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Section2.3Exercises

1

Prove that

12+22++n2=n(n+1)(2n+1)6\begin{equation*} 1^2 + 2^2 + \cdots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \end{equation*}

for nN.n \in {\mathbb N}\text{.}

Hint

The base case, S(1):[1(1+1)(2(1)+1)]/6=1=12S(1): [1(1 + 1)(2(1) + 1)]/6 = 1 = 1^2 is true. Assume that S(k):12+22++k2=[k(k+1)(2k+1)]/6S(k): 1^2 + 2^2 + \cdots + k^2 = [k(k + 1)(2k + 1)]/6 is true. Then

12+22++k2+(k+1)2=[k(k+1)(2k+1)]/6+(k+1)2=[(k+1)((k+1)+1)(2(k+1)+1)]/6,\begin{align*} 1^2 + 2^2 + \cdots + k^2 + (k + 1)^2 & = [k(k + 1)(2k + 1)]/6 + (k + 1)^2\\ & = [(k + 1)((k + 1) + 1)(2(k + 1) + 1)]/6, \end{align*}

and so S(k+1)S(k + 1) is true. Thus, S(n)S(n) is true for all positive integers n.n\text{.}

2

Prove that

13+23++n3=n2(n+1)24\begin{equation*} 1^3 + 2^3 + \cdots + n^3 = \frac{n^2(n + 1)^2}{4} \end{equation*}

for nN.n \in {\mathbb N}\text{.}

3

Prove that n!>2nn! \gt 2^n for n4.n \geq 4\text{.}

Hint

The base case, S(4):4!=24>16=24S(4): 4! = 24 \gt 16 =2^4 is true. Assume S(k):k!>2kS(k): k! \gt 2^k is true. Then (k+1)!=k!(k+1)>2k2=2k+1,(k + 1)! = k! (k + 1) \gt 2^k \cdot 2 = 2^{k + 1}\text{,} so S(k+1)S(k + 1) is true. Thus, S(n)S(n) is true for all positive integers n.n\text{.}

4

Prove that

x+4x+7x++(3n2)x=n(3n1)x2\begin{equation*} x + 4x + 7x + \cdots + (3n - 2)x = \frac{n(3n - 1)x}{2} \end{equation*}

for nN.n \in {\mathbb N}\text{.}

5

Prove that 10n+1+10n+110^{n + 1} + 10^n + 1 is divisible by 3 for nN.n \in {\mathbb N}\text{.}

6

Prove that 4102n+9102n1+54 \cdot 10^{2n} + 9 \cdot 10^{2n - 1} + 5 is divisible by 99 for nN.n \in {\mathbb N}\text{.}

7

Show that

a1a2ann1nk=1nak.\begin{equation*} \sqrt[n]{a_1 a_2 \cdots a_n} \leq \frac{1}{n} \sum_{k = 1}^{n} a_k. \end{equation*}
8

Prove the Leibniz rule for f(n)(x),f^{(n)} (x)\text{,} where f(n)f^{(n)} is the nnth derivative of f;f\text{;} that is, show that

(fg)(n)(x)=k=0n(nk)f(k)(x)g(nk)(x).\begin{equation*} (fg)^{(n)}(x) = \sum_{k = 0}^{n} \binom{n}{k} f^{(k)}(x) g^{(n - k)}(x). \end{equation*}
Hint

Follow the proof in Example 2.4.

