2=1

G

galactus

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Here is a proof that 2=1. No arguments. I don't want to hear it!. I am right and that's all there is too it!!!!.

:wink:


For x>0:

\(\displaystyle x=\underbrace{1+1+1+.....+1}_{\text{x times}}\)

\(\displaystyle x^{2}=\underbrace{x+x+x+....+x}_{\text{x times}}\)

\(\displaystyle D(x^{2})=D\left(\underbrace{x+x+x+...+x}_{\text{x times}}\right)\)

\(\displaystyle D(x^{2})=\underbrace{D(x)+D(x)+D(x)+...+D(x)}_{\text{x times}}\)

\(\displaystyle 2x=\underbrace{1+1+1+....+1}_{\text{x times}}\)

\(\displaystyle 2x=x\)

\(\displaystyle 2=1\)
 
There's a free court martial on offer for anyone that doesn't get this joke!
 
Thank you galactus for your excellent proof showing 2 = 1.

I always knew that 1 can't possibly = 1, but did you know that 0 does not equal 0 but actually 0 is < 0?

For that obnoxious individual who always says prove it, I will do so accordingly.

0 < 1-1/2+1/3-1/4+1/5-1/6+...

= (1+1/3+1/5+1/7+...)-(1/2+1/4+1/6+1/8+...)

=[(1+1/3+1/5+1/7+...)+(1/2+1/4+1/6+1/8+...)}-2(1/2+1/4+1/6+1/8+...)

=(1+1/2+1/3+1/4+1/5+...)-(1+1/2+1/3+1/4+1/5+...) = 0

So 0 < 0 QED
 
You're a true Master of Disguise, Glenn,
or a Master of Illusion.

If I have x empty bottles,
where the capacity of a bottle is 1 litre,
and I've drunk 5 litres,
what is the value of x ?
 
I have two current American coins that are used everyday as legal tender, here (USA), and other realms.

The two coins total 55 cents and one is not a nickel. What are the two coins?
 
chrisr, whether you drank 5 liters or fifty liters, what has that to do with x empty liter bottles?

A classic example of comparing apples to oranges.
 
No,
I never said I drank anything from those bottles,
or how many of us were drinking.
 
\(\displaystyle \int \frac{1}{x}dx\)

By parts let \(\displaystyle u=\frac{1}{x}, \;\ dv=dx\)

\(\displaystyle du=\frac{-1}{x^{2}}dx, \;\ v=x\)

\(\displaystyle \int\frac{1}{x}dx=\frac{x}{x}-\int\frac{-1}{x^{2}}\cdot xdx\)

\(\displaystyle \int\frac{1}{x}dx=1+\int\frac{1}{x}dx\)

\(\displaystyle 0=1\)
 
It seems like the hidden mishap is assuming indefinite integrals are unique?
 
I know, daon. These little things are just fun. :D

Same as the derivative one I posted.
 

From the study of Infinite Series, we have this fact:

. . \(\displaystyle \ln 2 \;=\;1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \hdots\)


Regroup the terms:

. . \(\displaystyle \ln 2 \;=\;\left(1 + \frac{1}{3} + \frac{1}{5} + \hdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \hdots\right)\)


\(\displaystyle \text{Add and subtract }\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \hdots\right)\)

. . \(\displaystyle \ln 2\;=\;\left(1+ \frac{1}{3} + \frac{1}{5} + \hdots\right) \;+\;\overbrace{ \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \hdots\right)}^{\text{add}} \;-\;\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \hdots\right) \;-\; \overbrace{\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\hdots\right)}^{\text{subtract}}\)

. . \(\displaystyle \ln 2 \;=\;\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \hdots\right) - 2\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} +\hdots\right)\)

. . \(\displaystyle \ln 2 \;=\;\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \hdots\right) - \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \hdots\right)\)


Hence: .\(\displaystyle \ln 2 \:=\:0\)

Therefore: .\(\displaystyle e^0 \:=\:2 \quad\Rightarrow\quad\boxed{ 1 \:=\:2}\)

 
That's the problem with infinity.... let's stop monkeying with 0, 1 and infinity .....
 
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