Question

Proof by Contradiction : Show that √ 2 is irrational.

Answer 1

Answer:

$\sqrt2$ is irrational

Step-by-step explanation:

Let us assume that $\sqrt2$ is rational. Thus, it can be expressed in the form of fraction $\frac{x}{y}$, where x and y are co-prime to each other.

$\sqrt2$ = $\frac{x}{y}$

Squaring both sides,

$2 = \frac{x^2}{y^2}$

Now, it is clear that x is an even number. So, let us substitute x = 2u

Thus,

$2 = \frac{(2u)^2}{y^2}\\y^2 = 2u^2$

Thus, $y^2$is even, which follows the fact that y is also an even number. But this is a contradiction as x and y have a common factor that is 2 but we assumed that the fraction $\frac{x}{y}$  was in lowest form.

Hence, $\sqrt2$ is not a rational number. But $\sqrt2$ is a an irrational number.