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pftq
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« on: September 26, 2006, 10:06:24 PM »

Proving 1=0, 2=1, and so on...


  Here is a pretty tricky proof that I've come across several times. It basically proves that a+b=a, which would mean that 1+0=0, 1=0, 2=1, and so on.

  I know very well how to prove this wrong, but do you? Tongue

a = b


ab = b²


ab - a² = b² - a²


a(b - a) = (b + a)(b - a)


a = b + a


If a and b are 1, then 1 = 1 + 1


1 = 2


0 = 1

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george
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« Reply #1 on: December 29, 2008, 10:56:09 PM »

no
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Bobby
Guest
« Reply #2 on: March 16, 2009, 05:23:08 PM »

im probably a little late but at the part where a(b-a)=(b+a)(b-a)

since b=a
b-a=0 and if u multiply 0 each side you would end up with 0
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dave
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« Reply #3 on: March 28, 2009, 01:36:51 PM »

your dividing by zero.  you cant divide both sides by (b-a). dividing by zero is undefined
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kenneth
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« Reply #4 on: January 02, 2012, 07:03:16 PM »

a(b - a) = (b + a)(b - a)
-(b - a)         -(b - a)
       a = (b + a)
      
Its not equal
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kenneth
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« Reply #5 on: January 02, 2012, 07:15:06 PM »

However

"0 IS 1 number" on the number scale.
In the begining is- 0 we now have 1 number. Oh its now 2. or is it three? Now im counting 4! ah <><>
Its called Axiom of infinity i think?

Its like the liers paradox.
A war between what it is and what it means
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Avinash
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« Reply #6 on: February 08, 2014, 07:06:11 AM »

nice
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