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Find a 90% confidence interval for the true (population) mean of statistics exam scores. | Solution A\n\nTo find the confidence interval, you need the sample mean, \( \bar{x} \), and the EBM.\n\n\[ \bar{x} = {68} \]\n\n\( {EBM} = {z}_{\frac{\alpha }{2}} \cdot \left( \frac{\sigma }{\sqrt{n}}\right) \)\n\n\( \sigma = 3;n = {36} \) ; The confidence level is \( {90}\% \left( {\mathrm{{CL}} = {0.90}}\right) \)\n\... | Yes |
Suppose we know that a confidence interval is \( \left( {{67.18},{68.82}}\right) \) and we want to find the error bound. We may know that the sample mean is 68 . Or perhaps our source only gave the confidence interval and did not tell us the value of the the sample mean. | - If we know that the sample mean is \( {68} : {EBM} = {68.82} - {68} = {0.82} \)\n- If we don’t know the sample mean: \( {EBM} = \frac{\left( {68.82} - {67.18}\right) }{2} = {0.82} \) | Yes |
Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. 500 randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes - they own cell phones. Using a \( {95}\% ... | Solution A\n\nLet \( X = \) the number of people in the sample who have cell phones. \( X \) is binomial. \( X \sim \) \( B\left( {{500},\frac{421}{500}}\right) \) .\n\nTo calculate the confidence interval, you must find \( {p}^{\prime },{q}^{\prime } \), and EBP.\n\n\( n = {500}\;x = \) the number of successes \( = {4... | Yes |
We want to test whether the mean grade point average in American colleges is different from 2.0 (out of 4.0). | \[ {H}_{o} : \mu = {2.0}\;{H}_{a} : \mu \neq {2.0} \] | No |
We want to test if college students take less than five years to graduate from college, on the average. | \[ {H}_{o} : \mu \geq 5\;{H}_{a} : \mu < 5 \] | No |
Suppose the null hypothesis, \( {H}_{o} \), is: Frank’s rock climbing equipment is safe. | Type I error: Frank thinks that his rock climbing equipment may not be safe when, in fact, it really is safe. Type II error: Frank thinks that his rock climbing equipment may be safe when, in fact, it is not safe. \( \alpha = \) probability that Frank thinks his rock climbing equipment may not be safe when, in fact, it... | Yes |
Suppose the null hypothesis, \( {H}_{o} \), is: The victim of an automobile accident is alive when he arrives at the emergency room of a hospital. | Type I error: The emergency crew thinks that the victim is dead when, in fact, the victim is alive. Type II error: The emergency crew does not know if the victim is alive when, in fact, the victim is dead. \( \alpha = \) probability that the emergency crew thinks the victim is dead when, in fact, he is really alive \( ... | Yes |
\( {H}_{o} : \mu = 5\;{H}_{a} : \mu < 5 \) | Test of a single population mean. \( {H}_{a} \) tells you the test is left-tailed. The picture of the p-value is as follows:\n\n | No |
[{H}_{o} : p \leq {0.2}\;{H}_{a} : p > {0.2}] | This is a test of a single population proportion. \( {H}_{a} \) tells you the test is right-tailed. The picture of the p-value is as follows:\n\n | No |
\[ {H}_{o} : \mu = {16.43}\;{H}_{a} : \mu < {16.43} \] | Since the problem is about a mean, this is a test of a single population mean.\n\nDetermine the distribution needed:\n\nRandom variable: \( \bar{X} = \) the mean time to swim the 25-yard freestyle.\n\nDistribution for the test: \( \bar{X} \) is normal (population standard deviation is known: \( \sigma = {0.8} \) )\n\n\... | Yes |
Suppose a consumer group suspects that the proportion of households that have three cell phones is \( {30}\% \) . A cell phone company has reason to believe that the proportion is \( {30}\% \) . Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households wi... | Set up the Hypothesis Test:\n\n\[ \n{\mathrm{H}}_{o} : p = {0.30}\;{\mathrm{H}}_{a} : p \neq {0.30} \]\n\nDetermine the distribution needed:\n\nThe random variable is \( {P}^{\prime } = \) proportion of households that have three cell phones.\n\nThe distribution for the hypothesis test is \( {P}^{\prime } \sim N\left( ... | Yes |
Can you use the information as it appears in the charts to conduct the goodness-of-fit test? | No. Notice that the expected number of absences for the \ | No |
What are the degrees of freedom \( \left( {df}\right) \) ? | There are 4 \ | Yes |
Suppose \( A = \) a speeding violation in the last year and \( B = \) a cell phone user while driving. If \( A \) and \( B \) are independent then \( P\left( {A\left| {AND}\right| B}\right) = P\left( A\right) P\left( B\right) \) . A \( {AND}\bar{B} \) is the event that a driver received a speeding violation last year a... | Let \( y = \) expected number of drivers that use a cell phone while driving and received speeding violations.\n\nIf \( A \) and \( B \) are independent, then \( P\left( {A \cap {ANDB}}\right) = P\left( A\right) P\left( B\right) \) . By substitution,\n\n\( \begin{array}{l} \frac{y}{755} = \frac{70}{755} \cdot \frac{305... | Yes |
Do male and female college students have the same distribution of living conditions? | Solution\n\n\( {H}_{o} \) : The distribution of living conditions for male college students is the same as the distribution of living conditions for female college students.\n\n\( {H}_{a} \) : The distribution of living conditions for male college students is not the same as the distribution of living conditions for fe... | Yes |
Suppose a math instructor believes that the standard deviation for his final exam is 5 points. One of his best students thinks otherwise. The student claims that the standard deviation is more than 5 points. If the student were to conduct a hypothesis test, what would the null and alternate hypotheses be? | Even though we are given the population standard deviation, we can set the test up using the population variance as follows.\n\n\[ \text{-}{H}_{o} : {\sigma }^{2} = {5}^{2} \]\n\n\[ \text{-}{\mathrm{H}}_{a} : {\sigma }^{2} > {5}^{2} \] | No |
