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Find a 90% confidence interval for the true (population) mean of statistics exam scores.
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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\n\[ {CL} = {0.90}\text{so}\alpha = 1 - {CL} = 1 - {0.90} = {0.10} \]\n\n\[ \frac{\alpha }{2} = {0.05}\;{z}_{\frac{\alpha }{2}} = {z}_{.05} \]\n\nThe area to the right of \( {z}_{.05} \) is 0.05 and the area to the left of \( {z}_{.05} \) is \( 1 - {0.05} = {0.95} \)\n\n\[ {z}_{\frac{\alpha }{2}} = {z}_{.05} = {1.645} \]\n\nusing invNorm \( \left( {{0.95},0,1}\right) \) on the TI-83,83+,84+ calculators. This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the Standard Normal distribution.\n\n\[ {EBM} = {1.645} \cdot \left( \frac{3}{\sqrt{36}}\right) = {0.8225} \]\n\n\[ \bar{x} - {EBM} = {68} - {0.8225} = {67.1775} \]\n\n\[ \bar{x} + {EBM} = {68} + {0.8225} = {68.8225} \]\n\nThe 90% confidence interval is (67.1775, 68.8225).
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Yes
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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.
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- If we know that the sample mean is \( {68} : {EBM} = {68.82} - {68} = {0.82} \)\n- If we don’t know the sample mean: \( {EBM} = \frac{\left( {68.82} - {67.18}\right) }{2} = {0.82} \)
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Yes
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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 \( {95}\% \) confidence level, compute a confidence interval estimate for the true proportion of adults residents of this city who have cell phones.
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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 \( = {421} \)\n\n\[ \n{p}^{\prime } = \frac{x}{n} = \frac{421}{500} = {0.842} \n\]\n\n\( {p}^{\prime } = {0.842} \) is the sample proportion; this is the point estimate of the population proportion.\n\n\[ \n{q}^{\prime } = 1 - {p}^{\prime } = 1 - {0.842} = {0.158} \n\]\n\nSince \( \mathrm{{CL}} = {0.95} \), then \( \alpha = 1 - \mathrm{{CL}} = 1 - {0.95} = {0.05}\;\frac{\alpha }{2} = {0.025} \) .\n\nThen \( {z}_{\frac{\alpha }{2}} = {z}_{.025} = {1.96} \)\n\nUse the TI-83, 83+ or 84+ calculator command invNorm(0.975,0,1) to find \( {z}_{.025} \) . Remember that the area to the right of \( {z}_{.{025}} \) is 0.025 and the area to the left of \( {z}_{0.025} \) is 0.975 . This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.\n\n\[ \n{EBP} = {z}_{\frac{\alpha }{2}} \cdot \sqrt{\frac{{p}^{\prime } \cdot {q}^{\prime }}{n}} = {1.96} \cdot \sqrt{\frac{\left( {0.842}\right) \cdot \left( {0.158}\right) }{500}} = {0.032} \n\]\n\n\[ \n{p}^{\prime } - \mathrm{{EBP}} = {0.842} - {0.032} = {0.81} \n\]\n\n\[ \n{p}^{\prime } + \mathrm{{EBP}} = {0.842} + {0.032} = {0.874} \n\]\n\nThe confidence interval for the true binomial population proportion is \( \left( {{p}^{\prime } - \mathrm{{EBP}},{p}^{\prime } + \mathrm{{EBP}}}\right) = \left( {{0.810},{0.874}}\right) . \)
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Yes
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We want to test whether the mean grade point average in American colleges is different from 2.0 (out of 4.0).
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\[ {H}_{o} : \mu = {2.0}\;{H}_{a} : \mu \neq {2.0} \]
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No
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We want to test if college students take less than five years to graduate from college, on the average.
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\[ {H}_{o} : \mu \geq 5\;{H}_{a} : \mu < 5 \]
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No
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Suppose the null hypothesis, \( {H}_{o} \), is: Frank’s rock climbing equipment is safe.
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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 really is safe. \( \beta = \) probability that Frank thinks his rock climbing equipment may be safe when, in fact, it is not safe. Notice that, in this case, the error with the greater consequence is the Type II error. (If Frank thinks his rock climbing equipment is safe, he will go ahead and use it.)
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Yes
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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.
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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 \( = \mathrm{P} \) (Type I error). \( \beta = \) probability that the emergency crew does not know if the victim is alive when, in fact, the victim is dead \( = \mathrm{P} \) (Type II error). The error with the greater consequence is the Type I error. (If the emergency crew thinks the victim is dead, they will not treat him.)
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Yes
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\( {H}_{o} : \mu = 5\;{H}_{a} : \mu < 5 \)
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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
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No
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[{H}_{o} : p \leq {0.2}\;{H}_{a} : p > {0.2}]
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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
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No
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\[ {H}_{o} : \mu = {16.43}\;{H}_{a} : \mu < {16.43} \]
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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\( \overline{X} \sim N\left( {\mu ,\frac{{\sigma }_{X}}{\sqrt{n}}}\right) \;\mathrm{{Therefore}},\overline{X} \sim N\left( {{16.43},\frac{0.8}{\sqrt{15}}}\right) \)\n\n\( \mu = {16.43} \) comes from \( {H}_{0} \) and not the data. \( \sigma = {0.8} \), and \( n = {15} \).\n\nCalculate the p-value using the normal distribution for a mean:\n\n---\n\np-value \( = P\left( {\bar{x} < {16}}\right) = {0.0187} \) where the sample mean in the problem is given as 16 .\n\np-value \( = {0.0187} \) (This is called the actual level of significance.) The p-value is the area to the left of the sample mean is given as 16 .\n\nGraph:\n\n\n\nFigure 9.1\n\n\( \mu = {16.43} \) comes from \( {H}_{o} \) . Our assumption is \( \mu = {16.43} \).\n\nInterpretation of the p-value: If \( {H}_{o} \) is true, there is a 0.0187 probability \( \left( {{1.87}\% }\right) \) that Jeffrey’s mean time to swim the 25-yard freestyle is 16 seconds or less. Because a \( {1.87}\% \) chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.\n\nCompare \( \alpha \) and the p-value:
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Yes
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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 with the result that 43 of the households have three cell phones.
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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 \( {P}^{\prime } \sim N\left( {{0.30},\sqrt{\frac{\left( {0.30}\right) \cdot \left( {0.70}\right) }{150}}}\right) \)
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Yes
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Can you use the information as it appears in the charts to conduct the goodness-of-fit test?
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No. Notice that the expected number of absences for the \
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No
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What are the degrees of freedom \( \left( {df}\right) \) ?
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There are 4 \
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Yes
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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 and is also a cell phone user while driving. Suppose, in a study of drivers who received speeding violations in the last year and who uses cell phones while driving, that 755 people were surveyed. Out of the 755,70 had a speeding violation and 685 did not; 305 were cell phone users while driving and 450 were not.
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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\( \begin{array}{l} \frac{y}{755} = \frac{70}{755} \cdot \frac{305}{755} \end{array} \)\n\nSolve for \( y : y = \frac{{70} \cdot {305}}{755} = {28.3} \)
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Yes
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Do male and female college students have the same distribution of living conditions?
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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 female college students.\n\nDegrees of Freedom (df):\n\n\( \mathrm{{df}} = \) number of columns \( - 1 = 4 - 1 = 3 \)\n\nDistribution for the test: \( {\chi }_{3}^{2} \)\n\nCalculate the test statistic: \( {\chi }^{2} = {10.1287}\; \) (calculator or computer)\n\nProbability statement: p-value \( = P\left( {{\chi }^{2} > {10.1287}}\right) = {0.0175} \)\n\nTI-83+ and TI-84 calculator: Press the MATRX key and arrow over to EDIT. Press 1: [A]. Press 2 ENTER 4 ENTER. Enter the table values by row. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C: \( ◈ \) -TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 10.1287 and the p-value \( = {0.0175} \) . Do the procedure a second time but arrow down to Draw instead of calculate.\n\nCompare \( \alpha \) and the p-value: Since no \( \alpha \) is given, assume \( \alpha = {0.05} \) . p-value \( = {0.0175} \) . \( \alpha > \) p-value.\n\nMake a decision: Since \( \alpha > \) p-value, reject \( {H}_{o} \) . This means that the distributions are not the same.\n\nConclusion: At a \( 5\% \) level of significance, from the data, there is sufficient evidence to conclude that the distributions of living conditions for male and female college students are not the same.\n\nNotice that the conclusion is only that the distributions are not the same. We cannot use the Test for Homogeneity to make any conclusions about how they differ.
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Yes
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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?
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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\[ \text{-}{\mathrm{H}}_{a} : {\sigma }^{2} > {5}^{2} \]
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No
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Find the equation that expresses the total cost in terms of the number of hours required to finish the word processing job.
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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\[ y = {31.50} + {32}\mathrm{x} \]\n
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Yes
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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.
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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 one variable occurring with low values of the other variable.\n\nYou can determine the strength of the relationship by looking at the scatter plot and seeing how close the points are to a line, a power function, an exponential function, or to some other type of function.\n\nWhen you look at a scatterplot, you want to notice the \( \mathbf{o} \) verall pattern and any deviations from the pattern. The following scatterplot examples illustrate these concepts.
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No
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Can you predict the final exam score of a random student if you know the third exam score?
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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 \
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No
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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.
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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.
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Yes
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What would you predict the final exam score to be for a student who scored a 66 on the third exam?
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145.27
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Yes
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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 \( a \mid \left( {b + c}\right) \) ;\n\n(v) \( a\left| {b\text{and}b}\right| c \) implies \( a \mid c \) .
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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 as an easy exercise for the reader to verify the remaining properties.
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No
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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 \( a = \pm 1 \) .
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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, \( a \mid b \) implies \( \left| a\right| \leq \left| b\right| \), and \( b\left| {a\text{implies}}\right| b\left| \leq \right| a\left| \text{; thus,}\right| a\left| = \right| b \mid \), and so \( a = \pm b \) . That proves the first statement. The second statement follows from the first by setting \( b \mathrel{\text{:=}} 1 \), and noting that \( 1 \mid a \) .
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Yes
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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 reordering of the primes.
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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 (possibly empty) of primes. We may prove this by induction on \( n \) . If \( n = 1 \), the statement is true, as \( n \) is the product of zero primes. Now let \( n > 1 \), and assume that every positive integer smaller than \( n \) can be expressed as a product of primes. If \( n \) is a prime, then the statement is true, as \( n \) is the product of one prime. Assume, then, that \( n \) is composite, so that there exist \( a, b \in \mathbb{Z} \) with \( 1 < a < n,1 < b < n \), and \( n = {ab} \) . By the induction hypothesis, both \( a \) and \( b \) can be expressed as a product of primes, and so the same holds for \( n \).
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Yes
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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 \( 0 \leq r < b \) .
