 179.9 cm. What is his height in feet and inches? 14. Throughout the world, the majority of countries use the centigrade scale rather than the Fahrenheit scale. This can cause problems for travelers not familiar with what to expect at certain temperature levels. To alleviate this prob-lem, the following approximate conversion is typically used: ( ) ° = ° + ° F 2 C 30 Comparing to the exact formula of ° = ° + ° F 9 5 C 32 , we find the ratio 9/5 is approximated to equal 2, and the tem-perature of 32° is changed to 30° simply to make the num-bers easier to work with and slightly compensate for the fact that 2(°C) is more than 9/5(°C). a. The temperature of 20°C is commonly accepted as nor-mal room temperature. Using the approximate formula, determine (in your head) the equivalent Fahrenheit temperature. PROBLEMS \| \| \| 51 S S I 24. Perform the following operations to hundredths-place accuracy: a. 4000 0.00008 b. × 0.006 6 10 6 c. 0.000440 0.00005 d. × × − 88 10 8 10 18 8 25. Perform the following operations: a. (100)3 b. (0.0001)1/2 c. (10,000)8 d. (0.00000010)9 26. Perform the following operations to hundredths-place accuracy: a. ( ) 400 2 b. ( ) × − 6 10 4 4 c. ( ) ( ) × − 0.005 3 10 4 2 d. 2 10 0.8 10 0.0005 10 4 5 6 3 ( )( ) [ ] ( ) × × × − 27. Perform the following operations to hundredths-place accuracy: a. ( ) ( ) × 300 100 3 10 2 4 b. [ ][ ] ( ) ( )− 40,000 20 2 3 c. ( ) ( ) 60,000 0.02 2 2 d. ( ) 0.000027 200,000 1 3 e. 4000 300 2 10 2 4 [ ][ ] ( ) × − f. 0.000016 100,000 0.02 1 2 5 [ ][ ][ ] ( ) ( ) g. 0.003 0.00007 160 200 0.0008 a challenge 3 2 2 1 2 [ ] [ ][ ] [ ] ( ) ( ) ( ) ( )( ) − − SECTION 1.7 Fixed-Point, Floating-Point, Scientific, and Engineering Notation 28. Write the following numbers in scientific and engineering notation to hundredths place: a. 20.46 b. 50,420 c. 0.000674 d. 000.0460 29. Write the following numbers in scientific and engineering notation to tenths place: a. × − 5 10 2 b. × + 0.45 10 2 c. 1 32 d. π 30. Perform the following operations and leave the answer in engineering notation: a. ( ) ( ) × × = − 8 10 4 10 3 5 b. ( )( ) × × = 70 10 0.04 10 6 3 c. ( ) ( ) ( ) × × × = 0.002 10 600 10 5 10 7 5 4 d. ( ) ( ) ( ) × × × = − 6.2 10 82 10 4.02 10 4 2 3 3 b. Use the exact formula and determine the equivalent Fahrenheit temperature for 20°C. c. How do the results of parts (a) and (b) compare? Is the approximation valid as a first estimate of the Fahrenheit temperature? d. Repeat parts (a) and (b) for a high temperature of 30°C and a low temperature of 5°C. SECTION 1.5 Significant Figures, Accuracy, and Rounding Off 15. Write the following numbers to tenths-place accuracy. a. 14.6026 b. 056.0420 c. 1,046.06 d. 1/16 e. π 16. Repeat Problem 15 using hundredths-place accuracy. 17. Repeat Problem 15 using thousandths-place accuracy. SECTION 1.6 Powers of T en 18. Express the following numbers as powers of ten to hundredths-place accuracy: a. 10,000 b. 1,000,000 c. 1000 d. 0.001 e. 1 f. 0.1 19. Using only those powers of ten listed in Table 1.2, express the following numbers in what seems to you the most logi-cal form for future calculations: a. 15,000 b. 0.005 c. 2,400,000 d. 60,000 e. 0.00040200 f. 0.0000000002 20. Perform the following operations to hundredths-place accuracy: a. + 4300 47,000 b. × + × 8 10 4.6 10 4 5 c. × − × − − 0.6 10 6 10 3 6 d. × + × − − 2.6 10 60,000 10 500 3 3 21. Perform the following operations: a. (1000)(10000) b. (0.001)(100) c. (102)(107) d. (100)(0.000001) e. (10−8)(10,000,000) f. (10,000)(10–10) (1026) 22. Perform the following operations to hundredths-place accuracy: a. ( )( ) 20,000 0.003 b. × 8150 0.001 c. ( ) ( ) × 0.000059 2.2 10 5 d. ( ) ( ) ( ) × × − 21 10 0.03 5 10 4 7 23. Perform the following operations: a. 100 10,000 b. 0.010 1000 c. 10,000 0.001 d. 0.0000001 100 e. 10 0.000100 38 f. ( ) 100 0.01 1 2 52 \| \| \| Introductio S S I 41. Find the distance in meters that a mass traveling at 800 cm/s will cover in 0.048 h. 42. Each spring there is a race up 86 floors of the 102-story Empire State Building in New York City. If you were able to climb two steps/second, how long would it take in minutes to reach the 86th floor if each floor is 14 ft high and each step is about 9 in.? 43 The record for the race in Problem 42 is 10.22 min. What was the racer’s speed in min/mi for the race? 44. If the race of Problem 42 were a horizontal distance, how long would it take a runner who can run 5-min miles to cover the distance? Compare this with the record speed of Problem 43. Did gravity have a significant effect on the overall time? SECTION 1.11 Conversion T ables 45. Using Appendix A, determine the number of a. Btu in 5 J of energy. b. cubic meters in 24 oz of a liquid. c. seconds in 1.4 days. d. pints in 1 m3 of a liquid. SECTION 1.12 Calculators Perform the following operations using a single sequence of calculator keys: 46. ( ) × + = 6 4 2 8 47. + = 42 3 6 5 48. ( ) + = 5 2 3 2 2 49. ° = cos21.87 50. = − tan 3 4 1 51. + = 400 62 10 5 52. ( ) × × = − 8.2 10 0.04 10 in engineering notation 3 3 53. ( )( ) ( ) ( ) × × = 0.06 10 20 10 0.01 in engineering notation 5 3 2 54. × × + × + × − − − 4 10 2 10 400 10 1 2 10 4 3 5 6 ( ) = in engineering notation SECTION 1.13 Computer Analysis 55. Investigate the availability of computer courses and com-puter time in your curriculum. Which languages are com-monly used, and which software packages are popular? 56. Develop a list of three popular computer languages, including a few characteristics of each. Why do you think some languages are better for the analysis of electric circuits than others? SECTION 1.8 Conversion between Levels of Powers of Ten 31. Fill in the blanks of the following conversions: a. 6 10 10 4 6 × = × b. × = × − − 0.4 10 10 3 6 c. × = × = × 50 10 10 10 5 3 6 10 9 = × d. 12 10 10 10 7 3 6 × = × = × − − − 10 9 = × − 32. Perform the following conversions: a. 0.06 s to milliseconds b. 4000 µs to milliseconds c. 0.08 ms to microseconds d. 6400 ps to microseconds e. × 100 10 4 mm to kilometers SECTION 1.9 Conversion within and between Systems of Units 33. Perform the following conversions to tenths-place accuracy: a. 1.5 min to seconds b. × − 2 10 2 h to seconds c. 0.05 s to microseconds d. 0.16 m to millimeters e. 0.00000012 s to nanoseconds f. × 4 108 s to days 34. Perform the following metric conversions to tenths-place accuracy: a. 88 mm to centimeters b. 60 cm to kilometres c. × − 12 10 3 m to micrometers d. 60 sq cm (cm2) to square meters (m2) 35. Perform the following conversions between systems to hundredths-place accuracy: a. 100 in. to meters b. 4 ft to meters c. 6 lb to newtons d. 60,000 dyn to pounds e. 150,000 cm to feet f. 0.002 mi to meters (5280 ft = 1 mi) 36. What is a mile in feet, yards, meters, and kilometers? 37. Convert 60 mph to meters per second. 38. How long would it take a runner to complete a 15-km race if a pace of 8.5 min/mi were maintained? 39. Quarters are about 1 in. in diameter. How many would be required to stretch from one end of a football field to the other (100 yd)? 40. Compare the total time required to drive a long, tiring day of 500 mi at an average speed of 60 mph versus an average speed of 70 mph. Is the time saved for such a long trip worth the added risk of the higher speed? S S I GLOSSARY \| \| \| 53 Language A communication link between user and computer to define the operations to be performed and the results to be displayed or printed. Leyden jar One of the first charge-storage devices. Menu A computer-generated list of choices for the user to deter-mine the next operation to be performed. Meter (m) A unit of measure for length in the SI and MKS sys-tems. Equal to 1.094 yards in the English system and 100 cen-timeters in the CGS system. MKS system The system of units employing the Meter, Kilogram, and Second as its fundamental units of measure. Nanotechnology The production of integrated circuits in which the nanometer is the typical unit of measurement. Newton (N) A unit of measurement for force in the SI and MKS systems. Equal to 100,000 dynes in the CGS system. Pound (lb) A unit of measurement for force in the English sys-tem. Equal to 4.45 newtons in the SI or MKS system. Program A sequential list of commands, instructions, and so on, to perform a specified task using a computer. Scientific notation A method for describing very large and very small numbers through the use of powers of ten, which re-quires that the multiplier be a number between 1 and 10. Second (s) A unit of measurement for time in the SI, MKS, English, and CGS systems. SI system The system of units adopted by the IEEE in 1965 and the USASI in 1967 as the International System of Units (Système International d’Unités). Slug A unit of measure for mass in the English system. Equal to 14.6 kilograms in the SI or MKS system. Software package A computer program designed to perform specific analysis and design operations or generate results in a particular format. Static electricity Stationary charge in a state of equilibrium. Transistor The first semiconductor amplifier. Voltaic cell A storage device that converts chemical to electrical energy. GLOSSARY Artificial intelligence A broad term for any technological effort to replicate the brain’s functions. Cathode-ray tube (CRT) A glass enclosure with a relatively flat face (screen) and a vacuum inside that will display the light generated from the bombardment of the screen by electrons. CGS system The system of units employing the Centimeter, Gram, and Second as its fundamental units of measure. Difference engine One of the first mechanical calculators. Edison effect Establishing a flow of charge between two ele-ments in an evacuated tube. Electromagnetism The relationship between magnetic and elec-trical effects. Engineering notation A method of notation that specifies that all powers of ten used to define a number be multiples of 3 with a mantissa greater than or equal to 1 but less than 1000. ENIAC The first totally electronic computer. Fixed-point notation Notation using a decimal point in a partic-ular location to define the magnitude of a number. Fleming’s valve The first of the electronic devices, the diode. Floating-point notation Notation that allows the magnitude of a number to define where the decimal point should be placed. Integrated circuit (IC) A subminiature structure containing a vast number of electronic devices designed to perform a par-ticular set of functions. Joule (J) A unit of measurement for energy in the SI or MKS system. Equal to 0.7378 foot-pound in the English system and 107 ergs in the CGS system. Kelvin (K) A unit of measurement for temperature in the SI sys-tem. Equal to + ° 273.15 C in the MKS and CGS systems. Kilogram (kg) A unit of measure for mass in the SI and MKS systems. Equal to 1000 grams in the CGS system. This page is intentionally left blank e I V Voltage and Current 2.1 INTRODUCTION Now that the foundation for the study of electricity/electronics has been established, the concepts of voltage and current can be investigated. The term voltage is encountered practically every day. We have all replaced batteries in our flashlights, answering machines, calculators, automobiles, and so on, that had specific voltage ratings. We are aware that most outlets in our homes are 120 volts ac. Although current may be a less familiar term, we know what happens when we place too many appliances on the same outlet—the circuit breaker opens due to the excessive current that results. It is fairly common knowledge that current is something that moves through the wires and causes sparks and possibly fire if there is a “short circuit.” Current heats up the coils of an electric heater or the burners of an electric stove; it generates light when passing through the filament of a bulb; it causes twists and kinks in the wire of an electric iron over time due to the heat generated; and so on. All in all, the terms voltage and current are part of the vocabulary of most individuals. In this chapter, the basic impact of current and voltage and the properties of each are intro-duced and discussed in some detail. Hopefully, any mysteries surrounding the general charac-teristics of each will be eliminated, and you will gain a clear understanding of the impact of each on an electric/electronics circuit. 2.2 ATOMS AND THEIR STRUCTURE A basic understanding of the fundamental concepts of current and voltage requires a degree of familiarity with the atom and its structure. The simplest of all atoms is the hydrogen atom, made up of two basic particles, the proton and the electron, in the relative positions shown in Fig. 2.1(a). The nucleus of the hydrogen atom is the proton, a positively charged particle. The orbiting electron carries a negative charge equal in magnitude to the positive charge of the proton. • Become aware of the basic atomic structure of conductors such as copper and aluminum and understand why they are used so extensively in the field. • Understand how the terminal voltage of a battery or any dc supply is established and how it creates a flow of charge in the system. • Become aware of the various sources of dc power. • Understand how current is established in a circuit and how its magnitude is affected by the elements it encounters. • Become familiar with the factors that affect the terminal voltage of a battery and how long a battery will remain effective. • Be able to apply a voltmeter and ammeter correctly to measure the voltage and current at any point in the network. Objectives 2 e I V 56 \| \| \| Voltage and Current e I V In all other elements, the nucleus also contains neutrons, which are slightly heavier than protons and have no electrical charge. The helium atom, for example, has two neutrons in addition to two electrons and two protons, as shown in Fig. 2.1(b). In general, the atomic structure of any stable atom has an equal number of electrons and protons. Different atoms have various numbers of electrons in concentric orbits called shells around the nucleus. The first shell, which is closest to the nucleus, can contain only two electrons. If an atom has three elec-trons, the extra electron must be placed in the next shell. The number of electrons in each succeeding shell is determined by n 2 , 2 where n is the shell number. Each shell is then broken down into subshells where the number of electrons is limited to 2, 6, 10, and 14 in that order as you move away from the nucleus. Copper is the most commonly used metal in the electrical/electronics industry. An examination of its atomic structure will reveal why it has such widespread application. As shown in Fig. 2.2, it has 29 electrons in orbits around the nucleus, with the 29th electron appearing all by itself in the 4th shell. Note that the number of electrons in each shell and subshell + Electron Nucleus Proton – (a) Hydrogen atom + + – – Protons Electron Nucleus Neutrons Electron (b) Helium atom FIG. 2.1 Hydrogen and helium atoms. – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 3rd shell 18 electrons 4th shell (Maximum = 32 electrons) 29th electron 1st shell 2 electrons 2nd shell 8 electrons Nucleus 29 protons 29 neutrons + FIG. 2.2 The atomic structure of copper. atoMS and tHeIr StruCture \| \| \| 57 e I V is as defined above. There are two important things to note in Fig. 2. 2 . First, the 4th shell, which can have a total of = = n 2 2(4) 32 2 2 electrons, has only one electron. The outermost shell is incomplete and, in fact, is far from complete because it has only one electron. Atoms with complete shells (that is, a number of electrons equal to n 2 2 ) are usually quite stable. Those atoms with a small percentage of the defined number for the outermost shell are normally considered somewhat unsta-ble and volatile. Second, the 29th electron is the farthest electron from the nucleus. Opposite charges are attracted to each other, but the far-ther apart they are, the less the attraction. In fact, the force of attraction between the nucleus and the 29th electron of copper can be determined by Coulomb’s law developed by Charles Augustin Coulomb ( Fig. 2. 3 ) in the late 18th century: F k Q Q r newtons, N 1 2 2 ( ) = ( 2. 1) where F is in newtons (N), = = × k a constant 9.0 10 Nm /C , 9 2 2 Q1 and Q2 are the charges in coulombs (a unit of measure discussed in the next section), and r is the distance between the two charges in meters. At this point, the most important thing to note is that the distance between the charges appears as a squared term in the denominator. First, the fact that this term is in the denominator clearly reveals that as it increases, the force will decrease. However, since it is a squared term, the force will drop dramatically with distance. This relationship where a parameter of interest has a squared term in the denominator occurs frequently in this and related fields, so we will take a closer look at its impact. Defining ′ = k kQ Q , 1 2 Eq. ( 2. 1 ) will become the following: = = ′ F kQ Q r k r 1 2 2 2 For very small distances between charges, the force of attraction or repulsion will be very high. Consider = r 0.3 m. The resulting force is ( ) = ′ = ′ ≅ ′ F k r k k 0.3 11.1 2 2 For = r 1 m : ( ) = ′ = ′ = ′ F k r k k 1 2 2 For = r 2 m: ( ) = ′ = ′ = ′ = ′ F k r k k k 2 4 0.25 2 2 and for = r 10 m : ( ) = ′ = ′ = ′ = ′ F k r k k k 10 100 0.01 2 2 A plot of force versus distance is provided as Fig. 2. 4 . Clearly the squared term in the denominator has a dramatic effect on the level of attraction or repulsion between two charges. The result, therefore, is that the force of attraction between the 29th electron and the nucleus is significantly less than that between an electron in the first shell and F k Q Q r 1 2 Q Q 1 2 Q Q 2 F k = F k r (m) 11.1 10 2 1 1.0 0 0.25 0.3 F kQ1Q2 = k9 r2 r2 F(k9) k9 100 = FIG. 2. 4 Demonstrating the impact of a squared term in the denominator of an expression. INTERFOTO/Personalities/ Alamy Stock Photo FIG. 2. 3 Charles Augustin Coulomb. French (Angoulème, Paris) (1736–1806) Scientist and Inventor Military Engineer, West Indies Attended the engineering school at Mézières, the first such school of its kind. Formulated Coulomb’s law, which defines the force between two electrical charges and is, in fact, one of the principal forces in atomic reactions. Performed extensive research on the friction encountered in machinery and windmills and the elasticity of metal and silk fibers. 58 \| \| \| Voltage and Current e I V the nucleus. The result is that the 29th electron is loosely bound to the atomic structure and with a little bit of pressure from outside sources could be encouraged to leave the parent atom. If this 29th electron gains sufficient energy from the surrounding medium to leave the parent atom, it is called a free electron. In 1 cubic in. of copper at room temperature, there are approximately × 1.4 10 24 free electrons. Expanded, that is 1,400,000,000,000,000,000,000,000 free elec-trons in a 1–in. square cube. The point is that we are dealing with enormous numbers of electrons when we talk about the number of free electrons in a copper wire—not just a few that you could leisurely count. Further, the numbers involved are clear evidence of the need to become proficient in the use of powers of ten to represent numbers and use them in mathematical calculations. Other metals that exhibit the same properties as copper, but to a dif-ferent degree, are silver, gold, and aluminum, and some rarer metals such as tungsten. Additional comments on the characteristics of conduc-tors are in the following sections. 2.3 VOLTAGE If we separate the 29th electron in Fig. 2.2 from the rest of the atomic structure of copper by a dashed line as shown in Fig. 2.5(a), we cre-ate regions that have a net positive and negative charge as shown in Fig. 2.5(b) and (c). For the region inside the dashed boundary, the num-ber of protons in the nucleus exceeds the number of orbiting electrons by 1, so the net charge is positive as shown in both figures. This posi-tive region created by separating the free electron from the basic atomic structure is called a positive ion. If the free electron then leaves the vicinity of the parent atom as shown in Fig. 2.5(d), regions of positive and negative charge have been established. This separation of charge to establish regions of positive and nega-tive charge is the action that occurs in every battery. Through chemical action, a heavy concentration of positive charge (positive ions) is estab-lished at the positive terminal, with an equally heavy concentration of negative charge (electrons) at the negative terminal. In general, every source of voltage is established by simply creating a separation of positive and negative charges. It is that simple: If you want to create a voltage level of any magni-tude, simply establish regions of positive and negative charge. The more the required voltage, the greater is the quantity of positive and negative charge. In Fig. 2.6(a), for example, a region of positive charge has been estab-lished by a packaged number of positive ions, and a region of negative charge by a similar number of electrons, separated by a distance r. Since it would be inconsequential to talk about the voltage established by the separation of a single electron, a package of electrons called a coulomb (C) of charge was defined as follows: One coulomb of charge is the total charge associated with × 6.242 1018 electrons. Conversely, the negative charge associated with a single electron is = × = × − Q 1 6.242 10 C 0.1602 10 C e 18 18 (b) + – (c) + – (d) Positive ion Free electron – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – (a) + + Positive region equal in charge to the isolated electron FIG. 2.5 Defining the positive ion. Voltage \| \| \| 59 e I V so that = × − Q 1.602 10 C e 19 ( 2. 