Proof of the theorem that there does not exist any rational number $p$ such that $p^2=2$ When we prove the theorem that there does not exist any rational number $p$ such that $p^2=2$, why we choose $p=\frac{m}{n}$ that $m$ and $n$ have no common factor other than one?
 A: There are, I think, several reasons why this is convenient.
First, we want to explore things which must be true about $m$ and $n$. We can always divide through by common factors, and if there are no common factors we have an extra piece of information which may prove useful in the proof.
Second, strictly it would be sufficient to say that we can divide through the fraction by powers of $2$ until at least one of $m$ and $n$ is odd, but reducing so there is no common factor is always possible and is a more familiar procedure in problems involving fractions, and can sometimes be more economically expressed.
Third, as a model for other similar proofs, it eliminates any common prime factor and can therefore be generalised a title more smoothly to square roots of other primes (and later to the roots of polynomials where the coefficients have no common factor)
A: Let's try to do the proof without it
A rational number is a ratio between two integers if $p$ is rational there exist two integers $m, n$ so that $p = \frac mn$.
So suppose $p^2 = (\frac mn)^2 = \frac {m^2}{n^2} = 2$.
So $m^2 = 2n^2$ so $m^2$ is even.  If $m$ is odd then $m^2$ is odd so we must conclude $m$ is even.  So $m= 2m_1$ for some integer $m_1$.
Then $m^2 = 4m_1^2 = 2n^2$ and so $2m_1^2 = n^2$ and so $n^2$ and therefore $n$ is also even.  Let $n = 2n_1$ for som integer $n_1$
Then $2m_1^2 = 4n_1^2$ and $m_1^2 = 2n_1^2$ so $m_1$ is even.
We can see this will go on forever and we will never reach an end.
But the thing is this isn't just true for our first choice of $m$ and $n$.  It is true for every choice of $m$ and $n$.
Surely this is impossible.  After all our choice of $m$ was a finite number and if $m = 2m_1 = 4m_2 = 8m_3 =...... 2^km_k =....$ then $m$ itself must be at least as big as ANY possible $2^k$ and there is not such limit.
So that is a contradiction.  But it is a contradiction that requires an uncomfortable diversion into infinity and a lot of handwaving and "well, you know..." and it doesn't resonate well.
Now, let's try it again but this time we are forearmed with the knowledge:
PROPOSITION:  Any rational number $p$ may be written as ratio of two integers where $p = \frac mn$ AND that $m,n$ have no common factor.
IF we can accept the proposition then if $p = \frac mn$ and $m,n$ have no common factor, and $p^2 = \frac {m^2}{n^2} = 2$ then we get:
$m^2 = 2n^2$ so $m$ is even.  So $m = 2m_1$ for some $m_1$.  So $4m_1^2 = 2n^2$ and $n^2 = 2m_1^2$ and $n$ is even.  So both $m$ and $n$ have a common factor of $2$.
!!!!STOP!!!!!  That's a contradiction right there and then.  We are done.  There can be no such rational number.
...
BUT, we have to accept the proposition as true.  Which means we have to prove it first.
Unfortunately the proof of the proposition involves the same handwaving and infinite regress.  But that's okay.  We are relegating it to one place where it wont be distracting.
Proof of proposition.  If $p = \frac mn$ ( and $m,n$ have a factor $k > 1$ in common, then $m = km_1; n = kn_1$ for integers $m_1, n_1$ and $p = \frac mn = \frac {km_1}{kn_1} = \frac {m_1}{n_1}$. Now if $m_1$ and $n_2$ have a factor in common we will have to repeat this.
But $k > 1$ so  $|n| > |n_1| > 0$. So if we repeat this multiple times we get $|n| > |n_1| > |n_2|> ..... > 0$.  Now we can't repeat this infinitely because $n$ was a regular integer and not infinite.  So there must be some last $n_k$ and $m_k$ and $n_k$ and $m_k$ have no factors in common.
(Note:  I haven't proven every rational number has a unique pair of integers with no common factor.  I've only proven that all pairs of integers can be reduced to a pair with no common factor-- not that that pair is unique.  It is.  But that it s proof for a different time and is not needed.)
A: In fact we want $m$ and $n$ to have no factor $2$ in common. Because the steps are
$$m^2=2n^2$$ so that $$m^2$$ is even, so that
$$m$$ is even, so that $$m^2$$ is a multiple of $4$ so that $$n^2$$ is even, so that $$n$$ is even.
Hence $m$ and $n$ are both even but have no factor $2$ in common!?

The contradiction wouldn't work if we allowed $m$ and $n$ both even.
Other common factors are unimportant.
A: Because a fraction $\frac{m}{n}$  where $m$ and $n$ have a common factor other than 1 can be simplified further by cancelling the common factors. e.g. $\frac{16}{24}$ Here, 16 and 24 have factors in common . Their "largest common factor" is 8. We can  write  $\frac{16}{24}$ as $\frac{8*2}{8*3}$ = $\frac{2}{3}$, which can't be simplified further as 2 and 3 don't have any common factors. Note that $\frac{16}{24}$ and $\frac{2}{3}$ are not "different" they are the same thing. But $\frac{16}{24}$ is not "simplest form", while $\frac{2}{3}$ is the simplest form as it can't be simplified further. . You can write it as $(-2)/(-3)$. But that would be inconvenient. 
So, whenever we use a fraction in unknowns  like $\frac{m}{n}$ , then according to our convenience, we can say that we have picked $m$ and $n$ such that that our fraction is simplified to the last step, that is , $m$ and $n$ have nothing in common or that $m$ and $n$ have no common factors to cancel out. The reason is, We can write any fraction in the "most simplified form" only by cancelling all the common factors in the numerator and denominator. 
Note that 1 is always a factor of both m and n but, you can try cancelling 1 , and after deep thought, you will find that it did not change anything :). 
$\frac{m}{n}$ = $\frac{m*1}{n*1}$ = $\frac{m}{n}$ . In the "most simplified form" , the only common factor of $m$ and $n$ is 1, and cancelling 1 is not going to simplify it.
That is why in some texts, it is written as g.c.d($m$,$n$) = $1$, or  h.c.f.($m$,$n$) = $1$ which essentially means that the greatest common factor of $m$ and $n$ is 1, which means that the fraction $\frac{m}{n}$ is in it's simplest form.
