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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

- Thread starter ArmoSkater87
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- #1

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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

- #2

Galileo

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Oh dear, you didn`t happen to divide by zero did you?

- #3

Zurtex

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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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- #7

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[sarcasm] this forum needs more post about how 1=0 and how .99999… doesn’t = 1 [/sarcasm]

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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.Gza said:

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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...

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...

Last edited:

- #10

Ivan Seeking

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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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Try integrating [itex]e^x \cosh x[/itex] by parts: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...

[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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I assume you mean [itex]0.99\overline{9}=1[/itex] and there are many proofs of this.garytse86 said:there is a proof in the philosophy section about .999999999999999999999 = 1 using a geometric series

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- #15

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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.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]

- #16

cronxeh

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i donno if it was posted here.. anyone can take on this:

[tex]\int x^x dx[/tex]

[tex]\int x^x dx[/tex]

- #17

Zurtex

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Sorry I don't believe that is expressible in terms of elementary functions.cronxeh said:i donno if it was posted here.. anyone can take on this:

[tex]\int x^x dx[/tex]

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