Prove that 7sqrt{5} is an irrational number. | vedclass

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(B) Let us assume,to the contrary,that $7\sqrt{5}$ is a rational number.Then,there exist coprime integers $a$ and $b$ $(b 
eq 0)$ such that $7\sqrt{5

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Irrational

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(B) Let us assume,to the contrary,that $7\sqrt{5}$ is a rational number.Then,there exist coprime integers $a$ and $b$ $(b 
eq 0)$ 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 $\sqrt{5}$ is a rational number.However,this contradicts the fact that $\sqrt{5}$ is an irrational number.Therefore,our assumption is incorrect,and $7\sqrt{5}$ must be an irrational number.

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

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