Hi

sevenplusplusClarification 1: Now 1. above can only be true if a is not equal to b or (a-b) is not zero ..right? (this is not mentioned in the book)

How did we (algebraically) arrive at this equivalency?

- Here the book only gives the rule that can be applied to ratios. Incase a-b=0, a=b, ratio is 1, so there's no need to go for componendo dividendo. It is useful to apply in some problems directly.

\(\frac{a}{b}\) = \(\frac{c}{d}\) =>Adding and subtracting 1 from both sides;

\(\frac{a}{b}\)+1=\(\frac{c}{d}\)+1 = \(\frac{(a+b)}{b}\)=\(\frac{(c+d)}{d}\) -------(1)

\(\frac{a}{b}\)-1=\(\frac{c}{d}\)-1 = \(\frac{(a-b)}{b}\)=\(\frac{(c-d)}{d}\)--------(2)

Dividing (1) by (2) gives the desired result.Clarification 2: can we also claim that: \(\frac{(a-b)}{(a+b)}\) = \(\frac{(c-d)}{(c+d)}\) ?

Yes. As it is inverse of the ratio.Please give kudos, if u liked my post!
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