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blockcolder
 2 years ago
Best ResponseYou've already chosen the best response.4Except of course when a=b=0 but otherwise, yeah.

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.0@badreferences , @apoorvk , @dumbcow , @dpaInc

blockcolder
 2 years ago
Best ResponseYou've already chosen the best response.4If a=b, and they're not zero, you can divide both sides by ab to get 1/a=1/b.

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.0@EarthCitizen , @Ishaan94 , @Mani_Jha

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.0oh!! thats the answer i was looking for!!@blockcolder

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.0this seems a simple qn ..bt i had this doubt from my small classes

EarthCitizen
 2 years ago
Best ResponseYou've already chosen the best response.0a=b, so long as a \[ a \neq0\]

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1It's been answered, but a more rigorous way of answering it would be:\[\left\{a=b\mid\forall \left(ab\neq0\right)\right\}\]

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1Whoops, I mean:\[\left\{a=b\therefore\frac{1}{a}=\frac{1}{b}\mid\forall \left(ab\neq0\right)\right\}\]

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1As either \(a,b\) can be \(0\), but not necessarily both, for the implication to be demonstrably false. A simpler way of showing this is by determining that the product must not be zero.
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