# How to prove that the result of multiplying two number of the same set is in the same set?

I'm a chemist and I'm studying abstract algebra on my own so I don't know if this is a trivial question.

If I have for example two number $a$ and $b$ (where $a,b \in \mathbb{Q}$). How can I prove that $a \times b \in \mathbb{Q}$ ? Of course I'm interested even if the numbers $\in \mathbb{Z}$ or any set. What I would like to know is the demonstration of this assumption, and if is valid for every set ($\mathbb{N,Q,I,R} \ldots)$ and operation.

From what I'm reading seems that this is deduced because multiplication is an operation and so the set is closed under the operation $\times$. But why?

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In general, when we define in mathematics a strucuture $S$ this is (at least) a "couple" $\langle D, * \rangle$ made by a domain "$D$" of "objects" and an operation ("$*$", e.g. a binary one) defined on them, and we write :

$S = \langle D , * \rangle$.

In this case, "by definition" the domain is "closed under" the operation, i.e.

for all $a, b \in D$, we have that $a*b \in D$.

The simplest example is $\langle \mathbb N, + \rangle$.

But after having defined the structure, we can introduce new ("derived") operations, like "$-$" in $\mathbb N$. In this case, it is not true in general that the structure is still closed under the new operation.

If we put :

$a - b$ iff $\exists x (a = b+x)$

we have that, for example, $2 - 3$ is not defined in $\mathbb N$.

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If you use \langle instead of < in the TeX code, it looks better, for example $\langle D,* \rangle$ instead of $< D,* >$. –  Jeppe Stig Nielsen Mar 12 '14 at 11:08
@JeppeStigNielsen - tahnk you ! I'll edit it. –  Mauro ALLEGRANZA Mar 12 '14 at 11:11

This does hold for your examples $\mathbb Q$ and $\mathbb Z$.

If $a,b \in \mathbb Q$, then $a=\frac{p}{q}$ and b = $\frac{r}{s}$ where $p,q,r,s \in \mathbb Z$ and $q,s \ne 0$ Now $a\times b = \frac{pr}{qs}$. And if $pr,qs \in \mathbb Z$, then $ab \in \mathbb Q$. So the first problem reduces to the second, as an exercise, try to prove the second part for yourself.

It doesn't hold for all sets however. Consider set $A=\{3,4\}$. Now $3,4 \in A$, but $3\times4 \not \in A$

Also, this need not hold for all operations either. Consider subtraction. If $a,b \in \mathbb N$, $a-b \in \mathbb N$ need not be true.

Consider division, if $a,b \in \mathbb Z$, $\frac{a}{b} \in \mathbb Z$ need not be true either.

If you consider $\sqrt{x}$ to be an operation, this this isn't valid on $\mathbb Q$ either.

EDIT: Apparently I was mistaken in the definition of an 'operation'. According to one user, the original statement "is valid for every set and operation" is true by definition. See comment thread below.

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$q,s$ and $qs$ should be restricted to be from $\mathbb{Z}\setminus\{0\}$. –  ccorn Mar 12 '14 at 10:04
@GitGud that would make the proposition is valid for every set and operation true by definition wouldn't it? –  Sabyasachi Mar 12 '14 at 10:21
@GitGud with \ you mean $\mathbb{R}$ without 0 because in $\mathbb{R}$ division is not an operation because is not define for 0. E.g. 4/0 is undefined right? –  G M Mar 12 '14 at 10:27
@GitGud That is not how I would define an operation. Also, and I realize that this in no way the final word, it's not how wikipedia defines an operation. en.wikipedia.org/wiki/Operation_%28mathematics%29 –  Taemyr Mar 12 '14 at 12:25
@Taemyr That link doesn't even provide a definition, it just useless blah, blah, blah. In any case, here the word operation is used as shorthand for binary operation and you can check the definition on this wikipedia link. Edit: I see now your link does provide a definition and it happens to be in accordance with what I say. It's a generalization of binary operation. –  Git Gud Mar 12 '14 at 12:26

The set of rationals is closed under multiplication. You can prove it if you take representation $$a=\frac pq, b=\frac rs \implies a\times b= \frac {pr}{qs}$$, where $p,q,r,s$ are integers. Here we use the fact that $\mathbb{Z}$ is also closed under multiplication and has non zero divisors. However, for example irationals are not closed under multiplication. Note that $\sqrt{3}\sqrt{3}=3$.

