Exponents Astronomical distances are written using exponents. (credit: “Our Solar System (Artist's Concept)” by NASA/Jet Propulsion Laboratory-Caltech/Public Domain) Learning Objectives After completing this section, you should be able to: Apply the rules of exponents to simplifying expressions. Sometimes, we look for shorthand when writing or expressing something that simply takes too long. The use of LOL and tl;dr. This shorthand only works if everyone reading the shorthand knows what it stands for. Using exponents is a similar instance. Writing out a long string of a number times itself over and over takes too much time, and eventually one would forget how many of the value has been written or read. For example, 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 . There has to be a shorter and more efficient way to write 8 times itself 1, 2, 3….hmmmm, 19 times. And that’s the role that exponents play in mathematics. They are shorthand for multiplying a number by itself a number of times. Without it, calculations would become a mess and we’d have to write a lot more. Applying the Rules of Exponents to Simplify Expressions Squaring a number is multiplying it by itself, and has that name because it is the area of a square with that side length. Cubing a number is finding the volume of a cube with that length of sides. That’s why we refer to 5 2 as five squared, or 10 3 as ten cubed. Exponents represent that multiplication. Let’s remind ourselves of the terminology associated with exponents and what exponents represent. Suppose you want to multiply a number, let’s label that number a , by itself some number of times. Let’s label the number of times b . We denote that as a b . We say a raised to the b th power. When we write or see 7 5 , we call the 7 the base and we call 5 the exponent . What it represents is 7 multiplied by itself 5 times. This means exponents are used as a shorthand for repeated multiplications, where we write 7 5 = 7 × 7 × 7 × 7 × 7 . We would write 7 5 and say seven to the fifth power. Exponential Notation The definitions of base and exponent make it possible to understand the exponent rules. Product Rule for Exponents The first rule we examine is the product rule, a n a m = a n + m . This rule means that when we multiply a base raised to a power times the same base to another power, the result is the base raised to the sum of the powers. To demonstrate, consider 9 3 × 9 5 . If we apply the product rule to that we get 9 3 × 9 5 = 9 3 + 5 = 9 8 . This can be tested by looking at the multiplications that are represented. The 9 3 is 9 times itself 3 times, while 9 5 is 9 times itself 5 times. Substituting those into 9 3 × 9 5 we see 9 3 × 9 5 = ( 9 × 9 × 9 ) × ( 9 × 9 × 9 × 9 × 9 ) = 9 8 , which is what the formula told us would happen. Caution: The product rule only applies when the bases are the same. If the bases are different, we do not apply this rule. If a number, a , raised to a power, n , is then multiplied by a raised to another power, m , the result is a n a m = a n + m . Using the Product Rule for Exponents If possible, use the product rule to simplify the following: 21 9 × 21 15 5 9 × 8 4 We can apply the product rule to simplify the expression because the bases are the same and we are multiplying. 21 9 × 21 15 = 21 ( 9 + 15 ) = 21 24 Since the bases are not the same (one is 5, the other 8), this cannot be simplified using the product rule for exponents. These rules can be applied to unknowns too. Using the Product Rule for Exponents of Unknowns Use the product rule to simplify a 4 × a 10 . The bases are the same, and we are multiplying, so we apply the multiplication rule to simplify the expression. a 4 × a 10 = a ( 4 + 10 ) = a 14 Quotient Rule for Exponents The next rule we examine is the quotient, or division, rule. When a number, a , raised to a power, n , is divided by a raised to another power, m , then the result is a n a m = a ( n − m ) . This rule means that when we divide a base raised to a power by the same base to another power, the result is the base raised to the difference of the powers. To demonstrate, consider 14 13 14 6 . If we apply the quotient rule to that, we get 14 13 14 6 = 14 13 − 6 = 14 7 . This can be tested by looking at the division that is represented. Remember, 14 13 is 14 multiplied to itself 13 times, while 14 6 is 14 multiplied to itself 6 times. Substituting those into 14 13 14 6 gives the following: 4 13 4 6 = 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 4 × 4 × 4 × 4 × 4 × 4 We see here that there are a LOT of fours to be divided out. = 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 4 × 4 × 4 × 4 × 4 × 4 = 4 × 4 × 4 × 4 × 4 × 4 × 4 1 = 4 × 4 × 4 × 4 × 4 × 4 × 4 What remains is 4 to the 7th power, 4 × 4 × 4 × 4 × 4 × 4 × 4 = 4 7 . All of the work above confirmed what the formula told us would be the result. Caution: The quotient rule only applies when the bases are the same. If the bases are different, we do not apply this rule. Using the Quotient Rule for Exponents Use the quotient rule to simplify 5 19 5 11 . We can apply the quotient rule to simplify the expression since the bases are the same and we are dividing. 5 19 5 11 = 5 ( 19 − 11 ) = 5 8 Product and Quotient Rule for Exponents A natural consequence of the quotient rule is what it means to raise a non-zero number to the zeroth power. Let’s look at the simplification when the exponents are equal. 3 6 3 6 = 3 ( 6 − 6 ) = 3 0 We know that a number divided by itself is 1, so 3 6 3 6 = 1 . From that is must be that 3 6 3 6 = 3 0 = 1 . This provides the rule for a number raised to the power 0: a ≠ 0 . If you have a non-zero number a , then a 0 = 1 . Distributive Rule for Exponents The next rule we look to is a distributive rule for exponents. If you have a product, ( a × b ) , and raise it to an exponent, n , then ( a × b ) n = a n × b n . This means that when we have two numbers multiplied together, and that is raised to a power, it is the same as raising each of the numbers to the same power first, then multiplying. For example, ( 3 × 7 ) 4 = 3 4 × 7 4 . This can be explained using the definition of exponents and multiplying all the factors. ( 3 × 7 ) 4 = ( 3 × 7 ) × ( 3 × 7 ) × ( 3 × 7 ) × ( 3 × 7 ) We may change the order in which numbers are multiplied. This is the commutative property of the real numbers. This can be written as 3 × 3 × 3 × 3 × 7 × 7 × 7 × 7 . Using exponents, that shortens to 3 4 × 7 4 . This also works in the other direction, a n × b n = ( a × b ) n . Read this way, if we have one base raised to an exponent, and another base raised to the same exponent, we can multiply the bases and raise that product to the shared exponent. For instance, 7 8 × 11 8 = ( 7 × 11 ) 8 = 77 8 . Caution: The exponent distributive rule , a n × b n = ( a × b ) n , only works if the exponents are the same . Using the Distributive Rule for Exponents Use the exponent distributive rule to expand ( 6 × 13 ) 7 . Applying the distributive rule to the product, we get ( 6 × 13 ) 7 = 6 7 × 13 7 . Using the Distributive Rule for Exponents Use the exponent distributive rule to expand ( c × d ) 10 . Applying the distributive rule to the product, we get ( c × d ) 10 = c 10 × d 10 . This distribution also works for quotients. A fraction raised to an exponent equals the numerator raised to the exponent divided by the denominator raised to the exponent. For example, ( 3 5 ) 7 = 3 7 5 7 . Demonstrating this is similar to the previous rule. When you have a fraction, a b , raised to an exponent, n , then ( a b ) n = a n b n . Using the Distributive Rule for Exponents with Fractions Use the exponent distributive rule to expand the following: ( 4 9 ) 6 ( 3 b ) 11 Applying the distributive rule to the quotient, we get ( 4 9 ) 6 = 4 6 9 6 . Applying the distributive rule to the quotient, we get ( 3 b ) 11 = 3 11 b 11 . Fraction Raised to a Power Power Rule In the previous two sets of rules, we’ve seen exponents applied to products and quotients. Now we look to exponents applied to other exponents. For example, ( 3 6 ) 4 = 3 ( 6 × 4 ) = 3 24 . This can be explained by examining what the outer exponent does. We raise 3 6 to the fourth power, so we multiply 3 6 by itself 4 times, ( 3 6 ) 4 = 3 6 × 3 6 × 3 6 × 3 6 . Now if we apply the product rule for exponents, this becomes 3 ( 6 + 6 + 6 + 6 ) = 3 24 . If you raise a non-zero base, say a , to an exponent n , and raise that to another exponent, m , you get the base raised to the product of the exponents, which is ( a n ) m = a ( n × m ) . Raising an Exponent to an Exponent Expand the following: ( 6 7 ) 3 ( b 12 ) 4 Using the power rule of exponents, ( 6 7 ) 3 = 6 ( 7 × 3 ) = 6 21 . Using the power rule of exponents, ( b 12 ) 4 = b ( 12 × 4 ) = b 48 . Negative Exponent Rule Up until now, we’ve only looked at positive exponents. The last exponent rule we look at is what negative exponents represent. Recall the quotient rule: a n a m = a ( n + m ) . What would happen if the exponent in the denominator was larger than that in the numerator? For example, 4 5 4 7 . If we apply the quotient rule, we obtain 4 5 4 7 = 4 5 − 7 = 4 − 2 . We need to make sense of that negative exponent. To do so, we can expand the quotient and see what happens: 4 5 4 7 = 4 × 4 × 4 × 4 × 4 4 × 4 × 4 × 4 × 4 × 4 × 4 . When we divide out common factors, only two factors of 4 are left in the denominator, as we see here: 1 4 × 4 . Using exponent notation, this is 1 4 2 . Since 4 − 2 and 1 4 2 represent the same number, 4 5 4 7 , they are equal. This demonstrates how negative exponents are defined. a − n = 1 a n provided that a ≠ 0 . Similarly, 1 a − n = a n . Eliminating Negative Exponents Convert the following to expressions with no negative exponent: 3 4 × 5 − 8 a − 9 × b 5 7 c − 2 Using the negative exponent rule on the 5 − 8 and multiplying, 3 4 × 5 − 8 = 3 4 × 1 5 8 = 3 4 5 8 . Using the negative exponent rule on the a − 9 and multiplying, a − 9 × b 5 = 1 a 9 × b 5 = b 5 a 9 . Begin by rewriting the expression as 7 c − 2 = 7 1 × 1 c − 2 . Apply the negative exponent rule to 1 c − 2 in the expression, which becomes 7 1 × 1 c − 2 = 7 × c 2 , which has no negative exponents. Eliminating Denominators by Using Negative Exponents Use negative exponents to rewrite the following expressions with no denominator: 7 3 13 9 c 4 d 8 Rewrite the expression 7 3 13 9 as 7 3 1 × 1 13 9 . Then use the definition of negative