## Proof that i=0 (not really but still)

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Parralelex
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### Proof that i=0 (not really but still)

So I got bored and wondered what would happen if you ignored the fact that you can't compare imaginary numbers to real numbers in the traditional sense. What I got was this:

First, i is not less than 0. If i<0, than i is negative. Multiplying both sides by i (which we assume is negative) we get -1>0, clearly not true. Now, we can also prove that i is not greater than 0 by the same method: If i>0, than i*i=-1>0, also clearly not true. Since it is not less than nor greater than 0, clearly it must be 0.

This is, of course, clearly not true. As everyone knows, only 2*pi*i is equal to 0.
I put the "fun" in "mathematics".

And then I took it back out.

jestingrabbit
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### Re: Proof that i=0 (not really but still)

Parralelex wrote:2*pi*i is equal to 0.

No, its not. exp(2*pi*i) = 1 = exp(0), but 2*pi*i does not equal 0. Please, look up what a field is and what zero divisors are.
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Parralelex
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### Re: Proof that i=0 (not really but still)

jestingrabbit wrote:
Parralelex wrote:2*pi*i is equal to 0.

No, its not. exp(2*pi*i) = 1 = exp(0), but 2*pi*i does not equal 0. Please, look up what a field is and what zero divisors are.

That was my attempt at math humor.
I put the "fun" in "mathematics".

And then I took it back out.

Dopefish
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### Re: Proof that i=0 (not really but still)

I'm inclined to think intentionally wrong 'proofs' tend to be better when the error isn't obvious. When you start of with something along the lines of "A is true. Assuming not A, ..." it kind of ruins the fun since you can derive anything at that point.

Do some sneaky divisions by 0, or treat multivalued functions as single valued, or something along those lines.

e.g.

x^2 = (x + x + x + ... + x) {x is being summed x times}
d/dx(x^2) = d/dx(x + x + x + ... + x) {differentiating both sides wrt x}
2x = (1 + 1 + 1 + ... + 1) {power rule on each term on either side}
d/dx(2x) = d/dx(1 + 1 +1 + ... + 1) {Differentiating both sides wrt x again}
2 = (0 + 0 + 0 + ... + 0) {power rule again}
2 = 0 {finite sum of 0 terms is 0}

See, no division or multivalued functions in sight, so clearly i doesn't = 0, 2 does!

dudiobugtron
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### Re: Proof that i=0 (not really but still)

Dopefish wrote:Do some sneaky divisions by 0, or treat multivalued functions as single valued, or something along those lines.

So, like this?

i2 = -1
d/di (i2) = d/di (-1)
2i = 0
i = 0/2
i = 0
Spoiler:
The error, of course, is in the second-to-last line, where I divided by 2. (Since 2 = 0 as shown above.)

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### Re: Proof that i=0 (not really but still)

e.g.

x^2 = (x + x + x + ... + x) {x is being summed x times}
d/dx(x^2) = d/dx(x + x + x + ... + x) {differentiating both sides wrt x}
2x = (1 + 1 + 1 + ... + 1) {power rule on each term on either side}
d/dx(2x) = d/dx(1 + 1 +1 + ... + 1) {Differentiating both sides wrt x again}
2 = (0 + 0 + 0 + ... + 0) {power rule again}
2 = 0 {finite sum of 0 terms is 0}

See, no division or multivalued functions in sight, so clearly i doesn't = 0, 2 does!

Haha, I love this one! Especially when just leaving it at 2x=x ... I'm sure I'll manage to confuse someone with it.

edit: quote fix
Last edited by Heptadecagon on Fri Dec 28, 2012 5:32 am UTC, edited 1 time in total.

Meteoric
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### Re: Proof that i=0 (not really but still)

Parralelex wrote:As everyone knows, only 2*pi*i is equal to 0.

In case Heptadecagon didn't convince you, here's an alternate proof:
e^2*pi*i = e^0, so yes, 2*pi*i = 0. By the zero product property, we can conclude that either 2, pi, or i must equal zero. We know it isn't pi (because there are circles with nonzero circumference), and we know it isn't i (as discussed already in this thread), therefore 2 = 0.
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Dopefish
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### Re: Proof that i=0 (not really but still)

Heptadecagon wrote:Haha, I love this one! Especially when just leaving it at 2x=x ... I'm sure I'll manage to confuse someone with it.

