It easy to show that √2 is an irrational number

So is the square root of any non-square number.

Prove that √22 is an irrational number, providing a reasoning applicable to any random-chosen non-square integer.

Assume that the square root of 22
is rational.<o:p></o:p>

This means there exist integers a and b, with no common factor, such that: a/b = square root of
22.<o:p></o:p>

Squaring
both sides: (a/b)^2 = 22 <o:p></o:p>

So, a^2/b^2 = 22

<o:p></o:p>

Hence, a^2 = 22b^2 = (2)(11)b^2 .......... eq (1)<o:p></o:p>

Since
the RHS has a factor of 2, by definition it is even. Therefore, so is the LHS
and so a^2 must be even. <o:p></o:p>

Since any odd number times itself
is odd, it follows that if a^2 is even, then a is even.<o:p></o:p>

Then,
there exists an integer k, s.t. a = 2k ....... eq. (2) <o:p></o:p>

Substitute eq. (2) into (1): (2k)^2 = (2)(11)b^2<o:p></o:p>

4k^2 =(2)(11)b^2<o:p></o:p>

Or 2k^2 = 11b^2 dividing both sides by 2<o:p></o:p>

Since the LHS is even, so is the RHS. On the RHS, 11 is odd which implies that b^2 must be even. And, as above, this
means b is even.<o:p></o:p>

If a and b are both even, they have a common factor of 2. This contradicts the initial assumption above that they have no common factor. Therefore, the square root of 22 cannot be
rational.<o:p></o:p>

This is easily generalized to any non-square integer.