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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 contains a minimum, we define \( r \) to be the smallest element of \( S \) . By definition, \( r \) is of the form \( r = a - {bq} \) for some \( q \in \mathbb{Z} \), and \( r \geq 0 \) . Also, we must have \( r < b \), since otherwise, \( r - b \) would be an element of \( S \) smaller than \( r \), contradicting the minimality of \( r \) ; indeed, if \( r \geq b \) , then we would have \( 0 \leq r - b = a - b\left( {q + 1}\right) \) .\n\nThat proves the existence of \( r \) and \( q \) . For uniqueness, suppose that \( a = {bq} + r \) and \( a = b{q}^{\prime } + {r}^{\prime } \), where \( 0 \leq r < b \) and \( 0 \leq {r}^{\prime } < b \) . Then subtracting these two equations and rearranging terms, we obtain\n\n\[ \n{r}^{\prime } - r = b\left( {q - {q}^{\prime }}\right)\n\]\n\nThus, \( {r}^{\prime } - r \) is a multiple of \( b \) ; however, \( 0 \leq r < b \) and \( 0 \leq {r}^{\prime } < b \) implies \( \left| {{r}^{\prime } - r}\right| < b \) ; therefore, the only possibility is \( {r}^{\prime } - r = 0 \) . Moreover, \( 0 = b\left( {q - {q}^{\prime }}\right) \) and \( b \neq 0 \) implies \( q - {q}^{\prime } = 0 \) .
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Yes
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Consider the ideal \( 3\mathbb{Z} + 5\mathbb{Z} \).
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This ideal contains \( 3 \cdot 2 + 5 \cdot \left( {-1}\right) = 1 \) . Since it contains 1, it contains all integers; that is, \( 3\mathbb{Z} + 5\mathbb{Z} = \mathbb{Z} \).
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Yes
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Theorem 1.6. Let \( I \) be an ideal of \( \mathbb{Z} \). Then there exists a unique non-negative integer \( d \) such that \( I = d\mathbb{Z} \).
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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 in \( I \). We want to show that \( I = d\mathbb{Z} \).\n\nWe first show that \( I \subseteq d\mathbb{Z} \). To this end, let \( a \) be any element in \( I \). It suffices to show that \( d \mid a \). Using the division with remainder property, write \( a = {dq} + r \), where \( 0 \leq r < d \). Then by the closure properties of ideals, one sees that \( r = a - {dq} \) is also an element of \( I \), and by the minimality of the choice of \( d \), we must have \( r = 0 \). Thus, \( d \mid a \).\n\nWe have shown that \( I \subseteq d\mathbb{Z} \). The fact that \( d\mathbb{Z} \subseteq I \) follows from the fact that \( d \in I \). Thus, \( I = d\mathbb{Z} \).\n\nThat proves the existence part of the theorem. For uniqueness, note that if \( d\mathbb{Z} = e\mathbb{Z} \) for some non-negative integer \( e \), then \( d \mid e \) and \( e \mid d \), from which it follows by Theorem 1.2 that \( d = \pm e \); since \( d \) and \( e \) are non-negative, we must have \( d = e \).
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Yes
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For all \( a, b \in \mathbb{Z} \), there exists a unique greatest common divisor \( d \) of \( a \) and \( b \), and moreover, \( a\mathbb{Z} + b\mathbb{Z} = d\mathbb{Z} \).
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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.\n\nSince \( a \in I = d\mathbb{Z} \), we see that \( d\left| {a\text{; similarly,}d}\right| b \). So we see that \( d \) is a common divisor of \( a \) and \( b \).\n\nSince \( d \in I = a\mathbb{Z} + b\mathbb{Z} \), there exist \( s, t \in \mathbb{Z} \) such that \( {as} + {bt} = d \). Now suppose \( a = {a}^{\prime }{d}^{\prime } \) and \( b = {b}^{\prime }{d}^{\prime } \) for some \( {a}^{\prime },{b}^{\prime },{d}^{\prime } \in \mathbb{Z} \). Then the equation \( {as} + {bt} = d \) implies that \( {d}^{\prime }\left( {{a}^{\prime }s + {b}^{\prime }t}\right) = d \), which says that \( {d}^{\prime } \mid d \). Thus, any common divisor \( {d}^{\prime } \) of \( a \) and \( b \) divides \( d \).\n\nThat proves that \( d \) is a greatest common divisor of \( a \) and \( b \). For uniqueness, note that if \( e \) is a greatest common divisor of \( a \) and \( b \), then \( d \mid e \) and \( e \mid d \), and hence \( d = \pm e \); since both \( d \) and \( e \) are non-negative by definition, we have \( d = e \).
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Yes
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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 \( t \) such that \( {as} + {bt} = 1 \) .
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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 the first, setting \( r \mathrel{\text{:=}} 1 \) .
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Yes
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Theorem 1.9. Let \( a, b, c \in \mathbb{Z} \) such that \( c \mid {ab} \) and \( \gcd \left( {a, c}\right) = 1 \) . Then \( c \mid b \) .
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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 \( c \) divides \( {ab} \) by hypothesis, and since \( c \) clearly divides \( {cbt} \), it follows that \( c \) divides the left-hand side of (1.1), and hence that \( c \) divides \( b \) .
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Yes
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Theorem 1.10. Let \( p \) be prime, and let \( a, b \in \mathbb{Z} \) . Then \( p \mid {ab} \) implies that \( p \mid a \) or \( p \mid b \) .
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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 \( p \mid b \) .
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Yes
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Theorem 1.11. There are infinitely many primes.
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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 such prime \( p \), since \( N \geq 2 \), and every positive integer can be written as a product of primes. Clearly, \( p \) cannot equal any of the \( {p}_{i} \) ’s, since if it did, then \( p \) would divide \( M \), and hence also divide \( N - M = 1 \), which is impossible. Therefore, the prime \( p \) is not among \( {p}_{1},\ldots ,{p}_{k} \) , which contradicts our assumption that these are the only primes.
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Yes
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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 \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) .
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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\[ x \in \left\lbrack a\right\rbrack \Rightarrow x \sim a\text{ (by definition) } \]\n\n\[ \Rightarrow x \sim b\text{(by transitivity, and since}x \sim a\text{and}a \sim b\text{)} \]\n\n\[ \Rightarrow x \in \left\lbrack b\right\rbrack \text{.} \]\n\nThus, \( \left\lbrack a\right\rbrack \subseteq \left\lbrack b\right\rbrack \) . By symmetry, we also have \( b \sim a \), and reversing the roles of \( a \) and \( b \) in the above argument, we see that \( \left\lbrack b\right\rbrack \subseteq \left\lbrack a\right\rbrack \) .
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Yes
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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) \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \) and \( b \equiv c\left( {\;\operatorname{mod}\;n}\right) \) implies \( a \equiv c\left( {\;\operatorname{mod}\;n}\right) \) .
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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 \( \left( {a - b}\right) + \left( {b - c}\right) = a - c \) .
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Yes
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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( {\;\operatorname{mod}\;n}\right) \text{ and }a \cdot b \equiv {a}^{\prime } \cdot {b}^{\prime }\left( {\;\operatorname{mod}\;n}\right) .
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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\[ a + b = {a}^{\prime } + {b}^{\prime } + n\left( {x + y}\right) ,\]\n\nwhich proves the first congruence of the theorem, and\n\n\[ {ab} = \left( {{a}^{\prime } + {nx}}\right) \left( {{b}^{\prime } + {ny}}\right) = {a}^{\prime }{b}^{\prime } + n\left( {{a}^{\prime }y + {b}^{\prime }x + {nxy}}\right) ,\]\n\nwhich proves the second congruence.
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Yes
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Let us find the set of solutions \( z \) to the congruence\n\n\[ \n{3z} + 4 \equiv 6\left( {\;\operatorname{mod}\;7}\right) \n\]
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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 \( 5 \cdot 3 \equiv 1\left( {\;\operatorname{mod}\;7}\right) \), we can achieve the same effect by multiplying both sides of (2.2) by 5 . If we do this, and then replace \( 5 \cdot 3 \) by 1, and \( 5 \cdot 2 \) by 3, we obtain\n\n\[ \nz \equiv 3\left( {\;\operatorname{mod}\;7}\right) \n\]\n\nThus, if \( z \) is a solution to (2.1), we must have \( z \equiv 3\left( {\;\operatorname{mod}\;7}\right) \) ; conversely, one can verify that if \( z \equiv 3\left( {\;\operatorname{mod}\;7}\right) \), then (2.1) holds. We conclude that the integers \( z \) that are solutions to (2.1) are precisely those integers that are congruent to 3 modulo 7, which we can list as follows:\n\n\[ \n\ldots , - {18}, - {11}, - 4,3,{10},{17},{24},\ldots \n\]
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Yes
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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 \( z \in \mathbb{Z} \), we have \( {az} \equiv 0\left( {\;\operatorname{mod}\;n}\right) \) if and only if \( z \equiv 0\left( {{\;\operatorname{mod}\;n}/d}\right) \). (iii) For all \( z,{z}^{\prime } \in \mathbb{Z} \), we have \( {az} \equiv a{z}^{\prime }\left( {\;\operatorname{mod}\;n}\right) \) if and only if \( z \equiv {z}^{\prime }\left( {{\;\operatorname{mod}\;n}/d}\right) \).
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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 \( z, y \in \mathbb{Z} \) \( \Leftrightarrow d \mid b \) (by Theorem 1.8). For (ii), we have \[ n\left| {{az} \Leftrightarrow n/d}\right| \left( {a/d}\right) z \Leftrightarrow n/d \mid z. \] All of these implications follow rather trivially from the definition of divisibility, except that for the implication \( n/d \mid \left( {a/d}\right) z \Rightarrow n/d \mid z \), we use Theorem 1.9 and the fact that \( \gcd \left( {a/d, n/d}\right) = 1 \). For (iii), we have \[ {az} \equiv a{z}^{\prime }\left( {\;\operatorname{mod}\;n}\right) \Leftrightarrow a\left( {z - {z}^{\prime }}\right) \equiv 0\left( {\;\operatorname{mod}\;n}\right) \] \[ \Leftrightarrow z - {z}^{\prime } \equiv 0\left( {{\;\operatorname{mod}\;n}/d}\right) \text{ (by part (ii)) } \] \[ \Leftrightarrow z \equiv {z}^{\prime }\left( {{\;\operatorname{mod}\;n}/d}\right) \]
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Yes
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The following table illustrates what Theorem 2.5 says for \( n = {15} \) and \( a = 1,2,3,4,5,6 \) .
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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 \( \gcd \left( {2,{15}}\right) = 1 \) .\n\nIn the third row, the only numbers hit are the multiples of 3 , which follows from the fact that \( \gcd \left( {3,{15}}\right) = 3 \) . Also note that the pattern in this row repeats every five columns; that is, \( {3z} \equiv 3{z}^{\prime }\left( {\;\operatorname{mod}\;{15}}\right) \) if and only if \( z \equiv {z}^{\prime }\left( {\;\operatorname{mod}\;5}\right) \) .\n\nIn the fourth row, we again see a permutation of the first row, which follows from the fact that \( \gcd \left( {4,{15}}\right) = 1 \) .\n\nIn the fifth row, the only numbers hit are the multiples of 5 , which follows from the fact that \( \gcd \left( {5,{15}}\right) = 5 \) . Also note that the pattern in this row repeats every three columns; that is, \( {5z} \equiv 5{z}^{\prime }\left( {\;\operatorname{mod}\;{15}}\right) \) if and only if \( z \equiv {z}^{\prime }\left( {\;\operatorname{mod}\;3}\right) \) .\n\nIn the sixth row, since \( \gcd \left( {6,{15}}\right) = 3 \), we see a permutation of the third row. The pattern repeats after five columns, although the pattern is a permutation of the pattern in the third row.
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Yes
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Observe that\n\n\[ 5 \cdot 2 \equiv 5 \cdot \left( {-4}\right) \left( {\;\operatorname{mod}\;6}\right) \]
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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.
