$$\sqrt{2}=a/b$$
$$2=\frac {a^2}{b^2}$$
$$b^2=2a^2$$
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
Best Answer
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.