Answer:
Prove the following:
(a) Given a, b ∈ Z with b 6= 0, there exist x, y ∈ Z with gcd(x, y) = 1
and a
b
=
x
y
.
Solution: Let d = gcd(a, b). Let x = a/d and y = b/d. Then
from class, we know that gcd(x, y) = 1. And we also have that
a/b = (a/d)/(b/d) = x/y.
(b) If p is a prime and a is a positive integer and p|a
n
, then p
n
|a
n
.
Solution: Suppose that p is a prime and p divides a
n = a· a · · · a.
Recall that when a prime divides a product of integers then it
must divide at least one of the integers contained in the product.
Hence p|a. Therefore, pk = a for some integer k. Hence, a
n =
(pk)
n = p
nk
n
. Therefore p
n
|a
n
.
(c) √5
5 is irrational.
Solution: Suppose that √5
5 is rational. Then √5
5 = a/b where
a, b ∈ Z. We may always cancel common divisors in a fraction,
hence we may assume that gcd(a, b) = 1.
Taking the fifth power of both sides of √5
5 = a/b gives 5 = a
5/b5
.
Hence a
5 = 5b
5
. Therefore 5 divides the product a
5 = a·a·a·a·a.
Recall that when a prime divides a product of integers then it must
divide at least one of the integers contained in the product. Since
5 is prime we must have that 5 divides a. Therefore a = 5k where
k is an integer. Substituting this expression into a
5 = 5b
5 yields
5
5k
5 = 5b
5
. Hence 5(53k
5
) = b
5
. Therefore 5 divides b
5
. Since 5
is prime we must have that 5|b. But then 5 would be a common
divisor of a and b and hence gcd(a, b) ≥ 5. This contradicts our
assumption that gcd(a, b) = 1.
Therefore √5
5 is irrational.
(d) If p is a prime, then √p is irrational.
Solution: Suppose that √p is rational. Then √p = a/b where
a, b ∈ Z. We may always cancel common divisors in a fraction,
hence we may assume that gcd(a, b) = 1.
Squaring both sides of √p = a/b and then multiplying through by
b
2 gives us that pb2 = a
2
. Hence p|a
2
. Recall that when a prime