Give a proof that nth root is irrational for every natural number n that is not a perfect square?

Answer 1

There are many proofs of this fact. I think one way to see it it let n be a number that is not a perfect square
Assume √n is rational.
In other words assume that

n = p2/q2

where p and q are integers and p/q is in lowest terms. So

n q2 = p2

So p2 is divisible by n. That means that p must be as well, so p2 is
divisible by n2.

So

q2 = p2/n

and q^2 is divisible by n.

But that means that p and q are both divisible by n, so they weren't
in lowest terms, which is a contradiction.


Here is another way to look at it.

If √n is rational, then it can be expressed by some number a/b (in lowest terms). This would mean: (a/b)² = n. Squaring, a² / b² = n Multiplying by b², a² = nb². If a and b are in lowest terms which we assumed, then their squares would each have an even number of prime factors. nb² has one more prime factor than b², meaning it would have an odd number of prime factors. Every composite has a unique prime factorization so I cannot have both an even and odd number of prime factors. This contradiction violates are assumption that a√n is rational. It is therefore irrational.