Prove that 3 + 2sqrt{5} is irrational. | vedclass

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(B) Let us assume,to the contrary,that $3 + 2\sqrt{5}$ is rational.Then,there exist coprime integers $a$ and $b$ $(b 
eq 0)$ such that $3 + 2\sqrt{5}

Vedclass · Accepted Answer

Irrational

Vedclass · Answer

(B) Let us assume,to the contrary,that $3 + 2\sqrt{5}$ is rational.Then,there exist coprime integers $a$ and $b$ $(b 
eq 0)$ such that $3 + 2\sqrt{5} = \frac{a}{b}$.Subtracting $3$ from both sides,we get $2\sqrt{5} = \frac{a}{b} - 3$.$2\sqrt{5} = \frac{a - 3b}{b}$.Dividing by $2$,we get $\sqrt{5} = \frac{a - 3b}{2b}$.Since $a$ and $b$ are integers,$\frac{a - 3b}{2b}$ is a rational number.This implies that $\sqrt{5}$ is a rational number.However,this contradicts the fact that $\sqrt{5}$ is irrational.Therefore,our assumption is incorrect,and $3 + 2\sqrt{5}$ is irrational.

Prove that $3 + 2\sqrt{5}$ is irrational.

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