Irrational numbers proof by contradiction

Question

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

$$2=\frac {a^2}{b^2}$$

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

since $2a^2$ is even $b^2$ is even, so is $b$

let $b=2c$ we have $4c^2=2a^2$ and thus $a^2=2c^2$ since $2c^2$ is even and since $a^2$ is, even so, is $a$

so $ \sqrt{2}$ is irrational right.

Now HERE THE QUESTION?

same steps for $\sqrt{4}$

$$\sqrt {4}=a/b$$

$$4=\frac {a^2}{b^2}$$

$$4b=2a^2$$

since $4a^2$ is even $b^2$ is even, so is $b$

let $b=4c$ we have $16c^2=2a^2$ and thus $a^2=8c^2$ since $8c^2$ is even and since $a^2$ is, even so, is $a$

but we know $\sqrt {4}$ is a rational number. So?

how does Proof by contradiction suppose to prove the truth!

Also when an even number is multiplied by a number it will be even.

EDIT: the question is if I use the same steps for $\sqrt2$ for $\sqrt 4$ it doesn't work, am I doing it wrong? or...

EDIT 2: I read this thanks to one of the comments Prove the sqrt of 4 is irrational, where did I go wrong? can someone dumb down the answer so I can understand, what's going on

Answer 1

Your confusion, stems from not keeping variables straight. In the first proof, you actually hit:$$2b^2=a^2$$ which implies a is even, first. This, by substitution, and simplification, gives back:$$b^2=2c^2$$ implying b is also even. These combine, to contradict ${a\over b}$ ever being in lowest form.

For the second proof, you were supposed to get: $$4b^2=a^2$$ implying a is even. This then follows to: $$4b^2=4c^2$$ This goes to:$$b^2=c^2$$ which taking both sides principle(positive) root, gives:$$b=c$$ which then proves:$$a=2b$$ and leads to:$${2b\over b}=2$$ being our root.

Answer 2

The poduct of two not-divisible-by-two (aka. odd) numbers is odd.

Proof: If $n,m$ are odd, we can write $n=2k+1$, $m=2l+1$. Then $nm=4kl+2k+2l+1=2\cdot(2kl+k+l)+1$ is also odd. $\square$

However, the product of two not-divisible-by-four numbers may happen to be divisible by four.

Answer 3

The correct inference is that $b$ is of the form $2c$, since that will still square to a multiple of $4$. More generally, if $p^k|n^2$ with $p$ prime then $p^{\lceil k/2\rceil}|n$. You now have $4a^2=b^2=(2c)^2=4c^2$, giving $a=c$. So, there is no infinite regress.

Answer 4

Your step $$4b=2a^2$$ is double nonsense. It should be $$4b^2=a^2$$ Therefore $a$ is even; but we can’t deduce that $b$ is also even.