Question

Proof by Contradiction : Show that √ 2 is irrational.

Answer 1

Step-by-step explanation:

Let's assume that $\sqrt{2}$ is rational,

So, it should be expressed in the form of $\dfrac{a}{b}$  

where a and b are two prime integers.

So, we can write,

 $\sqrt{2}=\dfrac{a}{b}$

$=>\ 2=\dfrac{a^2}{b^2}$

$=>\ a^2=2b^2.$

As $2b^2$ represents a even number, $a^2$ must also be even.  

Since $a^2$ is even number, so a is also an even number.

Let assume again that a=2c

$=>\ a^2\ =\ 4c^2$

$=>\ 2b^2=\ 4c^2$

$=>\ b^2=2c^2.$  

As $2c^2$ is even, so $b^2$ is also an even number, and so is b.  

And since and b both are even number here, two even numbers can not be relatively prime, so $\sqrt{2}$ can not be expressed in the form of $\dfrac{a}{b}$. So, $\sqrt{2}$ is considered as an irrational number.