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mirhaque
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HongHu
(A - B)/(A + B) = [(A - B)*(A + B)]/[(A + B) ^2]
Since (A+B)^2 is always non negative, (A - B)*(A + B) and (A - B)/(A + B) always have the same sign. However, you need to be careful when you do the division since it is possible that (A+B)^2 =0


Isn’t that a contradiction HongHu if (A + B)^2 is always non negative yet it could be zero?
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Folaa3
HongHu
(A - B)/(A + B) = [(A - B)*(A + B)]/[(A + B) ^2]
Since (A+B)^2 is always non negative, (A - B)*(A + B) and (A - B)/(A + B) always have the same sign. However, you need to be careful when you do the division since it is possible that (A+B)^2 =0

Isn’t that a contradiction HongHu if (A + B)^2 is always non negative yet it could be zero?


No, because a square is always positive, but its range is (0, infinite).

Thus the point 0 is included in the range of squares.

This case happens when A and B are the same magnitude and opposite sign. Or even both 0.

Hope that Helps.
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Folaa3


Isn’t that a contradiction HongHu if (A + B)^2 is always non negative yet it could be zero?


Notice I have very carefully said non negative, instead of positive. Non negative means it could be zero or positive. Similarly non positive would mean zero and negative.



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