Question

prove √2 is irrational number​

Answer 1

Answer:

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Answer 2

$\sqrt{2}$ is an irrational number.

Step-by-step explanation:

Let $\sqrt{2}$ be a rational number. Which can be written in form of $\frac{p}{q}$ where, there is no any common factor between 'p' and 'q' other than 1.

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

squaring on both sides

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

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

$2q^{2} = p^{2}$                   .................(ii)

2 divides $p^{2}$

2 also divides p

Let p = 2k           ................(iii)

putting p = 2k in (ii)

$2q^{2} = (2k)^2$

$2q^{2} = 4k^{2}$

$q^{2} = 2k^{2}$

2 divides $q^{2}$

2 also divides q

Here, 2 is a common factor of 'p' and 'q' and it creates a contradict.

∴ $\sqrt{2}$ is an irrational number.