Why operation * not defined in Q

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Homework Help Overview

The discussion revolves around the operation defined as l/m * k/n = (l+k)/(m²+n²) and its validity within the set of rational numbers Q, where l and k are integers, and m and n are non-zero integers. Participants are exploring why this operation cannot be consistently defined in Q.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the implications of the operation producing different results based on the values of l and m, particularly when negated. There is a focus on whether the operation can yield consistent results across different inputs.

Discussion Status

Some participants have noted that the operation appears to produce different outcomes, suggesting that it may not be well-defined. There is an acknowledgment of the need to clarify the implications of negating variables within the operation.

Contextual Notes

Participants are considering the constraints of the operation and its definitions within the context of rational numbers, particularly the implications of using integers and non-zero denominators.

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Homework Statement



Show that l/m * k/n = (l+k)/(m2+n2) can not be defined as an operation in Q when l,k € Z and m, n € Z\{0}

I do not know what is the issue here? Should I know something about Q that this not fulfilled by the operation *?
 
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pakkanen said:

Homework Statement



Show that l/m * k/n = (l+k)/(m2+n2) can not be defined as an operation in Q when l,k € Z and m, n € Z\{0}

I do not know what is the issue here? Should I know something about Q that this not fulfilled by the operation *?

Hint: What happens if you negate l and m?
 
Ok.. So the same operation can produce two different results??

So that l/m * k/n = (l+k)/(m2+n2) ≠ (-l+k)/((-m)2+n2) = -l/-m * k/n = l/m * k/n
 
pakkanen said:
Ok.. So the same operation can produce two different results??

So that l/m * k/n = (l+k)/(m2+n2) ≠ (-l+k)/((-m)2+n2) = -l/-m * k/n = l/m * k/n
That's right, so the operation is not well defined.
 
Thank you very much jbunniii! Helped me a lot. I think we'll meet again.
 

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