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Prove that âˆš7 is an irrational number.

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$\large \\ \text{Let us assume that} \sqrt{7}$ is rational. Then, there exist co-prime positive integers $a$ and $b$ such that \\ $\sqrt{7}=\frac{a}{b}$ \\ $\Longrightarrow a=b \sqrt{7}$ \\ Squaring on both sides, we get \\ $a^{2}=7 b^{2}$ \\ Therefore, $a^{2}$ is divisible by 7 and hence, $a$ is also divisible by7 so, we can write $a=7 p,$ for some integer $p$ \\ Substituting for $a$, we get $49 p^{2}=7 b^{2} \Longrightarrow b^{2}=7 p^{2}$. \\ This means, $b^{2}$ is also divisible by 7 and $s o, b$ is also divisible by 7 \\ Therefore, $a$ and $b$ have at least one common factor, i.e., $7 .$$

$\large \\ \text{But, this contradicts the fact that} a$ and $b$ are co-prime. Thus, our supposition is wrong. Hence, $\sqrt{7}$ is irrational.$

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Deependra Verma

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