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No
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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.
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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 in the interval \( \{ 0,\ldots, n - 1\} \). More generally, if \( d \mathrel{\text{:=}} \gcd \left( {a, n}\right) \), then the congruence \( {az} \equiv b\left( {\;\operatorname{mod}\;n}\right) \) has a solution if and only if \( d \mid b \). So suppose that \( d \mid b \). In this case, if we set \( {a}^{\prime } \mathrel{\text{:=}} a/d,{b}^{\prime } \mathrel{\text{:=}} b/d \), and \( {n}^{\prime } \mathrel{\text{:=}} n/d \), then for each \( z \in \mathbb{Z} \), we have \( {az} \equiv b\left( {\;\operatorname{mod}\;n}\right) \) if and only if \( {a}^{\prime }z \equiv {b}^{\prime }\left( {\;\operatorname{mod}\;{n}^{\prime }}\right) \). Moreover, \( \gcd \left( {{a}^{\prime },{n}^{\prime }}\right) = 1 \), and therefore, if we set \( t \mathrel{\text{:=}} {\left( {a}^{\prime }\right) }^{-1}{\;\operatorname{mod}\;{n}^{\prime }} \) and \( {z}_{0} \mathrel{\text{:=}} t{b}^{\prime }{\;\operatorname{mod}\;{n}^{\prime }} \), then the solutions to the congruence \( {az} \equiv b\left( {\;\operatorname{mod}\;n}\right) \) that lie in the interval \( \{ 0,\ldots, n - 1\} \) are the \( d \) integers \( {z}_{0},{z}_{0} + {n}^{\prime },\ldots ,{z}_{0} + \left( {d - 1}\right) {n}^{\prime }. \)
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Yes
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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\[ a \equiv {a}_{i}\left( {\;\operatorname{mod}\;{n}_{i}}\right) \;\left( {i = 1,\ldots, k}\right) . \]\n\nMoreover, any \( {a}^{\prime } \in \mathbb{Z} \) is a solution to this system of congruences if and only if \( a \equiv {a}^{\prime }\left( {\;\operatorname{mod}\;n}\right) \), where \( n \mathrel{\text{:=}} \mathop{\prod }\limits_{{i = 1}}^{k}{n}_{i} \) .
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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\[ {e}_{j} \equiv \left\{ \begin{array}{ll} 1\left( {\;\operatorname{mod}\;{n}_{i}}\right) & \text{ if }j = i, \\ 0\left( {\;\operatorname{mod}\;{n}_{i}}\right) & \text{ if }j \neq i. \end{array}\right. \]\n\n(2.6)\n\nIf we do this, then setting\n\n\[ a \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i}{e}_{i} \]\n\none sees that for \( j = 1,\ldots, k \), we have\n\n\[ a \equiv \mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i}{e}_{i} \equiv {a}_{j}\left( {\;\operatorname{mod}\;{n}_{j}}\right) \]\n\nsince all the terms in this sum are zero modulo \( {n}_{j} \), except for the term \( i = j \), which is congruent to \( {a}_{j} \) modulo \( {n}_{j} \) .\n\nTo construct \( {e}_{1},\ldots ,{e}_{k} \) satisfying (2.6), let \( n \mathrel{\text{:=}} \mathop{\prod }\limits_{{i = 1}}^{k}{n}_{i} \) as in the statement of the theorem, and for \( i = 1,\ldots, k \), let \( {n}_{i}^{ * } \mathrel{\text{:=}} n/{n}_{i} \) ; that is, \( {n}_{i}^{ * } \) is the product of all the moduli \( {n}_{j} \) with \( j \neq i \) . From the fact that \( {\left\{ {n}_{i}\right\} }_{i = 1}^{k} \) is pairwise relatively prime, it follows that for \( i = 1,\ldots, k \), we have \( \gcd \left( {{n}_{i},{n}_{i}^{ * }}\right) = 1 \), and so we may define \( {t}_{i} \mathrel{\text{:=}} {\left( {n}_{i}^{ * }\right) }^{-1}{\;\operatorname{mod}\;{n}_{i}} \) and \( {e}_{i} \mathrel{\text{:=}} {n}_{i}^{ * }{t}_{i} \) . One sees that \( {e}_{i} \equiv 1\left( {\;\operatorname{mod}\;{n}_{i}}\right) \), while for \( j \neq i \) , we have \( {n}_{i} \mid {n}_{j}^{ * } \), and so \( {e}_{j} \equiv 0\left( {\;\operatorname{mod}\;{n}_{i}}\right) \) . Thus,(2.6) is satisfied.\n\nThat proves the existence of a solution \( a \) to the given system of congruences. If \( a \equiv {a}^{\prime }\left( {\;\operatorname{mod}\;n}\right) \), then since \( {n}_{i} \mid n \) for \( i = 1,\ldots, k \), we see that \( {a}^{\prime } \equiv a \equiv {a}_{i}\left( {\;\operatorname{mod}\;{n}_{i}}\right) \) for \( i = 1,\ldots, k \), and so \( {a}^{\prime } \) also solves the system of congruences.\n\nFinally, if \( {a}^{\prime } \) is a solution to the given system of congruences, then \( a \equiv {a}_{i} \equiv \) \( {a}^{\prime }\left( {\;\operatorname{mod}\;{n}_{i}}\right) \) for \( i = 1,\ldots, k \) . Thus, \( {n}_{i} \mid \left( {a - {a}^{\prime }}\right) \) for \( i = 1,\ldots, k \) . Since \( {\left\{ {n}_{i}\right\} }_{i = 1}^{k} \) is pairwise relatively prime, this implies \( n \mid \left( {a - {a}^{\prime }}\right) \), or equivalently, \( a \equiv {a}^{\prime }\left( {\;\operatorname{mod}\;n}\right) \) .
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Yes
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Example 2.6. The following table illustrates what Theorem 2.6 says for \( {n}_{1} = 3 \) and \( {n}_{2} = 5 \) .
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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.
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Yes
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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\[ \left\lbrack 3\right\rbrack = \{ \ldots , - 9, - 3,3,9,{15},\ldots \} \]\n\n\[ \left\lbrack 4\right\rbrack = \{ \ldots , - 8, - 2,4,{10},{16},\ldots \} \]\n\n\[ \left\lbrack 5\right\rbrack = \{ \ldots , - 7, - 1,5,{11},{17},\ldots \} . \]\n\nLet us write down the addition and multiplication tables for \( {\mathbb{Z}}_{6} \) .
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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>[5]</td><td>[0]</td></tr><tr><td>[2]</td><td>[2]</td><td>[3]</td><td>[4]</td><td>[5]</td><td>[0]</td><td>[1]</td></tr><tr><td>[3]</td><td>[3]</td><td>[4]</td><td>[5]</td><td>[0]</td><td>[1]</td><td>[2]</td></tr><tr><td>[4]</td><td>[4]</td><td>[5]</td><td>[0]</td><td>[1]</td><td>[2]</td><td>[3]</td></tr><tr><td>[5]</td><td>[5]</td><td>[0]</td><td>[1]</td><td>[2]</td><td>[3]</td><td>[4].</td></tr></table>\n\nThe multiplication 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>[0]</td><td>[0]</td><td>[0]</td><td>[0]</td><td>[0]</td></tr><tr><td>[1]</td><td>[0]</td><td>[1]</td><td>[2]</td><td>[3]</td><td>[4]</td><td>[5]</td></tr><tr><td>[2]</td><td>[0]</td><td>[2]</td><td>[4]</td><td>[0]</td><td>[2]</td><td>[4]</td></tr><tr><td>[3]</td><td>[0]</td><td>[3]</td><td>[0]</td><td>[3]</td><td>[0]</td><td>[3]</td></tr><tr><td>[4]</td><td>[0]</td><td>[4]</td><td>[2]</td><td>[0]</td><td>[4]</td><td>[2]</td></tr><tr><td>[5]</td><td>[0]</td><td>[5]</td><td>[4]</td><td>[3]</td><td>[2]</td><td>[1].</td></tr></table>
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Yes
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We list the elements of \( {\mathbb{Z}}_{15}^{ * } \), and for each \( \alpha \in {\mathbb{Z}}_{15}^{ * } \), we also give \( {\alpha }^{-1} \):
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<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>
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Yes
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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\[ \n\theta : \;{\mathbb{Z}}_{n} \rightarrow {\mathbb{Z}}_{{n}_{1}} \times \cdots \times {\mathbb{Z}}_{{n}_{k}}\n\]\n\n\[ \n{\left\lbrack a\right\rbrack }_{n} \mapsto \left( {{\left\lbrack a\right\rbrack }_{{n}_{1}},\ldots ,{\left\lbrack a\right\rbrack }_{{n}_{k}}}\right) .\n\]\n\n(i) The definition of \( \theta \) is unambiguous.\n\n(ii) \( \theta \) is bijective.\n\n(iii) For all \( \alpha ,\beta \in {\mathbb{Z}}_{n} \), if \( \theta \left( \alpha \right) = \left( {{\alpha }_{1},\ldots ,{\alpha }_{k}}\right) \) and \( \theta \left( \beta \right) = \left( {{\beta }_{1},\ldots ,{\beta }_{k}}\right) \), then:\n\n(a) \( \theta \left( {\alpha + \beta }\right) = \left( {{\alpha }_{1} + {\beta }_{1},\ldots ,{\alpha }_{k} + {\beta }_{k}}\right) \) ;\n\n(b) \( \theta \left( {-\alpha }\right) = \left( {-{\alpha }_{1},\ldots , - {\alpha }_{k}}\right) \) ;\n\n(c) \( \theta \left( {\alpha \beta }\right) = \left( {{\alpha }_{1}{\beta }_{1},\ldots ,{\alpha }_{k}{\beta }_{k}}\right) \) ;\n\n(d) \( \alpha \in {\mathbb{Z}}_{n}^{ * } \) if and only if \( {\alpha }_{i} \in {\mathbb{Z}}_{{n}_{i}}^{ * } \) for \( i = 1,\ldots, k \), in which case \( \theta \left( {\alpha }^{-1}\right) = \left( {{\alpha }_{1}^{-1},\ldots ,{\alpha }_{k}^{-1}}\right) .