exponents to rewrite the 1 13 9 as 13 − 9 . Last, multiply, yielding 7 3 1 × 1 13 9 = 7 3 × 13 − 9 . Rewrite the expression c 4 d 8 as c 4 1 × 1 d 8 . Then use the definition of negative exponents to rewrite the 1 d 8 as d − 8 . Last, multiply, yielding c 4 1 × 1 d 8 = c 4 × d −8 . The table below shows a summary of the exponent rules from this section. Rule Example In Words Product Rule a n a m = a n + m 8 2 × 8 5 = 8 7 A base raised to a power, times the same based raised to another power, is the base raised to the sum of the powers. Quotient Rule a n a m = a ( n − m ) 11 14 11 12 = 11 12 A base raised to a power, divided by the same based raised to another power, is the base raised to the difference of the powers. Zero Power Rule a 0 = 1 provided that a ≠ 1 412 0 = 1 Any non-zero number raised to the zeroth power equals 1. Distributive Rule, Multiplication ( a × b ) n = a n × b n ( 14 × 31 ) 9 = 14 9 × 31 9 Exponents distribute across multiplication. Distributive Rule, Division ( a b ) n = a n b n ( 62 91 ) 8 = 62 8 91 8 Exponents distribute across division. Power Rule ( a n ) m = a ( n × m ) ( 5 9 ) 15 = 5 135 A base raised to a power, raised to another power, is the base raised to the first power times the second power. Negative Exponent Rule a − n = 1 a n provided that a ≠ 0 6 − 8 = 1 6 8 1 12 7 = 12 − 7 A base raised to a negative exponent is 1 divided by the base raised to the positive power, and vice versa. These rules often occur in tandem with each other, but it requires that you carefully apply the rules. Simplifying Expressions Using Exponent Rules Simplify the following: ( 4 2 × 7 9 3 ) 5 ( 5 a 4 b 9 ) 6 Step 1: To simplify this, we start by distributing the power 5 across the quotient: ( 4 2 × 7 9 3 ) 5 = ( 4 2 × 7 ) 5 ( 9 3 ) 5 Step 2: We distribute the power 5 in the numerator across that multiplication: ( 4 2 × 7 9 3 ) 5 = ( 4 2 × 7 ) ( 9 3 ) 5 5 = ( 4 2 ) 5 × 7 5 ( 9 3 ) 5 Step 3: We apply the power rule where indicated: ( 4 2 × 7 9 3 ) 5 = ( 4 2 × 7 ) ( 9 3 ) 5 5 = ( 4 2 ) 5 × 7 5 ( 9 3 ) 5 = 4 ( 2 × 5 ) × 7 5 9 ( 3 × 5 ) = 4 10 × 7 5 9 15 Step 1: To simplify this, we start by distributing the power 6 across the quotient: ( 5 a 4 b 9 ) 9 = ( 5 × a 4 ) 6 ( b 9 ) 6 Step 2: We distribute the power 5 in the numerator across that multiplication: ( 5 × a 4 ) 6 ( b 9 ) 6 = ( 5 ) 6 × ( a 4 ) 6 ( b 9 ) 6 Step 3: We apply the power rule where indicated: ( 5 ) 6 × ( a 4 ) 6 ( b 9 ) 6 = 5 6 a 24 b 54 Simplifying Expressions with Exponents Check Your Understanding Key Terms base exponent Key Concepts Exponents are used to express multiplying a number by itself a number of times. The number being multiplied by itself is the base. The number of times it is multiplied by itself is the exponent, which is often referred to as the power. Understanding that exponents represent repeated multiplication of a base makes it possible to establish some rules for combining exponential expressions, using the product rule, the quotient rule, and the power rule. Additionally, it allows us to formulate distributive rules for exponents. Any non-zero number raised to the 0th power is 1. This makes the definition of the 0th power consistent with the division rule for exponents. For consistency, negative exponents represent the reciprocal of the base raised to the power, so that a − n = 1 a n , provided that a ≠ 0 . Formulas a n a m = a n + m a n a m = a ( n − m ) a 0 = 1 , provided that a ≠ 0 ( a × b ) n = a n × b n ( a b ) n = a n b n ( a n ) m = a ( n × m ) a − n = 1 a n , provided that a ≠ 0 Videos Exponential Notation Product and Quotient Rule for Exponents Fraction Raised to a Power Simplifying Expressions with Exponents

Math


Scientific Notation Calculations in the sciences often involve numbers in scientific notation form. Learning Objectives After completing this section, you should be able to: Write numbers in standard or scientific notation. Convert numbers between standard and scientific notation. Add and subtract numbers in scientific notation. Multiply and divide numbers in scientific notation. Use scientific notation in computing real-world applications. The amount of information available on the Internet is simply incomprehensible. One estimate for the amount of data that will be on the Internet by 2025 is 175 Zettabytes. A single zettabyte is one billion trillion. Written out, it is 1,000,000,000,000,000,000,000. One estimate is that we’re producing 2.5 quintillion bytes of data per day. A quintillion is a trillion trillion, or, written out, 1,000,000,000,000,000,000. To determine how many days it takes to increase the amount of information that is on the Internet by 1 zettabyte, divide these two numbers, a zettabyte being 1,000,000,000,000,000,000,000, and 2.5 quintillion, being 2,500,000,000,000,000,000, shows it takes 400 days to generate 1 zettabyte of information. But doing that calculation is awkward with a calculator. Keeping track of the zeros can be tedious, and a mistake can easily be made. On the other end of the scale, a human red blood cell has a diameter of 7.8 micrometers. One micrometer is one millionth of a meter. Written out, 7.8 micrometers is 0.0000078 meters. Smaller still is the diameter of a virus, which is about 100 nanometers in diameter, where a nanometer is a billionth of a meter. Written out, 100 nanometers is 0.0000001 meters. To compare that to engineered items, a single transistor in a computer chip can be 14 nanometers in size (0.000000014 meters). Smaller yet is the diameter of an atom, at between 0.1 and 0.5 nanometers. Sometimes we have numbers that are incredibly big, and so have an incredibly large number of digits, or sometimes numbers are incredibly small, where they have a large number of digits after the decimal. But using those representations of the names of the sizes makes comparing and computing with these numbers problematic. That’s where scientific notation comes in. Writing Numbers in Standard or Scientific Notation Form When we say that a number is in scientific notation , we are specifying the form in which that number is written. That form begins with an integer with an absolute value between 1 and 9, then perhaps followed the decimal point and then some more digits. This is then multiplied by 10 raised to some power. When the number only has one non-zero digit, the scientific notation form is the digit multiplied by 10 raised to an exponent. When the number has more than one non-zero digit, the scientific notation form is a single digit, followed by a decimal, which is then followed by the remaining digits, which is then multiplied by 10 to a power. The following numbers are written in scientific notation: 1.45 × 10 3 − 8.345 × 10 − 4 3 × 10 2 3.14159 × 10 0 The following numbers are not written in scientific notation: 1.45 because it isn't multiplied by 10 raised to a power − 50.053 × 10 7 because the absolute value of −50.053 is not at least 1 and less than 10 41.7 × 10 9 because 41.7 is not at least 1 and less than 10 0.036 × 10 − 3 because 0.036 is not at least one and less than 10 Identifying Numbers in Scientific Notation Which of the following numbers are in scientific notation? If the number is not in scientific notation, explain why it is not. − 9.67 × 10 20 145 × 10 − 8 1.45 The number − 9.67 × 10 20 is in scientific notation because the absolute value of −9.67 is at least 1 and less than 10. The number 145 × 10 − 8 is not in scientific notation because 145 is not at least 1 and less than 10. The number 1.45 is not in scientific notation form. Even though it is at least 1 but less than 10, it is not multiplied by 10 raised to a power. Some numbers are so large or so small that it is impractical to write them out fully. Avogadro’s number is important in chemistry. It represents the number of units in 1 mole of any substance. The substance many be electrons, atoms, molecules, or something else. Written out, the number is: 602,214,076,000,000,000,000,000. Another example of a number that is impractical to write out fully is the length of a light wave. The wavelength of the color blue is about 0.000000450 to 0.000000495 meters. Such numbers are awkward to work with, and so scientific notation is often used. We need to discuss how to convert numbers into scientific notation, and also out of scientific notation. Recall that multiplying a number by 10 adds a 0 to the end of the number or moves the decimal one place to the right, as in 43 × 10 = 430 or 3.89 × 10 = 38.9 . And if you multiply by 100, it adds two zeros to the end of the number or moves the decimal two places to the right, and so on. For example, 38 × 100,000 = 3,800,000 and 32.998 × 10,000 = 329,980 . Multiplying a number by 1 followed by some number of zeros just adds that many zeros to the end of the number or moves the decimal place that many places to the right. Numbers written as 1 followed by some zeros are just powers of 10, as in 10 1 = 10 , 10 2 = 100 , 10 3 = 1,000 , etc. Generally, 10 n = 1 0...0 ︸ n . We can use this to write very large numbers. For instance, Avogadro’s number is 602,214,076,000,000,000,000,000, which can be written as 6.02214076 × 10 23 . The multiplication moves the decimal 23 places to the right. Similarly, when we divide by 10, we move the decimal one place to the left, as in 46.7 10 = 4.67 . If we divide by 100, we move the decimal two places to the left, as in 3.456 100 = 0.03456 . In general, when you divide a number by a 1 followed by n zeros, you move the decimal n places to the left, as in 8,244.902 1,000,000 = 0.008244902 . This denominator could be written as 10 6 . If we use that in the expression and allow for negative exponents, rewrite the number as 8,244.902 1,000,000 = 8,244.902 10 6 = 8,244.902 × 10 − 6 . With this, we can write division by a 1 followed by n zeros as multiplication by 10 raised to ‒ n . Using that information, we can demonstrate how to convert from a number in standard form into scientific notation form. Case 1: The number is a single-digit integer. In this case, the scientific notation form of the number is d i g i t × 10 1 . Case 2: The absolute value of the number is less than 1. Follow the process below. Step 1: Count the number of zeros between the decimal and the first non-zero digit. Label this n . Step 2: Starting with the first non-zero digit of the number, write the digits. If the number was negative, include the negative sign. Step 3: If there is more