Hisevenplusplus
Clarification 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 =\(\frac{(a+b)}{b}\)=\(\frac{(c+d)}{d}\)
...
Clarification 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 =\(\frac{(a+b)}{b}\)=\(\frac{(c+d)}{d}\)
...
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