Find the equation that expresses the total cost in terms of the number of hours required to finish the word processing job. | Let \( x = \) the number of hours it takes to get the job done.\n\nLet \( y = \) the total cost to the customer.\n\nThe \( \$ {31.50} \) is a fixed cost. If it takes \( x \) hours to complete the job, then (32) \( \left( x\right) \) is the cost of the word processing only. The total cost is:\n\n\[ y = {31.50} + {32}\ma... | Yes |
For the years 2000 through 2004, was there a relationship between the year and the number of m-commerce users? Construct a scatter plot. Let \( x = \) the year and let \( y = \) the number of m-commerce users, in millions. | A scatter plot shows the direction and strength of a relationship between the variables. A clear direction happens when there is either:\n\n- High values of one variable occurring with high values of the other variable or low values of one variable occurring with low values of the other variable.\n\n- High values of on... | No |
Can you predict the final exam score of a random student if you know the third exam score? | The third exam score, \( x \), is the independent variable and the final exam score, \( y \), is the dependent variable. We will plot a regression line that best \ | No |
Suppose you computed \( r = {0.801} \) using \( n = {10} \) data points. df \( = n - 2 = {10} - 2 = 8 \) . The critical values associated with \( \mathrm{{df}} = 8 \) are \( - {0.632} \) and \( + {0.632} \) . If \( r < \) negative critical value or \( r > \) positive critical value, then \( r \) is significant. | Since \( r = {0.801} \) and \( {0.801} > {0.632}, r \) is significant and the line may be used for prediction. If you view this example on a number line, it will help you. | Yes |
What would you predict the final exam score to be for a student who scored a 66 on the third exam? | 145.27 | Yes |
Theorem 1.1. For all \( a, b, c \in \mathbb{Z} \), we have\n\n(i) \( a\left| {a,1}\right| a \), and \( a \mid 0 \) ;\n\n(ii) \( 0 \mid a \) if and only if \( a = 0 \) ;\n\n(iii) \( a \mid b \) if and only if \( - a \mid b \) if and only if \( a \mid - b \) ;\n\n(iv) \( a\left| {b\text{and}a}\right| c \) implies \( a \m... | Proof. These properties can be easily derived from the definition of divisibility, using elementary algebraic properties of the integers. For example, \( a \mid a \) because we can write \( a \cdot 1 = a;1 \mid a \) because we can write \( 1 \cdot a = a;a \mid 0 \) because we can write \( a \cdot 0 = 0 \) . We leave it... | No |
For all \( a, b \in \mathbb{Z} \), we have \( a \mid b \) and \( b \mid a \) if and only if \( a = \pm b \) . In particular, for every \( a \in \mathbb{Z} \), we have \( a \mid 1 \) if and only if \( a = \pm 1 \) . | Proof. Clearly, if \( a = \pm b \), then \( a\left| {b\text{and}b}\right| a \) . So let us assume that \( a \mid b \) and \( b \mid a \), and prove that \( a = \pm b \) . If either of \( a \) or \( b \) are zero, then the other must be zero as well. So assume that neither is zero. By the above observation, \( a \mid b ... | Yes |
Theorem 1.3 (Fundamental theorem of arithmetic). Every non-zero integer \( n \) can be expressed as\n\n\[ n = \pm {p}_{1}^{{e}_{1}}\cdots {p}_{r}^{{e}_{r}} \]\n\nwhere \( {p}_{1},\ldots ,{p}_{r} \) are distinct primes and \( {e}_{1},\ldots ,{e}_{r} \) are positive integers. Moreover, this expression is unique, up to a ... | To prove Theorem 1.3, we may clearly assume that \( n \) is positive, since otherwise, we may multiply \( n \) by -1 and reduce to the case where \( n \) is positive.\n\nThe proof of the existence part of Theorem 1.3 is easy. This amounts to showing that every positive integer \( n \) can be expressed as a product (pos... | Yes |
Theorem 1.4 (Division with remainder property). Let \( a, b \in \mathbb{Z} \) with \( b > 0 \) . Then there exist unique \( q, r \in \mathbb{Z} \) such that \( a = {bq} + r \) and \( 0 \leq r < b \) . | Proof. Consider the set \( S \) of non-negative integers of the form \( a - {bt} \) with \( t \in \mathbb{Z} \) . This set is clearly non-empty; indeed, if \( a \geq 0 \), set \( t \mathrel{\text{:=}} 0 \), and if \( a < 0 \), set \( t \mathrel{\text{:=}} a \) . Since every non-empty set of non-negative integers contai... | Yes |
Consider the ideal \( 3\mathbb{Z} + 5\mathbb{Z} \). | This ideal contains \( 3 \cdot 2 + 5 \cdot \left( {-1}\right) = 1 \) . Since it contains 1, it contains all integers; that is, \( 3\mathbb{Z} + 5\mathbb{Z} = \mathbb{Z} \). | Yes |
Theorem 1.6. Let \( I \) be an ideal of \( \mathbb{Z} \). Then there exists a unique non-negative integer \( d \) such that \( I = d\mathbb{Z} \). | Proof. We first prove the existence part of the theorem. If \( I = \{ 0\} \), then \( d = 0 \) does the job, so let us assume that \( I \neq \{ 0\} \). Since \( I \) contains non-zero integers, it must contain positive integers, since if \( a \in I \) then so is \( - a \). Let \( d \) be the smallest positive integer i... | Yes |
For all \( a, b \in \mathbb{Z} \), there exists a unique greatest common divisor \( d \) of \( a \) and \( b \), and moreover, \( a\mathbb{Z} + b\mathbb{Z} = d\mathbb{Z} \). | Proof. We apply the previous theorem to the ideal \( I \mathrel{\text{:=}} a\mathbb{Z} + b\mathbb{Z} \). Let \( d \in \mathbb{Z} \) with \( I = d\mathbb{Z} \), as in that theorem. We wish to show that \( d \) is a greatest common divisor of \( a \) and \( b \). Note that \( a, b, d \in I \) and \( d \) is non-negative.... | Yes |
Theorem 1.8. Let \( a, b, r \in \mathbb{Z} \) and let \( d \mathrel{\text{:=}} \gcd \left( {a, b}\right) \) . Then there exist \( s, t \in \mathbb{Z} \) such that \( {as} + {bt} = r \) if and only if \( d \mid r \) . In particular, \( a \) and \( b \) are relatively prime if and only if there exist integers \( s \) and... | Proof. We have\n\n\[ \n{as} + {bt} = r\text{ for some }s, t \in \mathbb{Z} \]\n\n\[ \n\Leftrightarrow r \in a\mathbb{Z} + b\mathbb{Z} \]\n\n\[ \n\Leftrightarrow r \in d\mathbb{Z}\text{(by Theorem 1.7)} \]\n\n\[ \n\Leftrightarrow d \mid r\text{. } \]\n\nThat proves the first statement. The second statement follows from ... | Yes |