than one digit, place the decimal after the first digit from Step 2. Step 4: Multiply the number from Step 3 by 10 n + 1 . Case 3: The absolute value of the number is 10 or larger. Follow the process below. Step 1: Count the number of digits that are to the left of the decimal point. Label this n . Step 2: Write the digits of the number without the decimal place, if one was present. If the number was negative, include the negative sign. Step 3: If there is more than one digit, place the decimal point after the first digit. Step 4: Multiply the number from Step 3 by 10 n − 1 . Writing a Number in Scientific Notation Write the following numbers in scientific notation form: 428.9 −0.00000981 8 Since the absolute value of 428.9 is 10 or larger, so we use the process from Case 3, above. Step 1: There are three digits to the left of the decimal point, so n = 3 . Step 2: Write the digits of the number without the decimal place, which is 4289. Step 3: Since there is more than one digit, place the decimal point after the first digit. We now have 4.289. Step 4: Since n = 3 , we multiply 4.289 by 10 raised to the second power, 4.289 × 10 2 . The scientific notation form of 428.9 is 4.289 × 10 2 . Since the absolute value of −0.00000981 is less than 1, we use the process from Case 2. Step 1: The number of zeros between the decimal and the first non-zero digit is 5, so n = 5 . Step 2: We write the non-zero digits, including the negative sign, yielding −981. Step 3: The decimal gets placed to the right of the first digit, resulting in −9.81. Step 4: Since n = 5 , we multiply −9.81 by 10 raised to the fourth power, − 9.81 × 10 − 6 . The scientific notation form of −0.00000981 is 9.81 × 10 − 6 . Since 8 is a single-digit integer, apply Case 1. The scientific notation form of 8 is 8 × 10 1 . When we write numbers in scientific notation form, we can manipulate the representation of the number by moving the decimal around, and making an appropriate change to the exponent of the 10. For instance, let’s look at 145.8141 × 10 8 . If we wanted to move the decimal one place to the left, we’d have to increase the power of 10, as shown here: 145.8141 × 10 8 = 14.58141 × 10 9 . Since we moved the decimal one to the left, we balance that with moving the exponent up by one. Similarly, if we move the decimal one place to the right, we have to balance that by moving the exponent one to the left, or subtracting one from the exponent, as shown here: 145.8141 × 10 8 = 1458.141 × 10 7 . Generally, for a number in the form n u m b e r × 10 n : If you move the decimal to the left by k digits, you increase the exponent by k . If you move the decimal to the right by k digits, you decrease the exponent by k digits. Increasing the Exponent Change 456.142 × 10 5 by moving the decimal two places to the left. Since we are moving the decimal to the left by two places, we increase the exponent of 10 by 2, so that the exponent is now 7. This gives us 456.142 × 10 5 = 4.56142 × 10 7 . Decreasing the Exponent Change 12.3 × 10 2 by moving the decimal five places to the right. Since we are moving the decimal to the right by five places, we decrease the exponent of 10 by 5, so that exponent is now −3. This give us 12.3 × 10 2 = 1230000.0 × 10 − 3 . Converting Numbers from Scientific Notation to Standard Form In the previous section, converting a number from standard form to scientific notation was explored. Now, we explore converting from scientific notation back into standard form. Doing so involves moving the decimal according to the power of the 10. The decimal is moved a number of steps equal to the exponent of the 10. As demonstrated previously, when the exponent of the 10 is negative, the decimal is moved to the left and when the exponent of the 10 is positive, the decimal is moved to the right. Converting from Scientific Notation to Standard Form Convert the following into standard form: 2.78 × 10 9 9.04 × 10 − 8 Since the exponent is positive, the decimal moves nine places to the right, so 2.78 × 10 9 is 2,780,000,000 . Since the exponent is negative, the decimal moves eight places to the left, so 9.04 × 10 − 8 is 0.0000000904 . Converting from Standard Form to Scientific Notation Form Converting from Scientific Notation Form to Standard Form Scientific Notation on a Calculator Most scientific and graphing calculators come with the ability to directly convert from standard form to scientific notation. On the TI-83, it is accessed through the MODE menus. For a commonly used, free phone scientific calculator, the calculator can be forced to work in scientific notation mode through its settings. Some calculators, such as the Desmos online calculator, display scientific notation as a number times 10 to a power as you’ve seen in this section. However, some calculators indicate scientific notation by replacing the × 10 n with an E (or EE) followed by the exponent. For example, shows what you may see on a TI-84. Calculator screens Adding and Subtracting Numbers in Scientific Notation To add or subtract numbers in scientific notation, the numbers first need to have the same exponent for the 10s. It is possible to add the following since the powers of 10 match: 4.5 × 10 4 + 3.15 × 10 4 = 7.65 × 10 4 Notice that the number parts were added, but the exponent part remained the same. This is due to the distributive property of the real numbers. The 10 4 is factored from the two terms, as shown: 4.5 × 10 4 + 3.15 × 10 4 = ( 4.5 + 3.15 ) × 10 4 = 7.65 × 10 4 Numbers in scientific notation can be added or subtracted directly using a calculator. Simply enter the values in scientific form and set your calculator to display scientific notation. Adding and Subtracting Numbers in Scientific Notation with the Same Powers of 10 Calculate the following: 3.8 × 10 − 3 + 1.006 × 10 − 3 9.61 × 10 8 − 3.85 × 10 8 Since the powers of 10 match, we use the distributive property of real numbers to factor 10 −3 from the numbers. We then add the number parts separately to get 4.806. 3.8 × 10 − 3 + 1.006 × 10 − 3 = ( 3.8 + 1.006 ) × 10 − 3 = 4.806 × 10 − 3 Since the powers of 10 match, we use the distributive property of real numbers to factor 10 8 from the numbers. We then subtract the number parts separately to get 5.76. 9.61 × 10 8 − 3.85 × 10 8 = ( 9.61 − 3.85 ) × 10 8 = 5.76 × 10 8 Adding and subtracting in scientific notation is straightforward when the exponents are the same. There are two issues that can arise. The first issue is what to do if after adding or subtracting the result is not in scientific notation. Correcting an Answer to Scientific Notation After Adding or Subtracting Calculate the following: 7.03 × 10 13 + 8.5 × 10 13 4.3 × 10 21 − 4.613 × 10 21 Since the powers of 10 match, we add the number parts and multiply that by 10 13 : 7.03 × 10 13 + 8.5 × 10 13 = ( 7.03 + 8.5 ) × 10 13 = 15.53 × 10 13 . However, 15.53 × 10 13 is not in scientific notation because the absolute value of 15.53 is more than 10. To put this number in scientific notation, the decimal needs to move one to the left. To balance that move, the power of 10 must be increased by 1. So, the answer in scientific notation is 1.553 × 10 14 . Since the powers of 10 match, we add the number parts: 4.3 × 10 21 − 4.613 × 10 21 = ( 4.3 − 4.613 ) × 10 21 = − 0.313 × 10 21 However, − 0.313 × 10 21 is not in scientific notation because it is less than 1. To put it in scientific notation, the decimal needs to move one to the right. To balance that move, the power of 10 must be decreased by 1. So, the answer in scientific notation is − 3.13 × 10 20 . The second issue that might be encountered when adding or subtracting is that the powers of 10 do not match. In that case, one of the numbers must be changed so that the powers of 10 match. It is easiest to make the smaller power of 10 larger to match the other power of 10. For example, to perform the following, 4.5 × 10 5 + 3.9 × 10 3 , we’d change the 3.9 × 10 3 so that the power of 10 is 5. To do so, we need to increase the power of 10 and move the decimal in the number part two places to the left. That would alter 3.9 × 10 3 into 0.039 × 10 5 . We would use 0.039 × 10 5 in the addition problem, so that the exponents match, allowing the addition to occur. 4.5 × 10 5 + 3.9 × 10 3 = 4.5 × 10 5 + 0.039 × 10 5 = ( 4.5 + 0.039 ) × 10 5 = 4.539 × 10 5 The steps to take when the exponents of the 10s are not equal are: Step 1: Increase the smaller exponent to equal the larger exponent. Label the amount increased as n . Step 2: For the number with the smaller power of 10, move the decimal point of the number part to the left n places. Step 3: Perform the addition or subtraction. Step 4: If the result is not in scientific notation, adjust the number to be in scientific notation. Adding Numbers in Scientific Notation with Different Powers of 10 Calculate the following: 6.1 × 10 4 + 4.8 × 10 5 Step 1: The lower exponent is 4. To make this equal to the larger exponent, we increased it by 1. Step 2: Since the smaller exponent was increased by 1, move the decimal one to the left, so the addition become 6.1 × 10 4 + 4.8 × 10 5 = 0.61 × 10 5 + 4.8 × 10 5 . Step 3: Now add the numbers, 0.61 × 10 5 + 4.8 × 10 5 = 5.41 × 10 5 Step 4: The result is in scientific notation, so no additional adjustment is necessary. 6.1 × 10 4 + 4.8 × 10 5 = 5.41 × 10 5 Subtracting Numbers in Scientific Notation with Different Powers of 10 Calculate the following: 7.9 × 10 − 15 − 6.8 × 10 − 13 Step 1: The lower exponent is −15 and the larger is −13. To make −15 equal to the larger exponent, we increased it by 2. Step 2: Since the smaller exponent increased by 2, move the decimal two to the left. The subtraction changes to 7.9 × 10 − 15 − 6.8 × 10 − 13 = 0.079 × 10 − 13 − 6.8 × 10 − 13 . Step 3: Subtract the numbers, 0.079 × 10 − 13 − 6.8 × 10 − 13 = − 6.721 × 10 − 13 . Step 4: The result is in scientific notation, so no additional adjustment is necessary. 