Is the square root of an irrational number rational?

Answer 1

No:

Let r be some irrational number; as such it cannot be represented as s/t where s and t are both non-zero integers.

Assume the square root of this irrational number r was rational.

Then it can be represented in the form of p/q where p and q are both non-zero integers, ie √r = p/q

As p is an integer, p² = p×p is also an integer, let y = p²

And as q is an integer, q² = q×q is also an integer, let x = q²

The number is the square of its square root, thus:

(√r)² = (p/q)² = p²/q² = y/x

but (√r)² = r, thus r = y/x and is a rational number.

But r was chosen to be an irrational number, which is a contradiction (r cannot be both rational and irrational at the same time, so it cannot exist).

Thus the square root of an irrational number cannot be rational.

However, the square root of a rational number can be irrational, eg for the rational number ½ its square root (√½) is not rational.