Theorem 1.9. Let \( a, b, c \in \mathbb{Z} \) such that \( c \mid {ab} \) and \( \gcd \left( {a, c}\right) = 1 \) . Then \( c \mid b \) . | Proof. Suppose that \( c \mid {ab} \) and \( \gcd \left( {a, c}\right) = 1 \) . Then since \( \gcd \left( {a, c}\right) = 1 \), by Theorem 1.8 we have \( {as} + {ct} = 1 \) for some \( s, t \in \mathbb{Z} \) . Multiplying this equation by \( b \), we obtain\n\n\[ \n{abs} + {cbt} = b.\n\]\n\n(1.1)\n\nSince \( c \) divid... | Yes |
Theorem 1.10. Let \( p \) be prime, and let \( a, b \in \mathbb{Z} \) . Then \( p \mid {ab} \) implies that \( p \mid a \) or \( p \mid b \) . | Proof. Assume that \( p \mid {ab} \) . If \( p \mid a \), we are done, so assume that \( p \nmid a \) . By the above observation, \( \gcd \left( {a, p}\right) = 1 \), and so by Theorem 1.9, we have \( p \mid b \) . | Yes |
Theorem 1.11. There are infinitely many primes. | Proof. By way of contradiction, suppose that there were only finitely many primes; call them \( {p}_{1},\ldots ,{p}_{k} \) . Then set \(M \mathrel{\text{:=}} \mathop{\prod }\limits_{{i = 1}}^{k}{p}_{i} \) and \( N \mathrel{\text{:=}} M + 1 \) . Consider a prime \( p \) that divides \( N \) . There must be at least one ... | Yes |
Theorem 2.1. Let \( \sim \) be an equivalence relation on a set \( S \), and for \( a \in S \), let \( \left\lbrack a\right\rbrack \) denote its equivalence class. Then for all \( a, b \in S \), we have:\n\n(i) \( a \in \left\lbrack a\right\rbrack \) ;\n\n(ii) \( a \in \left\lbrack b\right\rbrack \) implies \( \left\lb... | Proof. (i) follows immediately from reflexivity. For (ii), suppose \( a \in \left\lbrack b\right\rbrack \), so that \( a \sim b \) by definition. We want to show that \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) . To this end, consider any \( x \in S \) . We have\n\n\[ x \in \left\lbrack a\right\rbra... | Yes |
Theorem 2.2. Let \( n \) be a positive integer. For all \( a, b, c \in \mathbb{Z} \), we have:\n\n(i) \( a \equiv a\left( {\;\operatorname{mod}\;n}\right) \) ;\n\n(ii) \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \) implies \( b \equiv a\left( {\;\operatorname{mod}\;n}\right) \) ;\n\n(iii) \( a \equiv b\left( {... | Proof. For (i), observe that \( n \) divides \( 0 = a - a \) . For (ii), observe that if \( n \) divides \( a - b \), then it also divides \( - \left( {a - b}\right) = b - a \) . For (iii), observe that if \( n \) divides \( a - b \) and \( b - c \), then it also divides \( \left( {a - b}\right) + \left( {b - c}\right)... | Yes |
Theorem 2.3. Let \( a,{a}^{\prime }, b,{b}^{\prime }, n \in \mathbb{Z} \) with \( n > 0 \) . If\n\n\[ a \equiv {a}^{\prime }\left( {\;\operatorname{mod}\;n}\right) \text{ and }b \equiv {b}^{\prime }\left( {\;\operatorname{mod}\;n}\right) ,\]\n\nthen\n\n\[ a + b \equiv {a}^{\prime } + {b}^{\prime }\left( {\;\operatornam... | Proof. Suppose that \( a \equiv {a}^{\prime }\left( {\;\operatorname{mod}\;n}\right) \) and \( b \equiv {b}^{\prime }\left( {\;\operatorname{mod}\;n}\right) \) . This means that there exist integers \( x \) and \( y \) such that \( a = {a}^{\prime } + {nx} \) and \( b = {b}^{\prime } + {ny} \) . Therefore,\n\n\[ a + b ... | Yes |
Let us find the set of solutions \( z \) to the congruence\n\n\[ \n{3z} + 4 \equiv 6\left( {\;\operatorname{mod}\;7}\right) \n\] | Suppose that \( z \) is a solution to (2.1). Subtracting 4 from both sides of (2.1), we obtain\n\n\[ \n{3z} \equiv 2\left( {\;\operatorname{mod}\;7}\right) \n\]\n\nNext, we would like to divide both sides of this congruence by 3, to get \( z \) by itself on the left-hand side. We cannot do this directly, but since \( 5... | Yes |
Theorem 2.5. Let \( a, n \in \mathbb{Z} \) with \( n > 0 \), and let \( d \mathrel{\text{:=}} \gcd \left( {a, n}\right) \). (i) For every \( b \in \mathbb{Z} \), the congruence \( {az} \equiv b\left( {\;\operatorname{mod}\;n}\right) \) has a solution \( z \in \mathbb{Z} \) if and only if \( d \mid b \). (ii) For every ... | Proof. For (i), let \( b \in \mathbb{Z} \) be given. Then we have \[ {az} \equiv b\left( {\;\operatorname{mod}\;n}\right) \text{ for some }z \in \mathbb{Z} \] \( \Leftrightarrow {az} = b + {ny} \) for some \( z, y \in \mathbb{Z} \) (by definition of congruence) \( \Leftrightarrow {az} - {ny} = b \) for some \( z, y \in... | Yes |
The following table illustrates what Theorem 2.5 says for \( n = {15} \) and \( a = 1,2,3,4,5,6 \) . | In the second row, we are looking at the values \( {2z}{\;\operatorname{mod}\;{15}} \), and we see that this row is just a permutation of the first row. So for every \( b \), there exists a unique \( z \) such that \( {2z} \equiv b\left( {\;\operatorname{mod}\;{15}}\right) \) . This is implied by the fact that \( \gcd ... | Yes |
Observe that\n\n\[ 5 \cdot 2 \equiv 5 \cdot \left( {-4}\right) \left( {\;\operatorname{mod}\;6}\right) \] | Part (iii) of Theorem 2.5 tells us that since \( \gcd \left( {5,6}\right) = 1 \), we may cancel the common factor of 5 from both sides of (2.4), obtaining \( 2 \equiv - 4\left( {\;\operatorname{mod}\;6}\right) \), which one can also verify directly. | No |
Let \( a, b, n \in \mathbb{Z} \) with \( n > 0 \). We can describe the set of solutions \( z \in \mathbb{Z} \) to the congruence \( {az} \equiv b\left( {\;\operatorname{mod}\;n}\right) \) very succinctly in terms of modular inverses. | If \( \gcd \left( {a, n}\right) = 1 \), then setting \( t \mathrel{\text{:=}} {a}^{-1}{\;\operatorname{mod}\;n} \), and \( {z}_{0} \mathrel{\text{:=}} {tb}{\;\operatorname{mod}\;n} \), we see that \( {z}_{0} \) is the unique solution to the congruence \( {az} \equiv b\left( {\;\operatorname{mod}\;n}\right) \) that lies... | Yes |
Theorem 2.6 (Chinese remainder theorem). Let \( {\left\{ {n}_{i}\right\} }_{i = 1}^{k} \) be a pairwise relatively prime family of positive integers, and let \( {a}_{1},\ldots ,{a}_{k} \) be arbitrary integers. Then there exists a solution \( a \in \mathbb{Z} \) to the system of congruences\n\n\[ a \equiv {a}_{i}\left(... | Proof. To prove the existence of a solution \( a \) to the system of congruences, we first show how to construct integers \( {e}_{1},\ldots ,{e}_{k} \) such that for \( i, j = 1,\ldots, k \), we have\n\n\[ {e}_{j} \equiv \left\{ \begin{array}{ll} 1\left( {\;\operatorname{mod}\;{n}_{i}}\right) & \text{ if }j = i, \\ 0\l... | Yes |
