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# Prove that $\sqrt 2$ is an irrational number?

Last updated date: 20th Jun 2024
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Hint: In this question, we will use the contradiction method of solving the question. First assume that the given number is a rational number of the $\dfrac{a}{b}$ where a and b have no common factor other than 1 and b$\ne$ 0. After this we follow a procedure to contradict our assumption.

The given number is$\sqrt 2$.
Now suppose that the given number is a rational number of the form $\dfrac{{\text{a}}}{{\text{b}}}$, where a and b have no common factor other than 1 and b$\ne$ 0.
$\therefore$ .$\sqrt 2$ =$\dfrac{{\text{a}}}{{\text{b}}}$
On squaring both sides, we get:
${\left( {\sqrt 2 } \right)^2} = \dfrac{{{{\text{a}}^2}}}{{{{\text{b}}^2}}}$
On further solving, we have:
$\Rightarrow 2 = \dfrac{{{{\text{a}}^2}}}{{{{\text{b}}^2}}}$
$\Rightarrow 2{{\text{b}}^2} = {{\text{a}}^2}$
$\because$ 2 divides ${{\text{b}}^2}$.
$\therefore$ It will also divide ${{\text{a}}^2}$and hence it will divide ‘a’.
Now, let a =2k
Putting the value of ‘a’ in above equation, we get:
$2{{\text{b}}^2} = {\left( {2{\text{k}}} \right)^2}$
On further solving, we get:
$2{{\text{b}}^2} = 4{{\text{k}}^2}$
$\Rightarrow {{\text{b}}^2} = 2{{\text{k}}^2}$
$\because$ 2 divides ${{\text{k}}^2}$.
$\therefore$ It will also divide ${{\text{b}}^2}$and hence it will divide ‘b’.
Therefore, we can say that 2 is the common factor of both ‘a’ and ‘b’.
This contradicts our assumption that ‘a’ and ‘b’ have no common factor other than 1
Hence, the number $\sqrt 2$ is an irrational number.

Note: Solving such forms of problems requires a determined approach of making an assumption which is the exact opposite of what is being asked and then contradicting that assumption therefore reaching to the proof. You should know that the product of a rational and an irrational number is always irrational. For example- $2\sqrt 2$ is an irrational number.