Any odd integer squared is always odd, and likewise, any even integer squared is always even. Therefore, the square root of an odd number must be odd and the square root of an even number must be even. For it can never be the case the square root of an odd number is even, or the square root of an even number is odd. Furthermore, only integers can be said of as being either even or odd.
Suppose,
$${a \over b} = \sqrt 2 $$
Then,
$${{{a^2}} \over {{b^2}}} = 2$$
$${a^2} = 2{b^2}$$
${a^2}$ must be even owing to the two present in the equation.
Therefore, $a$ must be even.
Suppose $a = 2$ then,
$${2^2} = 2({b^2})$$
$${{{2^2}} \over 2} = {b^2}$$
$$2 = {b^2}$$
Therefore, $b$ must be even.
But, $$b = \sqrt 2 $$
The square root of two is not an even integer, because it's not an integer at all.
In more general terms where $n$ is any integer,
$$\sqrt {{{{n^2}} \over 2}} = n\sqrt {{1 \over 2}} $$
So am I right in saying the square root of a half is also irrational?
Answer
A square must be divisible by an even number of powers of $2$.
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