It's usually just a time saver and an assurence that an infinite regress is a contradiction.
If $\frac ab$ is such that $(\frac ab)^2 =2$ then $a^2 = 2b^2$ so $2|a^2$ and so $2|a$. Let $a = 2a_1$ so $2a_1^2 = b^2$ so $2|b$. Let $b = 2b_1$ and $a_1^2 =2b_1^2$ so $a_1$ is divisible by $2$. Let $a_1 = 2a_2$. and so on ad infinitim.
Well... so what is wrong with ad infinitim? And how do we know that just because we got the same result $3$ times we will get it forever? And why is that a contradiction?
For that matter why is that all rational numbers actually have "lowest terms"?
Those all actually have answers and the pertain to the well-ordering principle and they are not trivial (although they aren't hard either).
We have a series of $a > a_1 > a_2 > ...$ so that $a_i = 2a_{i+1}$ as we as $b > b_1 > b_2 > ....$ so that $b_i = 2b_{i+1}$. But the induction principle in natural numbers we know these $a_i, b_i$ exist for all $i \in \mathbb N$ so that $a_i^2 = 2b_i^2$ and $b_i^2 = 2a_{i+1}^2$.
But by the well-ordering principle, we know that for such collections of $a_i$ and of $b_i$ that they must each have a least element.
And that is a contradiction.
But that involved a bit of a diversion that we feel simply distracts the student. Surely it's easier to just avoid the issue by saying "Oh, we can assume $\frac ab$ is in lowest terms".
But actually knowing that if $a,b \in \mathbb Z, b\ne 0$ that there must be some $a',b'$ so so that $\gcd(a',b') = 1$ and $\frac ab = \frac {a'}{b'}$ requires a similar well-ordering principle argument.