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Smallest terms is not necessary when $\mathbb{Q}$ is defined by means of equivalence classes. $\frac{pr}{qs}$ need not be in smallest terms anyway. –  ccorn Mar 12 '14 at 10:07
For the denominators, we also need nonzero entries and the fact that $\mathbb{Z}$ has no zero divisors. –  ccorn Mar 12 '14 at 10:11
For "smallest term" I rush with using that term and you are wright for zero divisors. I edited it. –  user101521 Mar 12 '14 at 10:18

Some abstract algebra books assume that multiplication is closed for the natural numbers, integers, rational numbers, real numbers and complex numbers.
Multiplication in the complex numbers is defined at the expense of multiplication in the real numbers. In fact the complex numbers themselves are constructed at the expense of the real numbers.
Multiplication in the real numbers is defined at the expense of multiplication in the rational numbers. In fact the real numbers themselves are constructed at the expense of the rational numbers.
Multiplication in the rational numbers is defined at the expense of multiplication in the integers. Once this is done, you can prove that multiplication in the rational numbers is closed. In fact the rational numbers themselves are constructed at the expense of the integer numbers.
Multiplication in the integers in defined at the expense of multiplication in the natural numbers. In fact the integers themselves are constructed at the expense of the natural numbers.
It works just the same for the sum.

So it boils down to knowing how to define multiplication in the natural numbers and then constructing each of the steps above.

Multiplication in the natural numbers is defined, by the way, at the expense of the sum in the natural numbers.

Some abstract algebra books deal with part of the process above. I've never seen it all done in an abstract algebra book. Usually sum in the natural numbers is assumed and the rest is done from there.

To see how the sum is defined in the natural numbers, you should pick up an elementary set theory book.

It goes roughly like this. One starts with the set of natural numbers (which can also be defined) to which something called $0$ belongs and a successor function, i.e., a function that given a natural number, yields what one would intuitively call its successor. Given a natural number $n$, it's common to denote it's successor by $n^+$.

From this one defines $+_{\mathbb N}$, recursively, as follows: let $m,n\in \mathbb N$, then

1. $n+_{\mathbb N}0:=0$
2. $n+_{\mathbb N}m^+:=(n+_{\mathbb N}m)^+$

From here one can prove the expected properties such as $n+_{\mathbb N}1=n^+, m+_{\mathbb N}(n+_{\mathbb N}k)=(m+_{\mathbb N}n)+_{\mathbb N}k$, etc.

Then multiplication in $\mathbb N$ can defined and one goes up from there.

I didn't focus on actually proving that multiplication in $\mathbb Q$ is closed because such a problem only arose to the OP because of his unawareness of the process explained above.

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Thanks I was looking for a general approach like this! But I don't understand the last part what do you mean with $+_{\mathbb{N}}$? Then I'm not familiar with := notation: $n+0 :=0$ means that $n$ is 0? Thanks a lot sorry for my ignorance –  G M Mar 12 '14 at 14:10
The notation $:=$ means that what's on the left of $:=$ is defined as being what's on the right of $:=$. So $n+0$ is, by definition, $0$, I'm defining the symbol $n+0$. The symbol $+_{\mathbb N}$ is meant to denote addition in the natural numbers. The subscript $\mathbb N$ is often left out for simplicity. I should have used the same symbol in the definitions below. I'll fix that. Thanks. –  Git Gud Mar 12 '14 at 14:22
Thank you! I will try to reasoning about it! –  G M Mar 12 '14 at 17:18
@GM I only gave an idea of how things are done in my answer. You're not supposed to be able to do them just with what I have said. Are you aware of this? –  Git Gud Mar 12 '14 at 17:31
Ah okay, thanks however I will try to acquire the knowledge necessary! –  G M Mar 12 '14 at 17:34