Question

State true or false
 $ \sqrt 2 $ is not a rational number.

Answer 1

Hint: Any number is called rational when it is in the form of $ \dfrac{p}{q} $ . Where $ \sqrt 2 $ cannot be expressed.
Let's suppose $ \sqrt 2 $ is a rational number. Then we can write it $ \sqrt 2 = \dfrac{a}{b} $ where a, b are whole numbers, b not zero.
We additionally assume that this $ \dfrac{a}{b} $ is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for $ \dfrac{a}{b} $ to be in simplest terms, both a and b cannot be even. One or both must be odd. Otherwise, we could simplify $ \dfrac{a}{b} $ further.

Complete step-by-step answer:
From the equality
 $ \sqrt 2 = \dfrac{a}{b} $
it follows that
 $
  2 = \dfrac{{{a^2}}}{{{b^2}}} \\
  {a^2} = 2{b^2} \;
 $
So, the square of a is an even number since it is two times something.
From this we know that a itself is also an even number.
Why? Because it can't be odd;
if a itself was odd, then a · a would be odd too.
Odd number times odd number is always odd.
Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols,
 $ a = 2k $
where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.
If we substitute a = 2k into the original equation $ 2 = \dfrac{{{a^2}}}{{{b^2}}} $ , this is what we get:
 $
  2 = \dfrac{{{{(2k)}^2}}}{{{b^2}}} \\
  2 = \dfrac{{4{k^2}}}{{{b^2}}} \\
  2{b^2} = 4{k^2} \\
  {b^2} = 2{k^2} \;
 $
This means that $ {b^2} $ is even, from which follows again that $ b $ itself is even. And that is a contradiction!!!
Therefore $ \sqrt 2 $ cannot be rational
So, the correct answer is “TRUE”.

Note: WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus, our original assumption (that $ \sqrt 2 $ is rational) is not correct. Therefore $ \sqrt 2 $ cannot be rational.