1 = 2? whats going on here?

  • #1
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1 = 2?? whats going on here??

Haha, just wanted to show this, because its just so funny. I'm sure some of you have seen it before.

a = b
a^2 = b^2
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b(a-b)
a+b=b
b+b=b
2b=b
2=1
 

Answers and Replies

  • #2
Galileo
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Oh dear, you didn`t happen to divide by zero did you?
 
  • #3
Zurtex
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If a = b then a - b = 0.

So this step:

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

Is not valid as you are dividing by zero. For example:

1*0 = 2*0

This does not prove 1 = 2.
 
  • #4
HallsofIvy
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Not only have "some" of us seen it before, it has been posted on this board more times that I want to remember!
 
  • #5
Gza
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The invalid step in this proof looks relatively obvious to me. I've seen some more intricate ones that aren't nearly as such, and take some time to figure out. (We've already had a few posts like this in the past, so I won't post them.)
 
  • #6
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hah, alrite, u guys saw it right away, it took me a little longer when i first saw it. I just though it would be fun to see some people's responses. :D
 
  • #7
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[sarcasm] this forum needs more post about how 1=0 and how .99999… doesn’t = 1 [/sarcasm]
 
  • #8
Gza said:
The invalid step in this proof looks relatively obvious to me. I've seen some more intricate ones that aren't nearly as such, and take some time to figure out. (We've already had a few posts like this in the past, so I won't post them.)
Just out of personal interest, could someone post or email me one of these more intricate and less noticable ones? I feel it would be an amusing email to confuse a few friends with.
 
  • #10
Ivan Seeking
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Has anyone seen the, either 101, or 151, or 351 [?] page proof that 1 + 1 = 2?

I can't remember the page number for sure but the last page was posted in the office of a math professor of mine. I think it was done as a thesis to show that certain theories were consistent; like number theory and set theory, for example.
 
  • #11
Zurtex
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Muzza said:
There are a few other here:

http://mathforum.org/dr.math/faq/faq.false.proof.html
http://www.math.toronto.edu/mathnet/falseProofs/second1eq2.html

I've seen one which involves integrals, but I can't find it...
Try integrating [itex]e^x \cosh x[/itex] by parts:

[tex]u = \cosh x[/tex]
[tex]v' = e^x[/tex]

[tex]\int e^x \cosh x dx = e^x \cosh x - \int e^x \sinh x dx[/tex]

Using by parts method again on [itex]e^x \sinh x[/itex]

[tex]s = \sinh x[/tex]
[tex]t' = e^x[/tex]

[tex]\int e^x \cosh x dx = e^x \cosh x - e^x \sinh x + \int e^x \cosh x dx[/tex]

[tex]0 = e^x \cosh x - e^x \sinh x[/tex]

[tex]\cosh x = \sinh x[/tex]

And this reduces down to 1 = -1.
 
  • #12
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there is a proof in the philosophy section about .999999999999999999999 = 1 using a geometric series
 
  • #13
Zurtex
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garytse86 said:
there is a proof in the philosophy section about .999999999999999999999 = 1 using a geometric series
I assume you mean [itex]0.99\overline{9}=1[/itex] and there are many proofs of this.
 
  • #14
Interesting!, thanks for the links and integral 'proof'. I'll be sure to use that to confuse a few people ^^.
 
  • #15
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Zurtex said:
Try integrating [itex]e^x \cosh x[/itex] by parts:

[tex]u = \cosh x[/tex]
[tex]v' = e^x[/tex]

[tex]\int e^x \cosh x dx = e^x \cosh x - \int e^x \sinh x dx[/tex]
Count up the negative signs: one from subtracting the result of the integration by parts, and one for using the derivative of [itex]\cosh x[/itex], which is [itex]-\sinh x[/itex]. You should be adding the the antiderivative, not subtracting it.
 
  • #16
cronxeh
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i donno if it was posted here.. anyone can take on this:

[tex]\int x^x dx[/tex]
 
  • #17
Zurtex
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cronxeh said:
i donno if it was posted here.. anyone can take on this:

[tex]\int x^x dx[/tex]
Sorry I don't believe that is expressible in terms of elementary functions.
 

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