1 = 2?? whats going on here??

ArmoSkater87

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

Galileo

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

Zurtex

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.

HallsofIvy

Not only have "some" of us seen it before, it has been posted on this board more times that I want to remember!

Gza

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.)

ArmoSkater87

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

JonF

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

KnowledgeIsPower

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.

Muzza

There are a few other here:

http://mathforum.org/dr.math/faq/faq.false.proof.html
http://www.math.toronto.edu/mathnet/...econd1eq2.html

I've seen one which involves integrals, but I can't find it...

Ivan Seeking

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.

Zurtex

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.

garytse86

there is a proof in the philosophy section about .999999999999999999999 = 1 using a geometric series

Zurtex

I assume you mean [itex]0.99\overline{9}=1[/itex] and there are many proofs of this.

KnowledgeIsPower

Interesting!, thanks for the links and integral 'proof'. I'll be sure to use that to confuse a few people ^^.

Dburghoff

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.

cronxeh

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

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

Zurtex

Sorry I don't believe that is expressible in terms of elementary functions.