Answer to Question #348773 in Calculus for Puja

Question

State if the following statement is true or false. Give reasons for your answers in the form of a short proof or a counterexample.

(√2, 1, ½) E Q X Z X R.

Answer 1

The statement is false because $\sqrt{2}\notin\mathbb{Q}$.

Proof that $\sqrt{2}\notin\mathbb{Q}$:

Suppose $\sqrt{2}$ is rational. That mean it can be writen as the ratio of two inegers $p$ and $q$

$\sqrt{2}=\frac{p}{q}$ (1)

where we may assume that $p$ and $q$ have no common factors. (If there are any common factors we cancel tham in the numerator and denominator.) Squaring in (1) on both sides gives

$2=\frac{p^2}{q^2}$ (2)

which impiels

$p^2=2q^2$ (3)

Thus $p^2$ is even. The only way this can be true is that $p$ itself is even. But then $p^2$ is actually divisible by $4$. Hence $q^2$ and therefore $q$ must me even. So $p$ and $q$ are both even which is a contradiction to our assumption that they have no common factor. The square root of 2 cannot be rational.