7.9 × 10 − 15 − 6.8 × 10 − 13 = − 6.721 × 10 − 13 Multiplying and Dividing Numbers in Scientific Notation Multiplying and dividing numbers in scientific notation is somewhat easier than adding or subtracting, because the exponents of the 10s do not have to match. However, it is much more likely that the result will not be in scientific notation, and so that will have to be adjusted at the end. Generally, we multiply or divide the number parts of the two values, and then apply exponent rules to the 10 raised to the powers. To multiply two numbers in scientific notation: Step 1: Multiply the number parts. Step 2: Add the exponents of the 10s. Step 3: The result is the answer from Step 1 times 10 raised to the answer from Step 2. Step 4: If the number is not in scientific notation, adjust it appropriately. Multiplying Numbers in Scientific Notation Calculate the following: ( 4.3 × 10 3 ) × ( 1.8 × 10 7 ) ( 5 × 10 − 1 3 ) × ( 7.3 × 10 6 ) Step 1: Multiply the number parts to get 4.3 × 1.8 = 7.74 . Step 2: Add the exponents of the 10s to get 3 + 7 = 10 . Step 3: The result is then 7.74 × 10 10 . Step 4: This number is already in scientific notation, so no additional adjustment is necessary, ( 4.3 × 10 3 ) × ( 1.8 × 10 7 ) = 7.74 × 10 10 . Step 1: Multiply the number parts to get 5 × 7.3 = 36.5 . Step 2: Add the exponents of the 10s to get − 13 + 6 = − 7 . Step 3: The result then is 36.5 × 10 − 7 . Step 4: Since the number is not in scientific notation, it must be adjusted. To put 36.5 × 10 − 7 into scientific notation, the decimal moves one to the left, so the exponent would be increased by 1, giving 3.65 × 10 − 6 . ( 5 × 10 − 13 ) × ( 7.3 × 10 6 ) = 3.65 × 10 − 6 Multiplying Numbers in Scientific Notation Dividing Numbers in Scientific Notation To divide two numbers that are in scientific notation: Step 1: Divide the number parts. Step 2: Subtract the exponent of the denominator from the exponent of the numerator. Step 3: The answer is the result from Step 1 times 10 raised to the result from Step 2. Step 4: If the number is not in scientific notation, adjust it appropriately. Dividing Numbers in Scientific Notation Calculate the following: ( 8.4 × 10 31 ) / ( 2.1 × 10 7 ) ( 4.14 × 10 − 13 ) / ( 8.28 × 10 9 ) ( 8.4 × 10 31 ) / ( 2.1 × 10 7 ) Step 1: Divide the number parts to get 8.4 ÷ 2.1 = 4 . Step 2: Subtract the exponent of the denominator from the exponent of the numerator to get 37 − 7 = 24 . Step 3: The result is then 4 × 10 24 . Step 4: This number is already in scientific notation, so no adjustment is necessary. ( 8.4 × 10 31 ) / ( 2.1 × 10 7 ) = 4 × 10 24 ( 4.14 × 10 − 13 ) / ( 8.28 × 10 9 ) Step 1: Divide the number parts to get 4.14 ÷ 8.28 = 0.5 . Step 2: Subtract the exponent of the denominator from the exponent of the numerator to get − 13 − 9 = − 22 . Step 3: The result then is 0.5 × 10 − 22 . Step 4: Since this number is not in scientific notation, it must be adjusted. To put 0.5 × 10 − 22 into scientific notation, the decimal needs to move one to the right, so the exponent is decreased by 1, giving 5 × 10 − 23 . ( 4.14 × 10 − 13 ) / ( 8.28 × 10 9 ) = 5 × 10 − 23 Dividing Numbers in Scientific Notation Using Scientific Notation in Computing Real-World Applications As noted at the start of this section, scientific notation is useful when the standard representation of a number is awkward or impractical, which occurs when the numbers being used are extremely large or extremely small. For example, Venus is 67,667,000 miles from the sun. In scientific notation, this is 6.7667 × 10 7 . Planetary and galaxy distances is one set of numbers that is easier to express using scientific notation. Calculating Distances How much farther from the sun is Earth compared to Venus if Venus is 6.7667 × 10 7 miles from the sun and Earth is 9.1692 × 10 7 miles from the sun? To determine how much farther Earth is compared to Venus, we’d subtract the distances. 9.1692 × 10 7 − 6.7667 × 10 7 = 2.4025 × 10 7 . So, Earth is 2.4025 × 10 7 miles farther from the sun than Venus. Calculating Probability The probability of winning the Mega Millions lottery is published as 3.304693 × 10 − 9 . The probability of being hit by lightning is approximated to be 2 × 10 − 6 . How many times more likely are you to be hit by lightning than win the Mega Millions? To find out how many times more likely you are to be hit by lightning, divide the probability of being hit by lightning by the probability of winning the Mega Millions. ( 2 × 10 − 6 ) / ( 3.304693 × 10 − 9 ) = 6.609386 × 10 3 Step 1: Divide the number parts to get = 0.6052 (rounded to the fourth digit). Step 2: Subtract the exponent of the denominator from the exponent of the numerator to get − 6 − ( − 9 ) = 3 . Step 3: The result then is 0.6052 × 10 3 . Step 4: Since this number is not in scientific notation, it must be adjusted. To put 0.6052 × 10 3 into scientific notation, the decimal needs to move one place to the right, so the exponent is decreased by 1, giving 6.052 × 10 2 . You are 6.052 × 10 2 , or 605.2, times more likely to be hit by lightning than you are to win the Mega Millions. Calculating Time and Length Sometimes it is entertaining to determine the time it takes for something to happen. Fingernails grow about 8.032 × 10 − 11 km per minute. How many kilometers long would fingernails be after 6 × 10 4 minutes? To find the length of the fingernails after the specified time, we multiply their rate of growth and the time they’ve grown. ( 8.032 × 10 − 11 ) × ( 6 × 10 4 ) = 48.192 × 10 − 7 = 4.8192 × 10 − 6 So, after 6 × 10 4 minutes, the fingernails would be 4.8192 × 10 − 6 km long. To put this in perspective, 1 × 10 − 6 km is a millimeter, and 6 × 10 4 minutes is about 4.16 days. So, after about 4.16 days, fingernails have grown about 4.8 millimeters. Calculating Data Generated As mentioned in the opening to this section, it is estimated that we’re producing 2.5 quintillion bytes of data per day. A good estimate is that there are 7.674 billion people on the planet. Convert both of those numbers to scientific notation, and then determine how much data is being generated per person each day. Written in standard form, 2.5 quintillion is 2,500,000,000,000,000,000. Changing that to scientific notation, move the decimal 18 places, so 2.5 quintillion bytes = 2.5 × 10 18 bytes. Writing 7.647 billion in scientific notation would be 7.647 × 10 9 because a billion is 1,000,000,000 = 10 9 . So, to find out how much data is being produced daily per person, we would divide these two numbers. 2.5 × 10 18 7.647 × 10 9 = 0.327 × 10 9 = 3.27 × 10 8 In standard form, that’s 327,000,000 bytes per person, so 327 million bytes of data daily are being produced per person. Application of Scientific Notation What Numbers Could Be Considered “Too Big” or “Too Small”? One wonders when the numbers we represent become too large or small for consideration. Perhaps the following examples put limits on what is meaningful. The number of particles in the known universe has been estimated at 4 × 10 80 particles. The smallest distance that has been measured is 1 × 10 − 18 m , though the theoretical smallest measurable value is 1 × 10 − 35 m . The distance across the universe is 4.4 × 10 26 m . Considering what those numbers represent, the extreme largest and extreme smallest, they might be numbers that constrain what we should reasonably be expected to deal with. Check Your Understanding Key Terms scientific notation standard notation Key Concepts Some numbers are so large or so small that writing the number out is clumsy and make it difficult to determine the true size of the number. Scientific notation makes the number more readable and make the relative size of the number immediately apparent. A number written in scientific notation is a number at least 1 and smaller than 10 multiplied by 10 raised to an exponent. Converting between scientific notation and standard notation involves correctly applying multiplication and division by powers of 10, which in practice equates to understanding how moving the decimal point of a number impacts the exponent of 10. Adding and subtracting numbers in base 10 requires the exponent of 10 in each number be the same. Once the numbers are converted to have the same exponent with the ten, then the numbers are added or subtracted as indicated, with the power of 10 remaining the same. If the result is not in scientific notation (for instance, the number has exceeded 10), then then number must be converted into scientific notation. Multiplying and dividing numbers in scientific notation is done by multiplying or dividing the number parts, then multiplying or dividing the 10 raised to the power parts, then multiplying those two results. If the new number is not in scientific notation, then the result must be converted into scientific notation. Videos Converting from Standard Form to Scientific Notation Form Converting from Scientific Notation Form to Standard Form Multiplying Numbers in Scientific Notation Dividing Numbers in Scientific Notation Application of Scientific Notation

Math


Arithmetic Sequences Learning Objectives After completing this section, you should be able to: Identify arithmetic sequences. Find a given term in an arithmetic sequence. Find the n th term of an arithmetic sequence. Find the sum of a finite arithmetic sequence. Use arithmetic sequences to solve real- world applications As we saw in the previous section, we are adding about 2.5 quintillion bytes of data per day to the Internet. If there are 550 quintillion bytes of data today, then there will be 552.5 quintillion bytes tomorrow, and 555 quintillion bytes in 2 days. This is an example of an arithmetic sequence. There are many situations where this concept of fixed increases comes into play, such as raises or table arrangements. Identifying Arithmetic Sequences A sequence of numbers is just that, a list of numbers in order. It can be a short list, such as the number of points earned on each assignment in a class, such as {10, 10, 8, 9, 10, 6, 10}. Or it can be a longer list, even infinitely long, such as the list of prime numbers. For example, here’s a sequence of numbers, specifically, the squares of the first 12 natural numbers. {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144} Each value in the sequence is called a term . Terms in the list are often referred to by their location in the sequence, as in the n th term. For the sequence {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}, the first term of the sequence is 1, the fourth term is 16, and so on. In the sequence of assignment scores {10, 10, 8, 9, 10, 6, 10}, the first term is 10 and the third term is 8 ( ). Sequence showing first, second, and fifth terms The notation we use with sequences is a letter, which represents a term in the sequence, and a subscript, which indicates what place the term is in the sequence. For the sequence {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}, we will use the letter a as a value in the sequence, and so a 5 would be the term in the sequence at the fifth position. That number is 25, so we can write a 5 = 25 . In this section, we focus on a special kind of sequence, one referred to as an arithmetic sequence . Arithmetic sequences have terms that increase by a fixed number or decrease by a fixed number, called the constant difference (denoted by d ), provided that value is not 0. This means the next term is always the previous term plus or minus a specified, constant value. Another way to say this is that the difference between any consecutive terms of the sequence is always the same value. To see a constant difference, look at the following sequence: {7, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87}. illustrates that each term of the sequence is the previous term plus 8. Eight is the constant difference here. Sequence of numbers with 8 added to each term Identifying Arithmetic Sequences Determine if the following sequences are arithmetic sequences. Explain your reasoning. { 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , ... } { 20 , 40 , 80 , 160 , 320 , 640 } { 7 , 1 , − 5 , − 11 , − 17 , − 23 , − 29 , − 34 , − 40 } In the sequence { 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , ... } , every term is the previous term plus 3. The ellipsis indicates that the pattern continues, which means keep adding 3 to the previous term to get the new term. Therefore, this is an infinite arithmetic sequence. In the sequence { 20 , 40 , 80 , 160 , 320 , 640 } , terms increase by various amounts, for instance from term 1 to term 2, the sequence increases by 20, but from term 2 to term 3 the sequence increases by 40. So, this is not an arithmetic sequence. In the sequence { 7 , 1 , − 5 , − 11 , − 17 , − 23 , − 29 , − 34 , − 40 } , every term is the previous term minus 6, so this is an arithmetic sequence. Arithmetic sequences can be expressed with a formula. When we know the first term of an arithmetic sequence, which we label a 1 , and we know the constant difference, which is denoted d , we can find any other term of the arithmetic sequence. The formula for the i th term of an arithmetic sequence is a i = a 1 + d × ( i − 1 ) . If we have an arithmetic sequence with first term a 1 and constant difference d , then the i th term of the arithmetic sequence is a i = a 1 + d × ( i − 1 ) . Let’s examine the formula with this arithmetic sequence: { 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , ... } . In this sequence a 1 = 4 and d = 3 . The table below shows the values calculated. i , Place in Sequence a i , i t h Term Value of Term Term Written as a 1 + 3 × ( i − 1 ) 1 a 1 4 4 + 3 × 0 2 a 2 7 4 + 3 × 1 3 a 3 10 4 + 3 × 2 4 a 4 13 4 + 3 × 3 5 a 5 16 4 + 3 × 4 i a i 4 + 3 × ( i − 1 ) We can see how the i th term can be directly calculated. In this sequence, the formula is a 1 + 3 × ( i − 1 ) where the first term, a 1 , is 4 and the constant difference d is 3. We can then determine the 47 th term of this sequence: a 47 = 4 + 3 × ( 47 − 1 ) = 4 + 3 × 46 = 4 + 138 = 142 . Calculating a Term in an Arithmetic Sequence Identify a 1 and d for the following arithmetic sequence. Use this information to determine the 60 th term. { 18 , 31 , 44 , 57 , 70 , 83 , ... } Inspecting the sequence shows that a 1 = 18 and d = 13 . We use those values in the formula, with i = 60 . a 60 = a 1 + d × ( i − 1 ) = 18 + 13 × ( 60 − 1 ) = 18 + 13 × 59 = 18 + 767 = 785 Arithmetic Sequences If we know two terms of the sequence, it is possible to determine the general form of an arithmetic sequence, a i = a 1 + d × ( i − 1 ) . If we have the i th term of an arithmetic sequence, a i , and the j th term of the sequence, a j , then the constant difference is d = a j − a i j − i and the first term of the sequence is a 1 = a i − d ( i − 1 ) . Determining First Term and Constant Difference Using Two Terms A sequence is known to be arithmetic. Two of its terms are a 7 = 56 and a 19 = 104 . Use that information to find the constant difference, the first term, and then the 50 th term of the sequence. To find the constant difference, use d = a j − a i j − i . The location of the terms is given by the subscript of the two a terms, i = 7 and j = 19 . So, the constant difference can be calculated as such: d = 104 − 56 19 − 7 = 48 12 = 4 . The constant difference of 4 is then used to find a 1 . a 1 = a i − d ( i − 1 ) = a 7 − 4 ( 7 − 1 ) = 56 − 4 × 6 = 32 . So d = 4 and a 1 = 32 . With this information, the 50 th term can be found. a 50 = a 1 + d × ( i − 1 ) = 32 + 4 × ( 50 − 1 ) = 32 + 4 × 49 = 32 + 196 = 228 . The 50 th term is a 50 = 228 . Finding the First Term and Constant Difference for an Arithmetic Sequence Finding the Sum of a Finite Arithmetic Sequence Sometimes we want to determine the sum of the numbers of a finite arithmetic sequence. The formula for this is fairly straightforward. The sum of the first n terms of a finite arithmetic sequence, written s n , with first and last term a 1 and a n , respectively, is s n = n ( a 1 + a n 2 ) . Finding the Sum of a Finite Arithmetic Sequence What is the sum of the first 60 terms of an arithmetic sequence with a 1 = 4.5 and d = 2.5 ? The formula requires the first and last terms of the sequence. The first term is given, a 1 = 4.5 . The 60 th term is needed. Using the formula a 1 = a i + d ( i − 1 ) provides the value for the 60 th term. a 60 = 4.5 + 2.5 ( 60 − 1 ) = 4.5 + 2.5 × 59 = 4.5 + 147.5 = 152 . Applying the formula s n = n ( a 1 + a n 2 ) provides the sum of the first 60 terms. s 60 = 60 ( 4.5 + 152 2 ) = 60 × 156.5 2 = 4,695 . The sum of the first 60 terms is 4,695. Finding the Sum of a Finite Arithmetic Sequence Using Arithmetic Sequences to Solve Real-World Applications Applications of arithmetic sequences occur any time some quantity increases by a fixed amount at each step. For instance, suppose someone practices chess each week and increases the amount of time they study each week. The first week the person practices for 3 hours, and vows to practice 30 more minutes each week. Since the amount of time practicing increases by a fixed number each week, this would qualify as an arithmetic sequence. Applying an Arithmetic Sequence Jordan has just watched The Queen’s Gambit and decided to hone their skills in chess. To really improve at the game, Jordan decides to practice for 3 hours the first week, and increase their time spent practicing by 30 minutes each week. How many hours will Jordan practice chess in week 20? Jordan’s practice scheme is an arithmetic sequence, as it increases by a fixed amount each week. The first week there are 3 hours of practice. This means a 1 = 3 . Jordan increases the time spent practicing by 30 minutes, or half an hour, each week. This means d = 0.5 . Using those values, and that we want to know the amount of time Jordan will study in week 20, we determine the time in week 20 using a i = a 1 + d × ( i − 1 ) . a 20 = 3 + 0.5 × ( 20 − 1 ) = 3 + 0.5 × 19 = 3 + 9.5 = 12.5 So, Jordan will practice 12.5 hours in week 20. Finding the Sum of a Finite Arithmetic Sequence Let’s check back in on Jordan. Recall, Jordan had just watched The Queen’s Gambit and decided to hone their skills, practicing for 3 hours the first week, and increasing the time spent practicing by 30 minutes each week. How many hours total will Jordan have practiced chess after 30 weeks of practice? To calculate the total amount of time that Jordan practiced, we need to use s n = n ( a 1 + a n 2 ) . The formula requires the first and last terms of the sequence. Since Jordan practiced 3 hours in the first week, the first term is a 1 = 3 . Because we want the total practice time after 30 weeks, we need the 30 th term. Because the constant difference is d = 0.5 , the 30 th term is a 30 = 3 + 0.5 ( 30 − 1 ) = 3 + 0.5 × 29 = 3 + 14.5 = 17.5 . Applying the formula s n = n ( a 1 + a n 2 ) provides the sum of the first 30 terms. s 30 = 30 ( 3 + 17.5 2 ) = 60 × 20.5 2 = 615 . This means that Jordan practiced a total of 615 hours after 30 weeks. The Fibonacci Sequence Not all sequences are arithmetic. One special sequence is the Fibonacci sequence , which is the sequence that has as its first two terms 1 and 1. Every term thereafter is the sum of the previous two terms. The first nine terms of the Fibonacci sequence are 1, 1, 2, 3, 5, 8, 13, 21, and 34. This sequence is found in nature, architecture, and even music! In nature, the Fibonacci sequence describes the spirals of sunflower seeds, certain galaxy spirals, and flower petals. In music, the band Tool used the Fibonacci sequence in the song “Lateralus.” The Fibonacci sequence even relates to architecture, as it is closely related to the golden ratio. Fibonacci Sequence and “Lateralus” Check Your Understanding Key Terms sequence term of a sequence arithmetic sequence first term constant difference Key Concepts A sequence is a list of numbers. Any individual number in that list, or sequence, is a term of the sequence. A specific term of a sequence is denoted by the sequence symbol with a subscript indicating where the term in the sequence is. A special form of a sequence is an arithmetic sequence. Each arithmetic sequence is determined by its first term and its constant difference. Any term in an arithmetic sequence is determined by adding the constant difference to the preceding term. If the first term and the constant difference of an arithmetic sequence are known, then any term of the sequence can be found directly. Because arithmetic sequences follow such a strict pattern, the sum of the first n terms of an arithmetic sequence can be determined with the formula s n = n ( a 1 + a n 2 ) . Formulas a i = a 1 + d × ( i − 1 ) d = a j − a i j − i a 1 = a i − d ( i − 1 ) s n = n ( a 1 + a n 2 ) Videos Arithmetic Sequences Finding the First Term and Constant Difference for an Arithmetic Sequence Finding the Sum of a Finite Arithmetic Sequence Fibonacci Sequence and “Lateralus”

Math


