Dovall rational numbers under multiplication form a group.

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Homework Help Overview

The discussion revolves around whether the set of all rational numbers forms a group under multiplication, exploring the properties of identity, inverses, and associativity.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the existence of an identity element and inverses for all rational numbers, questioning the implications of including zero in the set.

Discussion Status

Some participants express confusion regarding the criteria for a group, particularly in relation to the inverse of zero. Guidance has been offered in the form of hints that prompt further consideration of the properties of rational numbers under multiplication.

Contextual Notes

There is an emphasis on the definitions of group properties, particularly regarding the existence of inverses and the role of zero in the set of rational numbers.

buzzmath
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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.
 
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buzzmath said:
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?
 
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.
 
buzzmath said:
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? :smile:
 
buzzmath said:
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?
 
Wow, I can't believe I didn't think about that. Thanks
 

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