Proof that i=0 (not really but still)

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

Parralelex
Posts: 74
Joined: Sun Dec 02, 2012 8:10 am UTC
Location: Location Location

Proof that i=0 (not really but still)

Postby Parralelex » Wed Dec 26, 2012 5:54 am UTC

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.

User avatar
jestingrabbit
Factoids are just Datas that haven't grown up yet
Posts: 5967
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney

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

Postby jestingrabbit » Wed Dec 26, 2012 6:24 am UTC

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.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

Parralelex
Posts: 74
Joined: Sun Dec 02, 2012 8:10 am UTC
Location: Location Location

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

Postby Parralelex » Wed Dec 26, 2012 9:13 am UTC

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.

User avatar
Dopefish
Posts: 855
Joined: Sun Sep 20, 2009 5:46 am UTC
Location: The Well of Wishes

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

Postby Dopefish » Wed Dec 26, 2012 6:11 pm UTC

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!

User avatar
dudiobugtron
Posts: 1098
Joined: Mon Jul 30, 2012 9:14 am UTC
Location: The Outlier

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

Postby dudiobugtron » Thu Dec 27, 2012 11:45 pm UTC

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.)
Image

User avatar
Heptadecagon
Posts: 36
Joined: Mon May 28, 2012 7:11 pm UTC

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

Postby Heptadecagon » Fri Dec 28, 2012 2:43 am UTC

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
Posts: 333
Joined: Wed Nov 23, 2011 4:43 am UTC

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

Postby Meteoric » Fri Dec 28, 2012 4:30 am UTC

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.
No, even in theory, you cannot build a rocket more massive than the visible universe.

User avatar
Dopefish
Posts: 855
Joined: Sun Sep 20, 2009 5:46 am UTC
Location: The Well of Wishes

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

Postby Dopefish » Fri Dec 28, 2012 5:42 am UTC

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
Posts: 74
Joined: Sun Dec 02, 2012 8:10 am UTC
Location: Location Location

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

Postby Parralelex » Sat Dec 29, 2012 7:19 am UTC

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.

User avatar
PM 2Ring
Posts: 3715
Joined: Mon Jan 26, 2009 3:19 pm UTC
Location: Sydney, Australia

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

Postby PM 2Ring » Sat Dec 29, 2012 7:32 am UTC

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.

User avatar
Xenomortis
Not actually a special flower.
Posts: 1455
Joined: Thu Oct 11, 2012 8:47 am UTC

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

Postby Xenomortis » Mon Dec 31, 2012 9:44 am UTC

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. :)
Image

Protoform
Posts: 43
Joined: Sun Dec 30, 2012 1:09 pm UTC
Location: The Nether

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

Postby Protoform » Mon Dec 31, 2012 3:07 pm UTC

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
Posts: 74
Joined: Sun Dec 02, 2012 8:10 am UTC
Location: Location Location

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

Postby Parralelex » Mon Dec 31, 2012 9:11 pm UTC

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.


Return to “Mathematics”

Who is online

Users browsing this forum: No registered users and 6 guests