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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 directly from the statement of the Chinese remainder theorem.\n\nFor (iii), let \( \alpha = {\left\lbrack a\right\rbrack }_{n} \) and \( \beta = {\left\lbrack b\right\rbrack }_{n} \), so that for \( i = 1,\ldots, k \), we have \( {\alpha }_{i} = {\left\lbrack a\right\rbrack }_{{n}_{i}} \) and \( {\beta }_{i} = {\left\lbrack b\right\rbrack }_{{n}_{i}} \) . Then we have\n\n\[ \n\theta \left( {\alpha + \beta }\right) = \theta \left( {\left\lbrack a + b\right\rbrack }_{n}\right) = \left( {{\left\lbrack a + b\right\rbrack }_{{n}_{1}},\ldots ,{\left\lbrack a + b\right\rbrack }_{{n}_{k}}}\right) = \left( {{\alpha }_{1} + {\beta }_{1},\ldots ,{\alpha }_{k} + {\beta }_{k}}\right) ,\n\]\n\n\[ \n\theta \left( {-\alpha }\right) = \theta \left( {\left\lbrack -a\right\rbrack }_{n}\right) = \left( {{\left\lbrack -a\right\rbrack }_{{n}_{1}},\ldots ,{\left\lbrack -a\right\rbrack }_{{n}_{k}}}\right) = \left( {-{\alpha }_{1},\ldots , - {\alpha }_{k}}\right) \text{, and}\n\]\n\n\[ \n\theta \left( {\alpha \beta }\right) = \theta \left( {\left\lbrack ab\right\rbrack }_{n}\right) = \left( {{\left\lbrack ab\right\rbrack }_{{n}_{1}},\ldots ,{\left\lbrack ab\right\rbrack }_{{n}_{k}}}\right) = \left( {{\alpha }_{1}{\beta }_{1},\ldots ,{\alpha }_{k}{\beta }_{k}}\right) .\n\]\n\nThat proves parts (a), (b), and (c). For part (d), we have\n\n\[ \n\alpha \in {\mathbb{Z}}_{n}^{ * } \Leftrightarrow \gcd \left( {a, n}\right) = 1\n\]\n\n\[ \n\Leftrightarrow \gcd \left( {a,{n}_{i}}\right) = 1\text{ for }i = 1,\ldots, k\n\]\n\n\[ \n\Leftrightarrow {\alpha }_{i} \in {\mathbb{Z}}_{{n}_{i}}^{ * }\text{ for }i = 1,\ldots, k\text{. }\n\]\n\nMoreover, if \( \alpha \in {\mathbb{Z}}_{n}^{ * } \) and \( \beta = {\alpha }^{-1} \), then\n\n\[ \n\left( {{\alpha }_{1}{\beta }_{1},\ldots ,{\alpha }_{k}{\beta }_{k}}\right) = \theta \left( {\alpha \beta }\right) = \theta \left( {\left\lbrack 1\right\rbrack }_{n}\right) = \left( {{\left\lbrack 1\right\rbrack }_{{n}_{1}},\ldots ,{\left\lbrack 1\right\rbrack }_{{n}_{k}}}\right) ,\n\]\n\nand so for \( i = 1,\ldots, k \), we have \( {\alpha }_{i}{\beta }_{i} = {\left\lbrack 1\right\rbrack }_{{n}_{i}} \), which is to say \( {\beta }_{i} = {\alpha }_{i}^{-1} \) .
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Yes
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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\[ \varphi \left( n\right) = \mathop{\prod }\limits_{{i = 1}}^{k}\varphi \left( {n}_{i}\right) \]
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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 \( {\mathbb{Z}}_{n}^{ * } \) and \( {\mathbb{Z}}_{{n}_{1}}^{ * } \times \cdots \times {\mathbb{Z}}_{{n}_{k}}^{ * } \). The theorem now follows immediately.
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Yes
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Theorem 2.10. Let \( p \) be a prime and \( e \) be a positive integer. Then\n\n\[ \varphi \left( {p}^{e}\right) = {p}^{e - 1}\left( {p - 1}\right) . \]
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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, \( \varphi \left( {p}^{e}\right) = {p}^{e} - {p}^{e - 1} = {p}^{e - 1}\left( {p - 1}\right) \) .
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Yes
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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 \( n \) .
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<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</td><td>1</td></tr><tr><td>\( {3}^{i}{\;\operatorname{mod}\;7} \)</td><td>3</td><td>2</td><td>6</td><td>4</td><td>5</td><td>1</td></tr><tr><td>\( {4}^{i}{\;\operatorname{mod}\;7} \)</td><td>4</td><td>2</td><td>1</td><td>4</td><td>2</td><td>1</td></tr><tr><td>\( {5}^{i}{\;\operatorname{mod}\;7} \)</td><td>5</td><td>4</td><td>6</td><td>2</td><td>3</td><td>1</td></tr><tr><td>\( {6}^{i}{\;\operatorname{mod}\;7} \)</td><td>6</td><td>1</td><td>6</td><td>1</td><td>6</td><td>1</td></tr></table>\n\nSo we conclude that modulo 7: 1 has order 1; 6 has order 2; 2 and 4 have order 3; and 3 and 5 have order 6 .
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Yes
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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 \( \varphi \left( n\right) \) .
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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 \
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No
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Theorem 2.14 (Fermat’s little theorem). For every prime \( p \), and every \( \alpha \in {\mathbb{Z}}_{p} \) , we have \( {\alpha }^{p} = \alpha \) .
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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 \( {\alpha }^{p} = \alpha \) .
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Yes
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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 \( k/\gcd \left( {m, k}\right) \) .
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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\[{\alpha }^{m\ell } = 1 \Leftrightarrow m\ell \equiv 0\left( {\;\operatorname{mod}\;k}\right) \text{ (applying Theorem 2.12 to }\alpha \text{ ) }\n\]\n\[ \Leftrightarrow \ell \equiv 0\left( {{\;\operatorname{mod}\;k}/\gcd \left( {m, k}\right) }\right) \text{(by part (ii) of Theorem 2.5).} \]
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Yes
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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 \( \alpha \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \) and \( \beta \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \), then \( {\alpha \beta } \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \) .\n\n(iii) If \( \alpha \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \) and \( \beta \notin {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \), then \( {\alpha \beta } \notin {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \) .
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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 \( \alpha \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m},\beta \notin {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \), and \( {\alpha \beta } \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \) . Then by (i), \( {\alpha }^{-1} \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \), and by (ii), \( \beta = {\alpha }^{-1}\left( {\alpha \beta }\right) \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \), a contradiction.
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Yes
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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 \( \alpha \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m} \) .
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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\[ \alpha = {\alpha }^{\ell s + {mt}} = {\alpha }^{\ell s}{\alpha }^{mt} = {\beta }^{ms}{\alpha }^{mt} = {\left( {\beta }^{s}{\alpha }^{t}\right) }^{m} \in {\left( {\mathbb{Z}}_{n}^{ * }\right) }^{m}. \]
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Yes
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Theorem 2.18. Let \( p \) be an odd prime and \( \beta \in {\mathbb{Z}}_{p} \) . Then \( {\beta }^{2} = 1 \) if and only if \( \beta = \pm 1 \) .
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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\[ p \mid \left( {{b}^{2} - 1}\right) = \left( {b - 1}\right) \left( {b + 1}\right) \]\n\nand since \( p \) is prime, we must have \( p\left| \left( {b - 1}\right) \right| \) or \( p \mid \left( {b + 1}\right) \) . This implies \( b \equiv \pm 1\left( {\;\operatorname{mod}\;p}\right) \), or equivalently, \( \beta = \pm 1 \) .
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Yes
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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 \( \gamma = \pm \beta \) .
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Proof. This follows from the previous theorem:\n\n\[ \n{\gamma }^{2} = {\beta }^{2} \Leftrightarrow {\left( \gamma /\beta \right) }^{2} = 1 \Leftrightarrow \gamma /\beta = \pm 1 \Leftrightarrow \gamma = \pm \beta .\n\]
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Yes
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Theorem 2.20. Let \( p \) be an odd prime. Then \( \left| {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2}\right| = \left( {p - 1}\right) /2 \) .
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Proof. By the previous theorem, the \
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No
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Example 2.10. Let \( p = 7 \) . We can list the elements of \( {\mathbb{Z}}_{p}^{ * } \) as \[ \left\lbrack {\pm 1}\right\rbrack ,\left\lbrack {\pm 2}\right\rbrack ,\left\lbrack {\pm 3}\right\rbrack \text{.} \]
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Squaring these, we see that \[ {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} = \{ {\left\lbrack 1\right\rbrack }^{2},{\left\lbrack 2\right\rbrack }^{2},{\left\lbrack 3\right\rbrack }^{2}\} = \{ \left\lbrack 1\right\rbrack ,\left\lbrack 4\right\rbrack ,\left\lbrack 2\right\rbrack \} . \]
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Yes
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Theorem 2.22 (Wilson’s theorem). Let \( p \) be an odd prime. Then \( \mathop{\prod }\limits_{{\beta \in {\mathbb{Z}}_{p}^{ * }}}\beta = - 1 \) .
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\[ \left( {p - 1}\right) ! \equiv - 1\left( {\;\operatorname{mod}\;p}\right) \]
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Yes
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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 \( {\alpha \beta } \in {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} \) .
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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\[ \n{\left( \alpha \beta \right) }^{\left( {p - 1}\right) /2} = {\alpha }^{\left( {p - 1}\right) /2} \cdot {\beta }^{\left( {p - 1}\right) /2} = \left\lbrack {-1}\right\rbrack \cdot \left\lbrack {-1}\right\rbrack = 1,\n\]\n\nwhich again by Euler’s criterion implies that \( {\alpha \beta } \in {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} \) .
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Yes
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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 \( \beta = \pm 1 \) .
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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\[ \n{p}^{e} \mid \left( {{b}^{2} - 1}\right) = \left( {b - 1}\right) \left( {b + 1}\right) .\n\]\n\nIn particular, \( p \mid \left( {b - 1}\right) \left( {b + 1}\right) \), and so \( p \mid \left( {b - 1}\right) \) or \( p \mid \left( {b + 1}\right) \) . Moreover, \( p \) cannot divide both \( b - 1 \) and \( b + 1 \), as otherwise, it would divide their difference \( \left( {b + 1}\right) - \left( {b - 1}\right) = 2 \), which is impossible (because \( p \) is odd). It follows that \( {p}^{e}\left| {\left( {b - 1}\right) \text{ or }{p}^{e}}\right| \left( {b + 1}\right) \), which means \( \beta = \pm 1 \) .
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Yes
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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 \( p \) .
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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 quadratic residue modulo \( p \) .\n\nSuppose that \( a \) is not a quadratic residue modulo \( {p}^{e} \) . If \( a \) is divisible by \( p \), then by definition \( a \) is not a quadratic residue modulo \( p \) . So suppose \( a \) is not divisible by \( p \) . By Theorem 2.27, we have\n\n\[ {a}^{{p}^{e - 1}\left( {p - 1}\right) /2} \equiv - 1\left( {\;\operatorname{mod}\;{p}^{e}}\right) . \]\n\nThis congruence holds modulo \( p \) as well, and by Fermat’s little theorem (applied\n\n\( e - 1 \) times),\n\n\[ a \equiv {a}^{p} \equiv {a}^{{p}^{2}} \equiv \cdots \equiv {a}^{{p}^{e - 1}}\left( {\;\operatorname{mod}\;p}\right) ,\]\n\nand so\n\n\[ - 1 \equiv {a}^{{p}^{e - 1}\left( {p - 1}\right) /2} \equiv {a}^{\left( {p - 1}\right) /2}\left( {\;\operatorname{mod}\;p}\right) . \]\n\nTheorem 2.21 therefore implies that \( a \) is not a quadratic residue modulo \( p \) .
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Yes
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Theorem 2.31. Let \( p \) be an odd prime. Then -1 is a quadratic residue modulo \( p \) if and only \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) .
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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 \( {\left( -1\right) }^{\left( {p - 1}\right) /2} = 1 \) . If \( p \equiv 3\left( {\text{mod }4}\right) \), then \( \left( {p - 1}\right) /2 \) is odd, and so \( {\left( -1\right) }^{\left( {p - 1}\right) /2} = - 1 \) . \( \square \)
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Yes
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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 \( {\beta }^{2} = - 1 \) .
