SOLUTION: How do you solve this false proof and what is the mistake in the proof? a=1,b=1 a=b a2=ab a2+a2=a2+ab 2a2=a2+ab 2a2-2ab=a2-ab 2(a2-ab)= 1(a2-ab) 2=1

Algebra ->  Geometry-proofs -> SOLUTION: How do you solve this false proof and what is the mistake in the proof? a=1,b=1 a=b a2=ab a2+a2=a2+ab 2a2=a2+ab 2a2-2ab=a2-ab 2(a2-ab)= 1(a2-ab) 2=1      Log On


   

Question 704504: How do you solve this false proof and what is the mistake in the proof?
a=1,b=1
a=b
a2=ab
a2+a2=a2+ab
2a2=a2+ab
2a2-2ab=a2-ab
2(a2-ab)= 1(a2-ab)
2=1
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You let/force a = b.

So because a = b, this means that

a^2 - ab

b^2 - b*b ... plug in a = b (ie replace all copies of 'a' with 'b')

b^2 - b^2

0

So if a = b, then a^2 - ab is equal to 0.

When you jump from 2(a^2-ab)= 1(a^2-ab) to 2 = 1 in the last two lines, you are dividing both sides by a^2-ab. But if a = b, then a^2-ab = 0, which means you're dividing both sides by zero...which is undefined.

So the last step is NOT a valid algebraic move because you can't divide by zero.

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