I used to leave it there, but unfortunately people just concluded that x was 0 and there was no problem.

(Or they'd go ahead and divide by x from there and go "See, it doesn't work since 2 != 1" and leave it at that, feeling content they'd solved my little problem. That's wrong on a number of levels, so I took to another round of differentiating so that in addition to there being no divisions/multivalued functions, there aren't even any variables left.)

Parralelex
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### Re: Proof that i=0 (not really but still)

Dopefish wrote:I'm inclined to think intentionally wrong 'proofs' tend to be better when the error isn't obvious. When you start of with something along the lines of "A is true. Assuming not A, ..." it kind of ruins the fun since you can derive anything at that point.

Do some sneaky divisions by 0, or treat multivalued functions as single valued, or something along those lines.

e.g.

x^2 = (x + x + x + ... + x) {x is being summed x times}
d/dx(x^2) = d/dx(x + x + x + ... + x) {differentiating both sides wrt x}
2x = (1 + 1 + 1 + ... + 1) {power rule on each term on either side}
d/dx(2x) = d/dx(1 + 1 +1 + ... + 1) {Differentiating both sides wrt x again}
2 = (0 + 0 + 0 + ... + 0) {power rule again}
2 = 0 {finite sum of 0 terms is 0}

See, no division or multivalued functions in sight, so clearly i doesn't = 0, 2 does!

That one actually got me for a bit, I had to look up why it's incorrect.

From what I can tell, the function on the right hand side doesn't have any meaning for non whole valued functions, and therefore taking its derivative doesn't actually make sense.
I put the "fun" in "mathematics".

And then I took it back out.

PM 2Ring
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### Re: Proof that i=0 (not really but still)

Yeah, if x isn't a whole number then (x + x + x + ... + x) {x is being summed x times} is more than a little misleading, and so is differentiating it term by term.

Also note that
2x = (1 + 1 + 1 + ... + 1)
says that 2x = x.

Xenomortis
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### Re: Proof that i=0 (not really but still)

It makes some obvious sense for non-integer x (e.g. 1.5 summed 1.5 times = 1.5 + 0.75 = 2.25 = 1.5^2).
I think the problem comes is that you're hiding an x term that you naively ignore when you differentiate.
i.e. You say "sum x, x times" but ignore the second x when you differentiate.

Of course, extending the sum to adapt to the reals may have further implications that I've ignored.

Protoform
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### Re: Proof that i=0 (not really but still)

Meteoric wrote:
Parralelex wrote:As everyone knows, only 2*pi*i is equal to 0.

In case Heptadecagon didn't convince you, here's an alternate proof:
e^2*pi*i = e^0, so yes, 2*pi*i = 0. By the zero product property, we can conclude that either 2, pi, or i must equal zero. We know it isn't pi (because there are circles with nonzero circumference), and we know it isn't i (as discussed already in this thread), therefore 2 = 0.

Alternatively, one can offer the following simple proof: 0^5 = 0^3, therefore 5 = 3.

Parralelex
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### Re: Proof that i=0 (not really but still)

Protoform wrote:
Meteoric wrote:
Parralelex wrote:As everyone knows, only 2*pi*i is equal to 0.

In case Heptadecagon didn't convince you, here's an alternate proof:
e^2*pi*i = e^0, so yes, 2*pi*i = 0. By the zero product property, we can conclude that either 2, pi, or i must equal zero. We know it isn't pi (because there are circles with nonzero circumference), and we know it isn't i (as discussed already in this thread), therefore 2 = 0.

Alternatively, one can offer the following simple proof: 0^5 = 0^3, therefore 5 = 3.

Also:

Since 3*2=2 mod 4, 3=1 mod 4, meaning 3+4n=1+4m for some n and m. Simply set m and n to 0 and subtract both sides by 1.
I put the "fun" in "mathematics".

And then I took it back out.