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Proof. This is a simple calculation, based on Euler's criterion:\n\n\[ \n{\beta }^{2} = {\gamma }^{\left( {p - 1}\right) /2} = - 1 \n\]
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Yes
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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\[ r \equiv {bt}\left( {\;\operatorname{mod}\;n}\right) ,\;\left| r\right| < {r}^{ * },\text{ and }0 < \left| t\right| < {t}^{ * }.\]
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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 is, for some \( \left( {{i}_{1},{j}_{1}}\right) \neq \left( {{i}_{2},{j}_{2}}\right) \), we have \( {v}_{{i}_{1}{j}_{1}} \equiv {v}_{{i}_{2}{j}_{2}}\left( {\;\operatorname{mod}\;n}\right) \) . Setting \( r \mathrel{\text{:=}} {i}_{1} \) - \( {i}_{2}\;\mathrm{{and}}\;t \mathrel{\text{:=}} {j}_{1} \) - \( {j}_{2},\mathrm{{this}}\;\mathrm{{implies}}\;r \equiv {bt}\left( {\mathrm{{mod}}\;n}\right) , \) \( \left| r\right| < {r}^{ * },\left| t\right| < {t}^{ * } \), and that either \( r \neq 0 \) or \( t \neq 0 \) . It only remains to show that \( t \neq 0 \) . Suppose to the contrary that \( t = 0 \) . This would imply that \( r \equiv 0\left( {\;\operatorname{mod}\;n}\right) \) and \( r \neq 0 \) , which is to say that \( r \) is a non-zero multiple of \( n \) ; however, this is impossible, since \( \left| r\right| < {r}^{ * } \leq n \) .
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Yes
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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 \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) .
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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 alone equal to \( p \) ).\n\nFor the other direction, suppose \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . We know that -1 is a quadratic residue modulo \( p \), so let \( b \) be an integer such that \( {b}^{2} \equiv - 1\left( {\;\operatorname{mod}\;p}\right) \) . Now apply Theorem 2.33 with \( n \mathrel{\text{:=}} p, b \) as just defined, and \( {r}^{ * } \mathrel{\text{:=}} {t}^{ * } \mathrel{\text{:=}} \lfloor \sqrt{p}\rfloor + 1 \) . Evidently, \( \lfloor \sqrt{p}\rfloor + 1 > \sqrt{p} \), and hence \( {r}^{ * }{t}^{ * } > p \) . Also, since \( p \) is prime, \( \sqrt{p} \) is not an integer, and so \( \lfloor \sqrt{p}\rfloor < \sqrt{p} < p \) ; in particular, \( {r}^{ * } = \lfloor \sqrt{p}\rfloor + 1 \leq p \) . Thus, the hypotheses of that theorem are satisfied, and therefore, there exist integers \( r \) and \( t \) such that\n\n\[ r \equiv {bt}\left( {\;\operatorname{mod}\;p}\right) ,\;\left| r\right| \leq \lfloor \sqrt{p}\rfloor < \sqrt{p},\text{ and }0 < \left| t\right| \leq \lfloor \sqrt{p}\rfloor < \sqrt{p}. \]\n\nIt follows that\n\n\[ {r}^{2} \equiv {b}^{2}{t}^{2} \equiv - {t}^{2}\left( {\;\operatorname{mod}\;p}\right) \]\n\nThus, \( {r}^{2} + {t}^{2} \) is a multiple of \( p \) and \( 0 < {r}^{2} + {t}^{2} < {2p} \) . The only possibility is that \( {r}^{2} + {t}^{2} = p \) .
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Yes
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Theorem 2.35. There are infinitely many primes \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) .
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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, since if it were, it would divide both \( N \) and \( 4{M}^{2} \), and so also \( N - 4{M}^{2} = 1 \) . Also, \( p \) is clearly odd, since \( N \) is odd. Moreover, \( {\left( 2M\right) }^{2} \equiv - 1\left( {\;\operatorname{mod}\;p}\right) \) ; therefore, \( - 1 \) is a quadratic residue modulo \( p \), and so \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \), contradicting the assumption that \( {p}_{1},\ldots ,{p}_{k} \) are the only such primes.
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Yes
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Theorem 2.36. There are infinitely many primes \( p \equiv 3\left( {\;\operatorname{mod}\;4}\right) \) .
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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 \( p \equiv 3\left( {\;\operatorname{mod}\;4}\right) \) dividing \( N \) (if all primes dividing \( N \) were congruent to 1 modulo 4, then so too would be their product \( N \) ). Evidently, \( p \) is not among the \( {p}_{i} \) ’s, since if it were, it would divide both \( N \) and \( {4M} \), and so also \( {4M} - N = 1 \) . This contradicts the assumption that \( {p}_{1},\ldots ,{p}_{k} \) are the only primes congruent to 3 modulo 4. \( ▱ \)
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Yes
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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 \( f\left( n\right) = f\left( {p}_{1}^{{e}_{1}}\right) \cdots f\left( {p}_{r}^{{e}_{r}}\right) \) .
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Proof. Exercise.
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No
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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\[ \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) f\left( d\right) = \left( {1 - f\left( {p}_{1}\right) }\right) \cdots \left( {1 - f\left( {p}_{r}\right) }\right) . \]
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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 \( {\left( -1\right) }^{\ell }f\left( {{p}_{{i}_{1}}\cdots {p}_{{i}_{\ell }}}\right) = \) \( {\left( -1\right) }^{\ell }f\left( {p}_{{i}_{1}}\right) \cdots f\left( {p}_{{i}_{\ell }}\right) \) . These are the same as the terms in the expansion of the product on the right-hand side of (2.9).
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Yes
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Theorem 2.39 (Möbius inversion formula). Let \( f \) and \( F \) be arithmetic functions. Then \( F = 1 \star f \) if and only if \( f = \mu \star F \) .
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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\[ I \star f = I \star \left( {\mu \star F}\right) = \left( {I \star \mu }\right) \star F = I \star F = F. \]
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Yes
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Theorem 2.40. For every positive integer \( n \), we have \( \mathop{\sum }\limits_{{d \mid n}}\varphi \left( d\right) = n \) .
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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 \( n = {p}_{1}^{{e}_{1}}\cdots {p}_{r}^{{e}_{r}} \) is the prime factorization of \( n \), we have\n\n\[ \n\left( {\mu \star N}\right) \left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) \left( {n/d}\right) = n\mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) /d \n\]\n\n\[ \n= n\mathop{\prod }\limits_{{i = 1}}^{r}\left( {1 - 1/{p}_{i}}\right) \text{ (applying Theorem 2.38 with }f \mathrel{\text{:=}} M\text{ ) } \n\]\n\n\[ \n= \varphi \left( n\right) \text{(by Theorem 2.11).} \n\]
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Yes
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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, \( f = \Theta \left( g\right) \) .
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Indeed, \( f = \Theta \left( g\right) \) .
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No
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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 \( \lfloor x/y\rfloor = \left\lfloor {{x}^{\prime }/{y}^{\prime }}\right\rfloor \) .
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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\[ \frac{x}{y} = \frac{{x}^{\prime }}{{y}^{\prime }} + \frac{s}{{y}^{\prime }{2}^{n}} < \frac{{x}^{\prime }}{{y}^{\prime }} + \frac{1}{{y}^{\prime }} \leq \left( {\left\lfloor \frac{{x}^{\prime }}{{y}^{\prime }}\right\rfloor + \frac{{y}^{\prime } - 1}{{y}^{\prime }}}\right) + \frac{1}{{y}^{\prime }} \leq \left\lfloor \frac{{x}^{\prime }}{{y}^{\prime }}\right\rfloor + 1. \]\n\nThus, we have \( x/y < \left\lfloor {{x}^{\prime }/{y}^{\prime }}\right\rfloor + 1 \), and hence, \( \lfloor x/y\rfloor \leq \left\lfloor {{x}^{\prime }/{y}^{\prime }}\right\rfloor \) .
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Yes
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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 \( 2{y}^{\prime } \geq x/y \) . Then\n\n\[ \lfloor x/y\rfloor \leq \left\lfloor {{x}^{\prime }/{y}^{\prime }}\right\rfloor \leq \lfloor x/y\rfloor + 2. \]
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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 inequality.\n\nFor the second inequality, first note that from the definitions, we have \( x/y \geq \) \( {x}^{\prime }/\left( {{y}^{\prime } + 1}\right) \), which implies \( {x}^{\prime }y - x{y}^{\prime } - x \leq 0 \) . Further, \( 2{y}^{\prime } \geq x/y \) implies \( {2y}{y}^{\prime } - x \geq 0 \) . So we have \( {2y}{y}^{\prime } - x \geq 0 \geq {x}^{\prime }y - x{y}^{\prime } - x \), which implies \( x/y \geq {x}^{\prime }/{y}^{\prime } - 2 \), and hence \( \lfloor x/y\rfloor \geq \left\lfloor {{x}^{\prime }/{y}^{\prime }}\right\rfloor - 2 \) .
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Yes
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Suppose \( e = {37} = {\left( {100101}\right) }_{2} \) . The above algorithm performs the following operations in this case:
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\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\[ \beta \leftarrow {\beta }^{2},\beta \leftarrow \beta \cdot \alpha \]\n\n\[ \text{ // 1001} \]\n\n\[ \beta \leftarrow {\beta }^{2} \]\n\n\[ \beta \leftarrow {\beta }^{2},\beta \leftarrow \beta \cdot \alpha \;\text{ //100101 }.\]
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Yes
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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\[ {r}_{0} = {r}_{1}{q}_{1} + {r}_{2}\;\left( {0 < {r}_{2} < {r}_{1}}\right) ,\]\n\n\[ \vdots \]\n\n\[ {r}_{i - 1} = {r}_{i}{q}_{i} + {r}_{i + 1}\;\left( {0 < {r}_{i + 1} < {r}_{i}}\right) ,\]\n\n\[ \vdots \]\n\n\[ {r}_{\lambda - 2} = {r}_{\lambda - 1}{q}_{\lambda - 1} + {r}_{\lambda }\;\left( {0 < {r}_{\lambda } < {r}_{\lambda - 1}}\right) ,\]\n\n\[ {r}_{\lambda - 1} = {r}_{\lambda }{q}_{\lambda }\;\left( {{r}_{\lambda + 1} = 0}\right) .\n\nNote that by definition, \( \lambda = 0 \) if \( b = 0 \), and \( \lambda > 0 \), otherwise. Then we have \( {r}_{\lambda } = \gcd \left( {a, b}\right) \) . Moreover, if \( b > 0 \), then \( \lambda \leq \log b/\log \phi + 1 \), where \( \phi \mathrel{\text{:=}} \left( {1 + \sqrt{5}}\right) /2 \approx {1.62}. \)
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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 \( \gcd \left( {{r}_{i - 1},{r}_{i}}\right) = \gcd \left( {{r}_{i},{r}_{i + 1}}\right) \) . From this, it follows that\n\n\[ \gcd \left( {a, b}\right) = \gcd \left( {{r}_{0},{r}_{1}}\right) = \cdots = \gcd \left( {{r}_{\lambda },{r}_{\lambda + 1}}\right) = \gcd \left( {{r}_{\lambda },0}\right) = {r}_{\lambda }.\n\nTo prove the second statement, assume that \( b > 0 \), and hence \( \lambda > 0 \) . If \( \lambda = 1 \), the statement is obviously true, so assume \( \lambda > 1 \) . We claim that for \( i = 0,\ldots ,\lambda - 1 \) , we have \( {r}_{\lambda - i} \geq {\phi }^{i} \) . The statement will then follow by setting \( i = \lambda - 1 \) and taking logarithms.\n\nWe now prove the above claim. For \( i = 0 \) and \( i = 1 \), we have\n\n\[ {r}_{\lambda } \geq 1 = {\phi }^{0}\text{ and }{r}_{\lambda - 1} \geq {r}_{\lambda } + 1 \geq 2 \geq {\phi }^{1}.\n\nFor \( i = 2,\ldots ,\lambda - 1 \), using induction and applying the fact that \( {\phi }^{2} = \phi + 1 \), we have\n\n\[ {r}_{\lambda - i} \geq {r}_{\lambda - \left( {i - 1}\right) } + {r}_{\lambda - \left( {i - 2}\right) } \geq {\phi }^{i - 1} + {\phi }^{i - 2} = {\phi }^{i - 2}\left( {1 + \phi }\right) = {\phi }^{i},\n\nwhich proves the claim.