Geometric Sequences Savings grows in a geometric sequence. (credit: modification of “A big part of financial freedom is having your heart and mind free from worry about the what-ifs of life. – Suze Orman” by Morgan/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify geometric sequences. Find a given term in a geometric sequence. Find the n th term of a geometric sequence. Find the sum of a finite geometric sequence. Use geometric sequences to solve real-world applications. One of the concerns when investing is the doubling time , which is length of time it takes for the value of the investment to be twice, or double, that of its starting value. A shorter doubling times means the investment gets bigger, sooner. For example, if you invest $200 in an account with an 8-year doubling time, then in 8 years the value of the account will be double the starting amount, or 2 × $ 200 = $ 400 . After another 8 years (for a total of 16 years) the investment would be twice its value after the first 8 years, or 2 × ( 2 × $ 400 ) = 2 × ( $ 400 ) = $ 800 . Every 8 years, the investment would double again, so after the third 8-year period, the investment would be worth 2 × 2 × ( 2 × $ 400 ) = $ 1,600 . This process exhibits exponential growth, an application of geometric sequences, which is explored in this section. Identifying Geometric Sequences We know what a sequence is, but what makes a sequence a geometric sequence? In an arithmetic sequence, each term is the previous term plus the constant difference. So, you add a (possibly negative) number at each step. In a geometric sequence , though, each term is the previous term multiplied by the same specified value, called the common ratio . In the sequence { 3 , 6 , 12 , 24 , 48 , 96 , 192 , 384 , 728 , 1456 } the common ratio is 2. To see the difference between an arithmetic sequence and geometric sequence, examine these two sequences ( Figures 3.52 and 3.53 ). Arithmetic sequence Each term in this arithmetic sequence is the previous term plus 5. Geometric sequence Each term in this geometric sequence is the previous term times 2. In the sequence { 3 , 6 , 12 , 24 , 48 , 96 , 192 , 384 , 728 , 1456 } , the numbers get big fairly quickly, and stay positive. However, that’s not always the case with geometric sequences. Depending on the value of the common ratio, the terms could increase each time (like in the one shown in ), or the terms can get smaller each time, or the terms can alternate between positive and negative values. It all depends on the value of the common ratio, r . Consider this geometric sequence: { 5 , 15 , 45 , 135 , 405 , 2025... } Each term is the previous term times 5, which means the common ratio is 5. This common ratio is larger than 1, and so the terms increase each time. Now, look at this geometric sequence: { 2 , − 6 , 18 , − 54 , 162 , − 486 , 1458... } Each term is the previous term times −3, and the sign of the terms alternate from positive to negative. Then, there’s this geometric sequence: { 9 , 3 , 1 , 1 3 , 1 9 , 1 27 ... } Each term is the previous term times 1 3 , and the terms decrease each time. What we should take away from these three examples is if the common ratio is a positive number larger than 1, then the sequence increases. If the common ratio is a negative number, then the sign of the terms alternates between positive and negative. If the common ratio is between 0 and 1, then the terms decrease. Two special cases of geometric sequences are when the constant ratio is 1 and when the common ratio is 0. When the constant ratio is 1, every term of the sequence is the same, as in { 3 , 3 , 3 , 3 , 3 , 3 , 3 , 3 , 3 } . This is referred to as a constant sequence . When the constant ratio is 0, the first term can be any number, but every term after the first term is 0, as in { − 43.2 , 0 , 0 , 0 , 0 , 0 , 0 , 0 } . Identifying Geometric Sequences For each sequence, determine if the sequence is a geometric sequence. If so, identify the common ratio. { 5 , 20 , 80 , 320 , 1,280 , 5,120 , 20,480 , ... } { − 3 , 6 , − 12 , 24 , 11 , 33 } { 4 , 2 , 1 , 1 2 , 1 4 , 1 8 , ... } In the sequence { 5 , 20 , 80 , 320 , 1,280 , 5,120 , 20,480 , ... } , the jump from 5 to 20 is a multiplication by 4, as is the next jump to 80, and the next to 320. Each term is 4 times the previous term. Since each term is 4 times the previous, this is a geometric sequence. The common ratio is 4. In the sequence { − 3 , 6 , − 12 , 24 , 11 , 33 } , notice that 6 is −3 times −2. The jump from 6 to −12 is another multiplication by negative. So, if this is a geometric sequence, each term should be the previous term times −2. But the change from 24 to 11 is not a multiplication by −2, This means the sequence is not a geometric sequence. In the sequence { 4 , 2 , 1 , 1 2 , 1 4 , 1 8 , ... } , the change from 4 to 2 is a multiplication by 1 2 , as is the next jump, from 2 to 1, as is the next from 1 to 1 2 . Each term is 1 2 times the previous term. Since each term is 1 2 times the previous, this is a geometric sequence. The common ratio is 1 2 . As with arithmetic sequences, the first term of a geometric sequence is labeled a 1 . The number that is multiplied by each term is called the common ratio and is denoted r . So, if the first term is known, a 1 , and the common ratio is known, r , then the n th term, a n , can be calculated with the formula a n = a 1 r n − 1 . The n th term of the geometric sequence, a n , with first term a 1 and common ratio r , is a n = a 1 r n − 1 . Return to the sequence { 3 , 6 , 12 , 24 , 48 , 96 , 192 , 384 , 728 , ... } . We observe that the first term is 3, so a 1 = 3 . We also found that the common ratio is 2, so r = 2 . The table below shows how any term can be calculated using just a 1 and r . i , Place in Sequence a i , i t h ,Term Value of Term Term Written as a 1 × r i − 1 1 a 1 3 3 × 2 0 2 a 2 6 3 × 2 1 3 a 3 12 3 × 2 2 4 a 4 24 3 × 2 3 5 a 5 48 3 × 2 4 i a i 3 × 2 i − 1 Determining the Value of a Specific Term in a Geometric Sequence In the following geometric sequences, determine the indicated term of the geometric sequence with a given first term and common ratio. Determine the 9 th term of the geometric sequence with a 1 × 6 and r = 3 . Determine the 11 th term of the geometric sequence with a 1 = 2 and r = − 5 . Using a n = a 1 r n − 1 with a 1 = 6 , r = 3 , and n = 9 , we calculate a 9 = a 1 r 9 − 1 = 6 × ( 3 ) 9 − 1 = 6 × ( 3 ) 8 = 6 × 6561 = 39366 . The 9 th term of the geometric sequence with a 1 = 6 and r = 3 is a 9 = 39366 . Using a n = a 1 r n − 1 with a 1 = 2 , r = − 5 , and n = 11 , we calculate a 11 = a 1 r 11 − 1 = 2 × ( − 5 ) 11 − 1 = 2 × ( − 5 ) 10 = 2 × 9,765,625 = 19,531,250 . Geometric Sequences Finding the Sum of a Finite Geometric Sequence As with arithmetic sequences, it is possible to add the terms of the geometric sequence. Like arithmetic sequences, the formula for the finite sum of the terms of a geometric sequence has a straightforward formula. The sum of the first n terms of a finite geometric sequence, written s n , with first term a 1 and common ratio r , is s n = a 1 ( 1 − r n 1 − r ) provided that r ≠ 1 . Calculating the Sum of a Finite Geometric Sequence What is the sum of the first 13 terms of the geometric sequence with first term a 1 = 5 and common ratio r = 3 ? What is the sum of the first 7 terms of the geometric sequence with first term a 1 = 16 and common ratio r = 1 8 ? Using a 1 = 5 , r = 3 , and n = 13 , we find that the sum is: S 13 = a 1 ( 1 − r 13 1 − r ) = 5 ( 1 − 3 13 1 − 3 ) = 5 ( 1 − 1,594,323 − 2 ) = 5 ( − 1,594,322 − 2 ) = 5 ( 797,161 ) = 3,985,805 The sum of the first 13 terms of this geometric sequence is 3,985,805. Using a 1 = 16 , r = 1 8 , and n = 7 , we find that the sum is: s 7 = a 1 ( 1 − r 7 1 − r ) = 16 ( 1 − ( 1 8 ) 7 1 − ( 1 8 ) ) = 16 ( 1 − ( 1 2,097,152 ) 7 8 ) = 16 ( 2,097,151 2,097,152 7 8 ) = 16 ( 299,593 262,144 ) = 299,593 16,384 = 18.2857 The sum of the first 7 terms of this geometric sequence is 18.2857 . Using Geometric Sequences to Solve Real-World Applications Geometric sequences have a multitude of applications, one of which is compound interest. Compound interest is something that happens to money deposited into an account, be it savings or an individual retirement account, or IRA. The interest on the account is calculated and added to the account at regular intervals. This means the interest that was earned later gains its own interest. This allows the money to grow faster. If that interest is added every month, we say it is compounded monthly. If the interest is added daily, then we say it is compounded daily. The amount of money that is deposited into the account is called the principal and is denoted P . The account earns money on that principal. The amount it earns is a percentage of the money in the account. The interest rate, expressed as a decimal, is denoted r . If you deposit P dollars in an account that earns interest compounded yearly, then the amount in the account, A , after t years is calculated with the formula: A = P ( 1 + r ) t . This is a geometric sequence, with constant ratio ( 1 + r ) and first term a 1 = P . Calculating Interest Compounded Yearly Daryl deposits $1,000 in an account earning 4 % interest compounded yearly. How much money is in the account after 25 years? Using A = P ( 1 + r ) t with P = 1000 , r = 0.04 , and t = 25 , we find that A = P ( 1 + r ) t = 1,000 × ( 1 + 0.04 ) 25 = 1,000 × ( 1.04 ) 25 = 1,000 × 2.66583633 = 2,665.85 . After 25 years, there is $ 2,665.84 in the account. Another application of geometric sequences is exponential growth. This arises in biology quite frequently, especially in relation to bacterial cultures, but also with other organism population models. In bacterial cultures, the time it takes the population to double is often recorded. This time to double is the same, regardless of how big the population gets. So, if the population doubles after 3 hours, it doubles again after another 3 hours, and again after another 3 hours, and so on. Put into geometric sequence language, it has a common ratio of 2. Doubling a Bacterial Culture When Escherichia coli ( E. coli ) is in a broth culture at 37°C, the population of E. coli doubles in number with 30 organisms, how many E. coli bacteria are present in the culture after 16 hours? Since the population is doubling every 20 minutes, this is a geometric sequence situation with common ratio r = 2 . The culture begins with 30 organisms, so a 1 = 30 . The time,16 hours, is 48 twenty-minute periods, so we’re looking for the 48th term in the sequence. Using these values in the geometric sequence formula gives a 48 = a 1 r n − 1 = 30 × 2 48 − 1 = 30 × 2 47 = 30 × ( 1.40737 × 10 14 ) = 4.22212 × 10 15 . So, after 16 hours, the culture contains 4.22212 × 10 15 E. coli organisms. That’s