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

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

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Irrational

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(B) Assume,to the contrary,that $\sqrt{7}$ is a rational number.Then,there exist coprime positive integers $a$ and $b$ $(b 
eq 0)$ such that $\sqrt{7} = \frac{a}{b}$.Squaring both sides,we get $7 = \frac{a^2}{b^2}$,which implies $a^2 = 7b^2$.This means that $a^2$ is divisible by $7$,and by the Fundamental Theorem of Arithmetic,$a$ must also be divisible by $7$.So,we can write $a = 7k$ for some integer $k$.Substituting this into the equation $a^2 = 7b^2$,we get $(7k)^2 = 7b^2$,which simplifies to $49k^2 = 7b^2$,or $b^2 = 7k^2$.This implies that $b^2$ is divisible by $7$,and therefore $b$ must also be divisible by $7$.Since both $a$ and $b$ are divisible by $7$,they have a common factor of $7$,which contradicts our assumption that $a$ and $b$ are coprime.Therefore,our assumption that $\sqrt{7}$ is rational is false,and $\sqrt{7}$ is an irrational number.

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

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