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Yes
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Suppose \( a = {100} \) and \( b = {35} \) . Then the numbers appearing in Theorem 4.1 are easily computed as follows:
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<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 \( \gcd \left( {a, b}\right) = {r}_{3} = 5 \) .
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Yes
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Theorem 4.2. Euclid’s algorithm runs in time \( O\left( {\operatorname{len}\left( a\right) \operatorname{len}\left( b\right) }\right) \) .
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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\[ T = \mathop{\sum }\limits_{{i = 1}}^{\lambda }\operatorname{len}\left( {r}_{i}\right) \operatorname{len}\left( {q}_{i}\right) \leq \operatorname{len}\left( b\right) \mathop{\sum }\limits_{{i = 1}}^{\lambda }\operatorname{len}\left( {q}_{i}\right) \]\n\n\[ \leq \operatorname{len}\left( b\right) \mathop{\sum }\limits_{{i = 1}}^{\lambda }\left( {\operatorname{len}\left( {r}_{i - 1}\right) - \operatorname{len}\left( {r}_{i}\right) + 1}\right) \text{ (see Exercise 3.24) } \]\n\n\[ = \operatorname{len}\left( b\right) \left( {\operatorname{len}\left( {r}_{0}\right) - \operatorname{len}\left( {r}_{\lambda }\right) + \lambda }\right) \text{ (telescoping the sum) } \]\n\n\[ \leq \operatorname{len}\left( b\right) \left( {\operatorname{len}\left( a\right) + \log b/\log \phi + 1}\right) \text{(by Theorem 4.1)} \]\n\n\[ = O\left( {\operatorname{len}\left( a\right) \operatorname{len}\left( b\right) }\right) \text{.} \]
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No
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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\[ \n{s}_{1} \mathrel{\text{:=}} 0,\;{t}_{1} \mathrel{\text{:=}} 1, \]\n\n\[ \n{s}_{i + 1} \mathrel{\text{:=}} {s}_{i - 1} - {s}_{i}{q}_{i},\;{t}_{i + 1} \mathrel{\text{:=}} {t}_{i - 1} - {t}_{i}{q}_{i}\;\left( {i = 1,\ldots ,\lambda }\right) . \]\n\nThen:\n\n(i) for \( i = 0,\ldots ,\lambda + 1 \), we have \( a{s}_{i} + b{t}_{i} = {r}_{i} \) ; in particular, \( a{s}_{\lambda } + b{t}_{\lambda } = \gcd \left( {a, b}\right) \) ;\n\n(ii) for \( i = 0,\ldots ,\lambda \), we have \( {s}_{i}{t}_{i + 1} - {t}_{i}{s}_{i + 1} = {\left( -1\right) }^{i} \) ;\n\n(iii) for \( i = 0,\ldots ,\lambda + 1 \), we have \( \gcd \left( {{s}_{i},{t}_{i}}\right) = 1 \) ;\n\n(iv) for \( i = 0,\ldots ,\lambda \), we have \( {t}_{i}{t}_{i + 1} \leq 0 \) and \( \left| {t}_{i}\right| \leq \left| {t}_{i + 1}\right| \) ; for \( i = 1,\ldots ,\lambda \), we have \( {s}_{i}{s}_{i + 1} \leq 0 \) and \( \left| {s}_{i}\right| \leq \left| {s}_{i + 1}\right| \) ;\n\n(v) for \( i = 1,\ldots ,\lambda + 1 \), we have \( {r}_{i - 1}\left| {t}_{i}\right| \leq a \) and \( {r}_{i - 1}\left| {s}_{i}\right| \leq b \) ;\n\n(vi) if \( a > 0 \), then for \( i = 1,\ldots ,\lambda + 1 \), we have \( \left| {t}_{i}\right| \leq a \) and \( \left| {s}_{i}\right| \leq b \) ; if \( a > 1 \) and \( b > 0 \), then \( \left| {t}_{\lambda }\right| \leq a/2 \) and \( \left| {s}_{\lambda }\right| \leq b/2 \) .
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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\[ = \left( {a{s}_{i - 2} + b{t}_{i - 2}}\right) - \left( {a{s}_{i - 1} + b{t}_{i - 1}}\right) {q}_{i - 1} \]\n\n\[ = {r}_{i - 2} - {r}_{i - 1}{q}_{i - 1}\;\text{(by induction)} \]\n\n\[ = {r}_{i} \]\n\n(ii) is also easily proved by induction on \( i \) . For \( i = 0 \), the statement is clear. For \( i = 1,\ldots ,\lambda \), we have\n\n\[ {s}_{i}{t}_{i + 1} - {t}_{i}{s}_{i + 1} = {s}_{i}\left( {{t}_{i - 1} - {t}_{i}{q}_{i}}\right) - {t}_{i}\left( {{s}_{i - 1} - {s}_{i}{q}_{i}}\right) \]\n\n\[ = - \left( {{s}_{i - 1}{t}_{i} - {t}_{i - 1}{s}_{i}}\right) \;\text{(after expanding and simplifying)} \]\n\n\[ = - {\left( -1\right) }^{i - 1}\;\text{(by induction)} \]\n\n\[ = {\left( -1\right) }^{i}\text{.} \]\n\n(iii) follows directly from (ii).\n\nFor (iv), one can easily prove both statements by induction on \( i \) . The statement involving the \( {t}_{i} \) ’s is clearly true for \( i = 0 \) . For \( i = 1,\ldots ,\lambda \), we have\n\n\( {t}_{i + 1} = {t}_{i - 1} - {t}_{i}{q}_{i} \) ; moreover, by the induction hypothesis, \( {t}_{i - 1} \) and \( {t}_{i} \) have opposite signs and \( \left| {t}_{i}\right| \geq \left| {t}_{i - 1}\right| \) ; it follows that \( \left| {t}_{i + 1}\right| = \left| {t}_{i - 1}\right| + \left| {t}_{i}\right| {q}_{i} \geq \left| {t}_{i}\right| \), and that the sign of \( {t}_{i + 1} \) is the opposite of that of \( {t}_{i} \) . The proof of the statement involving the \( {s}_{i} \) ’s is the same, except that we start the induction at \( i = 1 \) .\n\nFor (v), one considers the two equations:\n\n\[ a{s}_{i - 1} + b{t}_{i - 1} = {r}_{i - 1} \]\n\n\[ a{s}_{i} + b{t}_{i} = {r}_{i} \]\n\nSubtracting \( {t}_{i - 1} \) times the second equation from \( {t}_{i} \) times the first, and applying (ii), we get \( \pm a = {t}_{i}{r}_{i - 1} - {t}_{i - 1}{r}_{i} \) ; consequently, using the fact that \( {t}_{i} \) and \( {t}_{i - 1} \) have opposite sign, we obtain\n\n\[ a = \left| {{t}_{i}{r}_{i - 1} - {t}_{i - 1}{r}_{i}}\right| = \left| {t}_{i}\right| {r}_{i - 1} + \left| {t}_{i - 1}\right| {r}_{i} \geq \left| {t}_{i}\right| {r}_{i - 1} \]
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Yes
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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:
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So we have \( \gcd \left( {a, b}\right) = 5 = - a + {3b} \) .
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Yes
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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 \( {b}^{-1}{\;\operatorname{mod}\;n} \) .
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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 \( b \) does not have a multiplicative inverse modulo \( n \) . Otherwise, if \( d = 1 \), then \( t \) is a multiplicative inverse of \( b \) modulo \( n \) ; however, it may not lie in the range \( \{ 0,\ldots, n - 1\} \), as required. By part (vi) of Theorem 4.3, we have \( \left| t\right| \leq n/2 < n \) . Thus, if \( t \geq 0 \), then \( {b}^{-1}{\;\operatorname{mod}\;n} \) is equal to \( t \) ; otherwise, \( {b}^{-1}{\;\operatorname{mod}\;n} \) is equal to \( t + n \) . Based on Theorem 4.4, it is clear that all the computations can be performed in time \( O\left( {\operatorname{len}{\left( n\right) }^{2}}\right) \) .
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Yes
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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.
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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 \
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No
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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 \( i = 1,\ldots, k \) . Let \( n \mathrel{\text{:=}} \mathop{\prod }\limits_{{i = 1}}^{k}{n}_{i} \) . Then in time \( O\left( {\operatorname{len}{\left( n\right) }^{2}}\right) \), we can compute the unique integer \( a \) satisfying \( 0 \leq a < n \) and \( a \equiv {a}_{i}\left( {\;\operatorname{mod}\;{n}_{i}}\right) \) for \( i = 1,\ldots, k. \)
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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\[ {n}_{i}^{ * } \leftarrow n/{n}_{i},{b}_{i} \leftarrow {n}_{i}^{ * }{\;\operatorname{mod}\;{n}_{i}},{t}_{i} \leftarrow {b}_{i}^{-1}{\;\operatorname{mod}\;{n}_{i}},{e}_{i} \leftarrow {n}_{i}^{ * }{t}_{i} \]\n\n\[ a \leftarrow \left( {\mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i}{e}_{i}}\right) {\;\operatorname{mod}\;n} \]\n\nWe leave it to the reader to verify the running time bound.
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No
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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 \( j \) be the smallest index (among \( 0,\ldots ,\lambda + 1 \) ) such that \( {r}_{j} < {r}^{ * } \) . Then, setting \( r \mathrel{\text{:=}} {r}_{j} \) and \( t \mathrel{\text{:=}} {t}_{j} \), we have\n\n\[ r \equiv {bt}\left( {\;\operatorname{mod}\;n}\right) ,\;0 \leq r < {r}^{ * },\text{ and }0 < \left| t\right| < {t}^{ * }.\]
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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\[ < {t}^{ * }\text{(since}n < {r}^{ * }{t}^{ * }\text{).} \]\n\nSince \( j \geq 1 \), by part (iv) of Theorem 4.3, we have \( \left| {t}_{j}\right| \geq \left| {t}_{1}\right| > 0 \) . Finally, since \( {r}_{j} = n{s}_{j} + b{t}_{j} \), we have \( {r}_{j} \equiv b{t}_{j}\left( {\;\operatorname{mod}\;n}\right) \).
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Yes
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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 this to the power \( \left( {p - 1}\right) /4 = {252} \). One can calculate that \( {17}^{252} \equiv {469}\left( {\;\operatorname{mod}\;{1009}}\right) \), and \( {469}^{2} \equiv - 1\left( {\;\operatorname{mod}\;{1009}}\right) \). So we were lucky with our first try. Now we run the extended Euclidean algorithm on input \( p = {1009} \) and \( b = {469} \), obtaining the following data:
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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 \( {r}^{2} + {t}^{2} = {28}^{2} + {15}^{2} = {1009} = p \).
