Are all non-integer square roots irrational numbers?

Question

I know that √2 and √3 are irrational, but I got to wondering if ALL square roots of integers that do not have an integer as the answer are irrational numbers.

Answer 1

Yes they are.

 

Here's a way to prove it.  It's a little complicated and it has to do with factoring integers into its prime factors.  If that's something you're familiar with, then follow along!

 

Suppose there is an integer (call it u) that has a non-integer square root that is not irrational.  That means that the square root has to be a rational fraction -- something like p/q, where p is not a multiple of q.  (If p were a multiple of q, then p/q would be an integer, which would violate one of the assumptions.)

 

Let's also assume that p/q is in the lowest terms.  That is, there are no common factors in p and q.  If there were any, we could just divide them out to get to a fraction that is in the lowest terms.

 

So what we're saying is that (p/q)² = u, or

 

p²/q² = u

 

So p² = uq²

 

Factor p² into its prime factors, and factor uq² into its prime factors.  Since the numbers are the same, then the factorizations must be the same. 

 

Now, we already know that u is not a perfect square.  That means that in the prime factorization of u, there is at least one prime with an odd power.  That's because if all the primes in the factorization had even powers, you could just divide them all by 2 and get the factorization for another integer.  But the square of that integer would be u!  And we've already said that the square root of u is not an integer, so that can't be.

 

And we also know that in the prime factorization of q², all the primes have even powers.  That's for a similar reason as above.  The prime factorization of q² is the same as the prime factorization of q, except that all the powers are doubled.  But if all the powers are doubled in the prime factorization of q², then the powers are all even numbers.

 

Since the prime factorization of u has one prime with an odd power, and the prime factorization of q² has all even powers for its primes, then the prime factorization of uq² has to have at least one prime with an odd power, the same one that is odd in the factorization of u.

 

However, the prime factorization of p² has all even powers for its primes, for the same reason as above for the prime factorization of q².

 

So go back to the equation p² = uq².  We've just shown that the left side has all even powers in its prime factorization, but the right side has at least one prime with an odd power.  But that's impossible.  If two numbers are equal, then they have to have the same prime factorization!  So our assumption that there is a non-integer non-irrational square root of u has led to a false statement, the assumption must be incorrect.

 

I hope that's clear!