Why do we need to prove a fraction can always be written in lowest terms?

Question

I'm currently reading the notes of a preliminary Math course.

Section 3.1.1 contains some proofs using the Well Ordering Principle. One of them is about the always apparent possibility to write a fraction in shortest terms.

But why does this require a proof? If the fraction has no common factors it is already the result, and otherwise, I don't see any contradiction as to why this needs to be explicitly proved. Even if numerator and denominator have no GCD other than $1$, the fraction has a presentation in shortest terms, i.e., the fraction itsself!
So why prove this? What does it prove at all?

Furthermore, can any proof be superfluous at all?
I'm not that into mathematics and all (I like it nevertheless), so bear with me in case of any gross misunderstanding, please.

Answer 1

In the proof of irrationality of $\sqrt{2}$, which the section you're referencing plainly uses as an example, it is necessary:

Suppose $a$ and $b$ are integers such that $\frac{a}{b}=\sqrt{2}$.

Then $a^2=2b^2$.

Since $2$ is prime, $2$ divides $a$, say $a=2c$.

Going back to $a^2=2b^2$, we now have $4c^2=2b^2$, and then $2c^2=b^2$. Since $2$ divides $b^2$, $2$ divides $b$.

At this point in what has been written, there is no problem.

However, if you had additionally assumed that $a$ and $b$ have no common prime divisors, you would have reached a contradiction.

Otherwise, this argument does not get anywhere:

say $b=2d$. Then $2c^2=4d^2\implies c^2=2d^2\implies 2|c$

say $c=2e$. Then $4e=2d^2\implies 2e^2=d^2\implies 2|d$

say $d=2f$. Then $2e^2=4f^2\implies e^2=2f^2\implies 2|e$

...

Answer 2

It does require rigorous proof that fractions can be written in lowest form, (i.e. with coprime numerator and denominator), because this property is not always true for other types of numbers. Though a proof is obvious in the classical integer case, it can fail for fractions formed from other numbers. The classical proof depends crucially on the fact that $\,\Bbb N\,$ is well-ordered, so continually cancelling common factors must eventually terminate with a fraction in lowest terms (else the cancellations would yield an infinite decreasing sequence of denominators, contra $\,\Bbb N\,$ is well-ordered.

For other types of numbers there may exist $\,a,b,c\,$ where $\, c^k\,$ divides $\,a\,$ and $\,b\,$ for all $\,k\ge 0.\,$ Here the above proof breaks down for $\,a/b,\,$ since no matter how many times we cancel $\,c\,$ from the fraction, there will always remain a common factor of $\,c.$

In domains like elementary number theory where we have very strong intuition based on experience, it is especially important to be extra careful not to confuse empirical inference with logical inference. This occurred many times in the past.

For example, for many centuries no one noticed that uniqueness of prime factorizations required proof. Apparently either no one conceived of the possibility of nonuniqueness (or those who did thought that the proof was so "obvious" that it did not deserve mention). This was not corrected until $1801$ when Gauss plugged this gaping logical gap in his book Disquisitiones Arithmeticae, where he wrote "It is clear from elementary considerations that any composite number can be resolved into prime factors, but it is often wrongly taken for granted that this cannot be done in several different ways".

But even decades later one still finds mistakes made around unique factorization, even by leading number theorists. For example, circa $1850$ a few eminent mathematicians mistakenly thought they had proved FLT by erroneously assuming statements (e.g. Bezout arguments) that they did not realize were equivalent to unique factorization (which generally fails in the rings of cyclotomic integers studied). Even a century later rigor still was lacking in some expositions, e.g. Harold Davenport wrote that some British schoolbooks deemed uniqueness of prime factorization to be a "law of thought".

Answer 3

The question has to do with the concept of equivalence classes. A fraction is not an "operation", a division in 'progress'. This is a quite common intuitive picture for the beginners (and there is nothing bad in that). Instead, a fraction is an equivalence class of 'pairs' (the equivalent 'fractions'): so, for example $$ \frac{1}{2}=\left[\frac{1}{2},\frac{2}{4},\frac{3}{6},\ldots\right] $$ or, more correctly, $$ \frac{1}{2}=\left[(1,2),(2,4),(3,6),\ldots\right]. $$ The symbol '$1/2$' has to be considered a representative of its class. But if we want to use only the representatives and not the underlying classes, we need to specify what representative we choose. Taking the 'minimal' fraction is an effective way to designate a 'canonical' representative.