Can the square root of any positive integer that is not a perfect square be irrational?

Question

For example, square root of 3 and the square root of 5 yields an irrational number. If not, how would you prove that all numbers that are not perfect squares be irrational?

Answer 1

You mean, can it be RATIONAL? You could take the classic proof of √2's irrationality and expand it to this general case. Or start off with a fraction that is defined as not being an integer, and show that its square can't be an integer.

Assume that there is an integer whose square root is not an integer nor an irrational number. Therefore it has to be a non-integer, rational number. Call this root a/b (where "a" and "b" are coprime integers, and b is not "1"). This means the original integer is (a/b)^2, or (a^2) / (b^2).

If this is to be a fraction, then a^2 has to be a multiple of b^2. This means all of the prime factors of b^2 must also be prime factors of a^2, which in turn means that all of the prime factors of b must be prime factors of a. But this contradicts our initial definition that a and b must be coprime.

Answer 2

All integers are rational, whether or not they are perfect squares.

Ithink you meant to as are the square roots of all numbers that are not perfect squares irrational.
If so the answer is yes.
sqrt(98) = 7sqrt(2) which is irrational.
So the square root of any number (except a prime number)can be reduced to a rational number times an irrational number which of course is an irrational number. the suare root of a prime number has to be irrational by definition.