why is the co-prime part not mentioned in the definition of the rational number?

Question

Proving $\sqrt{2}$ an irrational number is a quite popular exercise, in precalculus courses, but if we look clearly the definition that is introduced, in the beginning of the course, it never mentions, that the $p$ and $q$ in rational number $\frac{p}{q}$ have to be co-primes, My question is why is this part considered understood.

By there definition alone $\frac{2}{4}$ is a rational, but if we use the same idea that we use in proofing $\sqrt{2}$ irrational, any fraction where numerator and denominator are not co-primes, will not be rational number. I find it quite absurd.

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Edit :

second question, should have been why while proofing the ${\sqrt{2}}$ irrational, we are just satisfied with the fact, that $\exists d \in \mathbb{Q}(d = \gcd(p,q))$, so its irrational - by looking this way $\frac{2}{4}$ is also irrational, because $d = 2$.

Speaking of dentition, I'm not saying the co-prime part is necessary, neither I'm questioning anything, but in proofing $\sqrt{2}$, irrational all we are doing is - proofing the denominator and numerator are not co-primes, hence they are irrational; which seems wrong to me because clearly, $\frac{2}{4}$ is a rational, but if we use same logic which we are using in $\sqrt{2}$ then it is not.

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After having some conversation, and some reading or answers , I realized, that I was wrong, what we are proofing when we are proofing $\sqrt{2}$ irrational is no matter how we write the number there will always be something in common, and such number cant really exist, hence it cant be reprensed as ratio of two integers.

Answer 1

The definition of a rational number is a ratio between two integers. There is no requirement that the integers be coprime (else $\frac 24$ wouldn't be rational when it is). Notice that from the definition alone there is no statement that all rational numbers can be written as a ratio of coprime numbers nor that if they can the choice of numbers are unique. If the numbers have factors in common they can be reduced out as a ratio but what if they have other factors in common. We can't argue to factor out the greatest common factor because we have no reason to assume to integers have a greatest common factor.

However many (maybe most?) will take that for granted or will prove it before they get around to proving there is no rational square root of two. Most texts I've seen do choose to assume $p$ and $q$ are relatively prime so I have no idea why you say "never".

However you also said "have to be relative prime". They don't have to be. But whatever you choose then $(\frac pq)^2 =2\implies p^2 = 2q^2\implies 2\mid p$ so $p = 2p'$ for some integer $p'$. But then $4p' =2q^2\implies 2p'=q^2\implies 2|q$. Okay... stop for a second.

What we HAVE proven is if $(\frac pq)^2$ then which ever we choose whether $p$ and $q$ are or are not relatively prime we must always have that $p$ and $q$ will always be both even..... Actually, we have just proven if $(\frac pq)^2 =2$ then $p$ and $q$ CAN'T be relatively prime! (because they must both be divisible by $2$. Always.)

Now... we go back to the proof. Why is that impossible because we can repeat the argument forever and get an infinite chain of numbers $p = 2p'; p'=2p_2; p_2 = 2p_3$ etc. And that means we have an infinite chain of numbers where $p > p' > p_2 > p_3 > ..... $. And that IS impossible as $p$ has to be finite but it is strictly larger than an infinite number of distinct smaller positive numbers.

....

Now I don't see why you say that argues that $\frac 24$ isn't rational. If we try that reasoning we get: If $\frac pq = 24$ then $2p =4q$ so $p =2q$ so $q$ must be divisible by $2$ and ... that's it. $p$ may or may not be even. But any pair of numbers where $2p =q$ will do. $p=1$ and $q= 2$ will do. So will $p=2$ and $q=4$. And so will $p = 3072$ and $q=6144$.

If we assume $p$ is even we can go a step further. But then we might not need to go any further. With $(\frac pq)^2 =2$ we proved we could never stop, but with $\frac pq =24$ (or $\frac qp = \frac {1024}{2048}$ we never have to go on forever.) So .. it's just fine. $\frac 24$ and $\frac {1024}{2048}$ are most certainly rational because they are ... ratios between integers.

Answer 2

You are right that $p,q$ doesn't have to be co-primes. Let $d=\gcd(p,q)$. Then there are coprime integers $p_1,q_1$ such that $p=dp_1,q=dq_1$, so that $p/q=p_1/q_1$. Thus we can always assume that $p,q$ are coprime.

Answer 3

For a more sophisticated perspective, one can consider the question of how the rationals can be constructed from the integers.

The usual approach is to identify the rationals with equivalence classes of ordered pairs of integers, in which the second integer is required to be nonzero (corresponding to the idea that "you can't divide by zero"), where the equivalence relation is given by $$(a,b) \sim (c,d) \iff ad = bc$$

Then the rational that we usually write as 1/2 is understood to be the equivalence class containing such pairs as (1,2), (-1,-2), (2,4),(-2,-4), (3,6), and so on.

As part of this rather laborious construction of the rationals, along with operations on them such as addition and multiplication, we reassure ourselves along the way that we can safely treat rationals, understood strictly as equivalence classes, as single representatives of those classes.

So roughly speaking, we guarantee that we will get the same results for addition, multiplication and so on whether we write 1/2 or -1/-2 or 2/4 for the single rational number that is identified with the class $\{(1,2), (-1,-2), (2,4),(-2,-4), (3,6), \dots\}$.

However, the main point of this kind of construction is to ensure that all our naive ideas about the rational numbers really do make sense. But as soon as one has done it, one basically forgets about it. Almost no practicing mathematician actually thinks of rationals as ordered pairs of integers, and we go back to thinking that 1/2 "just is" a rational number. Certainly these details are not appropriate for a pre-calculus course.

If you are still a bit uneasy about this, you can always write proofs in the following kind of form, as explained in more detail in the other answer:

Suppose that $r$ is a rational number such that $r^2 = 2$. Without loss of generality, we can write $r$ in the form $r= p/q$ where $p$ and $q$ are positive integers such that $p$ and $q$ are coprime.

Alternatively, if this really bothers you, you can construct a different (and very interesting!) proof of the irrationality of $\sqrt 2$ that simply bypasses this issue, as noted in the comments.

Answer 4

I think, after checking your bio, that I see where your question comes from. I presume you've not studied set theory. About 2-3 hours of study will make what I say next clear in a formal sense if you pick up Halmos's Naive Set Theory.

What @StevenStadnicki is referring to as an equivalence class is exactly what you were taught is a rational: a number $\frac{p}{q}$ where $p,q$ are integers. We all know that $\frac{1}{2}=\frac{2}{4}=\frac{3}{6}$ etc. In set theoretic terms, we define an equivalence relation where any pairs of numbers $(a,b)$ and $(p,q)$ that satisfy $aq=pb$ are considered equivalent. So, $(1,2)$ and $(2,4)$ are such pairs, as are $(1,2)$ and $(3,6)$. You are taught to recognize all rationals (i.e., pairs) that reduce to the same coprime form as "the same", which is also what the equivalence relation just mentioned formalizes within the "canonical" definition of the rationals.

We choose as a convention to represent the rational with the least possible sum, the numbers we are most likely to recognize (given they are smallest).