2) (a) r – – (b) (c) b a 1 V 1 coulomb of charge 1 joule of energy – – 1 coulomb of charge 2 joules of energy 2 V (d) – – 4.8 V 1 coulomb of charge 4.8 joules of energy + – + – + – + – FIG. 2. 6 Defining the voltage between two points. = × Q 1.602 = × 1.602 = × 10 C − 0 C − e 19 0 C 19 0 C In Fig. 2. 6(b) , if we take a coulomb of negative charge near the sur-face of the positive charge and move it toward the negative charge, we must expend energy to overcome the repulsive forces of the larger nega-tive charge and the attractive forces of the positive charge. In the process of moving the charge from point a to point b in Fig. 2. 6(b) , if a total of 1 joule (J) of energy is used to move the negative charge of 1 coulomb (C), there is a difference of 1 volt (V) between the two points. The defining equation is V W Q V W Q volts V joules J coulombs C ( ) ( ) ( ) = = = = ( 2. 3) Take particular note that the charge is measured in coulombs, the energy in joules, and the voltage in volts. The unit of measurement, volt , was chosen to honor the efforts of Alessandro Volta, who first demonstrated that a voltage could be established through chemical action ( Fig. 2. 7 ). If the charge is now moved all the way to the surface of the larger negative charge as shown in Fig. 2. 6(c) , using 2 J of energy for the whole trip, there are 2 V between the two charged bodies. If the package of pos-itive and negative charge is larger, as shown in Fig. 2. 6(d) , more energy will have to be expended to overcome the larger repulsive forces of the large negative charge and attractive forces of the large positive charge. As shown in Fig. 2. 6(d) , 4.8 J of energy were expended, resulting in a voltage of 4.8 V between the two points. We can therefore conclude that it would take 12 J of energy to move 1 C of negative charge from the positive terminal to the negative terminal of a 12 V car battery. Through algebraic manipulations, we can define an equation to deter-mine the energy required to move charge through a difference in voltage: W QV (joules, J) = ( 2. 4) V W Q = W QV W Q = W Q Bilwissedition Ltd. & Co. KG/Alamy Stock Photo FIG. 2. 7 Count Alessandro Volta. Italian (Como, Pavia) (1745–1827) Physicist Professor of Physics, Pavia, Italy Began electrical experiments at the age of 18 working with other European investigators. Major contribution was the development of an electrical energy source from chemical action in 1800. For the first time, elec-trical energy was available on a continuous basis and could be used for practical purposes. Developed the first condenser, known today as the capacitor. Was invited to Paris to demonstrate the voltaic cell to Napoleon. The International Electrical Congress meeting in Paris in 1881 honored his efforts by choosing the volt as the unit of measure for electromotive force. 60 \| \| \| Voltage and Current e I V Finally, if we want to know how much charge was involved, we use Q W V (coulombs, C) = ( 2. 5) EXAMPLE 2. 1 Find the voltage between two points if 60 J of energy are required to move a charge of 20 C between the two points. Solution: Eq. ( 2. 3 ): = = = V W Q 3 V 60 J 20 C EXAMPLE 2. 2 Determine the energy expended moving a charge of µ 50 C between two points if the voltage between the points is 6 V. Solution: Eq. ( 2. 4 ): µ ( )( ) = = × = × = − − W QV 300 J 50 10 C 6 V 300 10 J 6 6 There are a variety of ways to separate charge to establish the desired voltage. The most common is the chemical action used in car batteries, flashlight batteries, and, in fact, all portable batteries. Other sources use mechanical methods such as car generators and steam power plants or alternative sources such as solar cells and windmills. In total, however, the sole purpose of the system is to create a separation of charge. In the future, therefore, when you see a positive and a negative terminal on any type of battery, you can think of it as a point where a large concentration of charge has gathered to create a voltage between the two points. More important is to recognize that a voltage exists between two points—for a battery between the positive and negative terminals. Hooking up just the positive or the negative terminal of a battery and not the other would be meaningless. Both terminals must be connected to define the applied voltage. As we moved the 1 C of charge in Fig. 2. 6(b) , the energy expended would depend on where we were in the crossing. The position of the charge is therefore a factor in determining the voltage level at each point in the crossing. Since the potential energy associated with a body is defined by its position, the term potential is often applied to define volt-age levels. For example, the difference in potential is 4 V between the two points, or the potential difference between a point and ground is 12 V, and so on. The electron volt A unit of energy sometimes applied in a number of physics oriented investigations is the electron volt . It is the level of energy required to move an electron through a poten-tial difference of 1 volt. Applying the basic energy equation, W QV W 1.602 10 C 1 volt 1.602 10 J 19 19 ( )( ) = = × = × − − Q W V = Current \| \| \| 61 e I V and 1 eV 1.602 10 J 19 = × − ( 2. 6) Conversely, = × − 1 J 1 1.602 10 eV 19 and 1 J 2.24 10 eV 18 = × ( 2. 7) 2. 4 CURRENT The question, “Which came first—the chicken or the egg?” can be applied here also because the layperson has a tendency to use the terms current and voltage interchangeably as if both were sources of energy. It is time to set things straight: The applied voltage is the starting mechanism—the current is a reaction to the applied voltage. In Fig. 2. 8(a) , a copper wire sits isolated on a laboratory bench. If we cut the wire with an imaginary perpendicular plane, producing the circu-lar cross section shown in Fig. 2. 8(b) , we would be amazed to find that there are free electrons crossing the surface in both directions. Those free electrons generated at room temperature are in constant motion in random directions. However, at any instant of time, the number of elec-trons crossing the imaginary plane in one direction is exactly equal to that crossing in the opposite direction, so the net flow in any one direc-tion is zero. Even though the wire seems dead to the world sitting by itself on the bench, internally, it is quite active. The same would be true for any other good conductor. 1 eV 1.602 10 J 19 = × 1.602 = × 1.602 − 1 J 2.24 10 eV 18 = × 2.24 = × 2.24 ( ( ( e– e– e– e– e– e– (b) Imaginary plane Perpendicular surface cut by plane (a) Isolated copper wire Perpendicular plane for Fig. 2.8(b) FIG. 2. 8 There is motion of free carriers in an isolated piece of copper wire, but the flow of charge fails to have a particular direction. Now, to make this electron flow do work for us, we need to give it a direction and be able to control its magnitude. This is accomplished by simply applying a voltage across the wire to force the electrons to move toward the positive terminal of the battery, as shown in Fig. 2. 9 . The instant the wire is placed across the terminals, the free electrons in the wire drift toward the positive terminal. The positive ions in the cop-per wire simply oscillate in a mean fixed position. As the electrons pass 62 \| \| \| Voltage and Current e I V through the wire, the negative terminal of the battery acts as a supply of additional electrons to keep the process moving. The electrons arriving at the positive terminal are absorbed, and through the chemical action of the battery, additional electrons are deposited at the negative terminal to make up for those that left. To take the process a step further, consider the configuration in Fig. 2.10, where a copper wire has been used to connect a light bulb to a battery to create the simplest of electric circuits. The instant the final connection is made, the free electrons of negative charge drift toward the positive terminal, while the positive ions left behind in the copper wire simply oscillate in a mean fixed position. The flow of charge (the elec-trons) through the bulb heats up the filament of the bulb through friction to the point that it glows red-hot and emits the desired light. e– e– e– Copper wire Battery terminals Chemical action V e– e– e– e– FIG. 2.9 Motion of negatively charged electrons in a copper wire when placed across battery terminals with a difference in potential of volts (V). Ielectron Copper wire Iconventional e e e e Chemical activity Battery e Imaginary plane e e e e e e e e FIG. 2.10 Basic electric circuit. In total, therefore, the applied voltage has established a flow of elec-trons in a particular direction. In fact, by definition, if × 6.242 1018 electrons (1 coulomb) pass through the imaginary plane in Fig. 2.10 in 1 second, the flow of charge, or current, is said to be 1 ampere (A). Current \| \| \| 63 e I V The unit of current measurement, ampere , was chosen to honor the efforts of André Ampère in the study of electricity in motion ( Fig. 2. 11 ). Using the coulomb as the unit of charge, we can determine the cur-rent in amperes from the following equation: I Q t I Q t amperes A coulombs C time s ( ) ( ) ( ) = = = = ( 2. 8) The capital letter I was chosen from the French word for current, intensité. The SI abbreviation for each quantity in Eq. ( 2. 8 ) is provided to the right of the equation. The equation clearly reveals that for equal time intervals, the more charge that flows through the wire, the larger is the resulting current. Through algebraic manipulations, the other two quantities can be determined as follows: Q It (coulombs, C) = ( 2. 9) and t Q l (seconds, s) = ( 2. 10) EXAMPLE 2. 3 The charge flowing through the imaginary surface in Fig. 2. 10 is 0.16 C every 64 ms. Determine the current in amperes. Solution: Eq. ( 2. 8 ): = = × = × × = − − − I Q t 2 50 A 0.16 C 64 10 s 160 10 C 64 10 s . 3 3 3 EXAMPLE 2. 4 Determine how long it will take × 4 1016 electrons to pass through the imaginary surface in Fig. 2. 10 if the current is 5 mA. Solution: Determine the charge in coulombs: × ×          = × − 4 10 electrons 1 C 6.242 10 electrons 0.641 10 C 16 18 2 = 6.41 mC Eq. ( 2. 10 ): t Q I 1 28 s 6.41 10 C 5 10 A . 3 3 = = × × = − − In summary, therefore, the applied voltage (or potential difference) in an electrical/ electronics system is the “pressure” to set the system in motion, and the current is the reaction to that pressure. A mechanical analogy often used to explain this is the simple garden hose. In the absence of any pressure, the water sits quietly in the hose with no general direction, just as electrons do not have a net direction in the absence of an applied voltage. However, release the spigot, and the applied pressure forces the water to flow through the hose. Similarly, apply a voltage to the circuit, and a flow of charge or current results. I Q t = Q It Q It Q I Q I = Q I t Q l = Nickolae/Fotolia FIG. 2. 11 André Marie Ampère. French (Lyon, Paris) (1775–1836) Mathematician and Physicist Professor of Mathematics, École Polytechnique, Paris On September 18, 1820, introduced a new field of study, electrodynamics, devoted to the effect of electricity in motion, including the interaction between currents in adjoining conductors and the interplay of the surrounding magnetic fields. Constructed the first solenoid and demonstrated how it could behave like a magnet (the first electromagnet ). Suggested the name galvanometer for an instrument designed to measure current levels. 64 \| \| \| Voltage and Current e I V A second glance at Fig. 2.10 reveals that two directions of charge flow have been indicated. One is called conventional flow, and the other is called electron flow. This text discusses only conventional flow for a variety of reasons; namely, it is the most widely used at educational institutions and in industry, it defines the direction of the arrow in the design of all electronic device symbols, and it is the popular choice for all major computer software packages. The flow controversy is a result of an assumption made at the time electricity was discovered that the positive charge was the moving particle in metallic conductors. Be assured that the choice of conventional flow will not create great difficulty and con-fusion in the chapters to follow. Once the direction of I is established, the issue is dropped and the analysis can continue without confusion. Safety Considerations It is important to realize that even small levels of current through the human body can cause serious, dangerous side effects. Experimental results reveal that the human body begins to react to currents of only a few milliamperes. Although most individuals can withstand currents up to perhaps 10 mA for very short periods of time without serious side effects, any current over 10 mA should be considered dangerous. In fact, currents of 50 mA can cause severe shock, and currents of over 100 mA can be fatal. In most cases, the skin resistance of the body when dry is sufficiently high to limit the current through the body to relatively safe levels for voltage levels typically found in the home. However, if the skin is wet due to perspiration, bathing, and so on, or if the skin barrier is broken due to an injury, the skin resistance drops dramatically, and current levels could rise to dangerous levels for the same voltage shock. In general, therefore, simply remember that water and electricity don’t mix. Granted, there are safety devices in the home today [such as the ground fault circuit interrupt (GFCI) receptacle and breakers, discussed in Chapter 4] that are designed specifically for use in wet areas such as the bathroom and kitchen, but accidents happen. Treat electricity with respect—not fear. 2.5 VOLTAGE SOURCES The term dc, used throughout this text, is an abbreviation for direct current, which encompasses all systems where there is a unidirectional (one direction) flow of charge. This section reviews dc voltage supplies that apply a fixed voltage to electrical/electronics systems. The graphic symbol for all dc voltage sources is shown in Fig. 2.12. Note that the relative length of the bars at each end define the polarity of the supply. The long bar represents the positive side; the short bar, the negative. Note also the use of the letter E to denote voltage source. It comes from the fact that an electromotive force (emf) is a force that establishes the flow of charge (or current) in a system due to the application of a difference in potential. In general, dc voltage sources can be divided into three basic types: (1) batteries (chemical action or solar energy), (2) generators (electro-mechanical), and (3) power supplies (rectification—a conversion pro-cess to be described in your electronics courses). E 12 V FIG. 2.12 Standard symbol for a dc voltage source. VOLTAGE SOURCES \| \| \| 65 e I V Batteries General Information For the layperson, the battery is the most common of the dc sources. By definition, a battery (derived from the expression “battery of cells”) consists of a combination of two or more similar cells, a cell being the fundamental source of electrical energy developed through the conversion of chemical or solar energy. All cells can be divided into the primary or secondary types. The secondary is rechargeable, whereas the primary is not. That is, the chemical reaction of the secondary cell can be reversed to restore its capacity. The two most common rechargeable batteries are the lead-acid unit (used primar-ily in automobiles) and the nickel–metal hydride (NiMH) battery (used in calculators, tools, photoflash units, shavers, and so on). The obvious advantages of rechargeable units are the savings in time and money of not continually replacing discharged primary cells. All the cells discussed in this chapter (except the solar cell, which absorbs energy from incident light in the form of photons) establish a potential difference at the expense of chemical energy. In addition, each has a positive and a negative electrode and an electrolyte to complete the circuit between electrodes within the battery. The electrolyte is the con-tact element and the source of ions for conduction between the terminals. Primary Cells (Nonrechargeable) The popular alkaline primary battery uses a powdered zinc anode ( ) + ; a potassium (alkali metal) hydroxide electrolyte; and a manganese dioxide/carbon cathode ( ) − as shown in Fig. 2.13(a). In Fig. 2.13(b), note that for the cylin-drical types (AAA, AA, C, and D), the voltage is the same for each, but the ampere-hour (Ah) rating increases significantly with size. The ampere-hour rating is an indication of the level of current that the battery can provide for a specified period of time (to be discussed in detail in Section 2.6). In particular, note that for the large, lantern-type battery, the voltage is only 4 times that of the AAA battery, but the ampere-hour rating of 52 Ah is almost 42 times that of the AAA battery. Metal spur Positive cover: plated steel Electrolyte: potassium hydroxide/water Cathode: manganese dioxide and carbon Separator: nonwoven fabric Metal washer Current collector: brass pin Can: steel Metalized plastic film label Anode: powdered zinc Seal: nylon Inner cell cover: steel Negative cover: plated steel (a) [(b) photo by Robert Boylestad] FIG. 2.13 Alkaline primary cells: (a) cutaway of cylindrical cell; (b) various types of primary cells. Lantern 6 V battery 1.5 V D cell 18 Ah 1.5 V C cell 8350 mAh AA cell 1.5 V 2850 mAh 9 V 625 mAh AAA cell 1.5 V 1250 mAh AAAA cell 1.5 V 600 mAh (b) 66 \| \| \| Voltage and Current e I V Another type of popular primary cell is the lithium battery, shown in Fig. 2.14. Again, note that the voltage is the same for each, but the size increases substantially with the ampere-hour rating and the rated drain current. It is particularly useful when low temperature is encountered. In general, therefore, for batteries of the same type, the size is dictated primarily by the standard drain current or ampere-hour rating, not by the terminal voltage rating. Secondary Cells (Rechargeable) Lead–Acid: The 12 V of Fig. 2.15, typically used in automobiles, has an electrolyte of sulfuric acid and electrodes of spongy lead (Pb) and lead peroxide ( ) PbO 2 . When a load is applied to the battery termi-nals, there is a transfer of electrons from the spongy lead electrode to the lead peroxide electrode through the load. This transfer of electrons 3 V 165 mAh Standard drain: 30 mA 3 V 1000 mAh Standard drain: 200 mA 3 V 1200 mAh Standard drain: 2.5 mA 3 V 5000 mAh Standard drain: 150 mA FIG. 2.14 Lithium primary batteries. Cells (each 2.1 V) Extrusion-fusion intercell connection Positive terminal Separator envelope Electrolyte reservoir Negative terminal Wrought lead–calcium grid Heat-sealed cover (Clive Streeter/DK Images) FIG. 2.15 Maintenance-free 12 V (actually 12.6 V) lead-acid battery. VOLTAGE SOURCES \| \| \| 67 e I V will continue until the battery is completely discharged. The discharge time is determined by how diluted the acid has become and how heavy the coating of lead sulfate is on each plate. The state of discharge of a lead storage cell can be determined by measuring the specific gravity of the electrolyte with a hydrometer. The specific gravity of a substance is defined to be the ratio of the weight of a given volume of the substance to the weight of an equal volume of water at ° 4 C. For fully charged batteries, the specific gravity should be somewhere between 1.28 and 1.30. When the specific gravity drops to about 1.1, the battery should be recharged. Since the lead storage cell is a secondary cell, it can be recharged at any point during the discharge phase simply by applying an external dc current source across the cell that passes current through the cell in a direction opposite to that in which the cell supplied current to the load. This removes the lead sulfate from the plates and restores the