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Prove that $\sqrt{6}$ is irrational.
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Suppose that $\sqrt{6}$ was rational. We show that this leads to a contradiction. We may write $\sqrt{6}=x / y$ where $x$ and $y$ are integers with $y \neq 0$ and $\operatorname{gcd}(x, y)=1$. Squaring this equation and cross-multiplying we get that $6 y^{2}=x^{2}$ or $2 \cdot 3 \cdot y^{2}=x^{2}$. Therefore, 2 divides $x^{2}=x \cdot x$. Since 2 is prime we must have that 2 divides $x$. Similarly, 3 divides $x^{2}=x \cdot x$.

And since 3 is prime we must have that 3 divides $x$. Since $2 \mid x$ and $3 \mid x$ and $\operatorname{gcd}(2,3)=1$, by the first part of this problem, we have that $6=2 \cdot 3$ must divide $x$. So $x=6 u$ where $u$ is a non-zero integer. Subbing this into $6 y^{2}=x^{2}$ gives us that $6 y^{2}=6^{2} u^{2}$. Thus $y^{2}=6 u^{2}$. Following the same reasoning as above, this forces that 6 must divide $y$. Therefore, 6 is a common divisor of $x$ and $y$ which contradicts the fact that $\operatorname{gcd}(x, y)=1$.
by Diamond (71,587 points)

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