Prove that $7 \sqrt{5}$ is an irrational number.

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(N/A) Assume that $7 \sqrt{5}$ is a rational number.
Therefore,we can find two integers $a$ and $b$ $(b \neq 0)$ such that $7 \sqrt{5} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{5} = \frac{a}{7b}$.
Since $a$ and $b$ are integers,$\frac{a}{7b}$ is a rational number.
This implies that $\sqrt{5}$ must also be a rational number.
However,this contradicts the established fact that $\sqrt{5}$ is an irrational number.
Therefore,our initial assumption that $7 \sqrt{5}$ is rational is incorrect.
Hence,$7 \sqrt{5}$ is an irrational number.

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