concentra-tion of sulfuric acid. The output of a lead storage cell over most of the discharge phase is about 2.1 V. In the commercial lead storage batteries used in automo-biles, 12.6 V can be produced by six cells in series, as shown in Fig. 2.15. In general, lead-acid storage batteries are used in situations where a high current is required for relatively short periods of time. At one time, all lead-acid batteries were vented. Gases created during the discharge cycle could escape, and the vent plugs provided access to replace the water or electrolyte and to check the acid level with a hydrometer. The use of a grid made from a wrought lead–calcium alloy strip, rather than the lead– antimony cast grid commonly used, has resulted in maintenance-free batteries, shown in Fig. 2.15. The lead–antimony structure was suscep-tible to corrosion, overcharge, gasing, water usage, and self-discharge. Improved design with the lead–calcium grid has either eliminated or substantially reduced most of these problems. It would seem that with so many advances in technology, the size and weight of the lead–acid battery would have decreased significantly in recent years, but even today it is used more than any other battery in automobiles and all forms of machinery. However, things are beginning to change with interest in nickel–metal hydride and lithium-ion batteries, which both pack more power per unit size than the lead–acid variety. Both will be described in the sections to follow. Nickel–Metal Hydride (NiMH): The nickel–metal hydride re- charge able battery has been receiving enormous interest and develop-ment in recent years. The Toyota Prius and two other hybrids would use NiMH batteries rather than the lead–acid variety. For applications such as flashlights, shavers, portable televisions, power drills, and so on, rechargeable batteries such as the nickel–metal hydride (NiMH) batter-ies shown in Fig. 2.16 are often the secondary batteries of choice. These batteries are so well made that they can survive over 1000 charge/dis-charge cycles over a period of time and can last for years. It is important to recognize that if an appliance calls for a recharge-able battery such as a NiMH battery, a primary cell should not be used. The appliance may have an internal charging network that would be dys-functional with a primary cell. In addition, note that NiMH batteries are about 1.2 V per cell, whereas the common primary cells are typically 1.5 V per cell. There is some ambiguity about how often a secondary cell should be recharged. Generally, the battery can be used until there is some indication that the energy level is low, such as a dimming light from a flashlight, less 68 \| \| \| Voltage and Current e I V power from a drill, or a signal from low-battery indicator. Keep in mind that secondary cells do have some “memory.” If they are recharged contin-uously after being used for a short period of time, they may begin to believe they are short-term units and actually fail to hold the charge for the rated period of time. In any event, always try to avoid a “hard” discharge, which results when every bit of energy is drained from a cell. Too many hard- discharge cycles will reduce the cycle life of the battery. Finally, be aware that the charging mechanism for nickel–cadmium cells is quite different from that for lead–acid batteries. The nickel–cadmium battery is charged by a constant–current source, with the terminal voltage staying fairly steady through the entire charging cycle. The lead–acid battery is charged by a constant voltage source, permitting the current to vary as determined by the state of the battery. The capacity of the NiMH battery increases almost linearly throughout most of the charging cycle. Nickel–cadmium batteries become relatively warm when charging. The lower the capacity level of the battery when charging, the higher is the temperature of the cell. As the battery approaches rated capacity, the temperature of the cell approaches room temperature. Lithium-ion (Li-ion): The battery receiving the most research and development in recent years is the lithium-ion battery. It carries more energy in a smaller space than either the lead–acid or NiMH rechargeable batteries. Its range of applications includes computers, a host of con-sumer products, power tools, and recently the sleek Tesla roadster with its battery pack composed of more than 6800 3.7 V Li-ion cells the size of a typical AA battery. It can travel some 265 miles but the battery pack costs between $10,000 and $15,000. Another problem is shelf life. Once manufactured, these batteries begin to slowly die even though they may go through normal charge/discharge cycles, which makes them similar to a normal primary cell, so lifetime is a major concern. The very popular iPhone 4S appearing in Fig. 2.17 has a 3.7 V, 5.45 Wh lithium-ion polymer battery as its power source. Note that it takes about 40% by volume of the internal structure and contributes a great deal to its weight. Industry is aware of the numerous positive characteristics of this power source and is pouring research money in at a very high rate. Recent D cell 1.2 V 2500 mAh @ 500 mA C cell 1.2 V 2500 mAh @ 500 mA AA cell 1.2 V 2300 mAh @ 460 mA AAA cell 1.2 V 700 mAh @ 140 mA (photo by Robert Boylestad) FIG. 2.16 Nickel–metal hydride (NiMH) rechargeable batteries. ((a) STANCA SANDA/Alamy Stock Photo (b) Brent Lewin/Bloomberg/Getty Images) FIG. 2.17 iPhone. Camera lens Volume control Charging port 3.7 V, 5.3 Wh Lithium-ion polymer battery VOLTAGE SOURCES \| \| \| 69 e I V use of nanotechnology and microstructures along with improved design has alleviated many of the concerns regarding safety and longevity. Solar Cell The use of solar cells as part of the effort to generate “clean” energy has grown exponentially in the last few years. At one time the cost and the low conversion efficiencies were the main stumbling blocks to widespread use of the solar cell. However, the company Nanosolar has significantly reduced the cost of solar panels by using a printing pro-cess that uses a great deal less of the expensive silicon material in the manufacturing process. Whereas the cost of generating solar electricity is about 4 to 10¢/kWh, compared to an average of 22 ¢/kWh using a local coal/oil utililty, this new printing process has had a significant impact on reducing the cost level. Another factor that will reduce costs is the improving level of efficiency being obtained by manufacturers. At one time the accepted efficiency level of conversion was between 10% and 14%. Recently, however, the company Semprius, Inc., set a new record for commercially available solar panels with a 35.5% efficiency level—almost twice the typical commercial level. The Fraunhofer Institute for Solar Energy Systems in Germany reached a laboratory level of 44.7% conversion efficiency. The two factors that affect the cost the most are the cost of the materials and the efficiency level. Silicon is the most frequently used material but it is expensive. Recent trends have been to find improved materials and to try innovative maneuvers like stacking the cells or imbedding lenses to focus the incident light. In general, it appears that there is a strong thrust to make solar cells a very important option for generating clean energy in the near future. Given that the maximum available wattage on an average bright, sunlit day is 100 mW/cm , 2 the efficiency is an important element in any future plans for the expansion of solar power. At 10% to 14% efficiency the maximum available power from a 1-m 2 panel (approximately ′ × ′ 3 3 ) would be 100 to 140 W. However, if the efficiency could be raised to 30%, the return could exceed 300 W per panel. When we consider the wide range of appliances that can be run on 300 W or less, this is a sig-nificant improvement. Even though the efficiency may not be as high as desired, the import-ant thing to remember when it comes to solar energy is that there is no cost associated with providing the energy in the first place. It is ready to be used and, hopefully, will never end. In cold climates with snow cover and long cloudy days, one might wonder about the feasibility or logic in installing solar panels. However, take a look at the fishing/hunting lodge in Fig. 2.18(a) located in northern Maine with twelve 265 W solar pan-els that provide a total of 3.18 kW under optimum conditions. Consider how many appliances, required in a camp of this kind, can operate with a power source of this magnitude. Keep in mind that the bank of 16 bat-teries will store unused energy on the sunny days for times when there is cloud cover or the panels are covered with snow. The system includes a solar charger and a 6 kW inverter to convert the dc to ac for use with appliances requiring 120V ac. During a recent visit, the owners revealed that the system works so well that they only turn on the generator about once a month for 3–5 hours, which usually occurs when a few rainy days coincide with a large group that uses the lodge late into the evening. They figure that they save about $3500 a year on diesel and maintenance costs. Most importantly the solar system permits the use of the kitchen, laundry, 70 \| \| \| Voltage and Current e I V and lights later in the evening, whereas with the generator as their sole source of power the supply of electricity normally ended at 10 p.m. The basic system operates as shown in Fig. 2.18(b). The solar panels (1) convert sunlight into dc electric power. An inverter (2) converts the dc power into the standard ac power for use in the home (6). The batteries (3) can store energy from the sun for use if there is insufficient sunlight or a power failure. At night or on dark days when the demand exceeds the solar panel and battery supply, the generators (4) can provide power to the appliances (6) through a special hookup in the electrical panel (5). (a) [(a) Courtesy of Red River Camps] FIG. 2.18 Red River Camps in Portage, Maine: (a) twelve 265 W panels on roof; (b) system operations. (b) VOLTAGE SOURCES \| \| \| 71 e I V There is no question that the use of solar panels is growing by leaps and bounds in recent years as manufacturers provide panels that meet the physical requirements of a wide range of applications. Consider the RV vehicle pictured in Fig. 2.19 designed to permit travel to all points of the globe. It has three panels of 240 W each to provide a total of 720 W of power that can be stored in a battery system with a 510 Ah capacity. The panels are obviously of a durability and strength to permit traveling in some of the harshest conditions. Generators The dc generator is quite different from the battery, both in construction (Fig. 2.20) and in mode of operation. When the shaft of the generator is rotating at the nameplate speed due to the applied torque of some external source of mechanical power, a voltage of rated value appears across the external terminals. The terminal voltage and power-handling capabilities of the dc generator are typically higher than those of most batteries, and its lifetime is determined only by its construction. Commercially used dc generators typically have an output voltage of 120 V or 240 V. For the pur-poses of this text, the same symbols are used for a battery and a generator. Power Supplies The dc supply encountered most frequently in the laboratory uses the rectification and filtering processes as its means toward obtaining a steady dc voltage. Both processes will be covered in detail in your basic electronics courses. In total, a time-varying voltage (such as ac voltage available from a home outlet) is converted to one of a fixed magnitude. A dc laboratory supply of this type is shown in Fig. 2.21. Most dc laboratory supplies have a regulated, adjustable volt-age output with three available terminals, as indicated horizontally at the bottom of Fig 2.21 and vertically in Fig 2.22(a). The symbol for ground or zero potential (the reference) is also shown in Fig. 2.22(a). (Virrage Images/Shutterstock) FIG. 2.19 EarthRoamer XV-LT . “Output” voltage Applied torque “Input” 120 V FIG. 2.20 dc generator. (Courtesy of B+K Precision.) FIG. 2.21 A 0 V to 60 V, 0 to 1.5 A digital display dc power supply. 72 \| \| \| Voltage and Current e I V If 10 V above ground potential are required, the connections are made as shown in Fig. 2.22(b). If 15 V below ground potential are required, the connections are made as shown in Fig. 2.22(c). If connections are as shown in Fig. 2.22(d), we say we have a “floating” voltage of 5 V since the reference level is not included. Seldom is the configuration in Fig. 2.22(d) used since it fails to protect the operator by providing a direct low-resistance path to ground and to establish a common ground for the system. In any case, the positive and negative terminals must be part of any circuit configuration. Fuel Cells One of the most exciting developments in recent years has been the steadily rising interest in fuel cells as an alternative energy source. Fuel cells are now being used in small stationary power plants, transportation (buses), and a wide variety of applications where portability is a major factor, such as the space shuttle. A great deal of money is now being spent in an effort to design affordable fuel-cell vehicles. Fuel cells have the distinct advantage of operating at efficiencies of 70% to 80% rather than the typical 20% to 25% efficiency of current internal combustion engine of today’s automobiles. They also have few moving parts, produce little or no pollution, generate very little noise, and use fuels such as hydrogen and oxygen that are readily available. Fuel cells are considered primary cells (of the continuous-feed variety) because they cannot be recharged. They hold their characteristics as long as the fuel (hydrogen) and oxygen are supplied to the cell. The only by-products of the conversion process are small amounts of heat (which is often used elsewhere in the system design), water (which may also be reused), and negligible levels of some oxides, depending on the com-ponents of the process. Overall, fuel cells are environmentally friendly. The operation of the fuel cell is essentially opposite to that of the chemical process of electrolysis. Electrolysis is the process whereby electric current is passed through an electrolyte to break it down into its fundamental components. An electrolyte is any solution that will permit conduction through the movement of ions between adjoining electrodes. For instance, passing current through water results in a hydrogen gas by the cathode (negative terminal) and oxygen gas at the anode (positive (–15 V) Gnd (0 V) Jumper 15 V 15 V Jumper 10 V 5 V 10 V 5 V (+10 V) (“Floating”) (a) (c) (b) (d) FIG. 2.22 dc laboratory supply: (a) available terminals; (b) positive voltage with respect to (w.r.t.) ground; (c) negative voltage w.r.t. ground; (d) floating supply. VOLTAGE SOURCES \| \| \| 73 e I V terminal). In 1839, Sir William Grove believed this process could be reversed and demonstrated that the proper application of the hydrogen gas and oxygen results in a current through an applied load connected to the electrodes of the system. The first commercial unit was used in a tractor in 1959, followed by an energy pack in the 1965 Gemini Space program. In 1996, the first small power plant was designed, and today it is an important component of the shuttle program. The basic components of a fuel cell are depicted in Fig. 2.23(a) with details of the construction in Fig. 2.23(b). Hydrogen gas (the fuel) is supplied to the system at a rate proportional to the current required by the load. At the opposite end of the cell, oxygen is supplied as needed. The net result is a flow of electrons through the load and a discharge of water with a release of some heat developed in the process. The amount of heat is minimal, although it can also be used as a component in the design to improve the efficiency of the cell. The water (very clean) can simply be discharged or used for other applications such as cooling in the overall application. If the source of hydrogen or oxygen is removed, the system breaks down. The flow diagram of the system is relatively simple, as shown in Fig. 2.23(a). In an actual cell, shown in Fig. 2.23(b), the hydrogen gas is applied to a porous electrode called the anode that is coated with a platinum catalyst. The catalyst on the anode serves to speed up the process of breaking down the hydrogen atom into positive hydrogen ions and free electrons. The electrolyte between the electrodes is a solution or membrane that permits the passage of positive hydro-gen ions but not electrons. Facing this wall, the electrons choose to pass through the load and light up the bulb, while the positive hydrogen ions migrate toward the cathode. At the porous cathode (also coated with the catalyst), the incoming oxygen atoms combine with the arriving hydro-gen ions and the electrons from the circuit to create water (H2O) and heat. The circuit is, therefore, complete. The electrons are generated and then absorbed. If the hydrogen supply is cut off, the source of electrons is shut down, and the system is no longer an operating fuel cell. In some fuel cells, either a liquid or molten electrolyte membrane is used. Depending on which the system uses, the chemical reactions will change slightly but not dramatically from that described above. Anode H2 (Hydrogen) Electrolyte Cathode Water (H2O) heat O2 (Oxygen) e– e– e– H+ e– H+ e– H+ e– H+ H+ e– e– e– e– O2 O2 O2 O2 H+ e– + – (a) (b) dc power Hydrogen FUEL CELL Oxygen Water and heat FIG. 2.23 Fuel cell (a) functional block diagram; (b) basic construction. 74 \| \| \| Voltage and Current e I V The phosphoric acid fuel cell is a popular cell using a liquid electrolyte, while the PEM uses a polymer electrolyte membrane. The liquid or mol-ten type is typically used in stationary power plants, while the membrane type is favored for vehicular use. The output from a single fuel cell is a low-voltage, high-current dc output. Stacking the cells in series or parallel increases the output volt-age or current level. For a number of years buses, trucks, motorcycles, forklifts, and many other vehicles have been running on fuel-cell technology. However, it was not until the summer of 2008 that Honda offered the FCX Clarity automobile, the first to be available for consumer use. The basic com-ponents of the vehicle are pointed out in Fig. 2.24. Basically, it is an electric car because there is no combustion and the source of power is the lithium batteries that are charged by the fuel cells. As long as oxygen and hydrogen are fed to the fuel cells, a dc voltage will be generated by the fuel cell to charge or maintain the voltage. This voltage is fed to an inverter, which in turn will provide the ac power for the drivetrain. Note in Fig. 2.24 that the vehicle also has the unique ability to capture braking energy and turn it into energy stored by a high-output battery for supple-mental use. The fuel-cell car currently has a range of about 300 miles on a full hydrogen tank with fueling times as low as 3 minutes to match that of a typical gasoline tank. Of course, the most pleasing characteristic of the vehicle is that the only discharge is some residual water that can be used for cooling. In addition, sealed batteries will eventually deteriorate but fuel cells have a longer life because the “chemicals” (oxygen and hydrogen) are continually being replaced. Hydrogen Storage Tank Stores hydrogen gas compressed at extremely high pressure to increase driving range Power Control Unit Governs the flow of electricity Electric Motor Propels the vehicle much more quietly, smoothly, and efficiently than an internal combustion engine and requires less maintenance Fuel-Cell Stack Converts hydrogen gas and oxygen into electricity to power the electric motor High-Output Battery Stores energy generated from regenerative braking and provides supplemental power to the electric motor (Courtesy of American Honda Motor Co. Inc.) FIG. 2.24 Honda FCX Clarity fuel-cell automobile. 2.6 AMPERE-HOUR RATING The most important piece of data for any battery (other than its voltage rating) is its ampere-hour (Ah) rating. You have probably noted in the photographs of batteries in this chapter that both the voltage and the ampere-hour rating have been provided for each battery. BatterY lIFe FaCtorS \| \| \| 75 e I V The ampere-hour (Ah) rating provides an indication of how long a battery of fixed voltage will be able to supply a particular current. A battery with an ampere-hour rating of 100 will theoretically pro-vide a current of 1 A for 100 hours, 10 A for 10 hours, or 100 A for 1 hour. Quite obviously, the greater the current, the shorter is the time. An equation for determining the length of time a battery will supply a par-ticular current is the following: Life hours ampere-hour Ah rating amperes drawn A ( ) ( ) ( ) = ( 2. 