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.

\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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2^{nd} way

We know that ,

\frac{1}{4}>\frac{1}{8}

\implies(\frac{1}{2})^2>(\frac{1}{2})^3

\implies2\log_{10}\frac{1}{2}>3\log_{10}\frac{1}{2}

\implies{2>3}

Here is a mistake. Find it.

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}