Bogus prove of the irrationality of $\sqrt{\frac{1}{2}}$.

Question

I need help figuring out my mistake. Proof. Proving by contradiction that $\sqrt{\frac{1}{2}}$ is irrational. Suppose $\sqrt{\frac{1}{2}}$ is rational so: $\sqrt{\frac{1}{2}}=\frac{m}{n}$. Where $m/n$ are in lowest terms.

Squaring both sides and solving for $n$ we have $n^{2}=2m^{2}$.

So $n^2$ is even therefore $n$ is also even therefore $n=2k$ and $n^{2}=4k^{2}$.

Now, replacing $n^2$ in $n^{2}=2m^{2}$, we have $4k^{2}=2m^{2}$. Which means $m$ is also even.

So we are left with an even numerator and an even denominator which contradicts the initial assumption that m/n are in lowest terms. Therefore $\sqrt{\frac{1}{2}}$ is irrational. (which we know is wrong)

Answer 1

$$\sqrt { \frac 12 }=\frac{1}{\sqrt2}=\frac{1}{\sqrt2} \times \frac{\sqrt2}{\sqrt2}=\frac{\color{red}{ \sqrt{2}} ~(\text{Irrational)}}{2~(\text{Rational)}} :=~~~\text{Is Irrational}$$

Answer 2

** hint to be continued**

Let us prove that $a=\sqrt {2} $ is irrational.$\frac {1}{a} $ will be also irrational.

at first we know that $$1 <\sqrt {2}<2$$ which means that $a\notin \mathbb N $.

assume $a=\frac {p}{q} $ with $p>q\geq 2$.

consider the set $$A=\{n>1 : na\in \mathbb N\}. $$ $$q\in A\implies A\neq\emptyset $$

$\implies A $ has a smallest element $b$.

We check easily that $$c=b\sqrt{2}-b\in A$$ and $$c<b$$ which is a contradiction.

Answer 3

Your proof is actually correct. We know$\sqrt{\frac12} = \frac{1}{\sqrt{2}}$. So, if we prove $\frac{1}{\sqrt{2}}$ is irrational, then so is $\sqrt{\frac12}$ since they are equal. Let $\sqrt{2} = \frac{m}{n}$. So that means $\frac{1}{\sqrt{2}} = \frac{n}{m}$. However, we know that $\sqrt{2}$ is irrational thus there is no $m,n$ such that $\sqrt{2} = \frac{m}{n}$. Since there is no such $m,n$, that means there are no solutions to $\frac{1}{\sqrt{2}} = \frac{n}{m}$, either. Thus, $\frac{1}{\sqrt{2}}$ is irrational, so your proof is actually fine, since there was no error in your logic and conclusion was correct.