Example 2.6. The following table illustrates what Theorem 2.6 says for \( {n}_{1} = 3 \) and \( {n}_{2} = 5 \) . | We see that as \( a \) ranges from 0 to 14, the pairs \( \left( {a{\;\operatorname{mod}\;3}, a{\;\operatorname{mod}\;5}}\right) \) range over all pairs \( \left( {{a}_{1},{a}_{2}}\right) \) with \( {a}_{1} \in \{ 0,1,2\} \) and \( {a}_{2} \in \{ 0,\ldots ,4\} \), with every pair being hit exactly once. | Yes |
Consider the residue classes modulo 6. These are as follows:\n\n\[ \left\lbrack 0\right\rbrack = \{ \ldots , - {12}, - 6,0,6,{12},\ldots \} \]\n\n\[ \left\lbrack 1\right\rbrack = \{ \ldots , - {11}, - 5,1,7,{13},\ldots \} \]\n\n\[ \left\lbrack 2\right\rbrack = \{ \ldots , - {10}, - 4,2,8,{14},\ldots \} \]\n\n\[ \left\l... | The addition table looks like this:\n\n<table><thead><tr><th>\( + \)</th><th>[0]</th><th>[1]</th><th>[2]</th><th>[3]</th><th>[4]</th><th>[5]</th></tr></thead><tr><td>[0]</td><td>[0]</td><td>[1]</td><td>[2]</td><td>[3]</td><td>[4]</td><td>[5]</td></tr><tr><td>[1]</td><td>[1]</td><td>[2]</td><td>[3]</td><td>[4]</td><td>[... | Yes |
We list the elements of \( {\mathbb{Z}}_{15}^{ * } \), and for each \( \alpha \in {\mathbb{Z}}_{15}^{ * } \), we also give \( {\alpha }^{-1} \): | <table><tr><td>\( \alpha \)</td><td>[1]</td><td>[2]</td><td>[4]</td><td>[7]</td><td>[8]</td><td>[11]</td><td>[13]</td><td>[14]</td></tr><tr><td>\( {\alpha }^{-1} \)</td><td>[1]</td><td>[8]</td><td>[4]</td><td>[13]</td><td>[2]</td><td>[11]</td><td>[7]</td><td>[14]</td></tr></table> | Yes |
Theorem 2.8 (Chinese remainder map). Let \( {\left\{ {n}_{i}\right\} }_{i = 1}^{k} \) be a pairwise relatively prime family of positive integers, and let \( n \mathrel{\text{:=}} \mathop{\prod }\limits_{{i = 1}}^{k}{n}_{i} \) . Define the map\n\n\[ \n\theta : \;{\mathbb{Z}}_{n} \rightarrow {\mathbb{Z}}_{{n}_{1}} \times... | Proof. For (i), note that \( a \equiv {a}^{\prime }\left( {\;\operatorname{mod}\;n}\right) \) implies \( a \equiv {a}^{\prime }\left( {\;\operatorname{mod}\;{n}_{i}}\right) \) for \( i = 1,\ldots, k \) , and so the definition of \( \theta \) is unambiguous (it does not depend on the choice of \( a \) ).\n\n(ii) follows... | Yes |
Theorem 2.9. Let \( {\left\{ {n}_{i}\right\} }_{i = 1}^{k} \) be a pairwise relatively prime family of positive integers, and let \( n \mathrel{\text{:=}} \mathop{\prod }\limits_{{i = 1}}^{k}{n}_{i} \). Then\n\n\[ \varphi \left( n\right) = \mathop{\prod }\limits_{{i = 1}}^{k}\varphi \left( {n}_{i}\right) \] | Proof. Consider the map \( \theta : {\mathbb{Z}}_{n} \rightarrow {\mathbb{Z}}_{{n}_{1}} \times \cdots \times {\mathbb{Z}}_{{n}_{k}} \) in Theorem 2.8. By parts (ii) and (iii.d) of that theorem, restricting \( \theta \) to \( {\mathbb{Z}}_{n}^{ * } \) yields a one-to-one correspondence between \( {\mathbb{Z}}_{n}^{ * } ... | Yes |
Theorem 2.10. Let \( p \) be a prime and \( e \) be a positive integer. Then\n\n\[ \varphi \left( {p}^{e}\right) = {p}^{e - 1}\left( {p - 1}\right) . \] | Proof. The multiples of \( p \) among \( 0,1,\ldots ,{p}^{e} - 1 \) are\n\n\[ 0 \cdot p,1 \cdot p,\ldots ,\left( {{p}^{e - 1} - 1}\right) \cdot p, \]\n\nof which there are precisely \( {p}^{e - 1} \) . Thus, \( \varphi \left( {p}^{e}\right) = {p}^{e} - {p}^{e - 1} = {p}^{e - 1}\left( {p - 1}\right) \) . | Yes |
Example 2.9. Let \( n = 7 \) . For each value \( a = 1,\ldots ,6 \), we can compute successive powers of \( a \) modulo \( n \) to find its multiplicative order modulo \( n \) . | <table><thead><tr><th>\( i \)</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th></tr></thead><tr><td>\( {1}^{i}{\;\operatorname{mod}\;7} \)</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>\( {2}^{i}{\;\operatorname{mod}\;7} \)</td><td>2</td><td>4</td><td>1</td><td>2</td><td>4<... | Yes |
Theorem 2.13 (Euler’s theorem). Let \( n \) be a positive integer and \( \alpha \in {\mathbb{Z}}_{n}^{ * } \) . Then \( {\alpha }^{\varphi \left( n\right) } = 1 \) . In particular, the multiplicative order of \( \alpha \) divides \( \varphi \left( n\right) \) . | Proof. Since \( \alpha \in {\mathbb{Z}}_{n}^{ * } \), for every \( \beta \in {\mathbb{Z}}_{n}^{ * } \) we have \( {\alpha \beta } \in {\mathbb{Z}}_{n}^{ * } \), and so we may define the \ | No |
Theorem 2.14 (Fermat’s little theorem). For every prime \( p \), and every \( \alpha \in {\mathbb{Z}}_{p} \) , we have \( {\alpha }^{p} = \alpha \) . | Proof. If \( \alpha = 0 \), the statement is obviously true. Otherwise, \( \alpha \in {\mathbb{Z}}_{p}^{ * } \), and by Theorem 2.13 we have \( {\alpha }^{p - 1} = 1 \) . Multiplying this equation by \( \alpha \) yields \( {\alpha }^{p} = \alpha \) . | Yes |
Theorem 2.15. Suppose \( \alpha \in {\mathbb{Z}}_{n}^{ * } \) has multiplicative order \( k \) . Then for every \( m \in \mathbb{Z} \) , the multiplicative order of \( {\alpha }^{m} \) is \( k/\gcd \left( {m, k}\right) \) . | Proof. Applying Theorem 2.12 to \( {\alpha }^{m} \), we see that the multiplicative order of \( {\alpha }^{m} \) is the smallest positive integer \( \ell \) such that \( {\alpha }^{m\ell } = 1 \) . But we have\n\n\[{\alpha }^{m\ell } = 1 \Leftrightarrow m\ell \equiv 0\left( {\;\operatorname{mod}\;k}\right) \text{ (appl... | Yes |
Theorem 2.16. Let \( n \) be a positive integer, let \( \alpha ,\beta \in {\mathbb{Z}}_{n}^{ * } \), and let \( m \) be any integer.\n\n(i) If \( \alpha \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \), then \( {\alpha }^{-1} \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \) .\n\n(ii) If \( \alpha \in {\left( {\math... | Proof. For (i), if \( \alpha = {\gamma }^{m} \), then \( {\alpha }^{-1} = {\left( {\gamma }^{-1}\right) }^{m} \).\n\nFor (ii), if \( \alpha = {\gamma }^{m} \) and \( \beta = {\delta }^{m} \), then \( {\alpha \beta } = {\left( \gamma \delta \right) }^{m} \).\n\nFor (iii), suppose that \( \alpha \in {\left( {\mathbb{Z}}_... | Yes |
