Proving that 0 = 1 (Part 1)

I've always enjoyed these kinds of "proofs." I saw my first one in my first algebra class, where you start with a^2 = b^2, manipulate the equalities until you can divide by a non-obvious zero, and come up with with 0 = 1. That's slight of hand trickery, and not very cool.More fun are other methods that show more of mathematical interest.

For example (this also involves some slight of hand) let's start out with the simple equation
x = 1 + 2x. No big deal, most could solve this with one hand tied behind their backs. But of these, some balk when I say that since this is an equality, I can substitute 1 + 2x for x however I like, and write x = 1 + 2 ( 1 + 2x ) = 1 + 2 + 4x. Of course, this is a perfectly valid operation, and I can do it again, as in x = 1 + 2 + 4 ( 1 + 2x ) = 1 + 2 + 4 + 8x. See the pattern? So I can continue doing this a and obtain x = 1 + 2 + 4 + 8 + 16 + 32 + 64 + ..... Clearly, the sum of all the powers of two diverges, and so is infinite. Therefore x equals infinity. On the other hand, we can solve the original equation for x = -1. Therefore -1 = infinity. Not exactly 0 = 1, but close enough, don't you think? Cute, but not exactly fair, since I hid a large factor of x off to the right of the ellipses. That's the slight of hand, which is cheating. Let's see if we can show that 0 = 1 without cheating.

Think about how taking the square root of a large number makes it smaller. For example, the square root of 2.0 is 1.414... And the square root of 1.414... is 1.189... Look at the following recursive formula (again, forgive the poor notation)

x = root ( 2 + root ( 2 + root ( 2 + root ( 2 + ....

Can you solve for x? It's actually easy, once you see the symmetry of the formula. As above, the left and right sides of the equals side are freely interchangeable, so you can write

x = root ( 2 + x )
x^2 = 2 + x
x^2 - x - 2 = 0
x = 2 (and of course also -1 but we'll only look at positive roots)

We can readily generalize this to solve for any x = root ( y + root ( y + root ( y+ ...
For example, root ( 3 + root ( 3 + root ( 3 + ... yields (1 + root 13 ) / 2 and root ( 1 + root ( 1 + root ( 1 + ... yield phi, the golden ration, ( 1 + root 5 ) / 2, or 1.618.... In general, from the quadratic formula, x = ( 1 + root ( 1 + 4y) ) / 2.

Consider, now, how taking the root of a small number makes it bigger. The square root of a quarter is a half,and the square root of a hundredth is a tenth. So what do we get when we ask for
x = root ( 1/2 + root ( 1/2 + root ( 1/2 + ... ? Well, this yields ( 1 + root 3 ) / 2 which is about 1.366... As y gets smaller, x approaches 1, from above. After all, y = 1 gave phi, well above 1. Think about it for a bit and I'm sure you'll agree it's intuitive.

So we can conclude that if we used an infinitely small y we would get 1 for our answer.
1 = root ( 0 + root ( 0 + root ( 0 + .... But of course the square root of zero is zero, and no matter how many zeroes you add up it's still zero. So root ( 0 + root ( 0 + root ( 0 + ... = 0 and there you are, once again, 0 = 1.

Where is the issue here? It's not in using only positive roots. What are your thoughts? I'll post more here myself later.