Proof that $\sqrt{2}$ is irrational using that $a=2^km$

Question

There is the proposition

Every integer $n$ can be represented uniquely as the product $2^km$ where $m$ is odd and $k\ge0$.

I have to use this to prove that $\sqrt{2}$ is irrational. As is it usual, assume the contrapositive

$$ \sqrt{2}=a/b, $$

where $a,b$ are integers and $b\neq 0$. Therefore

$$ a^2=2b^2. $$

Using the proposition, $b^2=2^jn$, with $n$ odd and $j\ge0$. Therefore $a^2$ is even (what the hell do I need this, since $a^2$ is clearly even, but I have to use the proposition...). Then I can continue in the usual way and finish the proof.

At this point, this proof seems very lame. Any ideas how to improve it?

Answer 1

A easy way to show $\sqrt{2}$ is not a rational number:

Suppose $\sqrt{2}$ is rational. note that $\sqrt{2}>1$, so w.l.o.g. I can take $a,b\in \mathbb{N}$ such that $\sqrt{2}=\frac{a}{b}$

Then $a^2=2b^2$. Note that $a^2\equiv 0,1(\mod 3)$

Case 1:

If $a^2\equiv 1(\mod 3)$, then $b^2\equiv 1(\mod 3)\Rightarrow 2b^2\equiv 2(\mod 3)$, but $a^2=2b^2$, then $a^2\equiv 2(\mod 3)\Rightarrow 1\equiv 2(\mod 3)$, which is impossible.

Case 2:

If $a^2\equiv 0(\mod 3)$, then $b^2\equiv 0(\mod 3)$, then $a=3c$, $b=3d$ where $c,d \in \mathbb{N}$.

Now we again get $c^2=2d^2$, since $9c^2=2\times9d^2$. This lead to a loop. The loop is we can not go to case 1(since that case is impossible). Then we have stay in case 2, which will lead to $3\mid c$ and $3\mid d$ and this process will continue until we end up to $1=2$ which is again impossible.

So there does not exist $a,b\in \mathbb{N}$ such that $a^2=2b^2$. Hence $\sqrt{2}$ is not a rational number$.\space\space\space\blacksquare$

Answer 2

Try applying the Proposition before squaring: consider $\sqrt{2} = a/b$ an irreducible fraction, write $a = 2^i n$ and $b = 2^j m$. We have multiple cases:

• If $n=1$ or $m=1$, then respectively we must have $j=0$ or $i=0$ because of $a/b$ being irreducible, then (squaring the original equality) respectively $m^2=4^i/2$ or $n^2 = 2\cdot 4^j m^2$. Thus either $m^2$ is even or $n^2$ is even, but $m$ and $n$ are odd and the square of an odd number is an odd number, a contradiction.

Assume $n\neq 1 \neq m$. Since $a/b$ is reducible, we must have either $i=0$ or $j=0$.

• If $i=0$ (notice that this contains the case $j = 0 = i$), squaring we obtain that:

$$2 = \frac{n^2}{4^j m^2} \text{ so } 2\cdot 4^j m^2 = n^2 \text{ is even, which is again a contradiction because $n$ is odd.}$$

• If $j=0$ and $i\neq 0$, squaring we obtain that:

$$2 = \frac{4^i n^2}{m^2} \text{ so } m^2 = \frac{4^i n^2}{2} \text{ is even ($i>0$), which is again a contradiction because $n$ is odd.}$$

Notice how the fact that $n$ and $m$ can be $1$ forces you to consider extra cases (alternatively, you could say that the integers do not have multiplicative inverses and include the first case in the other two).