Theorem 2.17. Let \( n \) be a positive integer. For each \( \alpha \in {\mathbb{Z}}_{n}^{ * } \), and all \( \ell, m \in \mathbb{Z} \) with \( \gcd \left( {\ell, m}\right) = 1 \), if \( {\alpha }^{\ell } \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \), then \( \alpha \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} ... | Proof. Suppose \( {\alpha }^{\ell } = {\beta }^{m} \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \) . Since \( \gcd \left( {\ell, m}\right) = 1 \), there exist integers \( s \) and \( t \) such that \( \ell s + {mt} = 1 \) . We then have\n\n\[ \alpha = {\alpha }^{\ell s + {mt}} = {\alpha }^{\ell s}{\alpha }^{mt} = {\... | Yes |
Theorem 2.18. Let \( p \) be an odd prime and \( \beta \in {\mathbb{Z}}_{p} \) . Then \( {\beta }^{2} = 1 \) if and only if \( \beta = \pm 1 \) . | Proof. Clearly, if \( \beta = \pm 1 \), then \( {\beta }^{2} = 1 \) . Conversely, suppose that \( {\beta }^{2} = 1 \) . Write \( \beta = \left\lbrack b\right\rbrack \), where \( b \in \mathbb{Z} \) . Then we have \( {b}^{2} \equiv 1\left( {\;\operatorname{mod}\;p}\right) \), which means that\n\n\[ p \mid \left( {{b}^{2... | Yes |
Theorem 2.19. Let \( p \) be an odd prime and \( \gamma ,\beta \in {\mathbb{Z}}_{p}^{ * } \) . Then \( {\gamma }^{2} = {\beta }^{2} \) if and only if \( \gamma = \pm \beta \) . | Proof. This follows from the previous theorem:\n\n\[ \n{\gamma }^{2} = {\beta }^{2} \Leftrightarrow {\left( \gamma /\beta \right) }^{2} = 1 \Leftrightarrow \gamma /\beta = \pm 1 \Leftrightarrow \gamma = \pm \beta .\n\] | Yes |
Theorem 2.20. Let \( p \) be an odd prime. Then \( \left| {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2}\right| = \left( {p - 1}\right) /2 \) . | Proof. By the previous theorem, the \ | No |
Example 2.10. Let \( p = 7 \) . We can list the elements of \( {\mathbb{Z}}_{p}^{ * } \) as \[ \left\lbrack {\pm 1}\right\rbrack ,\left\lbrack {\pm 2}\right\rbrack ,\left\lbrack {\pm 3}\right\rbrack \text{.} \] | Squaring these, we see that \[ {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} = \{ {\left\lbrack 1\right\rbrack }^{2},{\left\lbrack 2\right\rbrack }^{2},{\left\lbrack 3\right\rbrack }^{2}\} = \{ \left\lbrack 1\right\rbrack ,\left\lbrack 4\right\rbrack ,\left\lbrack 2\right\rbrack \} . \] | Yes |
Theorem 2.22 (Wilson’s theorem). Let \( p \) be an odd prime. Then \( \mathop{\prod }\limits_{{\beta \in {\mathbb{Z}}_{p}^{ * }}}\beta = - 1 \) . | \[ \left( {p - 1}\right) ! \equiv - 1\left( {\;\operatorname{mod}\;p}\right) \] | Yes |
Theorem 2.23. Let \( p \) be an odd prime and \( \alpha ,\beta \in {\mathbb{Z}}_{p}^{ * } \) . If \( \alpha \notin {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} \) and \( \beta \notin {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} \) , then \( {\alpha \beta } \in {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} \) . | Proof. Suppose \( \alpha \notin {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} \) and \( \beta \notin {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} \) . Then by Euler’s criterion, we have\n\n\[ \n{\alpha }^{\left( {p - 1}\right) /2} = - 1\text{ and }{\beta }^{\left( {p - 1}\right) /2} = - 1.\n\]\n\nTherefore,\n\n\[ \n{\left... | Yes |
Theorem 2.24. Let \( p \) be an odd prime, \( e \) be a positive integer, and \( \beta \in {\mathbb{Z}}_{{p}^{e}} \) . Then \( {\beta }^{2} = 1 \) if and only if \( \beta = \pm 1 \) . | Proof. Clearly, if \( \beta = \pm 1 \), then \( {\beta }^{2} = 1 \) . Conversely, suppose that \( {\beta }^{2} = 1 \) . Write \( \beta = \left\lbrack b\right\rbrack \), where \( b \in \mathbb{Z} \) . Then we have \( {b}^{2} \equiv 1\left( {\;\operatorname{mod}\;{p}^{e}}\right) \), which means that\n\n\[ \n{p}^{e} \mid ... | Yes |
Theorem 2.30. Let \( p \) be an odd prime, \( e \) be a positive integer, and \( a \) be any integer. Then \( a \) is a quadratic residue modulo \( {p}^{e} \) if and only if \( a \) is a quadratic residue modulo \( p \) . | Proof. Suppose that \( a \) is a quadratic residue modulo \( {p}^{e} \) . Then \( a \) is not divisible by \( p \) and \( a \equiv {b}^{2}\left( {\;\operatorname{mod}\;{p}^{e}}\right) \) for some integer \( b \) . It follows that \( a \equiv {b}^{2}\left( {\;\operatorname{mod}\;p}\right) \), and so \( a \) is a quadrat... | Yes |
Theorem 2.31. Let \( p \) be an odd prime. Then -1 is a quadratic residue modulo \( p \) if and only \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . | Proof. By Euler’s criterion,-1 is a quadratic residue modulo \( p \) if and only if \( {\left( -1\right) }^{\left( {p - 1}\right) /2} \equiv 1\left( {\;\operatorname{mod}\;p}\right) \) . If \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \), then \( \left( {p - 1}\right) /2 \) is even, and so \( {\left( -1\right) ... | Yes |
Theorem 2.32. Let \( p \) be a prime with \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) ,\gamma \in {\mathbb{Z}}_{p}^{ * } \smallsetminus {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} \), and \( \beta \mathrel{\text{:=}} {\gamma }^{\left( {p - 1}\right) /4} \) . Then \( {\beta }^{2} = - 1 \) . | Proof. This is a simple calculation, based on Euler's criterion:\n\n\[ \n{\beta }^{2} = {\gamma }^{\left( {p - 1}\right) /2} = - 1 \n\] | Yes |
Theorem 2.33 (Thue’s lemma). Let \( n, b,{r}^{ * },{t}^{ * } \in \mathbb{Z} \), with \( 0 < {r}^{ * } \leq n < {r}^{ * }{t}^{ * } \) . Then there exist \( r, t \in \mathbb{Z} \) with\n\n\[ r \equiv {bt}\left( {\;\operatorname{mod}\;n}\right) ,\;\left| r\right| < {r}^{ * },\text{ and }0 < \left| t\right| < {t}^{ * }.\] | Proof. For \( i = 0,\ldots ,{r}^{ * } - 1 \) and \( j = 0,\ldots ,{t}^{ * } - 1 \), we define the number \( {v}_{ij} \mathrel{\text{:=}} i - {bj} \) . Since we have defined \( {r}^{ * }{t}^{ * } \) numbers, and \( {r}^{ * }{t}^{ * } > n \), two of these numbers must lie in the same residue class modulo \( n \) ; that i... | Yes |
Theorem 2.34 (Fermat’s two squares theorem). Let \( p \) be an odd prime. Then \( p = {r}^{2} + {t}^{2} \) for some \( r, t \in \mathbb{Z} \) if and only if \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . | Proof. One direction is easy. Suppose \( p \equiv 3\left( {\;\operatorname{mod}\;4}\right) \) . It is easy to see that the square of every integer is congruent to either 0 or 1 modulo 4 ; therefore, the sum of two squares is congruent to either 0,1 , or 2 modulo 4 , and so can not be congruent to \( p \) modulo 4 (let ... | Yes |