9

Use induction to prove that 1+2+22++2n=2n+111 + 2 + 2^2 + \cdots + 2^n = 2^{n + 1} - 1 for nN.n \in {\mathbb N}\text{.}

10

Prove that

12+16++1n(n+1)=nn+1\begin{equation*} \frac{1}{2}+ \frac{1}{6} + \cdots + \frac{1}{n(n + 1)} = \frac{n}{n + 1} \end{equation*}

for nN.n \in {\mathbb N}\text{.}

11

If xx is a nonnegative real number, then show that (1+x)n1nx(1 + x)^n - 1 \geq nx for n=0,1,2,.n = 0, 1, 2, \ldots\text{.}

Hint

The base case, S(0):(1+x)01=00=0xS(0): (1 + x)^0 - 1 = 0 \geq 0 = 0 \cdot x is true. Assume S(k):(1+x)k1kxS(k): (1 + x)^k -1 \geq kx is true. Then

(1+x)k+11=(1+x)(1+x)k1=(1+x)k+x(1+x)k1kx+x(1+x)kkx+x=(k+1)x,\begin{align*} (1 + x)^{k + 1} - 1 & = (1 + x)(1 + x)^k -1\\ & = (1 + x)^k + x(1 + x)^k - 1\\ & \geq kx + x(1 + x)^k\\ & \geq kx + x\\ & = (k + 1)x, \end{align*}

so S(k+1)S(k + 1) is true. Therefore, S(n)S(n) is true for all positive integers n.n\text{.}

12Power Sets

Let XX be a set. Define the of X,X\text{,} denoted P(X),{\mathcal P}(X)\text{,} to be the set of all subsets of X.X\text{.} For example,

P({a,b})={,{a},{b},{a,b}}.\begin{equation*} {\mathcal P}( \{a, b\} ) = \{ \emptyset, \{a\}, \{b\}, \{a, b\} \}. \end{equation*}

For every positive integer n,n\text{,} show that a set with exactly nn elements has a power set with exactly 2n2^n elements.

13

Prove that the two principles of mathematical induction stated in Section 2.1 are equivalent.

14

Show that the Principle of Well-Ordering for the natural numbers implies that 1 is the smallest natural number. Use this result to show that the Principle of Well-Ordering implies the Principle of Mathematical Induction; that is, show that if SNS \subset {\mathbb N} such that 1S1 \in S and n+1Sn + 1 \in S whenever nS,n \in S\text{,} then S=N.S = {\mathbb N}\text{.}

15

For each of the following pairs of numbers aa and b,b\text{,} calculate gcd(a,b)\gcd(a,b) and find integers rr and ss such that gcd(a,b)=ra+sb.\gcd(a,b) = ra + sb\text{.}

  1. 14 and 39

  2. 234 and 165

  3. 1739 and 9923

  4. 471 and 562

  5. 23,771 and 19,945

  6. 4357-4357 and 3754

16

Let aa and bb be nonzero integers. If there exist integers rr and ss such that ar+bs=1,ar + bs =1\text{,} show that aa and bb are relatively prime.

17Fibonacci Numbers

The Fibonacci numbers are

1,1,2,3,5,8,13,21,.\begin{equation*} 1, 1, 2, 3, 5, 8, 13, 21, \ldots. \end{equation*}

We can define them inductively by f1=1,f_1 = 1\text{,} f2=1,f_2 = 1\text{,} and fn+2=fn+1+fnf_{n + 2} = f_{n + 1} + f_n for nN.n \in {\mathbb N}\text{.}

  1. Prove that fn<2n.f_n \lt 2^n\text{.}

  2. Prove that fn+1fn1=fn2+(1)n,f_{n + 1} f_{n - 1} = f^2_n + (-1)^n\text{,} n2.n \geq 2\text{.}

  3. Prove that fn=[(1+5)n(15)n]/2n5.f_n = [(1 + \sqrt{5}\, )^n - (1 - \sqrt{5}\, )^n]/ 2^n \sqrt{5}\text{.}

  4. Show that limnfn/fn+1=(51)/2.\lim_{n \rightarrow \infty} f_n / f_{n + 1} = (\sqrt{5} - 1)/2\text{.}

  5. Prove that fnf_n and fn+1f_{n + 1} are relatively prime.

Hint

For (a) and (b) use mathematical induction. (c) Show that f1=1,f_1 = 1\text{,} f2=1,f_2 = 1\text{,} and fn+2=fn+1+fn.f_{n + 2} = f_{n + 1} + f_n\text{.} (d) Use part (c). (e) Use part (b) and Exercise 2.3.16.

