Question
Prove that $\sqrt{2}$ is an irrational number.
✓
Answer
Let assume on the contrary that $\sqrt{2}$ is arational number.
Then, there exists positive integer $a$ and $b$ such that $\sqrt{2}=\frac{ a }{ b }$ where, $a$ and $b$ are co primes i.e. their $\text {HCF}$ is $1.$
$\Rightarrow \sqrt{2}=\left(\frac{a}{b}\right)^2$
$\Rightarrow 2=\frac{a^2}{b^2}$
$\Rightarrow a^2=2 b^2$
$\Rightarrow a^2 \text { is multiple of } 2$
$\Rightarrow a \text { is a multiple of } 2$
$\Rightarrow a=2 c \text { for some integer } c .$
$\Rightarrow a^2=4 c^2$
$\Rightarrow 2 b^2=4 c^2$
$\Rightarrow b^2=2 c^2$
$\Rightarrow b^2 \text { is a multiple of } 2$
$b \text { is a multiple of 2...(ii) }$
From $(i)$ and $(ii)$, a and b have at least $2$ as a common factor. But this contradicts the fact that a and b are co$-$prime. This means that $\sqrt{2}$ is an irrational number.
Then, there exists positive integer $a$ and $b$ such that $\sqrt{2}=\frac{ a }{ b }$ where, $a$ and $b$ are co primes i.e. their $\text {HCF}$ is $1.$
$\Rightarrow \sqrt{2}=\left(\frac{a}{b}\right)^2$
$\Rightarrow 2=\frac{a^2}{b^2}$
$\Rightarrow a^2=2 b^2$
$\Rightarrow a^2 \text { is multiple of } 2$
$\Rightarrow a \text { is a multiple of } 2$
$\Rightarrow a=2 c \text { for some integer } c .$
$\Rightarrow a^2=4 c^2$
$\Rightarrow 2 b^2=4 c^2$
$\Rightarrow b^2=2 c^2$
$\Rightarrow b^2 \text { is a multiple of } 2$
$b \text { is a multiple of 2...(ii) }$
From $(i)$ and $(ii)$, a and b have at least $2$ as a common factor. But this contradicts the fact that a and b are co$-$prime. This means that $\sqrt{2}$ is an irrational number.
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