11) EXAMPLE 2. 5 How long will a 9 V transistor battery with an ampere-hour rating of 520 mAh provide a current of 20 mA? Solution: Eq. ( 2. 11 ): = = = 26 h Life 520 mAh 20 mA 520 20 h EXAMPLE 2. 6 How long can a 1.5 V flashlight battery provide a current of 250 mA to light the bulb if the ampere-hour rating is 16 Ah? Solution: Eq. ( 2. 11 ): = = × = − 64 h Life 16 Ah 250 mA 16 250 10 h 3 2. 7 BATTERY LIFE FACTORS The previous section made it clear that the life of a battery is directly related to the magnitude of the current drawn from the supply. However, there are factors that affect the given ampere-hour rating of a battery, so we may find that a battery with an ampere-hour rating of 100 can supply a current of 10 A for 10 hours but can supply a current of 100 A for only 20 minutes rather than the full 1 hour calculated using Eq. ( 2. 11 ). In other words, the capacity of a battery (in ampere-hours) will change with change in current demand. This is not to say that Eq. ( 2. 11 ) is totally invalid. It can always be used to gain some insight into how long a battery can supply a particular cur-rent. However, be aware that there are factors that affect the ampere-hour rating. Just as with most systems, including the human body, the more we demand, the shorter is the time that the output level can be main-tained. This is clearly verified by the curves in Fig. 2. 25 for the Eveready Energizer D cell. As the constant-current drain increased, the ampere-hour rating decreased from about 18 Ah at 25 mA to around 12 Ah at 300 mA. Another factor that affects the ampere-hour rating is the temperature of the unit and the surrounding medium. In Fig. 2. 26 , the capacity of the same battery plotted in Fig. 2. 25 shows a peak value near the common room temperature of 68°F. At very cold temperatures and very warm temperatures, the capacity drops. Clearly, the ampere-hour rating will be provided at or near room temperature to give it a maximum value, but be aware that it will drop off with an increase or decrease in temperature. Most of us have noted that the battery in a car, radio, two-way radio, cellphone, flashlight, and so on seems to have less power in really cold weather. It would seem, then, that the battery capacity would increase Life hour ampere-hour Ah rating amperes drawn A ( ) hour ( ) hours ( ) s ( ) A ( ) Ah ( ) h ( ) A ( ) A = ( ( ( 76 \| \| \| Voltage and Current e I V with higher temperatures—which, however, is not always the case. In general, therefore, the ampere-hour rating of a battery will decrease from the room-temperature level with very cold and very warm temperatures. Another interesting factor that affects the performance of a battery is how long it is asked to supply a particular voltage at a continuous drain current. Note the curves in Fig. 2.27, where the terminal voltage dropped at each level of drain current as the time period increased. The lower the current drain, the longer it could supply the desired current. At 100 mA, it was limited to about 100 hours near the rated voltage, but at 25 mA, it did not drop below 1.2 V until about 500 hours had passed. That is an increase in time of 5:1, which is significant. The result is that the terminal voltage of a battery will eventually drop (at any level of current drain) if the time period of continuous discharge is too long. 0 mA 10 20 Ah 25 100 200 300 400 Ampere-hour rating I (constant current drain) 18@25 mA 17@100 mA 15@200 mA 12@300 mA 9.5@400 mA FIG. 2.25 Ampere-hour rating (capacity) versus drain current for an Energizer® D cell. 8F 10 20 Ah 0 10 20 30 40 15.5@328F 18@688F 50 60 70 80 90 130 120 110 100 Freezing Room temperature (688F) –10 FIG. 2.26 Ampere-hour rating (capacity) versus temperature for an Energizer® D cell. CONDUCTORS AND INSULATORS \| \| \| 77 e I V Hours 1.5 Voltage (V) 100 mA 800 700 600 500 400 300 200 100 50 mA 25 mA 1.4 1.3 1.2 1.1 1.0 Terminal voltage Discharge time 0.9 FIG. 2.27 Terminal voltage versus discharge time for specific drain currents for an Energizer® D cell. 2.8 CONDUCTORS AND INSULATORS Different wires placed across the same two battery terminals allow dif-ferent amounts of charge to flow between the terminals. Many factors, such as the density, mobility, and stability characteristics of a material, account for these variations in charge flow. In general, however, conductors are those materials that permit a generous flow of electrons with very little external force (voltage) applied. In addition, good conductors such as copper typically have only one electron in the valence (most distant from the nucleus) ring. Since copper is used most frequently, it serves as the standard of comparison for the relative conductivity in Table 2.1. Note that alumi-num, which has seen some commercial use, has only 61% of the conduc-tivity level of copper. The choice of material must be weighed against the cost and weight factors, however. Insulators are those materials that have very few free electrons and require a large applied potential (voltage) to establish a measurable current level. A common use of insulating material is for covering current- carrying wire, which, if uninsulated, could cause dangerous side effects. Power line workers wear rubber gloves and stand on rubber mats as safety measures when working on high-voltage transmission lines. A few different types of insulators and their applications appear in Fig. 2.28. Be aware, however, that even the best insulator will break down (per-mit charge to flow through it) if a sufficiently large potential (voltage) is applied across it. The breakdown strengths of some common insula-tors are listed in Table 2.2. According to this table, for insulators with the same geometric shape, it would require = 270/30 9 times as much potential to pass current through rubber as through air and approximately 67 times as much voltage to pass current through mica as through air. TABLE 2.1 Relative conductivity of various materials. Metal Relative Conductivity (%) Silver 105 Copper 100 Gold 70.5 Aluminum 61 Tungsten 31.2 Nickel 22.1 Iron 14 Constantan 3.52 Nichrome 1.73 Calorite 1.44 TABLE 2.2 Breakdown strength of some common insulators. Material Average Breakdown Strength (kV/cm) Air 30 Porcelain 70 Oils 140 Bakelite® 150 Rubber 270 Paper (paraffin-coated) 500 Teflon® 600 Glass 900 Mica 2000 78 \| \| \| Voltage and Current e I V 2.9 SEMICONDUCTORS Semiconductors are a specific group of elements that exhibit characteristics between those of insulators and those of conductors. The prefix semi, included in the terminology, has the dictionary defini-tion of half, partial, or between, as defined by its use. The entire electron-ics industry is dependent on this class of materials since the electronic devices and integrated circuits (ICs) are constructed of semiconductor materials. Although silicon (Si) is the most extensively employed mate-rial, germanium (Ge) and gallium arsenide (GaAs) are also used in many important high-speed electronic switching devices. Semiconductor materials typically have four electrons in the outermost valence ring. Semiconductors are further characterized as being photoconductive and having a negative temperature coefficient. Photoconductivity is a phenomenon in which the photons (small packages of energy) from inci-dent light can increase the carrier density in the material and thereby the charge flow level. A negative temperature coefficient indicates that the resistance (a characteristic to be described in detail in the next chap-ter) decreases with an increase in temperature (opposite to that of most conductors). A great deal more will be said about semiconductors in the chapters to follow and in your basic electronics courses. 2.10 AMMETERS AND VOLTMETERS It is important to be able to measure the current and voltage levels of an operating electrical system to check its operation, isolate malfunctions, and investigate effects impossible to predict on paper. As the names imply, ammeters are used to measure current levels; voltmeters, the potential difference between two points. If the current levels are usually of the order of milliamperes, the instrument will typically be referred to as a milliammeter, and if the current levels are in the microampere range, as a microammeter. Similar statements can be made for voltage levels. Throughout the industry, voltage levels are measured more frequently than current levels, primarily because measurement of the later does not require that the network connections be disturbed. (b) (c) (a) FIG. 2.28 Various types of insulators and their applications. (a) Fi-Shock extender insulator; (b) Fi-Shock corner insulator; (c) Fi-Shock screw-in post insulator. AMMETERS AND VOLTMETERS \| \| \| 79 e I V The potential difference between two points can be measured by sim-ply connecting the leads of the meter across the two points, as indicated in Fig. 2.29. An up-scale reading is obtained by placing the positive lead of the meter to the point of higher potential of the network and the com-mon or negative lead to the point of lower potential. The reverse connec-tion results in a negative reading or a below-zero indication. Ammeters are connected as shown in Fig. 2.30. Since ammeters mea-sure the rate of flow of charge, the meter must be placed in the network such that the charge flows through the meter. The only way this can be accomplished is to open the path in which the current is to be measured and place the meter between the two resulting terminals. For the config-uration in Fig. 2.30, the voltage source lead ( ) + must be disconnected from the system and the ammeter inserted as shown. An up-scale read-ing will be obtained if the polarities on the terminals of the ammeter are such that the current of the system enters the positive terminal. The introduction of any meter into an electrical/electronics system raises a concern about whether the meter will affect the behavior of the system. This question and others will be examined in Chapters 5 and 6 after additional terms and concepts have been introduced. For the moment, let it be said that since voltmeters and ammeters do have internal components, they will affect the network when introduced for measurement purposes. The design of each, however, is such that the impact is minimized. There are instruments designed to measure just current or just volt-age levels. However, the most common laboratory meters include the volt-ohm-milliammeter (VOM) and the digital multimeter (DMM), shown in Figs. 2.31 and 2.32, respectively. Both instruments measure voltage and current and a third quantity, resistance (introduced in the next chapter). The VOM uses an analog scale, which requires interpreting the position of a pointer on a continuous scale, while the DMM provides a display of numbers with decimal-point accuracy determined by the chosen scale. The use of an analog continuous scale can take some practice, but analog scales still appear so frequently that one must become adept at reading the scale correctly. The laboratory experience is probably the best opportunity to practice the reading of such a scale, but for the moment 12 V E V + FIG. 2.29 Voltmeter connection for an up-scale ( ) + reading. A + 40 V E System I FIG. 2.30 Ammeter connection for an up-scale ( ) + reading. (Andrew Scheck/Fotolia) FIG. 2.31 Volt-ohm-milliammeter (VOM) analog meter. (Courtesy of Fluke Corporation) FIG. 2.32 Digital multimeter (DMM). 80 \| \| \| Voltage and Current e I V let us try to interpret the reading of the VOM shown in Fig. 2.33. First, you will note that the pointer passes over a number of scales. If reading the resistance (Chapter 3) of a resistive element, the top scale is used. If the meter is set on × R 1, the resistance should be read as indicated. If the meter is set on × R 10, the result must be increased by a factor of ten. Assuming × R 1 for our current discussion, we must first notice that the resistance scale increases from the right to the left. The pointer appears to be right over the heavy bar between 2 and 5. Since there are three distinct regions between the numbers 2 and 5, the first bar to the left of 2 must signify 3 and the next bar 4 to complete the sequence from 2 to 5. The needle is right over the bar representing 4 so the reading is obviously 4 ohms. If the meter were set to read dc voltages, the band below the top resis-tance scale must be used. Now this scale increases from left to right with 250 V being the maximum voltage if the meter is set on × V 1. For this scale the pointer is between 150 and 200 V. If we count the divisions between the two numbers, we find there are 10 divisions. If we divide the difference in voltage by the number of divisions, we obtain 50 V/10 div. or 5 V per division. Counting from 150 V up, we find the pointer is 7.5 divisions above the 150 V level. The result is that the reading is at ( )( ) + = + = 150V 7.5 div. 5 V/div. 150 V 37.5 V 187.5 V. It is a bit of struggle to make a simple reading but this is the process one should be able to perform with a high degree of confidence. If ac (Chapter 13) voltages were being measured on the × V 1 scale, then the red ac scale below the dc scale must be used. Again, the scale increases from left to right with the pointer between the 6- and 8-volt levels. Because the heavy bar between them must represent 7 V, the five divisions between 7 and 8 V must have divisions of = 1 V/5 div. 0.2 V/div. The pointer is very close to 2.5 divisions above 7 V, so the reading is approximately ( )( ) + = + = 7 V 2.5 div. 0.2 V/div. 7 V 0.5 V 7.5 V. Now in this case the pointer is really just to the left of the halfway point between the two bars, so the reading is closer to 7.45 V. With all the extra effort required to read analog scales, why do we use them at all if we have digital meters that can read to hundredths-place accuracy? We do so because there are times when analog meters have characteristics that make them more efficient and safer to use. Consider the airline pilot and the huge dashboard in the cockpit with a range of 250 10 200 150 100 50 0 8 6 4 2 0 D. C A. C 2.5 V.A.C D. C A. C 0 .5 1.0 1.5 2.0 2.5 10 + 8 6 4 2 0 2 4 10 20 0 0 2 5 10 15 20 30 40 50 70 100 200 500 2K 1K 150 10 20 30 40 50 OHMS OHMS – VOLT-OHM-MILLIAMMETER FIG. 2.33 Reading an analog scale. APPLICATIONS \| \| \| 81 e I V meters to monitor. The majority of the meters are analog and not digital. The reason for this choice is over time the mind can scan a range of meters and establish the correct position of each pointer in memory and, if one seems out of whack, it can be quickly identified. This task is not so easy if facing you are 20 digital meters with all kinds of numbers. In addition a swinging or bouncing pointer can quickly be identified. The pointer also quickly shows whether the undesirable movement is up or down—again not so easy when you have to remember the numbers asso-ciated with a stable condition. As indicated above, the laboratory experience will provide practice in the use of both types of instruments. 2.11 APPLICATIONS Throughout the text, Applications sections such as this one have been included to permit a further investigation of terms, quantities, or systems introduced in the chapter. The primary purpose of these Applications is to establish a link between the theoretical concepts of the text and the real, practical world. Although the majority of com-ponents that appear in a system may not have been introduced (and, in fact, some components will not be examined until more advanced studies), the topics were chosen very carefully and should be quite interesting to a new student of the subject matter. Sufficient comment is included to provide a surface understanding of the role of each part of the system, with the understanding that the details will come at a later date. Since exercises on the subject matter of the Applications do not appear at the end of the chapter, the content is designed not to chal-lenge the student but rather to stimulate his or her interest and answer some basic questions such as how the system looks inside, what role specific elements play in the system, and, of course, how the system works. In essence, therefore, each Applications section provides an opportunity to begin to establish a practical background beyond simply the content of the chapter. Do not be concerned if you do not under-stand every detail of each application. Understanding will come with time and experience. For now, take what you can from the examples and then proceed with the material. Flashlight Although the flashlight uses one of the simplest of electrical circuits, a few fundamentals about its operation do carry over to more sophisti-cated systems. First, and quite obviously, it is a dc system with a lifetime totally dependent on the state of the batteries and bulb. Unless it is the rechargeable type, each time you use it, you take some of the life out of it. For many hours, the brightness will not diminish noticeably. Then, however, as it reaches the end of its ampere-hour capacity, the light becomes dimmer at an increasingly rapid rate (almost exponentially). The standard two-battery flashlight is shown in Fig. 2.34(a) with its elec-trical schematic in Fig. 2.34(b). Each 1.5 V battery has an ampere-hour rating of about 18 as indicated in Fig. 2.13. The single-contact miniature flange-base bulb is rated at 2.5 V and 300 mA with good brightness and a lifetime of about 30 hours. Thirty hours may not seem like a long lifetime, but you have to consider how long you usually use a flashlight on each occasion. If we assume a 300 mA drain from the battery for the 82 \| \| \| Voltage and Current e I V bulb when in use, the lifetime of the battery, by Eq. (2.11), is about 60 hours. Comparing the 60-hour lifetime of the battery to the 30-hour life expectancy of the bulb suggests that we normally have to replace bulbs more frequently than batteries. However, most of us have experienced the opposite effect. We can change batteries two or three times before we need to replace the bulb. This is simply one example of the fact that one cannot be guided solely by the specifications of each component of an electrical design. The operating conditions, terminal characteristics, and details about the actual response of the system for short and long periods of time must be considered. As mentioned earlier, the battery loses some of its power each time it is used. Although the terminal voltage may not change much at first, its ability to provide the same level of current drops with each usage. Further, batteries slowly discharge due to “leakage currents” even if the switch is not on. The air surrounding the battery is not “clean” in the sense that moisture and other elements in the air can provide a conduction path for leakage currents through the air through the surface of the battery itself, or through other nearby surfaces, and the battery eventually discharges. How often have we left a flashlight with new bat-teries in a car for a long period of time only to find the light very dim or the batteries dead when we need the flashlight the most? An additional problem is acid leaks that appear as brown stains or corrosion on the cas-ing of the battery. These leaks also affect the life of the battery. Further, when the flashlight is turned on, there is an initial surge in current that drains the battery more than continuous use for a period of time. In other words, continually turning the flashlight on and off has a very detrimen-tal effect on its life. We must also realize that the 30 hour rating of the bulb is for continuous use, that is, 300 mA flowing through the bulb for a continuous 30 hours. Certainly, the filament in the bulb and the bulb itself will get hotter with time, and this heat has a detrimental effect on the filament wire. When the flashlight is turned on and off, it gives the bulb a chance to cool down and regain its normal characteristics, thereby avoiding any real damage. Therefore, with normal use we can expect the bulb to last longer than the 30 hours specified for continuous use. Even though the bulb is rated for 2.5 V operation, it would appear that the two batteries would result in an applied voltage of 3 V, which suggests poor operating conditions. However, a bulb rated at 2.5 V can (b) 3 V Bulb Switch Ibulb Bulb Reﬂector Metal support 1.5 V D battery 1.5 V D battery Spring Sliding switch Contact 1.5 V 1.5 V (a) FIG. 2.34 (a) Eveready® D cell flashlight; (b) electrical schematic of flashlight of part (a). APPLICATIONS \| \| \| 83 e I V easily handle 2.5 V to 3 V. In addition, as was pointed out in this chapter, the terminal voltage drops with the current demand and usage. Under normal operating conditions, a 1.5 V battery is considered to be in good condition if the loaded terminal voltage is 1.3 V to 1.5 V. When it drops to the range from 1 V to 1.1 V, it is weak, and when it drops to the range from 0.8 V to 0.9 V, it has lost its effectiveness. Be aware that the total supplied voltage of 3 V will be obtained only if the batteries are connected as shown in Fig. 2.34(b). Accidentally plac-ing the two positive terminals together will result in a total voltage of 0 V, and the bulb will not light at all. For the vast majority of systems with more than one battery, the positive terminal of one battery will always be connected to the negative terminal of another. For all low-voltage batteries, the end with the nipple is the positive terminal, and the end with the flat end is the negative terminal. In addition, the flat or nega-tive end of a battery is always connected to the battery casing with the helical coil to keep the batteries in place. The positive end of the battery is always connected to a flat terminal connection or the element to be operated. If you look carefully at the bulb, you will find that the nipple connected to the positive end of the battery is insulated from the jacket around the base of the bulb. The jacket is the second terminal of the bat-tery used to complete the circuit through the on/off switch. If a flashlight fails to operate properly, the first thing to check is the state of the batteries. It is best to replace both batteries at once. A sys-tem with one good battery and one nearing the end of its life will result in pressure on the good battery to supply the current demand, and, in fact, the bad battery will actually be a drain on the good battery. Next, check the condition of the bulb by checking the filament to see whether it has opened at some point because a long-term, continuous current level occurred or because the flashlight was dropped. If the battery and bulb seem to be in good shape, the next area of concern is the contacts between the positive terminal and the bulb and the switch. Cleaning both with emery cloth often eliminates this problem. 