Question
Without using division method show that $\sqrt{7}$ is an irrational numbers.
✓
Answer
Let $\sqrt{7}$ be a raised number.
$\therefore \sqrt{7}=\frac{a}{b}$
$\Rightarrow 7=\frac{ a ^2}{ b ^2}$
$\Rightarrow a ^2=7 b ^2$
Since $a^2$ is divisible by $7 , a$ is also divisible by $7 .$
Let $a =7 c$
$\Rightarrow a^2=49 c^2$
$\Rightarrow 7 b^2=49 c^2$
$\Rightarrow b^2=7 c^2$
Since $b^2$ is divisible by $7 , b$ is also divisible by $7 .$
From $(I)$ and $(II)$, we get $a$ and $b$ both divisible by $7 .$
i.e., $a$ and $b$ have a common factor $7 .$
This contradicts our assumption that $\frac{ a }{ b }$ is rational.
i.e. $a$ and $b$ do not have any common factor other than unity $(1).$
$\Rightarrow \frac{ a }{ b }$ is not rational
$\Rightarrow \sqrt{7}$ is not rational, i.e. $\sqrt{7}$ is irrational.
$\therefore \sqrt{7}=\frac{a}{b}$
$\Rightarrow 7=\frac{ a ^2}{ b ^2}$
$\Rightarrow a ^2=7 b ^2$
Since $a^2$ is divisible by $7 , a$ is also divisible by $7 .$
Let $a =7 c$
$\Rightarrow a^2=49 c^2$
$\Rightarrow 7 b^2=49 c^2$
$\Rightarrow b^2=7 c^2$
Since $b^2$ is divisible by $7 , b$ is also divisible by $7 .$
From $(I)$ and $(II)$, we get $a$ and $b$ both divisible by $7 .$
i.e., $a$ and $b$ have a common factor $7 .$
This contradicts our assumption that $\frac{ a }{ b }$ is rational.
i.e. $a$ and $b$ do not have any common factor other than unity $(1).$
$\Rightarrow \frac{ a }{ b }$ is not rational
$\Rightarrow \sqrt{7}$ is not rational, i.e. $\sqrt{7}$ is irrational.
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