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# Impossible math 1

4 replies to this topic

### #1 comperr

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Posted 24 June 2007 - 02:59 AM

For a certain class I had to prove 25-30 things in math that are impossible. I was the only one in my class to do so. Since then I thought of more and I was shown more.
Here is the first of a a number of them
given
a = b
a = 1
-----
a^2 = ab (where ^ = raised to the power)
a^2 -b^2 = ab - b^2
factor it out:
(a+B)(a-B) = b(a-B)
divide through by a-b
a+b = b
1 + 1 = 1
2 = 1

What is the flaw with this?
What is a-b?
When you divide through by a-b you are actually dividing through by 0 which is impossible
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### #2 mannu

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Posted 28 June 2007 - 06:20 AM

if a=1 and b=1, a-b becomes 0. U can't perform a "divide-by-zero" operation on an equation.
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### #3 mathemagister

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Posted 09 July 2007 - 01:41 PM

This is the famous mathematical fallacy known as the "Zero Proof".
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### #4 comperr

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Posted 09 July 2007 - 11:53 PM

Sort of.
The famous one works a bit differently with the same fundamental flaw.
1. a = b + 1
2. (a-B.)a = (a-B.)(b+1)
3. a2 - ab = ab + a - b2 - b
4. a2 - ab -a = ab + a -a - b2 - b
5. a(a - b - 1) = b(a - b - 1)
6. a = b
7. b + 1 = b
8. Therefore, 1 = 0
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### #5 mathemagister

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Posted 11 July 2007 - 11:03 AM

Well, the Zero Proof is a bit different:

a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(difference of two squares)
(a+B.)(a-B.) = ab - b^2
(a+B.)(a-B.) = b(a-B.)
a+b = b
a = 0

A does not have any fixed value (no a = 1 (or any other number) step), it is just an unknown. Making you wonder how ANY NUMBER AT ALL could equal 0. Of course the step dividing by (a-B) is wrong as a-b is 0 because a=b.
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