mathematics puzzle

the above solution is on my line but with a flaw that stands close to Kmaran's comment. It is in substituting a=b not for performing equal operation. We can not just replace 2=3 than it would have nothing to prove at all our puzzle; 1=2 from 1=1. But we can say performa mathematical equal operation like adding up a certain value to both of the sides. Now, thanks again for your response but may place another request? You have started with algebric notation that also stands equal though but with the basic flaw of all in this particular case, that you haven't started from 1=1. again thank you for your participation.

just to give a hint if nay interested. 1=1 -2=-2 (multiplying both wth -2) 1-3=4-6 and so on

(a^2)-(b^2)=(a+b)(a-b) holds only good for real and distinct values of a and b. so it's falwed:-)

the fourth line is not following rules of solving equation. you just cannot perform any unequal operation on both the side of an equation.

when you are multiplying by 1 on both sides then why you are using 2 on right hand side and 1 on left hand side.so it is not following the commutative law.if there is statement that 0 equals to any number mulitplied by 0 then it may be correct.

In the last step, you are dividing both sides by 0, which is not a real mathematical operation since the result is an INDETERMINATION or a mathematical incongrence (what will you give each when you share something between zero persons?).