Theorem 2.35. There are infinitely many primes \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . | Proof. Suppose there were only finitely many such primes, \( {p}_{1},\ldots ,{p}_{k} \) . Set \( M \mathrel{\text{:=}} \) \( \mathop{\prod }\limits_{{i = 1}}^{k}{p}_{i} \) and \( N \mathrel{\text{:=}} 4{M}^{2} + 1 \) . Let \( p \) be any prime dividing \( N \) . Evidently, \( p \) is not among the \( {p}_{i} \) ’s, sin... | Yes |
Theorem 2.36. There are infinitely many primes \( p \equiv 3\left( {\;\operatorname{mod}\;4}\right) \) . | Proof. Suppose there were only finitely many such primes, \( {p}_{1},\ldots ,{p}_{k} \) . Set \( M \mathrel{\text{:=}} \) \( \mathop{\prod }\limits_{{i = 1}}^{k}{p}_{i} \) and \( N \mathrel{\text{:=}} {4M} - 1 \) . Since \( N \equiv 3\left( {\;\operatorname{mod}\;4}\right) \), there must be some prime \( p \equiv 3\lef... | Yes |
Theorem 2.37. If \( f \) is a multiplicative arithmetic function, and if \( n = {p}_{1}^{{e}_{1}}\cdots {p}_{r}^{{e}_{r}} \) is the prime factorization of \( n \), then \( f\left( n\right) = f\left( {p}_{1}^{{e}_{1}}\right) \cdots f\left( {p}_{r}^{{e}_{r}}\right) \) . | Proof. Exercise. | No |
Theorem 2.38. Let \( f \) be a multiplicative arithmetic function. If \( n = {p}_{1}^{{e}_{1}}\cdots {p}_{r}^{{e}_{r}} \) is the prime factorization of \( n \), then\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) f\left( d\right) = \left( {1 - f\left( {p}_{1}\right) }\right) \cdots \left( {1 - f\left( {p}_... | Proof. The only non-zero terms appearing in the sum on the left-hand side of (2.9) are those corresponding to divisors \( d \) of the form \( {p}_{{i}_{1}}\cdots {p}_{{i}_{\ell }} \), where \( {p}_{{i}_{1}},\ldots ,{p}_{{i}_{\ell }} \) are distinct; the value contributed to the sum by such a term is \( {\left( -1\right... | Yes |
Theorem 2.39 (Möbius inversion formula). Let \( f \) and \( F \) be arithmetic functions. Then \( F = 1 \star f \) if and only if \( f = \mu \star F \) . | Proof. If \( F = 1 \star f \), then\n\n\[ \mu \star F = \mu \star \left( {I \star f}\right) = \left( {\mu \star I}\right) \star f = I \star f = f, \]\n\nand conversely, if \( f = \mu \star F \), then\n\n\[ I \star f = I \star \left( {\mu \star F}\right) = \left( {I \star \mu }\right) \star F = I \star F = F. \] | Yes |
Theorem 2.40. For every positive integer \( n \), we have \( \mathop{\sum }\limits_{{d \mid n}}\varphi \left( d\right) = n \) . | Proof. Let us define the arithmetic functions \( N\left( n\right) \mathrel{\text{:=}} n \) and \( M\left( n\right) \mathrel{\text{:=}} 1/n \) . Our goal is to show that \( N = 1 \star \varphi \), and by Möbius inversion, it suffices to show that \( \mu \star N = \varphi \) . If \( n = {p}_{1}^{{e}_{1}}\cdots {p}_{r}^{{... | Yes |
Let \( f\left( x\right) \mathrel{\text{:=}} {x}^{2} \) and \( g\left( x\right) \mathrel{\text{:=}} 2{x}^{2} - {10x} + 1 \) . Then \( f = O\left( g\right) \) and \( f = \Omega \left( g\right) \) . Indeed, \( f = \Theta \left( g\right) \) . | Indeed, \( f = \Theta \left( g\right) \) . | No |
Theorem 3.1. Let \( x \) and \( y \) be integers such that\n\n\[ 0 \leq x = {x}^{\prime }{2}^{n} + s\text{ and }0 < y = {y}^{\prime }{2}^{n} \]\n\nfor some integers \( n, s,{x}^{\prime },{y}^{\prime } \), with \( n \geq 0 \) and \( 0 \leq s < {2}^{n} \) . Then \( \lfloor x/y\rfloor = \left\lfloor {{x}^{\prime }/{y}^{\p... | Proof. We have\n\n\[ \frac{x}{y} = \frac{{x}^{\prime }}{{y}^{\prime }} + \frac{s}{{y}^{\prime }{2}^{n}} \geq \frac{{x}^{\prime }}{{y}^{\prime }} \]\n\nIt follows immediately that \( \lfloor x/y\rfloor \geq \left\lfloor {{x}^{\prime }/{y}^{\prime }}\right\rfloor \) .\n\nWe also have\n\n\[ \frac{x}{y} = \frac{{x}^{\prime... | Yes |
Theorem 3.2. Let \( x \) and \( y \) be integers such that\n\n\[ 0 \leq x = {x}^{\prime }{2}^{n} + s\text{ and }0 < y = {y}^{\prime }{2}^{n} + t \]\n\nfor some integers \( n, s, t,{x}^{\prime },{y}^{\prime } \) with \( n \geq 0,0 \leq s < {2}^{n} \), and \( 0 \leq t < {2}^{n} \) . Further, suppose that \( 2{y}^{\prime ... | Proof. We have \( x/y \leq x/{y}^{\prime }{2}^{n} \), and so \( \lfloor x/y\rfloor \leq \left\lfloor {x/{y}^{\prime }{2}^{n}}\right\rfloor \), and by the previous theorem, \( \left\lfloor {x/{y}^{\prime }{2}^{n}}\right\rfloor = \left\lfloor {{x}^{\prime }/{y}^{\prime }}\right\rfloor \) . That proves the first inequalit... | Yes |
Suppose \( e = {37} = {\left( {100101}\right) }_{2} \) . The above algorithm performs the following operations in this case: | \n\n\[ \text{// computed exponent (in binary)} \]\n\n\[ \beta \leftarrow \left\lbrack 1\right\rbrack \;//0 \]\n\n\[ \beta \leftarrow {\beta }^{2},\beta \leftarrow \beta \cdot \alpha \;//1 \]\n\n\[ \beta \leftarrow {\beta }^{2}\;\text{//10} \]\n\n\[ \beta \leftarrow {\beta }^{2} \]\n\n\[ \text{// 100} \]\n\n\[ \beta \le... | Yes |
Theorem 4.1. Let \( a, b \) be integers, with \( a \geq b \geq 0 \) . Using the division with remainder property, define the integers \( {r}_{0},{r}_{1},\ldots ,{r}_{\lambda + 1} \) and \( {q}_{1},\ldots ,{q}_{\lambda } \), where \( \lambda \geq 0 \) , as follows:\n\n\[ a = {r}_{0} \]\n\n\[ b = {r}_{1} \]\n\n\[ {r}_{0}... | Proof. For the first statement, one sees that for \( i = 1,\ldots ,\lambda \), we have \( {r}_{i - 1} = \) \( {r}_{i}{q}_{i} + {r}_{i + 1} \), from which it follows that the common divisors of \( {r}_{i - 1} \) and \( {r}_{i} \) are the same as the common divisors of \( {r}_{i} \) and \( {r}_{i + 1} \), and hence \( \g... | Yes |
Suppose \( a = {100} \) and \( b = {35} \) . Then the numbers appearing in Theorem 4.1 are easily computed as follows: | <table><tr><td>\( i \)</td><td>0</td><td>1</td><td>2</td><td>3</td><td>4</td></tr><tr><td>\( {r}_{i} \)</td><td>100</td><td>35</td><td>30</td><td>5</td><td>0</td></tr><tr><td>\( {q}_{i} \)</td><td></td><td>2</td><td>1</td><td>6</td><td></td></tr></table>\n\n\nSo we have \( \gcd \left( {a, b}\right) = {r}_{3} = 5 \) . | Yes |