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Yes
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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\[ r \equiv {bt}\left( {\;\operatorname{mod}\;n}\right) ,\;\left| r\right| \leq {r}^{ * },\;0 < \left| t\right| \leq {t}^{ * },\]\n\n(4.7)\n\n\[ {r}^{\prime } \equiv b{t}^{\prime }\left( {\;\operatorname{mod}\;n}\right) ,\;\left| {r}^{\prime }\right| \leq {r}^{ * },\;0 < \left| {t}^{\prime }\right| \leq {t}^{ * }.\]\n\n(4.8)\n\nThen \( r/t = {r}^{\prime }/{t}^{\prime } \) .
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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\[ r{t}^{\prime } - {r}^{\prime }t \equiv 0\left( {\;\operatorname{mod}\;n}\right) \]\n\nHowever, we also have\n\n\[ \left| {r{t}^{\prime } - {r}^{\prime }t}\right| \leq \left| r\right| \left| {t}^{\prime }\right| + \left| {r}^{\prime }\right| \left| t\right| \leq 2{r}^{ * }{t}^{ * } < n.\]\n\nThus, \( r{t}^{\prime } - {r}^{\prime }t \) is a multiple of \( n \), but less than \( n \) in absolute value; the only possibility is that \( r{t}^{\prime } - {r}^{\prime }t = 0 \), which means \( r/t = {r}^{\prime }/{t}^{\prime } \) .
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Yes
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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 \( j \) be the smallest index (among \( 0,\ldots ,\lambda + 1 \) ) such that \( {r}_{j} \leq {r}^{ * } \), and set \[ {r}^{\prime } \mathrel{\text{:=}} {r}_{j},\;{s}^{\prime } \mathrel{\text{:=}} {s}_{j},\text{ and }{t}^{\prime } \mathrel{\text{:=}} {t}_{j}. \] Finally, suppose that there exist \( r, s, t \in \mathbb{Z} \) such that \[ r = {ns} + {bt},\left| r\right| \leq {r}^{ * },\text{ and }0 < \left| t\right| \leq {t}^{ * }. \] Then we have: (i) \( 0 < \left| {t}^{\prime }\right| \leq {t}^{ * } \) ; (ii) if \( n > 2{r}^{ * }{t}^{ * } \), then for some non-zero integer \( q \) , \[ r = {r}^{\prime }q,\;s = {s}^{\prime }q,\;\text{ and }t = {t}^{\prime }q. \]
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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 \[ 0 \leq {r}_{j} \leq {r}^{ * } < {r}_{j - 1},0 < \left| {t}_{j}\right| ,\left| r\right| \leq {r}^{ * },\text{ and }0 < \left| t\right| \leq {t}^{ * }, \] (4.9) along with the identities \[ {r}_{j - 1} = n{s}_{j - 1} + b{t}_{j - 1} \] (4.10) \[ {r}_{j} = n{s}_{j} + b{t}_{j} \] (4.11) \[ r = {ns} + {bt}\text{.} \] (4.12) We now turn to part (i) of the theorem. Our goal is to prove that \[ \left| {t}_{j}\right| \leq {t}^{ * } \] (4.13) This is the hardest part of the proof. To this end, let \[ \varepsilon \mathrel{\text{:=}} {s}_{j}{t}_{j - 1} - {s}_{j - 1}{t}_{j},\;\mu \mathrel{\text{:=}} \left( {{t}_{j - 1}s - {s}_{j - 1}t}\right) /\varepsilon ,\;\nu \mathrel{\text{:=}} \left( {{s}_{j}t - {t}_{j}s}\right) /\varepsilon . \] Since \( \varepsilon = \pm 1 \), the numbers \( \mu \) and \( v \) are integers; moreover, one may easily verify that they satisfy the equations \[ {s}_{j}\mu + {s}_{j - 1}v = s \] (4.14) \[ {t}_{j}\mu + {t}_{j - 1}v = t. \] (4.15) We now use these identities to prove (4.13). We consider three cases: (i) Suppose \( v = 0 \) . In this case,(4.15) implies \( {t}_{j} \mid t \), and since \( t \neq 0 \), this implies \( \left| {t}_{j}\right| \leq \left| t\right| \leq {t}^{ * } \) . (ii) Suppose \( {\mu v} < 0 \) . In this case, since \( {t}_{j} \) and \( {t}_{j - 1} \) have opposite sign,(4.15) implies \( \left| t\right| = \left| {{t}_{j}\mu }\right| + \left| {{t}_{j - 1}v}\right| \geq \left| {t}_{j}\right| \), and so again, we have \( \left| {t}_{j}\right| \leq \left| t\right| \leq {t}^{ * } \) . (iii) The only remaining possibility is that \( v \neq 0 \) and \( {\mu v} \geq 0 \) . We argue that this is impossible. Adding \( n \) times (4.14) to \( b \) times (4.15), and using the identities (4.10), (4.11), and (4.12), we obtain \[ {r}_{j}\mu + {r}_{j - 1}v = r. \] If \( \nu \neq 0 \) and \( \mu \) and \( \nu \) had the same sign, we would have \( \left| r\right| = \left| {{r}_{j}\mu }\right| + \left| {{r}_{j - 1}\nu }\right| \geq \) \( {r}_{j - 1} \), and hence \( {r}_{j - 1} \leq \left| r\right| \leq {r}^{ * } \) ; however, this contradicts the fact that \( {r}_{j - 1} > {r}^{ * } \) . That proves the inequality (4.13). We now turn to the proof of part (ii) of the theorem, which relies critically on this inequality. Assume that \[ n > 2{r}^{ * }{t}^{ * } \] (4.16) From (4.11) and (4.12), we have \[ {r}_{j} \equiv b{t}_{j}\left( {\;\operatorname{mod}\;n}\right) \text{ and }r \equiv {bt}\left( {\;\operatorname{mod}\;n}\right) . \] Combining this with the inequalities (4.9), (4.13), and (4.16), we see that the hypotheses of Theorem 4.8 are satisfied, and so we may conclude that \[ r{t}_{j} - {r}_{j}t = 0 \] (4.17) Subtracting \( {t}_{j} \) times (4.12) from \( t \) times (4.11), and using the identity (4.17), we obtain \( n\left( {s{t}
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Yes
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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 \( s/t = {0.7197183098591549}\cdots \) . Bob receives the digits \( 7,1,9,7,1,8,3 \), and computes \( n = {10}^{7} \) and \( b = {7197183} \) .
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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 \( z = - {s}^{\prime }/{t}^{\prime } = {511}/{710} \).
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Yes
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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.
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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 \( k \mathrel{\text{:=}} {71} \), we select \( k \) primes, \( {n}_{1},\ldots ,{n}_{k} \), each 16-bits in length. In fact, let us choose \( {n}_{1},\ldots ,{n}_{k} \) to be the largest \( k \) primes under \( {2}^{16} \) . If we do this, then the smallest prime among the \( {n}_{i} \) ’s turns out to be 64717, which is greater than \( {2}^{15.98} \) . We may set \( M \mathrel{\text{:=}} {2}^{1024} \), and since we want to correct up to 3 errors, we may set \( P \mathrel{\text{:=}} {2}^{3 \cdot {16}} \) . Then with \( n \mathrel{\text{:=}} \mathop{\prod }\limits_{i}{n}_{i} \), we have\n\n\[ n > {2}^{{71} \cdot {15.98}} = {2}^{1134.58} > {2}^{1121} = {2}^{1 + {1024} + 6 \cdot {16}} = {2M}{P}^{2}. \]\n\nThus, with these parameter settings, the above scheme will correct up to 3 corrupted blocks. This comes at a cost of increasing the length of the message from 1024 bits to \( {71} \cdot {16} = {1136} \) bits, an increase of about \( {11}\% \) .
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Yes
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Theorem 5.1 (Chebyshev's theorem). We have\n\n\[ \pi \left( x\right) = \Theta \left( {x/\log x}\right) . \]
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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):
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No
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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
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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}}^{{{2m} - 1}}\left( \begin{matrix} {2m} \\ i \end{matrix}\right) + 1 \leq 2 + \left( {{2m} - 1}\right) \left( \begin{matrix} {2m} \\ m \end{matrix}\right) \leq {2m}\left( \begin{matrix} {2m} \\ m \end{matrix}\right) .\n\nThe proves the first inequality. For the second, observe that the binomial coefficient \( \left( \begin{matrix} {2m} + 1 \\ m \end{matrix}\right) \) occurs twice in the binomial expansion of \( {\left( 1 + 1\right) }^{{2m} + 1} \), and is therefore less than \( {2}^{{2m} + 1}/2 = {2}^{2m} \) .
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Yes
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Lemma 5.3. Let \( n \) be a positive integer. For every prime \( p \), we have\n\n\[ \n{v}_{p}\left( {n!}\right) = \mathop{\sum }\limits_{{k \geq 1}}\left\lfloor {n/{p}^{k}}\right\rfloor \n\]
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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 sufficiently large \( k \) ). So we have\n\n\[ \n{v}_{p}\left( {n!}\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{v}_{p}\left( j\right) = \mathop{\sum }\limits_{{j = 1}}^{n}\mathop{\sum }\limits_{{k \geq 1}}{d}_{jk} = \mathop{\sum }\limits_{{k \geq 1}}\mathop{\sum }\limits_{{j = 1}}^{n}{d}_{jk}. \n\]\n\nFinally, note that \( \mathop{\sum }\limits_{{j = 1}}^{n}{d}_{jk} \) is equal to the number of multiples of \( {p}^{k} \) among the integers \( 1,\ldots, n \), which by Exercise 1.3 is equal to \( \left\lfloor {n/{p}^{k}}\right\rfloor \) .
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Yes
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Theorem 5.4. \( \pi \left( n\right) \geq \frac{1}{2}\left( {\log 2}\right) n/\log n \) for every integer \( n \geq 2 \) .
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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 to the identity \( N = \left( {2m}\right) !/{\left( m!\right) }^{2} \), we have\n\n\[ {v}_{p}\left( N\right) = \mathop{\sum }\limits_{{k \geq 1}}\left( {\left\lfloor {{2m}/{p}^{k}}\right\rfloor - 2\left\lfloor {m/{p}^{k}}\right\rfloor }\right) .\n\nEach term in this sum is either 0 or 1 (see Exercise 1.4), and for \( k > \log \left( {2m}\right) /\log p \) , each term is zero. Thus, \( {v}_{p}\left( N\right) \leq \log \left( {2m}\right) /\log p \) . So we have\n\n\[ \pi \left( {2m}\right) \log \left( {2m}\right) = \mathop{\sum }\limits_{{p \leq {2m}}}\frac{\log \left( {2m}\right) }{\log p}\log p \]\n\n\[ \geq \mathop{\sum }\limits_{{p \leq {2m}}}{v}_{p}\left( N\right) \log p = \log N \]\n\nwhere the summations are over the primes \( p \) up to \( {2m} \) . By Lemma 5.2, we have \( N \geq {2}^{2m}/{2m} \geq {2}^{m} \), and hence\n\n\[ \pi \left( {2m}\right) \log \left( {2m}\right) \geq m\log 2 = \frac{1}{2}\left( {\log 2}\right) \left( {2m}\right) .\n\nThat proves the theorem for even \( n \) . Now consider odd \( n \geq 3 \), so \( n = {2m} - 1 \) for some \( m \geq 2 \) . It is easily verified that the function \( x/\log x \) is increasing for \( x \geq 3 \) ; therefore,\n\n\[ \pi \left( {{2m} - 1}\right) = \pi \left( {2m}\right) \]\n\n\[ \geq \frac{1}{2}\left( {\log 2}\right) \left( {2m}\right) /\log \left( {2m}\right) \]\n\n\[ \geq \frac{1}{2}\left( {\log 2}\right) \left( {{2m} - 1}\right) /\log \left( {{2m} - 1}\right) .\n\nThat proves the theorem for odd \( n \) .