more than 4,000 trillion bacteria. Applying the Sum of a Finite Geometric Sequence A player places one grain of rice on the first square of a chess board. On the second square, the player places 2 grains of rice. On the third square, the player places 4 grains of rice. On each successive square of the board, the player doubles the number of grains of rice placed on the chess board. When the player places the last rice on the 64th square, how many total grains of rice have been placed on the board? Since the number of grains of rice is doubled at each step, this is a geometric sequence with first term a 1 = 1 and common ratio r = 2 . Rice is placed on 64 total squares, so we want the sum of the first 64 terms. Using this information and the formula, the total number of grains of rice on the board will be: s 64 = a 1 ( 1 − r n − 1 1 − r ) = 1 × ( 1 − 2 64 − 1 1 − 2 ) = ( 1 − 2 63 − 1 ) = − ( 1 − 2 63 ) = − ( − 9.2233720369 × 10 18 ) = 9.2233720369 × 10 18 That’s a 20-digit number! Sum of a Finite Geometric Sequence Check Your Understanding Key Terms geometric sequence common ratio Key Concepts A special form of a sequence is a geometric sequence. Each geometric sequence is determined by its first term and its constant ratio. Any term in a geometric sequence is determined by multiplying the constant ratio to the preceding term. If the first term and the constant ratio of a geometric sequence are known, then any term of the sequence can be found directly. Because geometric sequences follow such a strict pattern, the sum of the first n terms of a geometric sequence can be determined with the formula s n = a 1 ( 1 − r n − 1 1 − r ) . Finding the sum of a finite geometric sequence Applying arithmetic sequences Formulas a n = a 1 r n − 1 s n = a 1 ( 1 − r n − 1 1 − r ) Videos Geometric Sequences Sum of a Finite Geometric Sequence Encryption Throughout History Encryption began at least as far back as the Roman Empire. During the reign of Caesar, a particular cypher was used, fittingly named the Caesar Cypher. This encryption process granted the Romans a great tactical advantage. Even if a message was intercepted, it would not make sense to the person intercepting the message. Find four instances when encryption was used and cracked over the course of history. Projects The Golden Ratio in Art and Architecture The golden ratio has been used in art and architecture as far back as ancient Greece (possibly further). It also appears in South America (Incan architecture). Find five instances of the use of the golden ratio in art or architecture and describe its use in each of those instances. Your Budget Budgeting either is, or will shortly be, an important aspect of your life. Managing money well reduces stress in your life, and provides space for planning for future expenses, such as vacations or home improvements. Imagine your life 10 years from now. Estimate your monthly income. Identify expenses you will encounter monthly (mortgage or rent, car payment, insurance, entertainment, etc.). Decide on an amount you plan to save monthly (this is treated as an expense). Create a spreadsheet with those values. Record your monthly net income (your income minus your expenses). Determine how much money you will have saved over the course of 5 years (ignore interest). Write a reflection on your anticipated financial health. Estimating Pi The value of pi is the ratio of the circumference of a circle to the diameter of the circle. It is also equal to the ratio of the area of the circle to the square of the radius of the square. Research three ways to physically estimate pi. Estimate pi using all three processes you found. Present your process and solutions in class. Design Your Own Shift Cypher A cypher is a message written in such a way as to mask its contents. Changing a message into its cypher form is called encryption. Decryption or deciphering is the process of changing a cyphertext message back into the original (legible) message. One process of encryption is to scramble the letters, symbols, and punctuation of a message according to a mathematical rule. One rule that could be used for such a cypher is addition in a chosen modulus. In this project, you will create such a cypher, encrypt a message, and then decrypt the message. Step 1: Choose the letters, symbols, and punctuation marks you want to allow in your messages. This should include at least the uppercase letters and a space character. This is your character set. Step 2: Count the number of characters you will use. Label this number n . Step 3: Pair each character of your character set an integer from 0 to ( n − 1 ) . Do not assign more than one character to an integer. Step 4: Choose an integer between 1 and ( n − 1 ) . This will be the number used to create the cypher. Label this number s . Step 5: Write a message using your character set. Step 6: Replace every character in your message by the integer with which it was paired in Step 3. Step 7: For every number, x , from Step 7, perform the addition x + s (mod n ). Step 8: Replace every number found in Step 7 with the character with which it was paired in Step 3. This is your cyphertext. To decrypt your cyphertext, reverse the steps above. Step 1: Replace the cyphertext characters with the paired values. Step 2: For each value x , perform the subtraction x – s (mod n ). Step 3: Replace the numbers from Step 2 with their paired characters from the character set. The message is then deciphered. Design Your Own Cypher Using Multiplication A cypher is a message written in such a way as to mask its contents. Changing a message into its cypher form is called encryption. Decryption or deciphering is the process of changing a cyphertext message back into the original (legible) message. One process of encryption is to scramble the letters, symbols, and punctuation of a message according to a mathematical rule. One rule that could be used for such a cypher is multiplication in a chosen modulus. In this project, you will create such a cypher, encrypt a message, then decrypt the message. Step 1: Choose the letters, symbols, and punctuation marks you want to allow in your messages. This should include at least the uppercase letters and a space character. This is your character set. Step 2: Count the number of characters you will use. Label this number n . Step 3: Pair each character of your character set an integer from 0 to ( n − 1 ) . Do not assign more than one character to an integer. Step 4: Choose an integer, labeled s , between 1 and ( n − 1 ) so that GCF ( n , s ) = 1 . This will be the number used to create the cypher. Step 5: Write a message using your character set. Step 6: Replace every character in your message by the integer with which it was paired in Step 3. Step 7: For every number, x , from Step 6, perform the multiplication x − s (mod n ). Step 8: Replace every number found in Step 7 with the character with which it was paired in Step 3. This is your cyphertext. Before beginning to decrypt in this cypher, you need to know the multiplicative inverse of the value you chose as s . Step 1: The multiplicative inverse of s is the number that, when multiplied by s in your modulus, equals 1. To find this, you will have to multiply s and every number between 2 and ( n – 1 ) until the product is 1 (mod n ). Once this number is found, the message can be decrypted. Call this number r . Step 2: To decrypt your cyphertext, replace the cyphertext characters with the paired values. Step 3: For each of the value, x , perform the multiplication x × r ( mod n ) . Step 4: Replace the numbers from Step 3 with their paired characters from the character set. The message is then deciphered. Key Terms common ratio geometric sequence Key Concepts Geometric sequence. Finding an arbitrary term in a geometric sequence. Constant ratio. Finding the sum of a finite geometric sequence. Applying arithmetic sequences. Formulas a n = a 1 r n − 1 s n = a 1 ( 1 − r n − 1 1 − r ) Chapter Review Prime and Composite Numbers The Integers Order of Operations Rational Numbers Irrational Numbers Real Numbers Clock Arithmetic Exponents Scientific Notation Arithmetic Sequences Geometric Sequences Chapter Test

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Introduction Different cultures developed different ways to record quantity. (credit: modification of work "Tally sticks from the Swiss Alps" by Sandstein, Swiss Alpine Museum permanent collection/Wikimedia Commons, CC BY 3.0) Right now, almost all cultures use the familiar Hindu-Arabic numbering system, which uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 along with place values based on powers of ten. This is a relatively recent development. The system didn’t develop until the 6th or 7th century C.E. and took some time to spread across the world, which means other cultures at other times had to develop their own methods of counting and recording quantity. Being different cultures and different times means there were significant differences in counting systems. Cultures needed to count and measure time for agriculture and for religious observations. It was needed for trade. Some languages had words only for one, two, and many. Other cultures developed more complex ways to represent quantity, with the Oksapmin people of New Guinea using an astonishing 27 words for their system. Representing these quantities in a recorded form likely began with a simple marking system, where one scratch on a stick or bone represented one of whatever was being counted. We still see this today with tally marks. These systems use repeated symbols to represent more than one. We also have systems where different symbols represent different quantities but still use some repetition, such as in Roman numerals. Other systems were devised that rely on place values, like the Hindu-Arabic system in use today. Place value systems needed a zero, though, and weren't immediately recognized and took time to develop. And within these positional systems there is variation. Some systems counted in twenties, others in tens, and some in a mix (adding another reason to visit Hawaii). Even now, though we all use and think using tens, computers are designed to work in groups of two, which requires a different perspective on numbering. In this chapter, we explore different numbering systems and grouping systems, eventually discussing base 2, the language of computers.