12 V Car Battery Charger Battery chargers are a common household piece of equipment used to charge everything from small flashlight batteries to heavy-duty, marine, lead–acid batteries. Since all are plugged into a 120 V ac outlet such as found in the home, the basic construction of each is quite similar. In every charging system, a transformer (Chapter 23) must be included to reduce the ac voltage to a level appropriate for the charging dc level to be established. A diode (also called rectifier) arrangement must be included to convert the ac voltage, which varies with time, to an average dc level such as described in this chapter. Diodes and/or rectifiers will be discussed in detail in your first electronics course. Some dc chargers also include a regulator to provide an improved dc level (one that varies less with time or load). The car battery charger, one of the most common, is described here. The outside appearance and the internal construction of a DieHard 2 A/10 A AMP Manual Battery Charger are provided in Fig. 2.35. Note in Fig. 2.35(b) that the transformer (as in most chargers) takes up most of the internal space. The additional air space and the holes in the casing are there to ensure an outlet for the heat that will develop due to the resulting current levels. 84 \| \| \| Voltage and Current e I V The schematic in Fig. 2.36 includes all the basic components of the charger. Note first that the 120 V ac from the outlet are applied directly across the primary of the transformer. The charging rate of 2 A or 10 A is determined by the switch, which simply controls how many turns of the primary will be in the circuit for the chosen charging rate. If the battery is charging at the 2 A level, the full primary will be in the circuit, and the ratio of the turns in the primary to the turns in the secondary will be a maximum. If it is charging at the 10 A level, fewer turns of the primary are in the circuit, and the ratio drops. When you study transformers, you will find that the voltage at the primary and secondary is directly related to the turns ratio. If the ratio from primary to secondary drops, the volt-age drops also. The reverse effect occurs if the turns on the secondary exceed those on the primary. The general appearance of the waveforms appears in Fig. 2.36 for the 10 A charging level. Note that so far, the ac voltage has the same Charge level control Laminated iron-core of transformer Chasis ground Meter leads Full charge indicator Transformer Circuit board with diode elements 2-25 A Fuses Heat sink 470 F Capacitor (1 of 3) (b) 120 V ac 2 A 10 A Transformer (step-down) 13 V Diodes (rectifiers) Circuit Primary Current meter Negative clamp of charger Positive clamp of charger Peak = 18 V 12 V 12 V dc Secondary FIG. 2.36 Electrical schematic for the battery charger of Fig. 2.35. (a) (Don Johnson Photo) FIG. 2.35 Battery charger: (a) external appearance; (b) internal construction. APPLICATIONS \| \| \| 85 e I V wave shape across the primary and secondary. The only difference is in the peak value of the waveforms. Now the diodes take over and convert the ac waveform, which has zero average value (the waveform above equals the waveform below), to one that has an average value (all above the axis) as shown in the same figure. For the moment simply recog-nize that diodes are semiconductor electronic devices that permit only conventional current to flow through them in the direction indicated by the arrow in the symbol. Even though the waveform resulting from the diode action has a pulsing appearance with a peak value of about 18 V, it charges the 12 V battery whenever its voltage is greater than that of the battery, as shown by the shaded area. Below the 12 V level, the battery cannot discharge back into the charging network because the diodes per-mit current flow in only one direction. In particular, note in Fig. 2.35(b) the large plate that carries the cur-rent from the rectifier (diode) configuration to the positive terminal of the battery. Its primary purpose is to provide a heat sink (a place for the heat to be distributed to the surrounding air, such as a piece of aluminum with fins screwed to the diode) for the diode configuration. Otherwise, the diodes could eventually melt down and self-destruct due to the result-ing high current levels. Each component of Fig. 2.36 has been carefully labeled in Fig. 2.35(b) for reference. When current is first applied to a battery at the 10 A charge rate, the current demand as indicated by the meter on the face of the instrument may rise to 11 A or almost 12 A. However, the level of current decreases as the battery charges until it drops to a level of 2 A or 3 A. For units such as this that do not have an automatic shutoff, it is important to disconnect the charger when the current drops to the fully charged level; otherwise, the battery becomes overcharged and may be damaged. A battery that is at its 50% level can take as long as 10 hours to charge, so don’t expect it to be a 10-minute operation. In addition, if a battery is in very bad shape with a lower-than-normal voltage, the initial charging current may be too high for the design. To protect against such situations, the circuit breaker opens and stops the charging process. Because of the high current levels, it is important that the directions provided with the charger be carefully read and applied. The 50 A option on the face of the charger is for starting situations. This option can be used for limited periods of time if the battery voltage is too low. It can only be used for 10-second periods spaced by 3-minute periods for cooldown of the charger and battery. Answering Machines/Phones dc Supply A wide variety of systems in the home and office receive their dc operating voltage from an ac/dc conversion system plugged right into a 120 V ac outlet. Laptop computers, answering machines/phones, radios, clocks, cellular phones, CD players, and so on, all receive their dc power from a packaged system such as shown in Fig. 2.37. The conversion from ac to dc occurs within the unit, which is plugged directly into the outlet. The dc voltage is available at the end of the long wire, which is designed to be plugged into the operating unit. As small as the unit may be, it contains basically the same components as in the battery charger in Fig. 2.35. In Fig. 2.38, you can see the transformer used to cut the voltage down to appropriate levels (again the largest component of the system). Note that two diodes establish a dc level, and a capacitive filter (Chapter 10) is added to smooth out the dc voltage as shown. The system can be FIG. 2.37 Answering machine/phone 9 V dc supply. 86 \| \| \| Voltage and Current e I V relatively small because the operating current levels are quite small, permitting the use of thin wires to construct the transformer and limit its size. The lower currents also reduce the concerns about heating effects, permitting a small housing structure. The unit in Fig. 2.38, rated at 9 V at 200 mA, is commonly used to provide power to answering machines/ phones. Further smoothing of the dc voltage is accomplished by a reg-ulator built into the receiving unit. Its placement in the receiving unit reduces the possibility of picking up noise and oscillations along the long wire from the conversion unit to the operating unit, and it ensures that the full rated voltage is available at the unit itself, not a lesser value due to losses along the line. Transformer Capacitor Diodes (2) 120 V ac plug 9 V dc output FIG. 2.38 Internal construction of the 9 V dc supply in Fig. 2.37 . 1 C 2 C Q1 Q2 r FIG. 2.39 Problem 2. 8 C 40 C Q1 Q2 r m m FIG. 2.40 Problem 3. PROBLEMS SECTION 2.2 Atoms and Their Structure 1. a. The numbers of orbiting electrons in aluminum and sil-ver are 13 and 47, respectively. Draw the electronic con-figuration for each, and discuss briefly why each is a good conductor. b. Using the Internet, find the atomic structure of gold and explain why it is an excellent conductor of electricity. 2. Find the force of attraction in newtons between the charges Q1 and Q2 in Fig. 2.39 when a. = r 1 m b. = r 3 m c. = r 10 m d. Did the force drop off quickly with an increase in distance? 4. a. Plot the force of attraction (in newtons) versus separa-tion (in inches) between two unlike charges of µ 2 C. Use a range of 1 in. to 10 in. in 1 in. increments. Com-ment on the shape of the curve. Is it linear or nonlinear? What does it tell you about plotting a function whose magnitude is affected by a squared term in the denomi-nator of the equation? b. Using the plot of part (a), find the force of attraction at a 2.5 in. separation. c. Calculate the force of attraction with a 2.5 in. separation and compare with the result of part (b). 5. For two similar charges the force F1 exists for a separation of r meters. If the distance is increased to 2r, find the new level of force F2 in terms of the original force and the distance involved. 6. Determine the distance between two charges of µ 30 C if the force between the two charges is × 4.5 10 N 4 . 7. Two charged bodies Q1 and Q , 2 when separated by a dis-tance of 2 m, experience a force of repulsion equal to 1.8 N. a. What will the force of repulsion be when they are 10 m apart? b. If the ratio = Q Q / 1/2 1 2 , find Q1 and Q2 ( = r 10 m). 3. Find the force of repulsion in newtons between Q1 and Q2 in Fig. 2.40 when a. = r 1 ft b. = r 10 ft c. = r 100 yd d. Comment on the change in magnitude of the force as the distance between the charges was dramatically increased. e I V PROBLEMS \| \| \| 87 SECTION 2.6 Ampere-Hour Rating 26. What current will a battery with an Ah rating of 120 theoretically provide for 10 h? 27. What is the Ah rating of a battery that can provide 0.64 A for 80 h? 28. For how many hours will a battery with an Ah rating of 72 theoretically provide a current of 1.80 A? 29. A standard 12 V car battery has an ampere-hour rating of 45 Ah, whereas a heavy-duty battery has a rating of 75 Ah. How would you compare the energy levels of each and the available current for starting purposes? 30. At what current drain does the ampere-hour rating of the Energizer D Cell of Fig. 2.25 drop to 75% of its value at 25 mA? 31. What is the percentage loss in ampere-hour rating from room temperature to freezing for the Energizer D Cell of Fig. 2.26? 32. Using the graph of Fig. 2.27, how much longer can you maintain 1.2 V at a discharge rate of 25 mA compared to discharging at 100 mA? 33. A portable television using a 12 V, 4 Ah rechargeable battery can operate for a period of about 8 h. What is the average current drawn during this period? What is the energy expended by the battery in joules? SECTION 2.8 Conductors and Insulators 34. Discuss two properties of the atomic structure of copper that make it a good conductor. 35. Explain the terms insulator and breakdown strength. 36. List three uses of insulators not mentioned in Section 2.8. 37. a. Using Table 2.2, determine the level of applied voltage necessary to establish conduction through 1/2 in. of air. b. Repeat part (a) for 1/2 in. of rubber. c. Compare the results of parts (a) and (b). SECTION 2.9 Semiconductors 38. What is a semiconductor? How does it compare with a con-ductor and an insulator? 39. Consult a semiconductor electronics text and note the extensive use of germanium and silicon semiconductor materials. Review the characteristics of each material. SECTION 2.10 Ammeters and Voltmeters 40. What are the significant differences in the way ammeters and voltmeters are connected? 41. Compare analog and digital scales: a. Which are you more comfortable with? Why? b. Which can usually provide a higher degree of accuracy? c. Can you think of any advantages of the analog scale over a digital scale? Be aware that the majority of scales in a plane’s cockpit or in the control room of major power plants are analog. d. Do you believe it is necessary to become proficient in reading analog scales? Why? SECTION 2.3 Voltage 8. What is the voltage between two points if 3.4 J of energy are required to move 12 μC between the two points? 9. If the potential difference between two points is 50 V, how much energy is expended to bring 4 mC from one point to the other? 10. Find the charge in coulombs that requires µ 400 J of energy to be moved through a potential difference of 40 mV. 11. How much charge passes through a radio battery of 9 V if the energy expended is 620 mJ? 12. a. How much energy in electron volts is required to move 1 trillion (1 million million) electrons through a poten-tial difference of 40 V? b. How many joules of energy does the result of part (a) represent? c. Compare results (a) and (b). What can you say about the use of joules and electron volts as a unit of measure. Under what conditions should they be applied? SECTION 2.4 Current 13. Find the current in amperes if 96 mC of charge pass through a wire in 8.4 s. 14. If 600 C of charge pass through a wire in 4 min, find the current in amperes. 15. If a current of 40 mA exists for 1.2 min, how many cou-lombs of charge have passed through the wire? 16. How many coulombs of charge pass through a lamp in 1.2 min if the current is constant at 250 mA? 17. If the current in a conductor is constant at 2 mA, how much time is required for 6 mC to pass through the conductor? 18. If × + 21.847 10 18 electrons pass through a wire in 12 s, find the current. 19. How many electrons pass through a conductor in 5 min and 30 s if the current is 4 mA? 20. Will a fuse rated at 1 A “blow” if 86 C pass through it in 1.2 min? 21. If × + 0.92 10 16 electrons pass through a wire in 50 ms, find the current. 22. Which would you prefer? a. A penny for every electron that passes through a wire in µ 0.01 s at a current of 2 mA, or b. A dollar for every electron that passes through a wire in 1.5 ns if the current is µ 100 A. 23. If a conductor with a current of 300 mA passing through it converts 60 J of electrical energy into heat in 40 s, what is the potential drop across the conductor? 24. Charge is flowing through a conductor at the rate of 420 C/ min. If 742 J of electrical energy are converted to heat in 30 s, what is the potential drop across the conductor? 25. The potential difference between two points in an electric circuit is 24 V. If 0.8 J of energy were dissipated in a period of 6 ms, what would the current be between the two points? 88 \| \| \| Voltage and Current e I V Free electron An electron unassociated with any particular atom, relatively free to move through a crystal lattice structure under the influence of external forces. Fuel cell A nonpolluting source of energy that can establish cur-rent through a load by simply applying the correct levels of hydrogen and oxygen. Insulators Materials in which a very high voltage must be applied to produce any measurable current flow. Neutron The particle having no electrical charge found in the nucleus of the atom. Nucleus The structural center of an atom that contains both pro-tons and neutrons. Positive ion An atom having a net positive charge due to the loss of one of its negatively charged electrons. Potential difference The algebraic difference in potential (or voltage) between two points in an electrical system. Potential energy The energy that a mass possesses by virtue of its position. Primary cell Sources of voltage that cannot be recharged. Proton The particle of positive polarity found in the nucleus of an atom. Rectification The process by which an ac signal is converted to one that has an average dc level. Secondary cell Sources of voltage that can be recharged. Semiconductor A material having a conductance value between that of an insulator and that of a conductor. Of significant importance in the manufacture of electronic devices. Solar cell Sources of voltage available through the conversion of light energy (photons) into electrical energy. Specific gravity The ratio of the weight of a given volume of a substance to the weight of an equal volume of water at 4°C. Volt (V) The unit of measurement applied to the difference in potential between two points. If 1 joule of energy is required to move 1 coulomb of charge between two points, the differ-ence in potential is said to be 1 volt. Voltage The term applied to the difference in potential between two points as established by a separation of opposite charges. Voltmeter An instrument designed to read the voltage across an element or between any two points in a network. GLOSSARY Ammeter An instrument designed to read the current through elements in series with the meter. Ampere (A) The SI unit of measurement applied to the flow of charge through a conductor. Ampere-hour (Ah) rating The rating applied to a source of energy that will reveal how long a particular level of current can be drawn from that source. Cell A fundamental source of electrical energy developed through the conversion of chemical or solar energy. Conductors Materials that permit a generous flow of electrons with very little voltage applied. Conventional flow The movement of charge through a conduc-tor defined by the positive charge. Copper A material possessing physical properties that make it particularly useful as a conductor of electricity. Coulomb (C) The fundamental SI unit of measure for charge. It is equal to the charge carried by 6.242 × 1018 electrons. Coulomb’s law An equation defining the force of attraction or repulsion between two charges. Current The flow of charge resulting from the application of a dif-ference in potential between two points in an electrical system. dc current source A source that will provide a fixed current level even though the load to which it is applied may cause its terminal voltage to change. dc generator A source of dc voltage available through the turn-ing of the shaft of the device by some external means. Direct current (dc) Current having a single direction (unidirec-tional) and a fixed magnitude over time. Electrolysis The process of passing a current through an electro-lyte to break it down into its fundamental components. Electrolytes The contact element and the source of ions between the electrodes of the battery. Electron The particle with negative polarity that orbits the nucleus of an atom. Electron flow The movement of charge through a conductor defined by the negative charge. Electron volt A unit of energy defined by the movement of an electron through a potential difference of one volt. Resistance 3.1 INTRODUCTION In the previous chapter, we found that placing a voltage across the simplest of circuits results in a flow of charge through the circuit. The question remains, however, What determines the level of current that results when a particular voltage is applied? Why is the current heavier in some circuits than in others? The answers lie in the fact that there is an opposition to the flow of charge in the system that depends on the components of the circuit. This opposition to the flow of charge through an electrical circuit, called resistance, has the units of ohms and uses the Greek letter omega ( ) Ω as its symbol. The graphic symbol for resistance, which resembles the cutting edge of a saw, is provided in Fig. 3.1. • Become familiar with the parameters that determine the resistance of an element and be able to calculate the resistance from the given dimensions and material characteristics. • Understand the effects of temperature on the resistance of a material and how to calculate the change in resistance with temperature. • Develop some understanding of superconductors and how they can affect future development in the industry. • Become familiar with the broad range of commercially available resistors and how to read the value of each from the color code or labeling. • Become aware of a variety of elements such as thermistors, photoconductive cells, and varistors and how their terminal resistance is controlled. Objectives 3 R G R FIG. 3.1 Resistance symbol and notation. This opposition, due primarily to collisions and friction between the free electrons and other electrons, ions, and atoms in the path of motion, converts the supplied electrical energy into heat that raises the temperature of the electrical component and surrounding medium. The heat you feel from an electrical heater is simply due to current passing through a high-resistance material. Each material with its unique atomic structure reacts differently to pressures to establish current through its core. Conductors that permit a generous flow of charge with little external pressure have low resistance levels, while insulators have high resistance characteristics. 