Multiplying by 1 adds a restriction?

[tahayassen]

$$(a-b)=(a-b)*1\\ =(a-b)\frac { a+b }{ a+b } \\ =\frac { { a }^{ 2 }-{ b }^{ 2 } }{ a+b }$$

But now we have the restriction that a+b≠0?

 

[arildno]

It is the identification of 1=(a+b)/(a+b) that introduces that restriction, not multiplying with 1 as such.

 

[tahayassen]

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?

 

[arildno]

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.

 

[tahayassen]

Basically because the case a+b=0 can usually be treated as an uninteresting special case.

Ah.

Just FYI, I have an unanswered thread over here: https://www.physicsforums.com/showthread.php?t=648782

 

[Mark44]

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.