I just watched a proof which showed that the square root of is irrational. It seemed to me that the reason it was asserted irrational was because it was reducible. Therefore, I thought, that this must mean that all rational numbers can be represented by an irreducible fraction. Is this just a part of the definition of rationality or is there a proof that such is true?
Thanks in advance.
On
The definition of $x$ being rational is that it can be written as a fraction, i.e. as $x=\frac{m}{n}$ where $m,n$ are integers and $n\neq 0$. Of course, there are many different ways to do this for any given rational $x$.
Suppose $x$ is rational and we choose a way to write $x=\frac mn$ for which $n>0$ but is as small as possible. (Note that we can always make $n>0$ by multiplying top and bottom by $-1$, if necessary, so such a choice always exists.) Now we claim $\frac mn$ must be irreducible. Suppose not, so there is some common factor $h>1$ which divides both $m$ and $n$. Then $\frac{m/h}{n/h}=x$, but the denominator is smaller, contradicting our choice of representation.
It's not part of the definition, but the proof is easy.
Suppose one has a rational number $\frac{m}{n}$ that is reducible. We can assume $n \ge 1$ (divide the numerator and denominator by $-1$ otherwise).
Take an integer $k \ge 2$ which divides both of $m$ and $n$, and so $m' = m/k$ and $n'=n/k$ are smaller integers satisfying $\frac{m'}{n'} = \frac{m}{n}$.
Repeat this process as long as the fraction remains reducible.
One cannot repeat forever, because the sequence of denominators is a sequence of positive integers that gets smaller and smaller, and no such sequence can continue infinitely (essentially one is using mathematical induction in this step).
When the process stops, one has an irreducible fraction representing the same rational number