Theorem 4.2. Euclid’s algorithm runs in time \( O\left( {\operatorname{len}\left( a\right) \operatorname{len}\left( b\right) }\right) \) . | Proof. We may assume that \( b > 0 \) . With notation as in Theorem 4.1, the running time is \( O\left( T\right) \), where\n\n\[ T = \mathop{\sum }\limits_{{i = 1}}^{\lambda }\operatorname{len}\left( {r}_{i}\right) \operatorname{len}\left( {q}_{i}\right) \leq \operatorname{len}\left( b\right) \mathop{\sum }\limits_{{i ... | No |
Theorem 4.3. Let \( a, b,{r}_{0},\ldots ,{r}_{\lambda + 1} \) and \( {q}_{1},\ldots ,{q}_{\lambda } \) be as in Theorem 4.1. Define integers \( {s}_{0},\ldots ,{s}_{\lambda + 1} \) and \( {t}_{0},\ldots ,{t}_{\lambda + 1} \) as follows:\n\n\[ \n{s}_{0} \mathrel{\text{:=}} 1,\;{t}_{0} \mathrel{\text{:=}} 0, \]\n\n\[ \n{... | Proof. (i) is easily proved by induction on \( i \) . For \( i = 0,1 \), the statement is clear. For \( i = 2,\ldots ,\lambda + 1 \), we have\n\n\[ \na{s}_{i} + b{t}_{i} = a\left( {{s}_{i - 2} - {s}_{i - 1}{q}_{i - 1}}\right) + b\left( {{t}_{i - 2} - {t}_{i - 1}{q}_{i - 1}}\right) \]\n\n\[ = \left( {a{s}_{i - 2} + b{t}... | Yes |
Example 4.2. We continue with Example 4.1. The \( {s}_{i} \) ’s and \( {t}_{i} \) ’s are easily computed from the \( {q}_{i} \) ’s: | So we have \( \gcd \left( {a, b}\right) = 5 = - a + {3b} \) . | Yes |
Theorem 4.5. Suppose we are given integers \( n, b \), where \( 0 \leq b < n \) . Then in time \( O\left( {\operatorname{len}{\left( n\right) }^{2}}\right) \), we can determine if \( b \) is relatively prime to \( n \), and if so, compute \( {b}^{-1}{\;\operatorname{mod}\;n} \) . | Proof. We may assume \( n > 1 \), since when \( n = 1 \), we have \( b = 0 = {b}^{-1}{\;\operatorname{mod}\;n} \) . We run the extended Euclidean algorithm on input \( n, b \), obtaining integers \( d, s \), and \( t \) , such that \( d = \gcd \left( {n, b}\right) \) and \( {ns} + {bt} = d \) . If \( d \neq 1 \), then ... | Yes |
Suppose we are given integers \( a, b, n \), where \( 0 \leq a < n \), and \( 0 \leq b < n \), and we want to compute a solution \( z \) to the congruence \( {az} \equiv b\left( {\;\operatorname{mod}\;n}\right) \) , or determine that no such solution exists. | Based on the discussion in Example 2.5, the following algorithm does the job:\n\n\( d \leftarrow \gcd \left( {a, n}\right) \)\n\nif \( d \nmid b \) then\n\noutput \ | No |
Theorem 4.6 (Effective Chinese remainder theorem). Suppose we are given integers \( {n}_{1},\ldots ,{n}_{k} \) and \( {a}_{1},\ldots ,{a}_{k} \), where the family \( {\left\{ {n}_{i}\right\} }_{i = 1}^{k} \) is pairwise relatively prime, and where \( {n}_{i} > 1 \) and \( 0 \leq {a}_{i} < {n}_{i} \) for \( i = 1,\ldots... | Proof. The algorithm is a straightforward implementation of the proof of Theorem 2.6, and runs as follows:\n\n\[ n \leftarrow \mathop{\prod }\limits_{{i = 1}}^{k}{n}_{i} \]\n\nfor \( i \leftarrow 1 \) to \( k \) do\n\n\[ {n}_{i}^{ * } \leftarrow n/{n}_{i},{b}_{i} \leftarrow {n}_{i}^{ * }{\;\operatorname{mod}\;{n}_{i}},... | No |
Theorem 4.7 (Effective Thue’s lemma). Let \( n, b,{r}^{ * },{t}^{ * } \in \mathbb{Z} \), with \( 0 \leq b < n \) and \( 0 < {r}^{ * } \leq n < {r}^{ * }{t}^{ * } \) . Further, let \( \operatorname{EEA}\left( {n, b}\right) = {\left\{ \left( {r}_{i},{s}_{i},{t}_{i}\right) \right\} }_{i = 0}^{\lambda + 1} \), and let \( j... | Proof. Since \( {r}_{0} = n \geq {r}^{ * } > 0 = {r}_{\lambda + 1} \), the value of the index \( j \) is well defined; moreover, \( j \geq 1 \) and \( {r}_{j - 1} \geq {r}^{ * } \) . It follows that\n\n\[ \left| {t}_{j}\right| \leq n/{r}_{j - 1}\text{(by part (v) of Theorem 4.3)} \]\n\n\[ \leq n/{r}^{ * } \]\n\n\[ < {t... | Yes |
Example 4.4. One can check that \( p \mathrel{\text{:=}} {1009} \) is prime and \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \). Let us express \( p \) as a sum of squares using the above algorithm. First, we need to find a square root of -1 modulo \( p \). Let us just try a random number, say 17, and raise thi... | The first \( {r}_{j} \) that falls below the threshold \( {r}^{ * } = \lfloor \sqrt{1009}\rfloor + 1 = {32} \) is at \( j = 4 \), and so we set \( r \mathrel{\text{:=}} {28} \) and \( t \mathrel{\text{:=}} - {15} \). One verifies that \( {r}^{2} + {t}^{2} = {28}^{2} + {15}^{2} = {1009} = p \). | Yes |
Theorem 4.8. Let \( n, b,{r}^{ * },{t}^{ * } \in \mathbb{Z} \) with \( {r}^{ * } \geq 0,{t}^{ * } > 0 \), and \( n > 2{r}^{ * }{t}^{ * } \) . Further, suppose that \( r, t,{r}^{\prime },{t}^{\prime } \in \mathbb{Z} \) satisfy\n\n\[ r \equiv {bt}\left( {\;\operatorname{mod}\;n}\right) ,\;\left| r\right| \leq {r}^{ * },\... | Proof. Consider the two congruences\n\n\[ r \equiv {bt}\left( {\;\operatorname{mod}\;n}\right) \]\n\n\[ {r}^{\prime } \equiv b{t}^{\prime }\left( {\;\operatorname{mod}\;n}\right) \]\n\nSubtracting \( t \) times the second from \( {t}^{\prime } \) times the first, we obtain\n\n\[ r{t}^{\prime } - {r}^{\prime }t \equiv 0... | Yes |
Theorem 4.9 (Rational reconstruction). Let \( n, b,{r}^{ * },{t}^{ * } \in \mathbb{Z} \) with \( 0 \leq b < n \) , \( 0 \leq {r}^{ * } < n \), and \( {t}^{ * } > 0 \) . Further, let \( \operatorname{EEA}\left( {n, b}\right) = {\left\{ \left( {r}_{i},{s}_{i},{t}_{i}\right) \right\} }_{i = 0}^{\lambda + 1} \), and let \(... | Proof. Since \( {r}_{0} = n > {r}^{ * } \geq 0 = {r}_{\lambda + 1} \), the value of \( j \) is well defined, and moreover, \( j \geq 1 \), and we have the inequalities \[ 0 \leq {r}_{j} \leq {r}^{ * } < {r}_{j - 1},0 < \left| {t}_{j}\right| ,\left| r\right| \leq {r}^{ * },\text{ and }0 < \left| t\right| \leq {t}^{ * },... | Yes |
Alice chooses integers \( s, t \), with \( 0 \leq s < t \leq {1000} \), and tells Bob the high-order seven digits in the decimal expansion of \( z \mathrel{\text{:=}} s/t \), from which Bob should be able to compute \( z \) . Suppose \( s = {511} \) and \( t = {710} \) . Then \( s/t = {0.7197183098591549}\cdots \) . Bo... | Running the extended Euclidean algorithm on input \( n, b \), Bob obtains the data in Fig. 4.1. The first \( {r}_{j} \) that meets the threshold \( {r}^{ * } = {1000} \) is at \( j = {10} \), and Bob reads off \( {s}^{\prime } = {511} \) and \( {t}^{\prime } = - {710} \), from which he obtains \( z = - {s}^{\prime }/{t... | Yes |
