Use the method of contradiction to prove that √2 is irrational. I don't understand how to prove that √2 is irrational using this method. And I feel difficult to form the contradiction.

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    $\begingroup$ Write the steps of the proof and highlight what you're having difficulty with. $\endgroup$ – Deepak Aug 26 '19 at 4:42
  • $\begingroup$ The contradiction would be: $\sqrt 2$ is rational. Or in other words that there is a rational number $q$ so that $q^2 = 2$. Or in other words there are two integers $m$ and $n$ in lowest terms so that $2 =\frac {m^2}{n^2}$. I'll give you a boost to get started: That would mean $2n^2 = m^2$. Can you get a contradiction from that? Second hint: Can you determine whether $m$ and/or $n$ are even or odd? Or not? $\endgroup$ – fleablood Aug 26 '19 at 6:01

If you want to proof $A\Rightarrow B$, a proof by contradiction reads as follows:

Suppose $B$ is false, then $A$ has to be false.

So suppose $\sqrt{2}$ is rational, then $\sqrt{2}=\frac{p}{q}$ for $p\in\mathbb{Z}$, $q\in\mathbb{N}$ where $q\neq 0$.

Without loss of generality we can assume that the fraction $\frac{p}q$ is completly reduced.That means $\operatorname{gcd}(p,q)=1$ (greatest common divisor) This is our assumption we take to the contradiction! We can assume that, because if the fraction is not completly reduced, we can reduce it. :)

Now it is a little bit of equivalent manipulations:

$\sqrt{2}=\frac{p}{q}\Leftrightarrow 2=\frac{p^2}{q^2}\Leftrightarrow 2q^2=p^2$.

From this equality we get that $2$ divides $p^2$. Since $2$ is prime we have $2$ divides $p$.

[Indeed: $2\mid n^2$ iff $2\mid n$.


When $2\mid n$ then $n=2m$ for some $m\in\mathbb{Z}$. Then $n^2=(2m)^2=4m^2=2\cdot 2m^2$ so $2\mid n^2$.


Let $2\mid n^2$. Since $2$ is prime 2 has to divide a factor of $n^2=n\cdot n$. So $2\mid n$.


From this little lemma we can conclude that $2\mid p$. So we can write $p=2l$ for some $l\in\mathbb{N}$.

Then $2q^2=p^2\Leftrightarrow 2q^2=4l^2\Leftrightarrow q^2=2l^2$. But now we can conclude from this equation that $2\mid q^2$ so $2\mid q$. Which is a contradiction, because $\operatorname{gcd}(p,q)=1$, but we just showed that $p$ and $q$ both contain the factor $2$!


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