90 \| \| \| Resistanc R G 3.2 RESISTANCE: CIRCULAR WIRES The resistance of any material is due primarily to four factors: 1. Material 2. Length 3. Cross-sectional area 4. T emperature of the material As noted in Section 3.1, the atomic structure determines how eas-ily a free electron will pass through a material. The longer the path through which the free electron must pass, the greater is the resistance factor. Free electrons pass more easily through conductors with larger cross-sectional areas. In addition, the higher the temperature of the con-ductive materials, the greater is the internal vibration and motion of the components that make up the atomic structure of the wire, and the more difficult it is for the free electrons to find a path through the material. The first three elements are related by the following basic equation for resistance: T CM-ft at 20 C ρ = Ω = ° ρ = R l A = l feet (3.1) ( ) = A area in circular mils CM with each component of the equation defined by Fig. 3.2. The material is identified by a factor called the resistivity, which uses the Greek letter rho ρ ( ) as its symbol and is measured in Ω CM-ft. Its value at a temperature of ° 20 C room temperature 68 F ( ) = ° is pro-vided in Table 3.1 for a variety of common materials. Since the larger the resistivity, the greater is the resistance to setting up a flow of charge, it appears as a multiplying factor in Eq. (3.1); that is, it appears in the numerator of the equation. The effect of higher and lower temperatures is considered in Section 3.4. It is important to realize at this point that since the resistivity is provided at room temperature, Eq. (3.1) is applicable only at that temperature. Since the resistivity is in the numerator of Eq. (3.1), the higher the resistivity, the greater is the resistance of a conductor as shown for two conductors of the same length in Fig. 3.3(a). Further, T (8C) A Material ( ) l r FIG. 3.2 Factors affecting the resistance of a conductor. TABLE 3.1 Resistivity ρ ( ) of various materials. Material CM-ft 20 C @ ρ Ω ° ( ) Silver 9.9 Copper 10.37 Gold 14.7 Aluminum 17.0 Tungsten 33.0 Nickel 47.0 Iron 74.0 Constantan 295.0 Nichrome 600.0 Calorite 720.0 Carbon 21,000.0 R1 Copper R2 Iron R1 Copper R2 Copper R1 Copper R2 Copper (b) R2 > R1 l1 > R2 > R1 (c) (a) R2 > R1 1 2 > l2 A2 > A1 r r FIG. 3.3 Cases in which R R . 2 1 > For each case, all remaining parameters that control the resistance level are the same. RESISTANCE: CIRCULAR WIRES \| \| \| 91 R G the longer the conductor, the greater is the resistance since the length also appears in the numerator of Eq. (3.1). Note Fig. 3.3(b). Finally, the greater the area of a conductor, the less is the resistance because the area appears in the denominator of Eq. (3.1). Note Fig. 3.3(c). Circular Mils (CM) In Eq. (3.1), the area is measured in a quantity called circular mils (CM). It is the quantity used in most commercial wire tables, and thus it needs to be carefully defined. The mil is a unit of measurement for length and is related to the inch by 1 mil 1 1000 in. = or 1000 mils 1 in. = In general, therefore, the mil is a very small unit of measurement for length. There are 1000 mils in an inch, or 1 mil is only 1/1000 of an inch. It is a length that is not visible with the naked eye, although it can be measured with special instrumentation. The phrase milling used in steel factories is derived from the fact that a few mils of material are often removed by heavy machinery such as a lathe, and the thickness of steel is usually measured in mils. By definition, a wire with a diameter of 1 mil has an area of 1 CM as shown in Fig. 3.4. An interesting result of such a definition is that the area of a circular wire in circular mils can be defined by the following equation: ( ) = A d CM mils 2 (3.2) Verification of this equation appears in Fig. 3.5, which shows that a wire with a diameter of 2 mils has a total area of 4 CM, and a wire with a diameter of 3 mils has a total area of 9 CM. Remember, to compute the area of a wire in circular mils when the diameter is given in inches, first convert the diameter to mils by simply writing the diameter in decimal form and moving the decimal point three places to the right. For example, = = 1 8 in. 0.125 in. 125 mils 3 places Then the area is determined by ( ) ( ) = = = A d 15 625 CM 125 mils , CM mils 2 2 Sometimes when you are working with conductors that are not circu-lar, you will need to convert square mils to circular mils, and vice versa. Applying the basic equation for the area of a circle and substituting a diameter of 1 mil results in by definition π π π ( ) = = = ≡ A d 4 4 1 mil 4 sq mils 1 CM 2 2 from which we can conclude the following: 1 mil 1 circular mil (CM) 1 square mil FIG. 3.4 Defining the circular mil (CM). A = (2 mils)2 = 4 CM 1 2 3 4 3 A = (3 mils)2 = 9 CM 2 1 4 5 7 8 6 9 d = 2 mils d = 3 mils FIG. 3.5 Verification of Eq. (3.2): mils A d CM 2 ( ) = 92 \| \| \| Resistanc R G π = 1 CM 4 sq mils (3.3) or π = 1sq mil 4 CM (3.4) EXAMPLE 3.1 What is the resistance of a 100 ft length of copper wire with a diameter of 0.020 in. at ° 20 C? Solution: ρ = Ω = 10.37 CM-ft and 0.020 in. 20 mils ( ) ( ) = = = A d 20 mils 400 CM CM mils 2 2 ρ ( )( ) = = Ω R l A 10.37 CM-ft 100 ft 400 CM = R 2 59 Ω . EXAMPLE 3.2 An undetermined number of feet of wire have been used from the ′ 500 spool of wire in Fig. 3.6. Find the length of the remain-ing copper wire if the diameter is 1/16 in. and the resistance is Ω 0.8 . Solution: A d R l A l RA l 301 35 ft 10.37 CM-ft and 1 16 in. 0.0625 in. 62.5 mils 62.5 mils 3906.25 CM 0.8 3906.25 CM 10.37 CM-ft 3125 10.37 . CM mils 2 2 ρ ρ ρ ( ) ( )( ) ( ) = Ω = = = = = = ⇒ = = Ω Ω = = EXAMPLE 3.3 What is the resistance of a copper bus-bar, as used in the power distribution panel of a high-rise office building, with the dimensions indicated in Fig. 3.7? Solution: π ( )( ) = = = = × = ×              = ×                              A A A 5.0 in. 5000 mils 1 2 in. 500 mils 5000 mils 500 mils 2.5 10 sq mils 2.5 10 sq mils 4 CM 1 sq mil 3.183 10 CM CM 6 6 6 ρ ( )( ) = = Ω × = × R l A 10.37 CM-ft 3 ft 3.183 10 CM 31.11 3.183 10 6 6 = × − R 9 774 10 6 Ω . ( ) Ω≅ Ω quitesmall,0.000009774 0 + RT (Don Johnson Photo) FIG. 3.6 Example 3.2. 3 ft 5 in. 1/2 in. FIG. 3.7 Example 3.3. WIRE TABLES \| \| \| 93 R G You will learn in the following chapters that the lower the resistance of a conductor, the lower are the losses in conduction from the source to the load. Similarly, since resistivity is a major factor in determining the resistance of a conductor, the lower the resistivity, the lower is the resistance for the same-size conductor. It would appear from Table 3.1 that silver, copper, gold, and aluminum would be the best conductors and the most common. In general, there are other factors, however, such as malleability (ability of a material to be shaped), ductility (ability of a material to be drawn into long, thin wires), temperature sensitivity, resistance to abuse, and, of course, cost, that must all be weighed when choosing a conductor for a particular application. In general, copper is the most widely used material because it is quite malleable, ductile, and available; has good thermal characteristics; and is less expensive than silver or gold. It is certainly not cheap, however. Contractors always ensure that the copper wiring has been removed be-fore leveling a building because of its salvage value. Aluminum was once used for general wiring because it is cheaper than copper, but its thermal characteristics created some difficulties. The heating due to current flow and the cooling that occurred when the circuit was turned off resulted in expansion and contraction of the aluminum wire to the point where connections eventually loosened, resulting in dangerous side effects. Aluminum is still used today, however, in areas such as integrated circuit manufacturing and in situations where the connections can be made se-cure. Silver and gold are, of course, much more expensive than copper or aluminum, but the cost is justified for certain applications. Silver has ex-cellent plating characteristics for surface preparations, and gold is used quite extensively in integrated circuits. Tungsten has a resistivity three times that of copper, but there are occasions when its physical character-istics (durability, hardness) are the overriding considerations. 3.3 WIRE TABLES The wire table was designed primarily to standardize the size of wire produced by manufacturers. As a result, the manufacturer has a larger market, and the consumer knows that standard wire sizes will always be available. The table was designed to assist the user in every way possi-ble; it usually includes data such as the cross-sectional area in circular mils, diameter in mils, ohms per 1000 feet at ° 20 C, and weight per 1000 feet. The American Wire Gage (AWG) sizes are given in Table 3.2 for solid, round copper wire. A column indicating the maximum allowable current in amperes, as determined by the National Fire Protection Association, has also been included. The most commonly used appear in boldface. The chosen sizes have an interesting relationship: The area is doubled for every drop in 3 gage numbers and increased by a factor of 10 for every drop of 10 gage numbers. Examining Eq. (3.1), we note also that doubling the area cuts the resistance in half, and increasing the area by a factor of 10 decreases the resistance of 1/10 the original, everything else kept constant. The actual sizes of some of the gage wires listed in Table 3.2 are shown in Fig. 3.8 with a few of their areas of application. A few exam-ples using Table 3.2 follow. 94 \| \| \| Resistanc R G TABLE 3.2 American Wire Gage (AWG) sizes. AWG # Area (CM) Ω1000 ft ° at 20 C Maximum Allowable Current for RHW Insulation (A) (4/0) 0000 211,600 0.0490 230 (3/0) 000 167,810 0.0618 200 (2/0) 00 133,080 0.0780 175 (1/0) 0 105,530 0.0983 150 1 83,694 0.1240 130 2 66,373 0.1563 115 3 52,634 0.1970 100 4 41,742 0.2485 85 5 33,102 0.3133 — 6 26,250 0.3951 65 7 20,816 0.4982 — 8 16,509 0.6282 50 9 13,094 0.7921 — 10 10,381 0.9989 30 11 8,234.0 1.260 — 12 6,529.9 1.588 20 13 5,178.4 2.003 — 14 4,106.8 2.525 15 15 3,256.7 3.184 16 2,582.9 4.016 17 2,048.2 5.064 18 1,624.3 6.385 19 1,288.1 8.051 20 1,021.5 10.15 21 810.10 12.80 22 642.40 16.14 23 509.45 20.36 24 404.01 25.67 25 320.40 32.37 26 254.10 40.81 27 201.50 51.47 28 159.79 64.90 29 126.72 81.83 30 100.50 103.2 31 79.70 130.1 32 63.21 164.1 33 50.13 206.9 34 39.75 260.9 35 31.52 329.0 36 25.00 414.8 37 19.83 523.1 38 15.72 659.6 39 12.47 831.8 40 9.89 1049.0 { {              Not more than three conductors in raceway, cable, or direct burial. Businesses, Large homes Most new homes Some electrical appliances Most house wiring Electronic projects As AWG incresses resistance incresses As AWG incresses area drops                                   WIRE TABLES \| \| \| 95 R G EXAMPLE 3.4 Find the resistance of 650 ft of #8 copper wire = ° T ( 20 C). Solution: For #8 copper wire (solid), Ω/1000 ft at ° = Ω 20 C 0.6282 , and Ω           = 0 41 Ω 650 ft 0.6282 1000 ft . EXAMPLE 3.5 What is the diameter, in inches, of a #12 copper wire? Solution: For #12 copper wire (solid), = A 6529.9 CM, and = = ≅ d A 6529.9 CM 80.81 mils mils CM ( ) ≅ d 0 08 in. . or close to1 12 in. EXAMPLE 3.6 For the system in Fig. 3.9, the total resistance of each power line cannot exceed Ω 0.025 , and the maximum current to be drawn by the load is 95 A. What gage wire should be used? Solution: ρ ρ ( )( ) = ⇒ = = Ω Ω = R l A A l R 41 480 CM 10.37 CM-ft 100 ft 0.025 , Using the wire table, we choose the wire with the next largest area, which is #4, to satisfy the resistance requirement. We note, however, that 95 A must flow through the line. This specification requires that #3 wire be used since the #4 wire can carry a maximum current of only 85 A. D = 0.365 in. ù 1/3 in. 00 (AWG) Power distribution Stranded for increased flexibility D = 0.0808 in. ù 1/12 in. D = 0.064 in. ù 1/16 in. 12 (AWG) 14 (AWG) Lighting, outlets, general home use D = 0.013 in. ù 1/75 in. 28 (AWG) Telephone, instruments D = 0.032 in. ù 1/32 in. D = 0.025 in. 5 1/40 in. 20 (AWG) 22 (AWG) Radio, television FIG. 3.8 Popular AWG sizes and some of their areas of application. Solid round copper wire Input 100 ft Load FIG. 3.9 Example 3.6. 96 \| \| \| Resistanc R G 3.4 TEMPERATURE EFFECTS Temperature has a significant effect on the resistance of conductors, semiconductors, and insulators. Conductors Conductors have a generous number of free electrons, and any introduc-tion of thermal energy will have little impact on the total number of free carriers. In fact, the thermal energy only increases the intensity of the random motion of the particles within the material and makes it increas-ingly difficult for a general drift of electrons in any one direction to be established. The result is that for good conductors, an increase in temperature results in an increase in the resistance level. Consequently, conductors have a positive temperature coefficient. The plot in Fig. 3.10(a) has a positive temperature coefficient. Semiconductors In semiconductors, an increase in temperature imparts a measure of ther-mal energy to the system that results in an increase in the number of free carriers in the material for conduction. The result is that for semiconductor materials, an increase in temperature results in a decrease in the resistance level. Consequently, semiconductors have negative temperature coefficients. The thermistor and photoconductive cell discussed in Sections 3.12 and 3.13, respectively, are excellent examples of semiconductor devices with negative temperature coefficients. The plot in Fig. 3.10(b) has a negative temperature coefficient. Insulators As with semiconductors, an increase in temperature results in a decrease in the resistance of an insulator. The result is a negative temperature coefficient. 3.5 CONDUCTORS AND THE INFERRED ABSOLUTE TEMPERATURE ( ) Ti Fig. 3.11 reveals that for copper (and most other metallic conductors), the resistance increases almost linearly (in a straight-line relationship) with an increase in temperature. Since temperature can have such a pronounced effect on the resistance of a conductor, it is important that we have some method of determining the resistance at any temperature within operating limits. An equation for this purpose can be obtained by approximating the curve in Fig. 3.11 by the straight dashed line that intersects the temperature scale at − ° 234.5 C. Although the actual curve extends to absolute zero ( ) − ° 273.15 C,or 0 K , the straight-line approx-imation is quite accurate for the normal operating temperature range. At two temperatures T1 and T , 2 the resistance of copper is R1 and R , 2 (a) Temperature R 0 Temperature coefficient (b) Temperature R 0 Temperature coefficient FIG. 3.10 Demonstrating the effect of a positive and a negative temperature coefficient on the resistance of a conductor. R G CONDUCTORS AND THE INFERRED ABSOLUTE TEMPERATURE (Ti) \| \| \| 97 respectively, as indicated on the curve. Using a property of similar tri-angles, we may develop a mathematical relationship between these val-ues of resistance at different temperatures. Let x equal the distance from − ° 234.5 C to T1 and y the distance from − ° 234.5 C to T , 2 as shown in Fig. 3.11. From similar triangles, = x R y R 1 2 or T R T R 234.5 234.5 1 1 2 2 + = + (3.5) The temperature of − ° 234.5 C is called the inferred absolute tempera-ture (Ti) of copper. For different conducting materials, the intersection of the straight-line approximation occurs at different temperatures. A few typical values are listed in Table 3.3. The minus sign does not appear with the inferred absolute tempera-ture on either side of Eq. (3.5) because x and y are the distances from − ° 234.5 C to T1 and T , 2 respectively, and therefore are simply magni-tudes. For T1 and T2 less than zero, x and y are less than − ° 234.5 C, and the distances are the differences between the inferred absolute tempera-ture and the temperature of interest. Eq. (3.5) can easily be adapted to any material by inserting the proper inferred absolute temperature. It may therefore be written as follows: + = + T T R T T R i i 1 1 2 2 (3.6) where Ti indicates that the inferred absolute temperature of the mate-rial involved is inserted as a positive value in the equation. In general, therefore, associate the sign only with T1 and T . 2 EXAMPLE 3.7 If the resistance of a copper wire is Ω 50 at ° 20 C, what is its resistance at ° 100 C (boiling point of water)? Solution: Eq. (3.5): ° + ° Ω = ° + ° R 234.5 C 20 C 50 234.5 C 100 C 2 Ω ( )( ) = Ω ° ° = R 65 72 50 334.5 C 254.5 C . 2 R1 T1 T2 8C 08C –234.58C –273.158C Absolute zero x Inferred absolute zero R2 R y FIG. 3.11 Effect of temperature on the resistance of copper. TABLE 3.3 Inferred absolute temperatures ( ) T . i Material °C Silver −243 Copper −234 5 . Gold −274 Aluminum −236 Tungsten −204 Nickel −147 Iron −162 Nichrome −2,250 Constantan −125,000 98 \| \| \| Resistanc R G EXAMPLE 3.8 If the resistance of a copper wire at freezing ( ° 0 C) is Ω 30 , what is its resistance at − ° 40 C? Solution: Eq. (3.5): ° + Ω = ° − ° R 234.5 C 0 30 234.5 C 40 C 2 R 24 88 30 194.5 C 234.5 C . 2 Ω ( )( ) = Ω ° ° = EXAMPLE 3.9 If the resistance of an aluminum wire at room tem-perature ( ° 20 C) is Ω 100 m (measured by a milliohmmeter), at what temperature will its resistance increase to Ω 120 m ? Solution: Eq. (3.5): ° + ° Ω = ° + Ω T 236 C 20 C 100 m 236 C 120 m 2 and = Ω ° Ω           − ° T 120 m 256 C 100 m 236 C 2 = ° T 71 2 . C 2 3.6 TEMPERATURE COEFFICIENT OF RESISTANCE There is a second popular equation for calculating the resistance of a conductor at different temperatures. Defining α ( ) = + ° Ω° Ω T 1 20 C C i 20 (3.7) as the temperature coefficient of resistance at a temperature of ° 20 C and R20 as the resistance of the sample at ° 20 C, we determine the resis-tance R1 at a temperature T1 by α ( ) [ ] = + − ° R R T 1 20 C 1 20 20 1 (3.8) The values of α20 for different materials have been evaluated, and a few are listed in Table 3.4. Eq. (3.8) can be derived by applying the roles for similar triangles to the plot of Fig. 3.11. Eq. (3.8) can be written in the following form: α = − − °           = ∆ ∆ R R T R R T R 20 C 20 1 20 1 20 20 (3.9) from which the units of Ω° Ω C for α20 are defined. Since ∆ ∆ R T is the slope of the curve in Fig. 3.11, we can conclude that the higher the temperature coefficient of resistance for a material, the more sensitive is the resistance level to changes in temperature. Referring to Table 3.4, we find that copper is more sensitive to temperature variations than is silver, gold, or aluminum, although the TABLE 3.4 Temperature coefficient of resistance for various conductors at ° 20 C. Material Temperature Coefficient 20 α ( ) Silver 0.0038 Copper 0.00393 Gold 0.0034 Aluminum 0.00391 Tungsten 0.005 Nickel 0.006 Iron 0.0055 Constantan 0.000008 Nichrome 0.00044 TYPES OF RESISTORS \| \| \| 99 R G differences are quite small. The slope defined by α20 for constantan is so small that the curve is almost horizontal. Since R20 of Eq. (3.8) is the resistance of the conductor at ° 20 C and − ° T 20 C 1 is the change in temperature from ° 20 C, Eq. (3.8) can be written in the following form: R l A T l 20 ρ α [ ] = + ∆ (3.10) providing an equation for resistance in terms of all the controlling parameters. 