Proof attempt at $\sqrt{4} \in \mathbb{Q}$ I am attempting to self study Real Analysis, Abstract Algebra, and Topology out of private interest.
I'm working on the exercises in the text, ' Understanding Analysis ' and will be posting my attempts at proving the various theorems in the text.
Hopefully this is ok:
Question 1: Prove that the $\sqrt{4} \in \mathbb{Q}$ 
Proof:
Assume that$ \sqrt{4} = \frac{a}{b}:a,b \in \mathbb{Z}$.
Now,
Clearly by algebraic manipulation, 
$\sqrt{4} = \frac{a}{b} \Rightarrow  4=\frac{a^{2}}{b^{2}}$: equation.1.2
Now, Since $\mathbb{Q}$ is the set of all numbers that can be reduced to the ratio of two integers,
$a,b \in K$; $K$ of course being the set of all odd integers.
Now,
Since $a,b$ are odd integers than they can be defined as,
$ a=2n+1$
$b=2m+1$
Where, $m,n \in \mathbb{Z}$
Substituting $2n+1,2m+1$ into equation 1.2.


*

*$4=\frac{(2n+1)^{2}}{(2m+1)^{2}} \Rightarrow 4(2m+1)^{2}=(2n+1)^2$
$\Rightarrow 16m^2+16m+4=4n^2+4n+1n\Rightarrow n=\frac{1}{2}(4m+1):m,n \in \mathbb{Z}$
Since, $\exists a,b \in \mathbb{Z}$  for $\frac{a}{b}=\sqrt{4}$.
$\sqrt{4} \in \mathbb{Q}$.
Q.E.D
 A: Your proof is incorrect. In particular, you assume $a$ and $b$ are odd, yet this is false - $a$ must be even if $\left(\frac{a}b\right)^2=4.$ However, the proof salvages itself when you write the definition of an odd number incorrectly - the property that $a=n+1$ is true of all integers, not just odd ones. For instance $4=3+1$, but $4$ is even. But, at the end, you reach the result that
$$\frac{-k-1}{2k+2}=\sqrt{4}.$$
for any $k$, which is pretty close to true (you get $m$ and $n$ mixed up somewhere - so the fraction is upside-down - and for some reason, you get the negative root, but not the positive). Also, as a structural issue, your proof actually shows that if $\sqrt{4}=\frac{a}b$, then $a=2k+2$ and $b=k+1$ for integer $k$, which is true - but you actually need the converse of that (i.e. that if $a=2k+2$ and $b=k+1$, then $\sqrt{4}=\frac{a}b$), since you've otherwise started with the assumption that $\sqrt{4}$ is rational.
Writing $2^2=4$ would be a far more convincing argument that $\sqrt{4}$ is rational.
A: The fundamental problem here appears to be that you are trying 
to use the mechanisms you saw in another proof 
(probably the proof that $\sqrt2 \not\in \mathbb Q$).
But since you are trying to prove that $\sqrt4$ is rational,
whereas the other proof was trying to prove that a number is not rational,
the mechanisms you need are mostly opposite the ones you found in the other proof.
For example, to prove that $\sqrt2 \not\in \mathbb Q$,
a good first step is to write $\sqrt2 = \frac ab$ where $a,b \in \mathbb Z$.
The idea is that you will derive a contradiction from this assumption,
thereby proving that the assumption was false.
But when you assume $\sqrt4 = \frac ab$ where $a,b \in \mathbb Z$,
that's a true statement; you will not be able to disprove it by
deriving a contradiction or by any other means.
To prove $\sqrt4 \in \mathbb Q$ by contradiction, you could assume first of all
that $\sqrt4 \neq \frac ab$ for every $a,b \in \mathbb Z$.
Conversely, you could assume that there do not exist any $a,b \in \mathbb Z$
such that $\sqrt4 = \frac ab$.
But in either case, you will be unable to proceed with the next step of the
$\sqrt2$ proof, which requires you to have an equation involving $a,b \in \mathbb Z$.
Or if you do proceed with the next step, you will be doing something incorrectly.
In a broader sense, proofs of existence and proofs of non-existence tend to
rely on very different mechanisms. The difference between $\in$ and $\not\in$ is vast.
