[Math] Integers under multiplication a closed operation

abstract-algebra

I'm watching YouTube videos to teach myself some abstract algebra, a woman claims that integers under multiplication has a closed operation. But I don't understand why.

For example: $$7x = 3$$

The 3rd element is not an integer.

Please advise

Best Answer

Closure (under multiplication) means: "An integer times an integer is also an integer". It does not mean:

"An integer times something else which results in an integer, means that something else is also an integer."

This is pretty obvious to see:

$2\cdot \frac{1}{2} = 1$.

Here, $2$ is an integer, and the "something else" is $\frac{1}{2}$. Our product is an integer, but it is not the case that we can conclude $\frac{1}{2}$ is an integer; in fact, it is not.

It can be proven that $\Bbb N$ is closed under multiplication (hint: use induction), and using the rules, for $a,b \in \Bbb N$:

$a(-b) = (-a)b = -ab\\ (-a)(-b) = ab$

(If you do not consider $0$ to be a natural number, you have a few more cases to consider, but these are easy)

it is easy to see that if $\Bbb N$ is closed under multiplication, so is $\Bbb Z$.

Perhaps this will be easier to process when you see the differences between rings, and fields. What often throws people off-guard in thinking about this, is that ordinary high-school arithmetic typically takes place in the field of rational numbers, where the non-zero numbers are closed under multiplication, and the equation:

$7x = 3$ has a rational solution, even though it is an integral equation.