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# Prove that $\sqrt 6$ is an irrational number.

Last updated date: 04th Mar 2024
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Hint: Here we have to represent $\sqrt 6$ as fraction of two integers, and we have to represent that these two integers have common factor at lowest form and both cannot be even.By contradiction method (i.e assuming negation statement and proving that statement wrong) have to prove it is an irrational number.

This problem can be solved by a contradiction method i.e assuming it is a rational number.
The following proof is of contradiction
Let us assume that $\sqrt 6$ is rational number
Then it can be represented as factor of two integers
Let the lowest terms representation be $\sqrt 6 = \dfrac{a}{b}$, where $b \ne 0$
$\therefore {a^2} = 6{b^2}$ …… (1)
From above ${a^2}$ is even, if it is even then ‘a’ should also be even
$\Rightarrow a = 2c$ (c is constant and 2c is an even number)
Squaring both the sides of the above equation
${a^2} = 4{c^2}$ …… (2)
From equation (1) and (2)
$4{c^2} = 6{b^2}$ and $2{c^2} = 3{b^2}$
From above $3{b^2}$ is even, if it is even then ${b^2}$ should be even and also ‘b’ again should be even
Therefore, a and b have some common factors
But a and b were in lowest form and both cannot be even.
Hence assumption was wrong and hence$\sqrt 6$ is an irrational number.

NOTE: $\sqrt 6 = \dfrac{a}{b}$ , this representation is in lowest terms and hence, a and b have no common factors.So it is an irrational number.