Prove that 5 - sqrt{3} is an irrational numbe | vedclass

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(B) Let us assume,to the contrary,that $5 - \sqrt{3}$ is a rational number.Then,there exist coprime integers $a$ and $b$ $(b 
eq 0)$ such that $5 - \

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Irrational

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(B) Let us assume,to the contrary,that $5 - \sqrt{3}$ is a rational number.Then,there exist coprime integers $a$ and $b$ $(b 
eq 0)$ such that $5 - \sqrt{3} = \frac{a}{b}$.Rearranging the equation,we get $5 - \frac{a}{b} = \sqrt{3}$,which simplifies to $\frac{5b - a}{b} = \sqrt{3}$.Since $a$ and $b$ are integers,$\frac{5b - a}{b}$ is a rational number.This implies that $\sqrt{3}$ is a rational number.However,this contradicts the fact that $\sqrt{3}$ is an irrational number.Therefore,our assumption is incorrect,and $5 - \sqrt{3}$ must be an irrational number.

Prove that $5 - \sqrt{3}$ is an irrational number.

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