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Hindu-Arabic Positional System This manuscript is an early example of Hindu numerals. (credit: modification of work “Bakshali manuscript”, Bodleian Libraries/ University of Oxford, public domain) Learning Objectives After completing this section, you should be able to: Evaluate an exponential expression. Convert a Hindu-Arabic numeral to expanded form. Convert a number in expanded form to a Hindu-Arabic numeral. The modern system of counting and computing isn’t necessarily natural. That different symbols are used to indicate different quantities or amounts is a relatively new invention. Simple marking by scratches or dots, one for each item being counted, was the norm long into human history. The modern system doesn’t use repeated symbols to indicate more than one of a thing. It uses the place of a digit in a numeral to determine what that digit represents. A numeral is a symbol used to represent a number. A number is an abstract idea that represents quantity or amount. Being clear about the difference between numeral and number is important. Just like a person can be called by various names, such as brother, father, husband, uncle, they are all representing the same person, John Smith. The person John Smith is the number, and the names brother, father, husband, and uncle are the numerals. Hindu-Arabic Numerals The numerals we currently use are referred to as Hindu-Arabic numerals, although they have changed as time has passed. Early forms of the numerals for 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 began in India, and passed through Persia to the Middle East. Place value was also employed in the early systems of India. Once this system was in north Africa and the Middle East, it spread to Europe, eventually replacing Roman numerals. Over time, the original symbols transformed into our modern ones. Read this article for another perspective on how the symbols began (based on the moon!) . The system we use for counting and computing uses place values based on powers of 10. In this section, we review exponents and our positional system. Evaluating Exponential Expressions Most modern numerical systems depend on place values, where the quantity represented depends not only on the digit, but also on where the digit is in the number. The place value is a power of some specific number, which means most numbering systems are actually exponential expressions. An exponential expression is any mathematical expression that includes exponents. So, evaluating such an expression means performing the calculation. In this chapter, we will be using exponents that are positive integer values. Before we do so, let’s remind ourselves about exponents and what they represent. Suppose you want to multiply a number. Let’s label that number a , by itself some number of times. Let’s label the number of times b . We denote that as a b . We say a , or the base , raised to the b th power, or the exponent . For example, if we are multiplying 13 by itself eight times, we write 13 8 and say 13 to the eighth power. When computing exponential expressions, we should be careful to remember the order of operations. Using the order of operation rules, calculations inside the parentheses are done first, then exponents are calculated, then multiplication and division calculations are performed, and then addition and subtraction. Exponential Notation Evaluating an Exponential Expression Evaluate the following exponential expressions. 4 × 5 2 + 2 × 6 3 6 × 8 2 + 3 × 8 1 + 4 × 8 0 3 × 10 2 + 0 × 10 1 + 6 × 10 0 To evaluate, or calculate, this expression, we use order of operations, which means the exponents are done first, then multiplications, and then additions. 4 × 5 2 + 2 × 6 3 = 4 × 5 × 5 + 2 × 6 × 6 × 6 = 4 × 25 + 2 × 216 = 100 + 432 = 532 To evaluate the expression, we use the order of operations, which means the exponents are done first, then the multiplications, then the additions. Remember that any base raised to the exponent 0 is 1. 6 × 8 2 + 3 × 8 1 + 4 × 8 0 = 6 × 8 × 8 + 3 × 8 + 4 × 1 = 6 × 64 + 3 × 8 + 4 × 1 = 384 + 24 + 4 = 412 To evaluate the expression, we use the order of operations, which means the exponents are done first, then the multiplications, and then the additions. Remember that any base raised to the exponent 0 is 1. 3 × 10 2 + 0 × 10 1 + 6 × 10 0 = 3 × 100 + 0 × 10 + 6 × 1 = 300 + 0 + 6 = 306 Converting Hindu-Arabic Numerals to Expanded Form When you see the number 738, and you speak the number out loud, what do you say? You probably said “seven hundred thirty-eight” while wondering what point could possibly be made by asking this. What you didn’t say was “seven, and three, and eight.” A pre-K student might say that. Which should make you wonder, why? The reason is that you’ve been taught place values , or the positions of digits in a number that determine the values of those digits. You know that in a three-digit number, the first digit is hundreds, the second digit is tens, and the last digit is ones. These place values rely on powers of 10, which makes this system a base 10 system . This sense of place value is what makes our system of numbers so useful. You’ve also been taught the Hindu-Arabic numeration system . This system, which uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9, and also employs place value based on powers of 10, is in use today. Writing a number using these place values is writing them in expanded form . For a number with n digits, the expanded form is the first digit times 10 raised to one less than n , plus each following digit times 10 raised to one less than the previous power of 10. For example, the number 738 would be written as 7 × 10 2 + 3 × 10 1 + 8 × 10 0 . What about a four-digit number, like 5,825? Out loud, we’d say five thousand, seven hundred twenty-five. In expanded form, it would be 5 × 10 3 + 8 × 10 2 + 2 × 10 1 + 5 × 10 0 . Notice that the largest exponent is one less than the number of digits, and that the exponents go down by one as we move through the number. Aryabhata of Kusumapura and Brahmagupta The Hindu-Arabic numeral system developed in India, and Aryabhata of Kusumapura is credited with the place value notation in the 5th century. However, the system wasn’t as complete as it could be, until. roughly a century later, when Brahmagupta introduced the symbol for 0. The 0 is necessary to indicate that a given place value has been skipped, as in 4,098. In 4,098, the 10 2 power is skipped. Without such a symbol, 4,098 and 498 look similar. The value of both the place value notation and the introduction of the symbol 0 cannot be overstated, for math and the sciences. Writing a Number in Expanded Form Write the following in expanded form. 563 4,821 903,786 Step 1: Since there are three digits in 563, n is 3. So, this is the first digit times 10 raised to the power of 2, so we start with 5 × 10 2 . Step 2: Then we add the next digit, 6, multiplied by 10 to a power one less than the previous, at which point we have 5 × 10 2 + 6 × 10 1 . Step 3: Finally, the last digit is multiplied by 10 to the zeroth power and added to the previous. This results in 5 × 10 2 + 6 × 10 1 + 3 × 10 0 . Step 1: Since there are four digits in 4,821, n is 4. We multiply the first digit, 4, by 10 raised to the power of 3, which is 4 × 10 3 . Step 2: Then we add the next digit, 8, multiplied by 10 to a power one less than the previous, at which point we have 4 × 10 3 + 8 × 10 2 . Step 3: We continue to the next digit, lowering the exponent of 10 by one. Now we have 4 × 10 3 + 8 × 10 2 + 2 × 10 1 . Step 4: Finally, the last digit is multiplied by 10 to the zeroth power and added to the previous. This results in 4 × 10 3 + 8 × 10 2 + 2 × 10 1 + 1 × 10 0 . Since there are six digits in 903,786, n is 6. So, we begin the process with 9 times 10 raised to the 5th power and continue through the numbers, reducing the exponent of 10 by one each time. This results in 9 × 10 5 + 0 × 10 4 + 3 × 10 3 + 7 × 10 2 + 8 × 10 1 + 6 × 10 0 . Converting Numbers in Expanded Form to Hindu-Arabic Numerals Converting from expanded form back into a Hindu-Arabic numeral is the reverse process of expanding a number, and is equivalent to evaluating the exponential expression. Converting a Number from Expanded Form to a Hindu-Arabic Numeral Convert the following into Hindu-Arabic numerals. 3 × 10 2 + 4 × 10 1 + 8 × 10 0 5 × 10 3 + 0 × 10 2 + 9 × 10 1 + 9 × 10 0 6 × 10 6 + 2 × 10 5 + 0 × 10 4 + 9 × 10 3 + 1 × 10 2 + 1 × 10 1 + 7 × 10 0 Evaluating the expression results in: 3 × 10 2 + 4 × 10 1 + 8 × 10 0 = 3 × 100 + 4 × 10 + 8 × 1 = 300 + 40 + 8 = 348 Evaluating the expression results in: 5 × 10 3 + 0 × 10 2 + 9 × 10 1 + 9 × 10 0 = 5 × 1000 + 0 × 100 + 9 × 10 + 9 × 1 = 5000 + 0 + 90 + 9 = 5099 Evaluating the expression results in: 6 × 10 6 + 2 × 10 5 + 0 × 10 4 + 9 × 10 3 + 1 × 10 2 + 1 × 10 1 + 7 × 10 0 = 6 × 1,000,000 + 2 × 100,000 + 0 × 10,000 + 9 × 1,000 + 1 × 100 + 1 × 10 + 7 × 1 = 6,000,000 + 200,000 + 0 + 9,000 + 100 + 10 + 7 = 6,209,117 Check Your Understanding Key Terms numeral number exponential expression base exponent place value base 10 system Hindu-Arabic numeration system expanded form Key Concepts Exponents are used to represent repeated multiplication of a base. In arithmetic, exponents are computed before multiplication, division, addition, and subtraction. Computing an exponent is done by multiplying the base by itself the number of times equal to the exponent. The system of numbers currently used is the Hindu-Arabic system. Digits in this system take on values based on their place in the number. The place values are determined by multiplying the digit by 10 raised to the appropriate power. The expanded form of a Hindu-Arabic number is the sum of each digit times 10 raised to the exponent for that place value. Videos Exponential Notation

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