can anyone proof that 2=0 if anyone solve it guve me the answer otherwise i have the answer
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Let a=b=2
So, a×b=(b)^2 [2×2=(2)^2]
On subtracting (a)^2 from both sides
ab-(a)^2=(b)^2-(a)^2
a(b-a) =(b+a) (b-a)
So, a=b+a
a-a=b on transposing
0=b
Or, 0=2
So, a×b=(b)^2 [2×2=(2)^2]
On subtracting (a)^2 from both sides
ab-(a)^2=(b)^2-(a)^2
a(b-a) =(b+a) (b-a)
So, a=b+a
a-a=b on transposing
0=b
Or, 0=2
tarun0001:
brilliant bro
ur method is thr 2nd1 which i got
thanks
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