In the proof of irrationality of $\sqrt{2}$ or $\sqrt{7}$, why do the numerator and denominator of a rational number have to be in their lowest term? I have been confused by this problem for a very long period of time, and I think I am personally opposed to this concept and refused to agree with it  in my introduction to mathematical proof course. 
It's pretty much the same thing, the problem arises in the proof that shows square root of 2 or square root of 7 is an irrational number. 
I agree with the beginning that assumes these two numbers are rational numbers for the sake of applying contradiction, but as the solution in example proceeds, I notice that they all assume that the numerator and denominator must be in their lowest term, which I think they are relatively prime? 
I mean seriously, why do they have to be in the lowest  term? I mean there are lots of rational numbers that don't satisfy this condition, like 4/2, 10/2 right? 
Can someone explain to me why this in lowest term must be satisfied in order to show a contradiction?
Appreciated.

The point is that all rational numbers can be expressed in lowest terms.  They don't have to be.  Yes, $\frac 24$ is a perfectly good rational number, but the same number can be expressed as $\frac 12$.  When we prove $\sqrt 2$ is irrational, we assume it is rational, then say to express that rational in lowest terms as $\frac ab$.  The reason to do that is that we then show both $a$ and $b$ are even, so $\frac ab$ is not in lowest terms.  If we had not assumed it was in lowest terms we would not have the contradiction we are seeking.