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Yes
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Theorem 5.5. We have\n\n\[ \vartheta \left( x\right) = \Theta \left( {\pi \left( x\right) \log x}\right) . \]
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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\[ \vartheta \left( x\right) = \mathop{\sum }\limits_{{p \leq x}}\log p \geq \mathop{\sum }\limits_{{{x}^{1/2} < p \leq x}}\log p \geq \frac{1}{2}\log x\mathop{\sum }\limits_{{{x}^{1/2} < p \leq x}}1 \]\n\n\[ = \frac{1}{2}\log x\left( {\pi \left( x\right) - \pi \left( {x}^{1/2}\right) }\right) = \frac{1}{2}\left( {1 - \pi \left( {x}^{1/2}\right) /\pi \left( x\right) }\right) \pi \left( x\right) \log x. \]\n\nIt will therefore suffice to show that \( \pi \left( {x}^{1/2}\right) /\pi \left( x\right) = o\left( 1\right) \) . Clearly, \( \pi \left( {x}^{1/2}\right) \leq {x}^{1/2} \) . Moreover, by the previous theorem, \( \pi \left( x\right) = \Omega \left( {x/\log x}\right) \) . Therefore,\n\n\[ \pi \left( {x}^{1/2}\right) /\pi \left( x\right) = O\left( {\log x/{x}^{1/2}}\right) = o\left( 1\right) ,\]\n\nand the theorem follows.
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Yes
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Theorem 5.6. \( \vartheta \left( x\right) < 2\left( {\log 2}\right) x \) for every real number \( x \geq 1 \) .
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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 on \( n \) .\n\nFor \( n = 1 \) and \( n = 2 \), this is clear, so assume \( n > 2 \) . If \( n \) is even, then using the induction hypothesis for \( n - 1 \), we have\n\n\[ \vartheta \left( n\right) = \vartheta \left( {n - 1}\right) < 2\left( {\log 2}\right) \left( {n - 1}\right) < 2\left( {\log 2}\right) n. \]\n\nNow consider the case where \( n \) is odd. Write \( n = {2m} + 1 \), where \( m \) is a positive integer, and consider the binomial coefficient\n\n\[ M \mathrel{\text{:=}} \left( \begin{matrix} {2m} + 1 \\ m \end{matrix}\right) = \frac{\left( {{2m} + 1}\right) \cdots \left( {m + 2}\right) }{m!}. \]\n\nObserve that \( M \) is divisible by all primes \( p \) with \( m + 1 < p \leq {2m} + 1 \) . Moreover, be Lemma 5.2, we have \( M < {2}^{2m} \) . It follows that\n\n\[ \vartheta \left( {{2m} + 1}\right) - \vartheta \left( {m + 1}\right) = \mathop{\sum }\limits_{{m + 1 < p \leq {2m} + 1}}\log p \leq \log M < 2\left( {\log 2}\right) m. \]\n\nUsing this, and the induction hypothesis for \( m + 1 \), we obtain\n\n\[ \vartheta \left( n\right) = \vartheta \left( {{2m} + 1}\right) - \vartheta \left( {m + 1}\right) + \vartheta \left( {m + 1}\right) \]\n\n\[ < 2\left( {\log 2}\right) m + 2\left( {\log 2}\right) \left( {m + 1}\right) = 2\left( {\log 2}\right) n. \]
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Yes
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Theorem 5.8 (Bertrand’s postulate). For every positive integer \( m \), we have\n\n\[ \pi \left( {2m}\right) - \pi \left( m\right) > \frac{m}{3\log \left( {2m}\right) }.\]
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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 \( N \) is divisible only by primes less than \( {2m} \), and that we have the identity\n\n\[ {v}_{p}\left( N\right) = \mathop{\sum }\limits_{{k \geq 1}}\left( {\left\lfloor {{2m}/{p}^{k}}\right\rfloor - 2\left\lfloor {m/{p}^{k}}\right\rfloor }\right) ,\]\n\n(5.1)\n\nwhere each term in the sum is either 0 or 1 . We can characterize the values \( {v}_{p}\left( N\right) \) a bit more precisely, as follows:
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Yes
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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\[ \n\text{if}{2m}/3 < p \leq m\text{, then}{v}_{p}\left( N\right) = 0\text{;} \n\]\n\n(5.4)\n\n\[ \n\text{if}m < p < {2m}\text{, then}{v}_{p}\left( N\right) = 1\text{.} \n\]\n\n(5.5)
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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 \( {2m}/3 < p \leq m \), then \( {2m}/p < 3 \), and we must also have \( p \geq 3 \) , since \( p = 2 \) implies \( m < 3 \) . We have \( {p}^{2} > p\left( {{2m}/3}\right) = {2m}\left( {p/3}\right) \geq {2m} \), and hence all terms with \( k > 1 \) in (5.1) vanish. The term with \( k = 1 \) also vanishes, since \( 1 \leq m/p < 3/2 \), from which it follows that \( 2 \leq {2m}/p < 3 \), and hence \( \lfloor m/p\rfloor = 1 \) and \( \lfloor {2m}/p\rfloor = 2 \).\n\nFor (5.5), if \( m < p < {2m} \), it follows that \( 1 < {2m}/p < 2 \), so \( \lfloor {2m}/p\rfloor = 1 \) . Also, \( m/p < 1 \), so \( \lfloor m/p\rfloor = 0 \) . It follows that the term with \( k = 1 \) in (5.1) is 1, and it is clear that \( {2m}/{p}^{k} < 1 \) for all \( k > 1 \), and so all the other terms vanish.
|
Yes
|
Theorem 5.10. We have\n\n\[ \mathop{\sum }\limits_{{p \leq x}}\frac{1}{p} = \log \log x + O\\left( 1\\right) \]
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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\[ \mathop{\sum }\limits_{{p \leq x}}\frac{\log p}{p} = \log x + O\left( 1\right) \]
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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\[ \log \left( {n!}\right) = \mathop{\sum }\limits_{{p \leq n}}\mathop{\sum }\limits_{{k \geq 1}}\left\lfloor {n/{p}^{k}}\right\rfloor \log p = \mathop{\sum }\limits_{{p \leq n}}\lfloor n/p\rfloor \log p + \mathop{\sum }\limits_{{k \geq 2}}\mathop{\sum }\limits_{{p \leq n}}\left\lfloor {n/{p}^{k}}\right\rfloor \log p. \]\n\nWe next show that the last sum is \( O\left( n\right) \) . We have\n\n\[ \mathop{\sum }\limits_{{p \leq n}}\log p\mathop{\sum }\limits_{{k \geq 2}}\left\lfloor {n/{p}^{k}}\right\rfloor \leq n\mathop{\sum }\limits_{{p \leq n}}\log p\mathop{\sum }\limits_{{k \geq 2}}{p}^{-k} \]\n\n\[ = n\mathop{\sum }\limits_{{p \leq n}}\frac{\log p}{{p}^{2}} \cdot \frac{1}{1 - 1/p} = n\mathop{\sum }\limits_{{p \leq n}}\frac{\log p}{p\left( {p - 1}\right) } \]\n\n\[ \leq n\mathop{\sum }\limits_{{k \geq 2}}\frac{\log k}{k\left( {k - 1}\right) } = O\left( n\right) \]\n\nThus, we have shown that\n\n\[ \log \left( {n!}\right) = \mathop{\sum }\limits_{{p \leq n}}\lfloor n/p\rfloor \log p + O\left( n\right) . \]\n\nSince \( \lfloor n/p\rfloor = n/p + O\left( 1\right) \), applying Theorem 5.6 (and Exercise 3.12), we obtain\n\n\[ \log \left( {n!}\right) = \mathop{\sum }\limits_{{p \leq n}}\left( {n/p}\right) \log p + O\left( {\mathop{\sum }\limits_{{p \leq n}}\log p}\right) + O\left( n\right) = n\mathop{\sum }\limits_{{p \leq n}}\frac{\log p}{p} + O\left( n\right) . \]\n\n(5.7)\n\nWe can also estimate \( \log \left( {n!}\right) \) by estimating a sum by an integral (see §A5):\n\n\[ \log \left( {n!}\right) = \mathop{\sum }\limits_{{k = 1}}^{n}\log k = {\int }_{1}^{n}\log {tdt} + O\left( {\log n}\right) = n\log n - n + O\left( {\log n}\right) . \]\n\n(5.8)\n\nCombining (5.7) and (5.8), and noting that \( \log x - \log n = o\left( 1\right) \) (see Exercise 3.11), we obtain\n\n\[ \mathop{\sum }\limits_{{p \leq x}}\frac{\log p}{p} = \log n + O\left( 1\right) = \log x + O\left( 1\right) \]\n\nwhich proves the theorem.
|
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 with a continuous derivative \( {f}^{\prime }\left( t\right) \) on the interval \( \left\lbrack {k, x}\right\rbrack \), where \( x \) is a real number, with \( x \geq k \) . Then\n\n\[ \n\mathop{\sum }\limits_{{k \leq i \leq x}}{c}_{i}f\left( i\right) = C\left( x\right) f\left( x\right) - {\int }_{k}^{x}C\left( t\right) {f}^{\prime }\left( t\right) {dt}. \n\]
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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\[ \n= \mathop{\sum }\limits_{{i = k}}^{{n - 1}}C\left( i\right) \left\lbrack {f\left( i\right) - f\left( {i + 1}\right) }\right\rbrack + C\left( n\right) f\left( n\right) \n\]\n\n\[ \n= \mathop{\sum }\limits_{{i = k}}^{{n - 1}}C\left( i\right) \left\lbrack {f\left( i\right) - f\left( {i + 1}\right) }\right\rbrack + C\left( n\right) \left\lbrack {f\left( n\right) - f\left( x\right) }\right\rbrack + C\left( x\right) f\left( x\right) . \n\]\n\nObserve that for \( i = k,\ldots, n - 1 \), we have \( C\left( t\right) = C\left( i\right) \) for all \( t \in \lbrack i, i + 1) \), and so\n\n\[ \nC\left( i\right) \left\lbrack {f\left( i\right) - f\left( {i + 1}\right) }\right\rbrack = - C\left( i\right) {\int }_{i}^{i + 1}{f}^{\prime }\left( t\right) {dt} = - {\int }_{i}^{i + 1}C\left( t\right) {f}^{\prime }\left( t\right) {dt} \n\]\n\nlikewise,\n\n\[ \nC\left( n\right) \left\lbrack {f\left( n\right) - f\left( x\right) }\right\rbrack = - {\int }_{n}^{x}C\left( t\right) {f}^{\prime }\left( t\right) {dt} \n\]\n\nfrom which the theorem directly follows.
|
Yes
|
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