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So I know that $\sqrt{4}=2$, and therefore is rational, but, it seems like I proved it to be irrational? Could I have some help as to where I went wrong? Thank you!

Prove $\sqrt4$ is irrational. Suppose per contradiction $\sqrt4$ is rational, that is, $\sqrt4 = \frac ab$ where $a$ and $b$ are integers. We assume $\frac ab$ is in reduced form.

$$\begin{align}\sqrt4^2 &= \left(\frac{a}{b}\right)^2\\ 4 &= \frac{a^2}{b^2}\\ 4b^2 &= a^2 \end{align}$$

This implies $4$ divides $a^2$, and thus $4|a$, and thus $2|a$. That is, $a = 2n$ for some integer $n$. The definition of an even integer is $2k$ for some integer $k$, so $a$ is even. Applying this new logic to our old statement we get:

$$\begin{align} 4b^2 &= (4n)^2\\ 4b^2 &= 16n^2\\ b^2 &= 4n^2 \end{align}$$

This implies $4|b^2$, and thus $4|b$, and thus $2|b$. Since we know $b = 2n$ for some integer $n$, and the definition of an even integer, $b$ is therefore an even integer. Since we know that both $a$ and $b$ cannot be even, as that would contradict the fact that $\frac ab$ is in reduced form, then we know that there is no $a$ and $b$ such that $\sqrt4 = \frac ab$. Therefore, $\sqrt4$ is irrational.

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    $\begingroup$ You have a $(4n)^2$ where it should be a $(2n)^2$... $\endgroup$ – Daniel Robert-Nicoud May 7 '17 at 21:32
  • $\begingroup$ Please use MathJax. Formatting tips here. $\endgroup$ – Em. May 7 '17 at 21:33
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So $4|a^2$ does not imply that $4|a$. But your real error comes from

$$4b^2 = (4n)^2 4b^2 = 16n^2 b^2 = 4n^2$$ and you conclude that $4|b^2$. You should divide by $4$ to get $$b^2=n^2$$ from which little can be concluded about divisibility by $4$.

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