Prove that 3 is an irrational number. - Vidyadip

Question

Prove that $\sqrt3$ is an irrational number.

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Answer

If possible, let $\sqrt{3}$ be rational and let its simplest form be $\frac{a}{b}$
The, a and b are integers having no common factor other than $1$ , and $b \neq 0$
Now, $\sqrt{3}=\frac{ a ^2}{b^2}$
$\Rightarrow 3=\frac{ a ^2}{b^2} \ldots$ (On squaring both sides)
$\Rightarrow 3 b^2=a^2$
$\Rightarrow 3$ divides $a ^2 \ldots\left[\because 3\right.$ divies $\left.3 b^2\right]$
$\Rightarrow 3$ divides a $\ldots\left[\because 3\right.$ is prime and 3 divides $a^2 \Rightarrow 3$ divides $\left.a\right]$
Let $a =3 c$ for some integers c .
Putting $a=3 c$ in (i), we get
$3 b^2=9 c^2$
$\Rightarrow b^2=3 c^2$
$\Rightarrow 3 \text { divides } b^2 \ldots\left[\because 3 \text { divies } 3 c^2\right]$
$\Rightarrow 3 \text { divides } b \ldots\left[\because 3 \text { is prime and } 3 \text { divides } a^2 \Rightarrow 3 \text { divides } a\right]$
Thus, $3$ is a common factor of $a$ and $b$
But, this contradicts the fact that $a$ and $b$ have no common factor other than $1$
The contradiction arises by assuming that $\sqrt{3}$ is rational.
Hence, $\sqrt{3}$ is irrational.

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