18

Let aa and bb be integers such that gcd(a,b)=1.\gcd(a,b) = 1\text{.} Let rr and ss be integers such that ar+bs=1.ar + bs =1\text{.} Prove that

gcd(a,s)=gcd(r,b)=gcd(r,s)=1.\begin{equation*} \gcd(a,s) = \gcd(r,b) = \gcd(r,s) = 1. \end{equation*}
19

Let x,yNx, y \in {\mathbb N} be relatively prime. If xyxy is a perfect square, prove that xx and yy must both be perfect squares.

Hint

Use the Fundamental Theorem of Arithmetic.

20

Using the division algorithm, show that every perfect square is of the form 4k4k or 4k+14k + 1 for some nonnegative integer k.k\text{.}

21

Suppose that a,b,r,sa, b, r, s are pairwise relatively prime and that

a2+b2=r2a2b2=s2.\begin{align*} a^2 + b^2 & = r^2\\ a^2 - b^2 & = s^2. \end{align*}

Prove that a,a\text{,} r,r\text{,} and ss are odd and bb is even.

22

Let nN.n \in {\mathbb N}\text{.} Use the division algorithm to prove that every integer is congruent mod nn to precisely one of the integers 0,1,,n1.0, 1, \ldots, n-1\text{.} Conclude that if rr is an integer, then there is exactly one ss in Z{\mathbb Z} such that 0s<n0 \leq s \lt n and [r]=[s].[r] = [s]\text{.} Hence, the integers are indeed partitioned by congruence mod n.n\text{.}

23

Define the of two nonzero integers aa and b,b\text{,} denoted by lcm(a,b),\lcm(a,b)\text{,} to be the nonnegative integer mm such that both aa and bb divide m,m\text{,} and if aa and bb divide any other integer n,n\text{,} then mm also divides n.n\text{.} Prove there exists a unique least common multiple for any two integers aa and b.b\text{.}

Hint

Use the Principle of Well-Ordering and the division algorithm.

24

If d=gcd(a,b)d= \gcd(a, b) and m=lcm(a,b),m = \lcm(a, b)\text{,} prove that dm=ab.dm = |ab|\text{.}

25

Show that lcm(a,b)=ab\lcm(a,b) = ab if and only if gcd(a,b)=1.\gcd(a,b) = 1\text{.}

26

Prove that gcd(a,c)=gcd(b,c)=1\gcd(a,c) = \gcd(b,c) =1 if and only if gcd(ab,c)=1\gcd(ab,c) = 1 for integers a,a\text{,} b,b\text{,} and c.c\text{.}

27

Let a,b,cZ.a, b, c \in {\mathbb Z}\text{.} Prove that if gcd(a,b)=1\gcd(a,b) = 1 and abc,a \mid bc\text{,} then ac.a \mid c\text{.}

Hint

Since gcd(a,b)=1,\gcd(a,b) = 1\text{,} there exist integers rr and ss such that ar+bs=1.ar + bs = 1\text{.} Thus, acr+bcs=c.acr + bcs = c\text{.}

28

Let p2.p \geq 2\text{.} Prove that if 2p12^p - 1 is prime, then pp must also be prime.

29

Prove that there are an infinite number of primes of the form 6n+5.6n + 5\text{.}

Hint

Every prime must be of the form 2, 3, 6n+1,6n + 1\text{,} or 6n+5.6n + 5\text{.} Suppose there are only finitely many primes of the form 6k+5.6k + 5\text{.}

30

Prove that there are an infinite number of primes of the form 4n1.4n - 1\text{.}

31

Using the fact that 2 is prime, show that there do not exist integers pp and qq such that p2=2q2.p^2 = 2 q^2\text{.} Demonstrate that therefore 2\sqrt{2} cannot be a rational number.