Some Model Answers for HW on Chapter 0
Notes: 1. This HW was graded as follows:
√
correct (3p)
(√)
imperfect, but judged `complete’ (3p)
Partial credit:
incomplete, or hard to read or both
2. Future HW will be graded more and more on correctness.
3. `if and only if’ statements usually require two proofs: one for each
direction of the assertion
#9: If a and b are integers and n is a positive integer, prove that a mod n= b
mod n if and only if n divides a-b.
Let’s write a=nq1+r1 and b=nq2+r2. Then a-b=n(q1-q2)+r1-r2
►
assume that a mod n= b mod n. Then r1=r2 and a-b=n(q1-q2).
Therefore n divides a-b.
◄
assume that n divides a-b. Then by the division algorithm r1-r2 has to
be 0. So r1=r2 and a mod n= b mod n.
#13: Let n and a be positive integers and let d=gcd(a,n). Show that the
equation ax mod n =1 has a solution if and only if d=1.
►
assume that ax mod n =1 has a solution. Then ax = nq+1. in other
words ax-nq=1. This means there is a linear combination of a and n
that yields 1. By the GCD Theorem, this means that the GCD of a and
n has to be 1.
◄
assume that gcd(a,n) =1. Then by the GCD Theorem there is a linear
combination that yields 1: ax+nq=1. Taking the mod n on both sides
we get that ax mod n + nq mod n =1 mod n. This implies that ax mod
n =1. So there is a solution of the equation ax mod n =1
#14: Show that 5n+3 and 7n+4 are relatively prime for all n.
By the GCD Theorem it suffices to show that there is a linear
combination of 5n+3 and 7n+4 that gives 1, because then 1 would
have to be the gcd, and the two numbers are relatively prime. Indeed,
taking
7 (5n+3) – 5 (7n+4) = 1,
shows that this linear combination exists.
#18: Let p1,p2,. . . pn be primes. Show that p1p2∙∙∙ pn +1 is not divisible by
any of these primes.
Proof. By the division algorithm p1p2∙∙∙ pn +1= p1(p2∙∙∙ pn) +1 is the
unique division of the number by p1. It has a remainder of 1, and
therefore is not divisible by p. Similarly, dividing the number by any
of the other primes always leaves a remainder of 1.
#19. Prove that there are infinitely many primes
Proof. Suppose that there are finitely many primes. Let’s enumerate
them by p1,p2,. . . pn, where pn is the largest prime. Then define the
number p1p2∙∙∙ pn +1. As #18 shows this number is not divisible by any
of the primes, so is itself a prime. And it is larger than pn. That is a
contradiction. So there must be infinitely many primes.
#20. For every positive integer n prove that 1+2+∙∙∙+n=½ n(n+1)
Proof by induction:
initial step: if n=1, then the formula is obviously true: 1=1.
Now assume that the formula is true for n. For n+1 the left side
becomes
1+2+∙∙∙+n+n+1=½ n(n+1) + (n+1) (here we used the assumption)
=½(n2+n+2n+2)
=½(n+1)(n+2)
which is the right hand side of the formula for n+1.
#30. Prove that for every integer n, n3 mod 6 = n mod 6.
Proof: It suffices (by exercise #9) to prove that 6 divides n 3 – n. But n3
– n factors into n(n2-1) which factors into n(n-1)(n+1). This means
that n3 – n factors into the product of three successive integers. At
least one of them has to be even, and at least one of them has to be
divisible by 3. This means the product is divisible by 6.