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AshleyyWhuddupp Group TitleBest ResponseYou've already chosen the best response.0
yes,
 2 years ago

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

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

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

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

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

AravindG Group TitleBest ResponseYou've already chosen the best response.0
@blockcolder
 2 years ago

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

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

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

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

badreferences Group TitleBest ResponseYou've already chosen the best response.1
As 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.
 2 years ago
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