A cute 2=1

[Antrax]

Not too difficult to figure out, but at least it's new (or so I think. First time I've seen it):

x^2 = x*x = x+x+...+x (x times)
take derivative of both sides of the equation:
2x=1+1+...+1 (x times) -> 2x=x
and we get either x=0 (which means all numbers are zero) or 1=2.
Antrax

------------------
"Look, that's why there's rules, understand? So that you think before you break 'em" - Lu-Tze, Thief of Time

[Tahnan]

For those puzzled (and on a puzzle board, who wouldn't be?), what goes wrong is that Antrax took the derivative of "x + x + ... + x" correctly, but the ellipsis corresponds to "x times," which is a variable, which you can't ignore when taking the derivative. (A simpler version: if x = (1 + 1 + ... + 1), x times, so 1 = (0 + 0 + ... + 0), x times. Clearly the slope of the graph at any given point depends on how many 1's there are, which makes that "x times" a, pardon the expression, integral part of the formula.)

[Bicho the Inhaler]

Here's another "cute" one (depending on your definition of "cute"). I saw this on some other message board.

It is a well-known fact that if you have a complex function f expressed as a power series that converges on the whole complex plane (also known as an "entire" function), then there can be at most one complex number missed by f, i.e., there can be at most one complex y such that f(x) != y for any complex x. For the purposes of this "proof," you can just take this on faith. There is a problem with the argument, of course, but it isn't here. Also, the composition of two entire functions is another entire function.

The function f(x) = e^x is an entire function that misses the point 0. The function g(x) = f(f(x)) is therefore another entire function. But since f misses 0, g misses both f(0) and 0, and f(0) = 1. Since g can only miss one point, it follows that 1 = 0. Add 1 to get that 1 = 2.

[Antrax]

[mith]

Nope, Ant:

It still doesn't hit 0. The problem is that just because 0 is not hit by e^x does not mean 1 is not hit by e^(e^x). We just need another number than 0 (call it y) such that e^y = 1, and then let x = ln y. Since e^pi*i = -1, e^2pi*i = 1, so x = ln(2pi*i) (or any number ln(2*n*pi*i)).

[kevinatilusa]

My favorite cutie.

Consider the sum S=1-1/2+1/3-1/4+1/5-1/6+1/7-1/8+... This is an alternating series whose terms decrease to 0, so converges by the alternating series test to a number betweeen its first 2 partial sums, i.e. S exists and 1/2<S<1. In fact you can prove (though it requires a bit of work) that S=ln(2)

Now that we've proven the existence of S we can rearrange its terms so long as we keep track of the signs:
S=1-1/2-1/4+1/3-1/6-1/8+1/5-1/10-1/12+1/7-1/14-1/16+...
=(1-1/2)-1/4+(1/3-1/6)-1/8+(1/5-1/10)-1/12+(1/7-1/14)-1/16...
=1/2-1/4+1/6-1/8+1/10-1/12+1/14-1/16...
=1/2(1-1/2+1/3-1/4+1/5-1/6+...)=1/2*S

Solving, S=0. Thus we have 1/2<0<1, and, furthermore, that since ln(2)=0, 1=2. (taking e to the power of both sides)

[Lepton]

~Hands Wave Frantically~

This doesn't work because of the re-arrangement of the terms in an alternating series. There is a Named Theorem that says that any alternating series that satisfies a few basic criteria can be re-arranged to give any natural number, if one tries hard enough.

[jadesmar]

x^2 = x*x = x+x+...+x (x times)

Differentiating:

d(x^2)/dx = 2x
d(x*x)/dx = (1)x+x(1)
d(x+x+...+ x(x times))/dx= 1+1+...+1(x times)+ x+x+...+x(1 times)

[i_h8_evil_stuff]

The "Let S=ln(2)" one doesn't quite work, my calculus teacher says, because you only have two-thirds as many terms, or something. I didn't fully understand what he said.

[starfruit]

How about this?
sqrt = square root

sqrt 1 = sqrt [(-1)(-1)] --> sqrt ab = sqrt a x sqrt b
= sqrt (-1) x sqrt (-1)
= i x i
= -1

Which results in 1 = -1 !

[Antrax]

sqrt(-1)=+-i, not i.

so we get on the left side +-1, and on the right side we get +-i times +-i. Looks fine to me.

[starfruit]

well...just a simple trick to snare others

[Antrax]

sqrt(-1)=(-1) ^1/2 =(-1) ^2/4 =( _(-1) 2) ^1/4 =1 ^1/4 =1

[mathgrant]

1 ^1/4 is not 1. It is 1 or i or -1 or -i.
_________________
My logic puzzle blog

[winterHLepsilon]

If a=b, a,b>0, then 1=2.

a,b>0
a=b
ab=b^2
ab-a^2=b^2-a^2
a(b-a)=(b+a)(b-a)
a=(b+a)
a=a+a
a=2a
1=2
_________________
*HL--^H^e^L^en *
=A girl from Guangzhou, China, Asia=

Ask, and it shall be given you;
Seek, and ye shall find;
Knock, and it shall be opened unto you.

[i_h8_evil_stuff]

This was posted earlier, in a different thread.

When you divide both sides by (a-b) to go from a(b-a)=(b+a)(b-a) to a=(b+a), you're dividing by zero. You're just cleverly disguising the fact that 1*0=2*0.
_________________
Space for sale. PM i_h8_evil_stuff for details.

[winterHLepsilon]

I think many paradoxes end up dividing something by zero. And disaster...
_________________
*HL--^H^e^L^en *
=A girl from Guangzhou, China, Asia=

Ask, and it shall be given you;
Seek, and ye shall find;
Knock, and it shall be opened unto you.

[CrystyB]

[kevinatilusa]

Another fun (calculus) one:

Suppose you couldn't remember that int(1/x)=ln(x), and tried to do it by parts

u=1/x dv=dx
du=-1/x^2(dx) v=x

int(udv)=uv-int(vdu), so int(1/x dx)=x(1/x)-int(x*-1/x^2 dx)=1+int(1/x dx). Cancelling the common term on both sides, 0=1.

[CrystyB]

Gotcha! I admit, i had to think for a good five minutes about this, and finally decided to see if the justification of the formula, d(uv)=udv+vdu still holds. (it does)

The answer is...

.

.

.

Cancelling indefinite integrals does not necessarily leave a zero in their place. After all, int(f(x)dx) = F(x) + C, so what we get is C = 1 + C, which is perfectly all right.