Suppose we want to encode a 1024-bit message as a sequence of 16- bit blocks, so that the above scheme can correct up to 3 corrupted blocks. Without any error correction, we would need just \( {1024}/{16} = {64} \) blocks. However, to correct this many errors, we need a few extra blocks; in fact, 7 will do. | Of course, a 1024-bit message can naturally be viewed as an integer \( a \) in the set \( \left\{ {0,\ldots ,{2}^{1024} - 1}\right\} \), and the \( i \) th 16-bit block in the encoding can be viewed as an integer \( {a}_{i} \) in the set \( \left\{ {0,\ldots ,{2}^{16} - 1}\right\} \) . Setting \( k \mathrel{\text{:=}} ... | Yes |
Theorem 5.1 (Chebyshev's theorem). We have\n\n\[ \pi \left( x\right) = \Theta \left( {x/\log x}\right) . \] | It is not too difficult to prove this theorem, which we now proceed to do in several steps. We begin with some elementary bounds on binomial coefficients (see §A2): | No |
Lemma 5.2. If \( m \) is a positive integer, then\n\n\[ \left( \begin{matrix} {2m} \\ m \end{matrix}\right) \geq {2}^{2m}/{2m}\text{ and }\left( \begin{matrix} {2m} + 1 \\ m \end{matrix}\right) < {2}^{2m} \]\n | Proof. As \( \left( \begin{matrix} {2m} \\ m \end{matrix}\right) \) is the largest binomial coefficient in the binomial expansion of \( {\left( 1 + 1\right) }^{2m} \), we have\n\n\[ {2}^{2m} = \mathop{\sum }\limits_{{i = 0}}^{{2m}}\left( \begin{matrix} {2m} \\ i \end{matrix}\right) = 1 + \mathop{\sum }\limits_{{i = 1}}... | Yes |
Lemma 5.3. Let \( n \) be a positive integer. For every prime \( p \), we have\n\n\[ \n{v}_{p}\left( {n!}\right) = \mathop{\sum }\limits_{{k \geq 1}}\left\lfloor {n/{p}^{k}}\right\rfloor \n\] | Proof. For all positive integers \( j, k \), define \( {d}_{jk} \mathrel{\text{:=}} 1 \) if \( {p}^{k} \mid j \), and \( {d}_{jk} \mathrel{\text{:=}} 0 \) , otherwise. Observe that \( {v}_{p}\left( j\right) = \mathop{\sum }\limits_{{k \geq 1}}{d}_{jk} \) (this sum is actually finite, since \( {d}_{jk} = 0 \) for all su... | Yes |
Theorem 5.4. \( \pi \left( n\right) \geq \frac{1}{2}\left( {\log 2}\right) n/\log n \) for every integer \( n \geq 2 \) . | Proof. Let \( m \) be a positive integer, and consider the binomial coefficient\n\n\[ N \mathrel{\text{:=}} \left( \begin{matrix} {2m} \\ m \end{matrix}\right) = \frac{\left( {2m}\right) !}{{\left( m!\right) }^{2}} \]\n\nIt is clear that \( N \) is divisible only by primes \( p \) up to \( {2m} \) . Applying Lemma 5.3 ... | Yes |
Theorem 5.5. We have\n\n\[ \vartheta \left( x\right) = \Theta \left( {\pi \left( x\right) \log x}\right) . \] | Proof. On the one hand, we have\n\n\[ \vartheta \left( x\right) = \mathop{\sum }\limits_{{p \leq x}}\log p \leq \log x\mathop{\sum }\limits_{{p \leq x}}1 = \pi \left( x\right) \log x. \]\n\nOn the other hand, we have\n\n\[ \vartheta \left( x\right) = \mathop{\sum }\limits_{{p \leq x}}\log p \geq \mathop{\sum }\limits_{... | Yes |
Theorem 5.6. \( \vartheta \left( x\right) < 2\left( {\log 2}\right) x \) for every real number \( x \geq 1 \) . | Proof. It suffices to prove that \( \vartheta \left( n\right) < 2\left( {\log 2}\right) n \) for every positive integer \( n \), since then \( \vartheta \left( x\right) = \vartheta \left( {\lfloor x\rfloor }\right) < 2\left( {\log 2}\right) \lfloor x\rfloor \leq 2\left( {\log 2}\right) x \) . We prove this by induction... | Yes |
Theorem 5.8 (Bertrand’s postulate). For every positive integer \( m \), we have\n\n\[ \pi \left( {2m}\right) - \pi \left( m\right) > \frac{m}{3\log \left( {2m}\right) }.\] | The proof uses Theorem 5.6, along with a more careful re-working of the proof of Theorem 5.4. The theorem is clearly true for \( m \leq 2 \), so we may assume that \( m \geq 3 \) . As in the proof of the Theorem 5.4, define \( N \mathrel{\text{:=}} \left( \begin{matrix} {2m} \\ m \end{matrix}\right) \), and recall that... | Yes |
Lemma 5.9. Let \( m \geq 3 \) and \( N \mathrel{\text{:=}} \left( \begin{matrix} {2m} \\ m \end{matrix}\right) \) . For all primes \( p \), we have:\n\n\[ \n{p}^{{v}_{p}\left( N\right) } \leq {2m} \n\]\n\n(5.2)\n\n\[ \n\text{if}p > \sqrt{2m}\text{, then}{v}_{p}\left( N\right) \leq 1\text{;} \n\]\n\n(5.3)\n\n\[ \n\text{... | Proof. For (5.2), all terms with \( k > \log \left( {2m}\right) /\log p \) in (5.1) vanish, and hence \( {v}_{p}\left( N\right) \leq \log \left( {2m}\right) /\log p \), from which it follows that \( {p}^{{v}_{p}\left( N\right) } \leq {2m} \).\n\n(5.3) follows immediately from (5.2).\n\nFor (5.4), if \( {2m}/3 < p \leq ... | Yes |
Theorem 5.10. We have\n\n\[ \mathop{\sum }\limits_{{p \leq x}}\frac{1}{p} = \log \log x + O\\left( 1\\right) \] | The proof of this theorem, while not difficult, is a bit technical, and we proceed in several steps. | No |
Theorem 5.11. We have\n\n\[ \mathop{\sum }\limits_{{p \leq x}}\frac{\log p}{p} = \log x + O\left( 1\right) \] | Proof. Let \( n \mathrel{\text{:=}} \lfloor x\rfloor \) . The idea of the proof is to estimate \( \log \left( {n!}\right) \) in two different ways. By Lemma 5.3, we have\n\n\[ \log \left( {n!}\right) = \mathop{\sum }\limits_{{p \leq n}}\mathop{\sum }\limits_{{k \geq 1}}\left\lfloor {n/{p}^{k}}\right\rfloor \log p = \ma... | Yes |
Theorem 5.12 (Abel’s identity). Let \( {\left\{ {c}_{i}\right\} }_{i = k}^{\infty } \) be a sequence of real numbers, and for each real number \( t \), define\n\n\[ \nC\left( t\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{k \leq i \leq t}}{c}_{i} \n\]\n\nFurther, suppose that \( f\left( t\right) \) is a function ... | Proof. Let \( n \mathrel{\text{:=}} \lfloor x\rfloor \) . We have\n\n\[ \n\mathop{\sum }\limits_{{i = k}}^{n}{c}_{i}f\left( i\right) = C\left( k\right) f\left( k\right) + \mathop{\sum }\limits_{{i = k + 1}}^{n}\left\lbrack {C\left( i\right) - C\left( {i - 1}\right) }\right\rbrack f\left( i\right) \n\]\n\n\[ \n= \mathop... | Yes |
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