# Q is closed under division or not?

1. Sep 10, 2011

### flyingpig

1. The problem statement, all variables and given/known data

Q = rational numbers

My professor proved that it is closed under addition yesterday. I kinda understood a bit...

How the heck does it work for when n_1 or n_2 = 0?

2. Sep 10, 2011

### micromass

Staff Emeritus
Take a look at the first line. There we take $r_1,r_2\in \mathbb{Q}$. Then we say that

$$r_i=\frac{m_i}{n_i}$$

but obviously this doesn't work for $n_i=0$ since division by zero is not defined in $\mathbb{Q}$.

That is $\frac{m}{n}\in \mathbb{Q}$ if and only if $n\neq 0$...

3. Sep 10, 2011

### flyingpig

I thought it is the same logic for integers, I take 1 and 0 and take 0 to be the bottom, can't do that.

If I take 0 and 1 and put 0 on top, that works, but it is an exception when it is the bottom?

4. Sep 10, 2011

### micromass

Staff Emeritus
Yes, you can never have 0 in the denominator. By definition of a rational number.

5. Sep 10, 2011

### flyingpig

But the whole close division thing i can take two integers (including 0)?

6. Sep 10, 2011

### micromass

Staff Emeritus
No, an element of $\mathbb{Q}$ is defined as a fraction $\frac{m}{n}$, where n is nonzero. So you can't take 0 is the denominator, by definition.

You can't divide by 0.

7. Sep 10, 2011

### SammyS

Staff Emeritus
No. The integers are not closed under division for reasons other than the fact that 1/0 is undefined.

Is 4/3 an integer?

Perhaps you should review the definition of a 'ring'.

8. Sep 10, 2011

### flyingpig

Thank you, you just added more work for me...

9. Sep 10, 2011

### SammyS

Staff Emeritus
Anytime! You're welcome.

To help out a bit more:

Whether we're discussing the Integers, the Rationals, the Reals, or Complex Numbers (all with the usual arithmetic operations):
It's always true that 0 times any element is 0, 0 being the identity element for the addition operation. Because of this, there is no multiplicative inverse for 0, and thus division (the operation that is the inverse of multiplication) by 0 is undefined. Therefore, when discussing whether division is closed, we exclude the case of division by zero.​