Dovall rational numbers under multiplication form a group.

[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.

 

[radou]

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.

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

 

[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.

 

[radou]

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.

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

 

[JasonRox]

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.

What's the inverse of 0?

 

[buzzmath]

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