Prove that sqrt{2} is irrational. | vedclass

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(B) Assume,to the contrary,that $\sqrt{2}$ is a rational number. Then,there exist coprime integers $a$ and $b$ $(b 
eq 0)$ such that $\sqrt{2} = \fra

Vedclass · Accepted Answer

It is an irrational number.

Vedclass · Answer

(B) Assume,to the contrary,that $\sqrt{2}$ is a rational number. Then,there exist coprime integers $a$ and $b$ $(b 
eq 0)$ such that $\sqrt{2} = \frac{a}{b}$. Squaring both sides,we get $2 = \frac{a^2}{b^2}$,which implies $a^2 = 2b^2$. This shows that $a^2$ is even,and therefore $a$ must also be even. Let $a = 2k$ for some integer $k$. Substituting this into the equation,we get $(2k)^2 = 2b^2$,so $4k^2 = 2b^2$,which simplifies to $b^2 = 2k^2$. This shows that $b^2$ is even,and therefore $b$ must also be even. Since both $a$ and $b$ are even,they have a common factor of $2$,which contradicts our assumption that $a$ and $b$ are coprime. Thus,our assumption is false,and $\sqrt{2}$ is irrational.

Prove that $\sqrt{2}$ is irrational.

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