# Dovall rational numbers under multiplication form a group.

1. Oct 24, 2006

### buzzmath

I know the set of positives rationals form a group under multiplication and that the negative irrationals do not form a group under multiplication because there is no identity or inverse. My question is does the set of all rational numbers under multiplication form a group.

2. Oct 24, 2006

### radou

Hint: does there for every $$a \in \textbf{Q}$$ exist a unique element $$a^{-1}\in\textbf{Q}$$ such that $$a\cdot a^{-1}=1$$?

3. Oct 24, 2006

### buzzmath

Yes, it would just be 1/a. That's what I put on my test and got it wrong. He wrote that the set of rational numbers under multiplication is not a group. 1 i rational, the product of any two rationals is rational, 1/a is the inverse, and it's associative. I don't understand.

4. Oct 24, 2006

### radou

Hint 2: does there exist 1/a for a = 0?

5. Oct 24, 2006

### JasonRox

What's the inverse of 0?

6. Oct 24, 2006

### buzzmath

Wow, I can't believe I didn't think about that. Thanks

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