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PreTeXt Sample Book: Abstract Algebra (SAMPLE ONLY)

Exercises 2.4 Exercises

1.

Prove that
12+22++n2=n(n+1)(2n+1)6
for nN.
Answer.
The base case, S(1):[1(1+1)(2(1)+1)]/6=1=12 is true.
Assume that S(k):12+22++k2=[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,
and so S(k+1) is true. Thus, S(n) is true for all positive integers n.

3.

Prove that n!>2n for n4.
Answer.
The base case, S(4):4!=24>16=24 is true. Assume S(k):k!>2k is true. Then (k+1)!=k!(k+1)>2k2=2k+1, so S(k+1) is true. Thus, S(n) is true for all positive integers n.

5.

Prove that 10n+1+10n+1 is divisible by 3 for nN.

6.

Prove that 4102n+9102n1+5 is divisible by 99 for nN.

8.

Use induction to prove that 1+2+22++2n=2n+11 for nN.

9.

Prove the Leibniz rule for f(n)(x), where f(n) is the nth derivative of f; that is, show that
(fg)(n)(x)=k=0n(nk)f(k)(x)g(nk)(x).
Hint.
Follow the proof in Example 2.1.4.

11.

If x is a nonnegative real number, then show that (1+x)n1nx for n=0,1,2,.
Hint.
The base case, S(0):(1+x)01=00=0x is true. Assume S(k):(1+x)k1kx 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,
so S(k+1) is true. Therefore, S(n) is true for all positive integers n.

12. Power Sets.

Let X be a set. Define the power set of X, denoted P(X), to be the set of all subsets of X. For example,
P({a,b})={,{a},{b},{a,b}}.
For every positive integer n, show that a set with exactly n elements has a power set with exactly 2n 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 SN such that 1S and n+1S whenever nS, then S=N.

15.

For each of the following pairs of numbers a and b, calculate gcd(a,b) and find integers r and s such that gcd(a,b)=ra+sb.
  1. 14 and 39
  2. 234 and 165
  3. 1739 and 9923
  4. 471 and 562
  5. 23,771 and 19,945
  6. 4357 and 3754

16.

Let a and b be nonzero integers. If there exist integers r and s such that ar+bs=1, show that a and b are relatively prime.

17. Fibonacci Numbers.

The Fibonacci numbers are
1,1,2,3,5,8,13,21,.
We can define them inductively by f1=1, f2=1, and fn+2=fn+1+fn for nN.
  1. Prove that fn<2n.
  2. Prove that fn+1fn1=fn2+(1)n, n2.
  3. Prove that fn=[(1+5)n(15)n]/2n5.
  4. Show that limnfn/fn+1=(51)/2.
  5. Prove that fn and fn+1 are relatively prime.
Hint.
For Item 2.4.17.a and Item 2.4.17.b use mathematical induction. Item 2.4.17.c Show that f1=1, f2=1, and fn+2=fn+1+fn. Item 2.4.17.d Use part Item 2.4.17.c. Item 2.4.17.e Use part Item 2.4.17.b and Exercise 2.4.16.

18.

Let a and b be integers such that gcd(a,b)=1. Let r and s be integers such that ar+bs=1. Prove that
gcd(a,s)=gcd(r,b)=gcd(r,s)=1.

19.

Let x,yN be relatively prime. If xy is a perfect square, prove that x and y 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 4k or 4k+1 for some nonnegative integer k.

21.

Suppose that a,b,r,s are pairwise relatively prime and that
a2+b2=r2a2b2=s2.
Prove that a, r, and s are odd and b is even.

22.

Let nN. Use the division algorithm to prove that every integer is congruent mod n to precisely one of the integers 0,1,,n1. Conclude that if r is an integer, then there is exactly one s in Z such that 0s<n and [r]=[s]. Hence, the integers are indeed partitioned by congruence mod n.

23.

Define the least common multiple of two nonzero integers a and b, denoted by lcm(a,b), to be the nonnegative integer m such that both a and b divide m, and if a and b divide any other integer n, then m also divides n. Prove that any two integers a and b have a unique least common multiple.
Hint.
Let S={sN:as, bs}. Then S, since |ab|S. By the Principle of Well-Ordering, S contains a least element m. To show uniqueness, suppose that an and bn for some nN. By the division algorithm, there exist unique integers q and r such that n=mq+r, where 0r<m. Since a and b divide both m, and n, it must be the case that a and b both divide r. Thus, r=0 by the minimality of m. Therefore, mn.

24.

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

25.

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

26.

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

27.

Let a,b,cZ. Prove that if gcd(a,b)=1 and abc, then ac.
Hint.
Since gcd(a,b)=1, there exist integers r and s such that ar+bs=1. Thus, acr+bcs=c. Since a divides both bc and itself, a must divide c.

28.

Let p2. Prove that if 2p1 is prime, then p must also be prime.

29.

Prove that there are an infinite number of primes of the form 6n+5.
Hint.
Every prime must be of the form 2, 3, 6n+1, or 6n+5. Suppose there are only finitely many primes of the form 6k+5.

30.

Prove that there are an infinite number of primes of the form 4n1.

31.

Using the fact that 2 is prime, show that there do not exist integers p and q such that p2=2q2. Demonstrate that therefore 2 cannot be a rational number.
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