Q is closed under division or not?

[flyingpig]

Homework Statement

Q = rational numbers

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

Proof

Let r_1, r_2, \in \mathbb{Q}

r_1 + r_2 = \frac{m_1}{n_1} + \frac{m_2}{n_2} = \frac{m_1 n_2 + n_1 m_2}{n_1 n_2} \in \mathbb{Q}

By letting m_1 n_2 + n_1 m_2 = m_3 and n_3 = n_1 n_2

Since m_1, m_2, n_1, n_2 are integers, m_1 n_2 + n_1 m_2, n_1 n_2 are also integers.

So that

\frac{m_1 n_2 + n_1 m_2}{n_1 n_2} \in \mathbb{Q}

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

 

[micromass]

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...

 

[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?

 

[micromass]

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

 

[flyingpig]

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

 

[micromass]

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

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.

 

[SammyS]

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

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'.

 

[flyingpig]

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'.

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

 

[SammyS]

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

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.​