3.7 PPM/°C For resistors, as for conductors, resistance changes with a change in temperature. The specification is normally provided in parts per million per degree Celsius (PPM C ° ), providing an immediate indication of the sensitivity level of the resistor to temperature. For resistors, a 5000 PPM level is considered high, whereas 20 PPM is quite low. A 1000 ° PPM C characteristic reveals that a ° 1 change in temperature results in a change in resistance equal to 1000 PPM, or = 1000 1,000,000 1 1000 of its nameplate value—not a significant change for most applications. However, a ° 10 change results in a change equal to 1/100 (1%) of its nameplate value, which is becoming significant. The concern, therefore, lies not only with the PPM level but also with the range of expected temperature variation. In equation form, the change in resistance is given by R R T 10 PPM nominal 6 ( ) ( ) ∆ = ∆ (3.11) where Rnominal is the nameplate value of the resistor at room temperature and ∆T is the change in temperature from the reference level of ° 20 C. EXAMPLE 3.10 For a 1 Ω k carbon composition resistor with a PPM of 2500, determine the resistance at ° 60 C. Solution: ( ) ( ) ∆ = Ω ° − ° R 1000 10 2500 60 C 20 C 6 = Ω 100 and = + ∆ = Ω+ Ω R R R 1000 100 nominal = 1100 Ω 3.8 TYPES OF RESISTORS Fixed Resistors Resistors are made in many forms, but all belong in either of two groups: fixed or variable. The most common of the low-wattage, fixed-type re-sistors is the film resistor shown in Fig. 3.12. It is constructed by depos-iting a thin layer of resistive material (typically carbon, metal, or metal 100 \| \| \| Resistanc R G oxide) on a ceramic rod. The desired resistance is then obtained by cut-ting away some of the resistive material in a helical manner to establish a long, continuous band of high-resistance material from one end of the resistor to the other. In general, carbon-film resistors have a beige body and a lower wattage rating. The metal-film resistor is typically a stron-ger color, such as brick red or dark green, with higher wattage ratings. The metal-oxide resistor is usually a softer pastel color, such as rating powder blue shown in Fig. 3.12(b), and has the highest wattage rating of the three. When you search through most electronics catalogs or visit a local electronics dealer to purchase resistors, you will find that the most common resistor is the film resistor. In years past, the carbon com-position resistor in Fig. 3.13 was the most common, but fewer and fewer companies are manufacturing this variety, with its range of ap-plications reduced to applications in which very high temperatures and inductive effects (Chapter 11) can be a problem. Its resistance is determined by the carbon composition material molded directly to each end of the resistor. The high-resistivity characteristics of carbon ρ ( ) = Ω 21,000 CM- /ft provide a high-resistance path for the current through the element. For a particular style and manufacturer, the size of a resistor increases with the power or wattage rating. The concept of power is covered in detail in Chapter 4, but for the moment recognize that increased power ratings are normally associated with the ability to handle higher current and temperature levels. Fig. 3.14 depicts the actual size of thin-film, metal-oxide resistors in the 1/4 W to 5 W rating range. All the resistors in Fig. 3.14 are 1 Ω M , revealing that the size of a resistor does not define its resistance level. A variety of other fixed resistors are depicted in Fig. 3.15. The wire-wound resistors of Fig. 3.15(a) are formed by winding a high-resistance wire around a ceramic core. The entire structure is then baked in a ceramic cement to provide a protective covering. Wire-wound resistors are typically used for larger power applications, although they are also available with very small wattage ratings and very high accuracy. Fig. 3.15(c) and (g) are special types of wire-wound resis-tors with a low percent tolerance. Note, in particular, the high power ratings for the wire-wound resistors for their relatively small size. Figs. 3.15(b), (d), and (f) are power film resistors that use a thicker layer of film material than used in the variety shown (a) Wire lead connected to continuous thin-film path of resistive material Spiral trimmed grooves End cap Molded insulating casing Ceramic core Thin film of high-resistance metal FIG. 3.12 Film resistors: (a) construction; (b) types. (b) Carbon-film (1/2 W) Metal-film (2 W) Metal-oxide film (2 W) Leads Color bands Insulation material Resistance material (Carbon composition) (a) FIG. 3.13 Fixed-composition resistors: (a) construction; (b) appearance. (b) ACTUAL SIZE 5 W 3 W 1 W 1 2 W / 1 4 W / FIG. 3.14 Fixed metal-oxide resistors of different wattage ratings. TYPES OF RESISTORS \| \| \| 101 R G 100 MV, 0.75 W Precision power film resistor (d) 25 kV, 5 W Silicon-coated, wire-wound resistor (g) 1 kV bussed (all connected on one side) single in-line resistor network (e) Resistive material Terminals Bakelite (insulator) coating 1 kV, 25 W Aluminum-housed, chassis-mount resistor–precision wire-mount (c) 470 V, 35 W Thick-film power resistor (b) Tinned alloy terminals Vitreous enamel coating Even uniform winding High-strength welded terminal Resilient mounting brackets Strong ceramic core Welded resistance wire junction Wire-wound resistors (a) 2 kV, 8 W 100 Ω, 25 W 22 kV, 1 W Surface mount thick-film chip resistors with gold electrodes (f) Electrodes (Terminals) Ceramic base Resistive material FIG. 3.15 Various types of fixed resistors. in Fig. 3.12. The chip resistors in Fig. 3.15(f) are used where space is a priority, such as on the surface of circuit board. Units of this type can be less than 1/16 in. in length or width, with thickness as small as 1/30 in., yet they can still handle 0.5 W of power with resistance levels as high as Ω 1000 M —clear evidence that size does not determine the resistance level. The fixed resistor in Fig. 3.15(e) has terminals applied to a layer of resistor material, with the resistance between the terminals a function of the dimensions of the resistive material and the placement of the terminal pads. 102 \| \| \| Resistanc R G Maximum Voltage Rating In addition to a wattage rating, every resistor has a maximum voltage rating. Both must be considered with the use of any resistor in any application. In some cases, the maximum voltage rating may seem so high from typical operating voltages that the power rating is the only one of any consequence. Consider, however, a commercially available carbon composition resistor rated for 1 W, a 10% tolerance, and a max-imum voltage rating of 500 V available in resistor values from Ω 2.2 to Ω 1 M . Keep in mind that the complete line of resistors with this 1 W power rating and maximum voltage rating all have the same size and construction. Internally, of course, the resistive material is altered to provide the desired resistance, but the external appearance is the same. Within this series of resistors, a 10 Ω resistor would need only 3.16 V applied across it to reach the 1 W rating. The calculations surrounding this conclusion will be discussed in Chapter 4. Certainly, the 3.16 V is magnitudes smaller than the rated voltage of 500 V and it may never be necessary to worry about this rating. However, we would also find that if we were using a Ω 250 k resistor from the package of resistors, it would require 500 V to reach the 1 W power dissipation level. In fact any resis-tor between Ω 250 k and Ω 1 M would require more than the rated 500 V to dissipate a power level of 1 W. For the Ω 1 M resistor, the power dis-sipated at 500 V is only 14 W. The result, therefore, is that the applied voltage is a very important factor because it is related to the maximum power dissipation capabilities of the resistor. Furthermore, if the applied voltage exceeds the maximum rated value, the resistive qualities of the resistor may deteriorate, high surface currents may develop, arcing may occur, or the resistor itself may open or cause a short circuit. Variable Resistors Variable resistors, as the name implies, have a terminal resistance that can be varied by turning a dial, knob, screw, or whatever seems appropriate for the application. They can have two or three terminals, but most have three terminals. If the two- or three-terminal device is used as a variable resistor, it is usually referred to as a rheostat. If the three-terminal device is used for controlling potential levels, it is then commonly called a potentiometer. Even though a three-terminal device can be used as a rheostat or a potentiometer (depending on how it is connected), it is typically called a potentiometer when listed in trade magazines or requested for a particular application. The symbol for a three-terminal potentiometer appears in Fig. 3.16(a). When used as a variable resistor (or rheostat), it can be hooked up in one of two ways, as shown in Figs. 3.16(b) and (c). In Fig. 3.16(b), points a and (d) R (c) a R Rab b, c (b) R a c b Rab (a) R b a c FIG. 3.16 Potentiometer: (a) symbol; (b) and (c) rheostat connections; (d) rheostat symbol. TYPES OF RESISTORS \| \| \| 103 R G a b c Rotating shaft (a) (Don Johnson Photo) FIG. 3.17 Potentiometer: (a) external, (b) internal, (c) circuit equivalent. a b (b) (c) c a b c b are hooked up to the circuit, and the remaining terminal is left hanging. The resistance introduced is determined by that portion of the resistive element between points a and b. In Fig. 3.16(c), the resistance is again between points a and b, but now the remaining resistance is “ shorted-out” (effect removed) by the connection from b to c. The universally accepted symbol for a rheostat appears in Fig. 3.16(d). Most potentiometers have three terminals in the relative positions shown in Fig. 3.17. The knob, dial, or screw in the center of the hous-ing controls the motion of a contact that can move along the resistive element connected between the outer two terminals. The contact is con-nected to the center terminal, establishing a resistance from movable contact to each outer terminal. The internal construction of the potentiometer is provided in Fig. 3.17(b). The resistive element is typically carbon or wire-wound. The center leg (b) is connected to the moveable arm (in blue) through the Philips screw without making contact with the moveable arm. Terminal c is connected directly to the bottom right of the resistive material with-out making contact with the moveable arm. The resistance between terminals c and b is then the major part of the resistive material as shown in Figs. 3.17(b) and (c). Contact is made at the point indicated by the dot in the moveable arm. The resistance between terminals b and a is then the smaller portion of the resistive material as shown in both figures. The contact point has the full range of the resistive material to control the level of resistance between the three points of the potentiometer. The resistance between the outside terminals a and c in Fig. 3.17 is always fixed at the full rated value of the potentiometer, regardless of the position of the wiper arm b. In other words, the resistance between terminals a and c in Fig. 3.18(a) for a Ω 1 M potentiometer will always be Ω 1 M , no matter how we turn the control element and move the contact. In Fig. 3.18(a), the center con-tact is not part of the network configuration. The resistance between the wiper arm and either outside terminal can be varied from a minimum of Ω 0 to a maximum value equal to the full rated value of the potentiometer. (b) + + 1 MV b a c (a) 1 MV b a c + FIG. 3.18 Resistance components of a potentiometer: (a) between outside terminals; (b) between wiper arm and each outside terminal. 104 \| \| \| Resistanc R G In Fig. 3.18(b), the wiper arm has been placed 1/4 of the way down from point a to point c. The resulting resistance between points a and b will therefore be 1/4 of the total, or Ω 250 k (for a Ω 1 M potentiometer), and the resistance between b and c will be 3/4 of the total, or Ω 750 k . The sum of the resistances between the wiper arm and each outside terminal equals the full rated resistance of the potentiometer. This is demonstrated in Fig. 3.18(b), where Ω+ Ω= Ω 250 k 750 k 1 M . Specifically, R R R ac ab bc = + (3.12) Therefore, as the resistance from the wiper arm to one outside contact increases, the resistance between the wiper arm and the other outside terminal must decrease accordingly. For example, if Rab of a Ω 1 k po-tentiometer is Ω 200 , then the resistance Rbc must be Ω 800 . If Rab is further decreased to Ω 50 , then Rbc must increase to Ω 950 , and so on. The molded carbon composition potentiometer is typically applied in networks with smaller power demands, and it ranges in size from Ω 20 to Ω 22 M (maximum values). A miniature trimmer (less than 1/4 in. in diameter) appears in Fig. 3.19(a), and a variety of potentiometers that use a cermet resistive material appear in Fig. 3.19(b). The contact point of the three-point wire-wound resistor in Fig. 3.19(c) can be moved to set the resistance between the three terminals. (a) FIG. 3.19 Variable resistors: (a) 4 mm in. ( ) ≈5 32 trimmer; (b) conductive plastic and cermet elements; (c) three-point wire-wound resistor. ((a) vlabo/ Shutterstock) (b) (c) a Vab c Vbc R b FIG. 3.20 Potentiometer control of voltage levels. When the device is used as a potentiometer, the connections are as shown in Fig. 3.20. It can be used to control the level of V V , , ab bc or both, depending on the application. Additional discussion of the potentiometer in a loaded situation can be found in later chapters. 3.9 COLOR CODING AND STANDARD RESISTOR VALUES A wide variety of resistors, fixed or variable, are large enough to have their resistance in ohms printed on the casing. Some, however, are too small to have numbers printed on them, so a system of color coding is used. For the thin-film resistor, four, five, or six bands may be used. The four-band scheme is described. Later in this section, the purpose of the fifth and sixth bands will be described. COLOR CODING AND STANDARD RESISTOR VALUES \| \| \| 105 R G 1 2 3 4 FIG. 3.21 Color coding for fixed resistors. For the four-band scheme, the bands are always read from the end that has a band closest to it, as shown in Fig. 3.21. The bands are num-bered as shown for reference in the discussion to follow. The first two bands represent the first and second digits, respectively. They are the actual first two numbers that define the numerical value of the resistor. The third band determines the power-of-ten multiplier for the first two digits (actually the number of zeros that follow the second digit for resistors greater than 10 Ω). The fourth band is the manufacturer’s tolerance, which is an indication of the precision by which the resistor was made. If the fourth band is omitted, the tolerance is assumed to be ±20%. The number corresponding to each color is defined in Fig. 3.22. The fourth band will be either ±5% or ±10% as defined by gold and silver, respectively. To remember which color goes with which percent, simply re-member that ±5% resistors cost more and gold is more valuable than silver. Remembering which color goes with each digit takes a bit of practice. In general, the colors start with the very dark shades and move toward the lighter shades. The best way to memorize is to simply repeat over and over that red is 2, yellow is 4, and so on. Simply practice with a friend or a fellow student, and you will learn most of the colors in short order. EXAMPLE 3.11 Find the value of the resistor in Fig. 3.23. Solution: Reading from the band closest to the left edge, we find that the first two colors of brown and red represent the numbers 1 and 2, respectively. The third band is orange, representing the number 3 for the power of the multiplier as follows: × Ω 12 10 3 resulting in a value of Ω 12 k . As indicated above, if Ω 12 k is written as Ω 12,000 , the third band reveals the number of zeros that follow the first two digits. Now for the fourth band of gold, representing a tolerance of 5%: ± To find the range into which the manufacturer has guaranteed the re-sistor will fall, first convert the 5% to a decimal number by moving the decimal point two places to the left: ⇒ 5% 0.05 Then multiply the resistor value by this decimal number: ( ) Ω = Ω 0.05 12 k 600 Finally, add the resulting number to the resistor value to determine the maximum value, and subtract the number to find the minimum value. That is, = Ω+ Ω= Ω Maximum 12,000 600 12.6 k = Ω− Ω= Ω Minimum 12,000 600 11.4 k = 11 4 Ωto 12 6 kΩ Range . k . 0 1 2 3 4 5 6 7 8 9 610% (0.01 multiplier if 3rd band) Black Brown Red Orange Yellow Green Blue Violet Gray White Gold Silver Number Color 65% (0.1 multiplier if 3rd band) FIG. 3.22 Color coding. FIG. 3.23 Example 3.11. 106 \| \| \| Resistanc R G The result is that the manufacturer has guaranteed with the 5% gold band that the resistor will fall in the range just determined. In other words, the manufacturer does not guarantee that the resistor will be ex-actly Ω 12 k , but rather that it will fall in a range as defined above. Using the above procedure, the smallest resistor that can be labeled with the color code is Ω 10 . However, the range can be extended to include resistors from 0.1 Ω to 10 Ω by simply using gold as a multiplier color (third band) to represent 0.1 and using silver to represent 0.01. This is demonstrated in the next example. EXAMPLE 3.12 Find the value of the resistor in Fig. 3.24. Solution: The first two colors are gray and red, representing the numbers 8 and 2, respectively. The third color is gold, representing a multiplier of 0.1. Using the multiplier, we obtain a resistance of ( ) ( ) Ω = Ω 0.1 82 8.2 The fourth band is silver, representing a tolerance of 10%. ± Converting to a decimal number and multiplying through yields ( ) ( ) = Ω = Ω 10% 0.10 and 0.1 8.2 0.82 = Ω+ Ω= Ω Maximum 8.2 0.82 9.02 = Ω− Ω= Ω Minimum 8.2 0.82 7.38 so that Ω Ω = 7 38 to 9 02 Range . . Although it will take some time to learn the numbers associated with each color, it is certainly encouraging to become aware that the same color scheme to represent numbers is used for all the important elements of electrical circuits. Later on, you will find that the numerical value associated with each color is the same for capacitors and inductors. Therefore, once learned, the scheme has repeated areas of application. Some manufacturers prefer to use a five-band color code. In such cases, as shown in the top portion of Fig. 3.25, three digits are provided before the multiplier. The fifth band remains the tolerance indicator. If the manufacturer decides to include the temperature coefficient, a sixth band will appear as shown in the lower portion of Fig. 3.25, with the color indicating the PPM level. For four, five, or six bands, if the tolerance is less than 5%, the fol-lowing colors are used to reflect the % tolerances: brown 1%, red 2%, green 0.5%, blue 0.25%, = ± = ± = ± = ± and violet 0.1%. brown 1%,red 2%,green 0.5%,blue 0.25%, = ± = ± = ± = ± = ± You might expect that resistors would be available for a full range of values such as Ω Ω Ω Ω Ω 10 ,20 ,30 ,40 ,50 , and so on. However, this is not the case, with some typical commercial values as 27 , 56 , Ω Ω and Ω 68 . There is a reason for the chosen values, which is best demon-strated by examining the list of standard values of commercially avail-able resistors in Table 3.5. The values in boldface are the most common and typically available with 5%, 10%, and 20% tolerances. FIG. 3.24 Example 3.12. Temperature coefficient Brown = 100 PPM Red = 50 PPM Orange = 15 PPM Yellow = 25 PPM 1st digit 470 kV 275 V 2nd digit 3rd digit Multiplier Tolerance FIG. 3.25 Five-band color coding for fixed resistors. COLOR CODING AND STANDARD RESISTOR VALUES \| \| \| 107 R G Examining the impact of the tolerance level will help explain the choice of numbers for the commercial values. Take the sequence Ω− Ω− Ω 47 68 100 , which are all available with 20% tolerances. In Fig. 3.26(a), the tolerance band for each has been determined and plotted on a single axis. Note that with this tolerance (which is all that the manu-facturer will guarantee), the full range of resistor values is available from TABLE 3.5 Standard values of commercially available resistors. Ohms Ω ( ) Kilohms kΩ ( ) Megohms MΩ ( ) 0.10 1.0 10 100 1000 10 100 1.0 10.0 0.11 1.1 11 110 1100 11 110 1.1 11.0 0.12 1.2 12 120 1200 12 120 1.2 12.0 0.13 1.3 13 130 1300 13 130 1.3 13.0 0.15 1.5 15 150 1500 15 150 1.5 15.0 0.16 1.6 16 160 1600 16 160 1.6 16.0 0.18 1.8 18 180 1800 18 180 1.8 18.0 0.20 2.0 20 200 2000 20 200 2.0 20.0 0.22 2.2 22 220 2200 22 220 2.2 22.0 0.24 2.4 24 240 2400 24 240 2.4 0.27 2.7 27 270 2700 27 270 2.7 0.30 3.0 30 300 3000 30 300 3.0 0.33 3.3 33 330 3300 33 330 3.3 0.36 3.6 36 360 3600 36 360 3.6 0.39 3.9 39 390 3900 39 390 3.9 0.43 4.3 43 430 4300 43 430 4.3 0.47 4.7 47 470 4700 47 470 4.7 0.51 5.1 51 510 5100 51 510 5.1 0.56 5.6 56 560 5600 56 560 5.6 0.62 6.2 62 620 6200 62 620 6.2 0.68 6.8 68 680 6800 68 680 6.8 0.75 7.5 75 750 7500 75 750 7.5 0.82 8.2 82 820 8200 82 820 8.2 0.91 9.1 91 910 9100 91 910 9.1 37.6 V 56.4 V 80 V 120 V 47 V 68 V 100 V 54.4 V 81.6 V (a) 50.4 V 61.6 V 90 V 110 V 47 V 68 V 100 V 90.2 V 73.8 V (b) 6 10% 100 V 6 10% 47 V 42.3 V 51.7 V 56 V 82 V 61.2 V 74.8 V 6 10% 82 V 6 10% 56 V 6 20% 68 V 6 20% 100 V 6 20% 47 V 6 20% 68 V FIG. 3.26 Guaranteeing the full range of resistor values for the given tolerance: (a) 20%; (b) 10%. 