Multiplying by 1 adds a restriction?

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Discussion Overview

The discussion revolves around the implications of multiplying by 1 in algebra, specifically when using the expression (a+b)/(a+b) to manipulate equations. Participants explore whether this introduces restrictions, particularly the condition a+b≠0, and how this affects the validity of algebraic manipulations.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that the restriction a+b≠0 arises from the identification of 1 as (a+b)/(a+b), rather than from the act of multiplying by 1 itself.
  • Another participant questions the validity of multiplying by the conjugate in algebra, implying it may not maintain equality under certain conditions.
  • A different viewpoint posits that the case where a+b=0 can often be considered an uninteresting special case, suggesting it may not require further consideration in typical algebraic contexts.
  • One participant clarifies that while multiplying by 1 generally preserves value, the specific expression used [(a+b)/(a+b)] is not identically 1, which introduces complications depending on the values of a and b.

Areas of Agreement / Disagreement

Participants express differing views on whether multiplying by 1 introduces restrictions. There is no consensus on the implications of using (a+b)/(a+b) in algebraic manipulations, and the discussion remains unresolved.

Contextual Notes

The discussion highlights the dependence on the definitions of equality and the conditions under which certain algebraic manipulations are valid. The implications of treating a+b=0 as a special case are also noted but not resolved.

tahayassen
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[tex](a-b)=(a-b)*1\\ =(a-b)\frac { a+b }{ a+b } \\ =\frac { { a }^{ 2 }-{ b }^{ 2 } }{ a+b }[/tex]

But now we have the restriction that a+b≠0?
 
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It is the identification of 1=(a+b)/(a+b) that introduces that restriction, not multiplying with 1 as such.
 
arildno said:
It is the identification of 1=(a+b)/(a+b) that introduces that restriction, not multiplying with 1 as such.

So how come we do that all the time in algebra (multiplying by the conjugate) to get rid of annoying radicals? Doesn't that make it not equal to the previous line?
 
tahayassen said:
So how come we do that all the time with binomial conjugates? Doesn't that make it not equal to the previous line?

Basically because the case a+b=0 can usually be treated as an uninteresting special case.
 
You can always multiply by 1 to get a new expression with the same value. The problem here is that what you multiplied by [ (a + b)/(a + b) ] isn't identically 1, and it is those values of a and b that cause a problem.
 

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