1^{st} way
2=1
ধরি, a=b
কিন্তু,a\ne0 b\ne0
So, a^2=ab
\implies{a^2-b^2}=ab-b^2
\implies(a+b)(a-b)=b (a-b)
\implies{a+b=b}
\implies{2b=b}
\implies{2=1}
Here is a mistake. Find it.
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\fbox{In the second line of the equation,}
\fbox{You can write-}
(a+b)(a-b)=b(a-b)
\implies{(a+a)(a-a)=b(a-a) [a=b]}
\implies{0=0}
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\fbox{Right answer}
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In any inequality, if you divide or multiply both sides with something negative, the < or > will be reversed.In the way to 2nd line from 3rd line, you have divide both sides by log1/2 that is negative. So the sign of > will be reversed. But you haven’t changed it.Here is the mistake.
\fbox {Right answer}
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