108 \| \| \| Resistanc R G Ω 37.6 to Ω 120 . In other words, the manufacturer is guaranteeing the full range, using the tolerances to fill in the gaps. Dropping to the 10% level introduces the Ω 56 and Ω 82 resistors to fill in the gaps, as shown in Fig. 3.26(b). Dropping to the 5% level would require additional resis-tor values to fill in the gaps. In total, therefore, the resistor values were chosen to ensure that the full range was covered, as determined by the tolerances employed. Of course, if a specific value is desired but is not one of the standard values, combinations of standard values often result in a total resistance very close to the desired level. If this approach is still not satisfactory, a potentiometer can be set to the exact value and then inserted in the network. Throughout the text, you will find that many of the resistor values are not standard values. This was done to reduce the mathematical com-plexity, which might interfere with the learning process. In the problem sections, however, standard values are frequently used to ensure that you start to become familiar with the commercial values available. 3.10 SURFACE MOUNT RESISTORS In general, surface mount resistors, such as appearing in Fig. 3.15(f), are marked in three ways: color coding, three symbols, and two symbols. The color coding is the same as just described earlier in this section for through-hole resistors. The three-symbol approach uses three digits. The first two define the first two digits of the value; the last digit defines the power of the power-of-ten multiplier. For instance: × Ω= Ω 82 820 is 82 10 0 × Ω= Ω= Ω 2 2 222 is 22 10 2200 . k 2 × Ω= Ω 1 010 is 1 10 0 The two-symbol marking uses a letter followed by a number. The letter defines the value as in the following list. Note that all the numbers of the commercially available list of Table 3.5 are included. A 1.0 = B 1.1 = C 1.2 = D 1.3 = E 1.5 = F 1.6 = G 1.8 = H 2 = J 2.2 = K 2.4 = L 2.7 = M 3 = N 3.3 = P 3.6 = Q 3.9 = R 4.3 = S 4.7 = T 5.1 = U 5.6 = V 6.2 = W 6.8 = X 7.5 = Y 8.2 = Z 9.1 = The second symbol is the power of the power-of-ten multiplier. For example, = × Ω= 1 2 kΩ C3 1.2 10 . 3 = × Ω= 5 1 Ω T0 5.1 10 . 0 = × Ω= 91 Ω Z1 9.1 101 Additional symbols may precede or follow the codes and may differ depending on the manufacturer. These may provide information on the internal resistance structure, power rating, surface material, tapping, and tolerance. CONDUCTANCE \| \| \| 109 R G 3.11 CONDUCTANCE By finding the reciprocal of the resistance of a material, we have a mea-sure of how well the material conducts electricity. The quantity is called conductance, has the symbol G, and is measured in siemens (S) (note Fig. 3.27). In equation form, conductance is ( ) = G R 1 siemens, S (3.13) A resistance of Ω 1 M is equivalent to a conductance of − 10 S, 6 and a resistance of Ω 10 is equivalent to a conductance of − 10 S. 1 The larger the conductance, therefore, the less is the resistance and the greater is the conductivity. In equation form, the conductance is determined by G A l S ρ ( ) = (3.14) indicating that increasing the area or decreasing either the length or the resistivity increases the conductance. EXAMPLE 3.13 a. Determine the conductance of a Ω 1 , a Ω 50 k , and a Ω 10 M resistor. b. How does the conductance level change with increase in resistance? Solution: Eq. (3.13): a. Ω = = Ω= G R 1 S 1 : 1 1 1 Ω = = Ω= × Ω= × = − G R 0 02 mS 50 k : 1 1 50 k 1 50 10 0.02 10 S . 3 3 µ Ω = = Ω= × Ω= × = − G R 0 1 S 10 M : 1 1 10 M 1 10 10 0.1 10 S . 6 6 b. The conductance level decreases rapidly with significant increase in resistance levels. EXAMPLE 3.14 What is the relative increase or decrease in conduc-tivity of a conductor if the area is reduced by 30% and the length is increased by 40%? The resistivity is fixed. Solution: Eq. (3.14): G R l A A l 1 1 i i i i i i i i ρ ρ = = = with the subscript i for the initial value. Using the subscript n for the new value, we obtain ρ ρ ρ ( ) = = = = G A l A l A l G 0.70 1.4 0.70 1.4 0.70 1.4 n n n n i i i i i i i and G 0.5G i n = INTERFOTO/Alamy Stock Photo FIG. 3.27 Werner von Siemens. German (Lenthe, Berlin) (1816–92) Electrical Engineer Telegraph Manufacturer, Siemens & Halske AG Developed an electroplating process during a brief stay in prison for acting as a second in a duel be-tween fellow officers of the Prussian army. Inspired by the electronic telegraph invented by Sir Charles Wheatstone in 1817, he improved on the design and with the help of his brother Carl proceeded to lay cable across the Mediterranean and from Europe to India. His inventions included the first self-excited generator, which depended on the residual magnet ism of its electromagnet rather than an inefficient permanent magnet. In 1888 he was raised to the rank of nobility with the addition of von to his name. The current firm of Siemens AG has manufacturing out-lets in some 35 countries and sales offices in some 125 countries. 110 \| \| \| Resistanc R G 3.12 OHMMETERS The ohmmeter is an instrument used to perform the following tasks and several other useful functions: 1. Measure the resistance of individual or combined elements. 2. Detect open-circuit (high-resistance) and short-circuit (low-resistance) situations. 3. Check the continuity of network connections and identify wires of a multilead cable. 4. T est some semiconductor (electronic) devices. For most applications, the ohmmeters used most frequently are the ohmmeter section of a VOM or DMM. The details of the internal circuitry and the method of using the meter will be left primarily for a laboratory exercise. In general, however, the resistance of a resistor can be measured by simply connecting the two leads of the meter across the resistor, as shown in Fig. 3.28. There is no need to be concerned about which lead goes on which end; the result is the same in either case since resistors offer the same resistance to the flow of charge (current) in either direction. If the VOM is used, a switch must be set to the proper resistance range, and a nonlinear scale (usually the top scale of the meter) must be properly read to obtain the resistance value. The DMM also requires choosing the best scale setting for the resistance to be measured, but the result appears as a numerical display, with the proper placement of the decimal point determined by the chosen scale. When measuring the resistance of a single resistor, it is usually best to remove the resistor from the network before making the measurement. If this is difficult or impossible, at least one end of the resistor must not be connected to the network, otherwise the reading may include the effects of the other elements of the system. If the two leads of the meter are touching in the ohmmeter mode, the resulting resistance is zero. A connection can be checked as shown in Fig. 3.29 by simply hooking up the meter to either side of the connection. If the resistance is zero, the connection is secure. If it is other than zero, the connection could be weak; if it is infinite, there is no connection at all. If one wire of a harness or ribbon cable is known, a second can be found as shown in Fig. 3.30. Simply connect the end of the known lead to the end of any other lead. When the ohmmeter indicates zero ohms (or very low resistance), the second lead has been identified. The above procedure can also be used to determine the first known lead by simply connecting the meter to any wire at one end and then touching all the leads at the other end until a zero ohm indication is obtained. Ribbon cables are typically used to electrically interconnect multiple printed cir-cuit board (PCB) functions. V FIG. 3.28 Measuring the resistance of a single element. V FIG. 3.29 Checking the continuity of a connection. V FIG. 3.30 Identifying the leads of a multilead ribbon cable. Preliminary measurements of the condition of some electronic devices such as the diode and the transistor can be made using the ohmmeter. The meter can also be used to identify the terminals of such devices. RESISTANCE: METRIC UNITS \| \| \| 111 R G One important note about the use of any ohmmeter: Never hook up an ohmmeter to a live circuit! The reading will be meaningless, and you may damage the instru-ment. The ohmmeter section of any meter is designed to pass a small sensing current through the resistance to be measured. A large external current could damage the movement and would certainly throw off the calibration of the instrument. In addition: Never store a VOM or a DMM in the resistance mode. If the two leads of the meter touch, the small sensing current could drain the internal battery. VOMs should be stored with the selector switch on the highest voltage range, and the selector switch of DMMs should be in the off position. 3.13 RESISTANCE: METRIC UNITS The design of resistive elements for various areas of application, includ-ing thin-film resistors and integrated circuits, uses metric units for the quantities of Eq. (3.1) introduced in Section 3.2. In SI units, the resistiv-ity would be measured in ohm-meters, the area in square meters, and the length in meters. However, the meter is generally too large a unit of mea-sure for most applications, and so the centimeter is usually employed. The resulting dimensions for Eq. (3.1) are therefore ρ l A : ohm-centimeters : centimeters : square centimeters The units for ρ can be derived from RA l -cm cm -cm 2 ρ = = Ω = Ω The resistivity of a material is actually the resistance of a sample such as that appearing in Fig. 3.31. Table 3.6 provides a list of values of ρ in ohm-centimeters. Note that the area now is expressed in square centi-meters, which can be determined using the basic equation π = A d 4, 2 eliminating the need to work with circular mils, the special unit of mea-sure associated with circular wires. EXAMPLE 3.15 Determine the resistance of 100 ft of #28 copper telephone wire if the diameter is 0.0126 in. Solution: Unit conversions: =                      = l 100 ft 12 in. 1 ft 2.54 cm 1 in. 3048 cm =           = d 0.0126 in. 2.54 cm 1 in. 0.032 cm Therefore, A d 4 3.1416 0.032 cm 4 8.04 10 cm 2 2 4 2 π ( ) ( ) = = = × − ρ ( )( ) = = × Ω × ≅ − − R l A 6 5 Ω 1.723 10 -cm 3048 cm 8.04 10 cm . 6 4 2 A = 1 cm2 l = 1 cm FIG. 3.31 Defining ρ in ohm-centimeters. TABLE 3.6 Resistivity (r) of various materials. Material -cm Ω Silver Copper Gold Aluminum Tungsten Nickel Iron Tantalum Nichrome Tin oxide Carbon × − 1.645 10 6 1 723 10 6 . × − × − 2.443 10 6 × − 2.825 10 6 × − 5.485 10 6 × − 7.811 10 6 × − 12.299 10 6 × − 15.54 10 6 × − 99.72 10 6 × − 250 10 6 × − 3500 10 6 112 \| \| \| Resistanc R G Using the units for circular wires and Table 3.2 for the area of a #28 wire, we find ρ ( )( ) = = Ω ≅ R l A 6 5 Ω 10.37 CM-/ft 100 ft 159.79 CM . EXAMPLE 3.16 Determine the resistance of the thin-film resistor in Fig. 3.32 if the sheet resistance Rs (defined by ρ = R d s ) is Ω 100 . Solution: For deposited materials of the same thickness, the sheet resistance factor is usually employed in the design of thin-film resistors. Eq. (3.1) can be written ρ ρ ρ ( )( ) = = = = R l A l dw d l w R l w s where l is the length of the sample and w is the width. Substituting into the above equation yields ( )( ) = = Ω = R R l w 200 Ω 100 0.6 cm 0.3 cm s as one might expect since = l w 2 . The conversion factor between resistivity in circular mil-ohms per foot and ohm-centimeters is the following: ρ ( ) ( ) ( ) Ω = × × Ω − -cm 1.662 10 value in CM-/ft 7 For example, for copper, ρ = Ω 10.37 CM-/ft: ρ ( ) ( ) Ω = × Ω − -cm 1.662 10 10.37 CM-/ft 7 = × Ω − 1.723 10 -cm 6 as indicated in Table 3.6. The resistivity in an integrated circuit design is typically in ohm- centimeter units, although tables often provide ρ in ohm-meters or microhm-centimeters. Using the conversion technique of Chapter 1, we find that the conversion factor between ohm-centimeters and ohm- meters is the following: [ ] × Ω        = × Ω − − 1.723 10 - cm 1 m 100 cm 1 100 1.723 10 -m 6 6 or the value in ohm-meters is 1/100 the value in ohm-centimeters, and ρ ( ) ( ) ( ) Ω = × Ω -m 1 100 value in -cm (3.15) Similarly, ρ µ ( ) ( ) ( ) Ω = × Ω -cm 10 value in -cm 6 (3.16) For comparison purposes, typical values of ρ in ohm-centimeters for conductors, semiconductors, and insulators are provided in Table 3.7. In particular, note the power-of-ten difference between conductors and insulators( ) 10 21 —a difference of huge proportions. There is a sig-nificant difference in levels of ρ for the list of semiconductors, but the power-of-ten difference between the conductor and insulator levels is at least 10 6 for each of the semiconductors listed. r 0.3 cm 0.6 cm d FIG. 3.32 Thin-film resistor. Example 3.16. THE FOURTH ELEMENT—THE MEMRISTOR \| \| \| 113 R G 3.14 THE FOURTH ELEMENT—THE MEMRISTOR In May 2008, researchers at Hewlett Packard Laboratories led by Dr. Stanley Williams had an amazing announcement—the discovery of the “missing” link in basic electronic circuit theory called a memristor (memory resis-tor), shown in Fig. 3.33. Up to this point the basic passive elements of circuit theory were the resistor, the capacitor, and the inductor, with the last two to be introduced later in this text. The presence of this fourth element was postulated in a seminal 1971 paper in the IEEE Transactions on Circuit Theory by Leon Chua of the University of California at Berkeley. However, it was not until this announcement that the device was actually constructed and found to function as predicted. Many attempts were made to build a memristor through the years, but it was not until work was done at the nanometer scale that success was obtained. It turns out that the smaller the structure, the more prominent is the memristance response. The level of memristance at the nanometer scale is a million times stronger than at the micrometer scale and is almost undetectable at the millimeter scale. However, this property can work to the advantage of current integrated cir-cuit (IC) designs that are already in the nanometer range. The four basic circuit quantities of charge, current, voltage, and magnetic flux can be related in six ways. Three relations derive from the basic elements of the resistor, the capacitor, and the inductor. The resistor provides a direct relationship between current and voltage, the capacitor provides a relationship between charge and voltage, and the inductor provides a relationship between current and magnetic flux. That leaves the relationship between the magnetic flux and the charge moving through an element. Chua sought a device that would define the relationship between magnetic flux and charge similar to that between the voltage and current of a resistor. In general, Chua was looking for a device whose resistance would be a function of how much charge has passed through it. In Chapter 11 the relationship between the movement of charge and the surrounding magnetic field will be described in keeping with the need to find a device relating charge flow and the surrounding magnetic field. The memristor is a device whose resistance increases with in-crease in the flow of charge in one direction and decreases as the flow of charge decreases in the reverse direction. Furthermore, and vastly important, it maintains its new resistance level when the exci-tation has been removed. This behavior in the nanometer range was discovered using the semi-conductor titanium dioxide ( ) TiO , 2 which is a highly resistive material but can be doped with other materials to make it very conductive. In this material, the dopants move in response to an applied electric field and drift in the direction of the resulting current. Starting out with a memristor with dopants only one side and pure TiO 2 on the other, one can apply a biasing voltage to establish a current in the memristor. The TABLE 3.7 Comparing levels of ρ in Ω-cm. Conductor -cm ( ) Ω Semiconductor -cm ( ) Ω Insulator -cm ( ) Ω × − Copper1.723 10 6 Ge 50 In general :1015 Si × 200 10 3 GaAs × 70 10 6 (a) [(a) Copyright © 2014 Hewlett-Packard Development Company, L.P. Reproduced with Permission.] FIG. 3.33 (a) An image of a circuit with 17 memristors captured by an atomic force microscope. Each memristor is composed of two layers of titanium dioxide sandwiched between a lower common wire and its own upper wire. As a voltage is applied across a memristor, the small signal resistance of one of the titanium dioxide layers is changed, which in turn is used as a method to register data. (b) Symbol. (b)
5899	https://arxiv.org/abs/2303.00828	We gratefully acknowledge support from the Simons Foundation, member institutions, and all contributors. Donate math > arXiv:2303.00828 Mathematics > Combinatorics arXiv:2303.00828 (math) [Submitted on 1 Mar 2023] Title:The structure of large sum-free sets in $\mathbb{F}_p^n$ Authors:Leo Versteegen View a PDF of the paper titled The structure of large sum-free sets in $\mathbb{F}_p^n$, by Leo Versteegen View PDF Abstract:A set $A\subset \mathbb{F}_p^n$ is sum-free if $A+A$ does not intersect $A$. If $p\equiv 2 \mod 3$, the maximal size of a sum-free in $\mathbb{F}_p^n$ is known to be $(p^n+p^{n-1})/3$. We show that if a sum-free set $A\subset \mathbb{F}_p^n$ has size at least $p^n/3-p^{n-1}/6+p^{n-2}$, then there exists subspace $V<\mathbb{F}_p^n$ of co-dimension 1 such that $A$ is contained in $(p+1)/3$ cosets of $V$. For $p=5$ specifically, we show the stronger result that every sum-free set of size larger than $1.2\cdot 5^{n-1}$ has this property, thus improving on a recent theorem of Lev. \| \| \| --- \| \| Comments: \| 15 pages \| \| Subjects: \| Combinatorics (math.CO); Number Theory (math.NT) \| \| Cite as: \| arXiv:2303.00828 [math.CO] \| \| \| (or arXiv:2303.00828v1 [math.CO] for this version) \| \| \| arXiv-issued DOI via DataCite \| Submission history From: Leo Versteegen [view email] [v1] Wed, 1 Mar 2023 21:28:18 UTC (19 KB) Full-text links: Access Paper: View a PDF of the paper titled The structure of large sum-free sets in $\mathbb{F}_p^n$, by Leo Versteegen View PDF TeX Source Other Formats view license Current browse context: math.CO < prev \| next > new \| recent \| 2023-03 Change to browse by: math math.NT References & Citations NASA ADS Google Scholar Semantic Scholar export BibTeX citation Loading... BibTeX formatted citation × Data provided by: Bookmark Bibliographic and Citation Tools Bibliographic Explorer (What is the Explorer?) Connected Papers (What is Connected Papers?) Litmaps (What is Litmaps?) scite Smart Citations (What are Smart Citations?) Code, Data and Media Associated with this Article alphaXiv (What is alphaXiv?) CatalyzeX Code Finder for Papers (What is CatalyzeX?) DagsHub (What is DagsHub?) Gotit.pub (What is GotitPub?) Hugging Face (What is Huggingface?) Papers with Code (What is Papers with Code?) ScienceCast (What is ScienceCast?) Demos Replicate (What is Replicate?) Hugging Face Spaces (What is Spaces?) TXYZ.AI (What is TXYZ.AI?) Recommenders and Search Tools Influence Flower (What are Influence Flowers?) CORE Recommender (What is CORE?) Author Venue Institution Topic arXivLabs: experimental projects with community collaborators arXivLabs is a framework that allows collaborators to develop and share new arXiv features directly on our website. Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. arXiv is committed to these values and only works with partners that adhere to them. Have an idea for a project that will add value for arXiv's community? Learn more about arXivLabs. Which authors of this paper are endorsers? \| Disable MathJax (What is MathJax?)

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