A small math problem I found myself with...

A spot for all things TestTube
Post Reply
User avatar
robly18
Posts: 413
Joined: Tue Jun 04, 2013 2:03 pm

A small math problem I found myself with...

Post by robly18 »

What you're about to see was found entirely by me. I did my research and found nothing on this.

We all know that every number has two square roots. But what if I told you any exponent of ANY kind has FOUR results? For instance, x^1 = x, -x, ix, -ix. Allow me to explain:

To put a number to the power of a fraction, we make the root of index denominator and then put it to the power of the numerator. i.e. 4^0.5 = 4^(1/2) = √(4^1) which equals obviously √4 = 2 and -2. However, there's more. 1/2 also equals 2/4. So putting four to that power equals 4√(4^2). That equals 4√16. However, root index four can also be two square roots: 4√16 = √(√16). That'll equal √4 and √-4. This equals 2, -2 AND i2 and -i2.

The plot thickens from hereon out, as every number can be written as a fraction of denominator being a multiple of four. For instance, x^1 can be written as x^(4/4). Therefor, x^1 = x, -x, ix and -ix.

This is where I fear I fell into some sort fallacy. Because if this is correct, then if x^1 = x then x = -x. Which obviously means that x^1 does not equal exclusively x.

So, what is YOUR opinion on this? Does every exponent have four results?
Convincing people that 0.9999... = 1 since 2012
A Random Player
Posts: 523
Joined: Mon Jun 03, 2013 4:54 pm

Re: A small math problem I found myself with...

Post by A Random Player »

I haven't looked carefully yet, but I think the problem is applying radical properties which only hold in the reals to complex numbers.
Ex. 1 = sqrt(1) = sqrt(-1 x -1) = sqrt(-1) x sqrt(-1) = i x i = -1

Edit: Or possibly extraneous roots - Can't tell which right now.
$1 = 100¢ = (10¢)^2 = ($0.10)^2 = $0.01 = 1¢ [1]
Always check your units or you will have no money!
User avatar
robly18
Posts: 413
Joined: Tue Jun 04, 2013 2:03 pm

Re: A small math problem I found myself with...

Post by robly18 »

No, I don't think that's what I'm doing. I'm taking numbers... Wait... This means.. i IS in fact 1! Holy dip! Did I revolutionize mathemathics?
(
Anyway, what I'm doing is taking into account that 1 = 4/4 and x^(4/4) equals √(√(1^4)). This equals √(√1) which equals √1 and √-1, which equals 1, -1, i and -1
Convincing people that 0.9999... = 1 since 2012
A Random Player
Posts: 523
Joined: Mon Jun 03, 2013 4:54 pm

Re: A small math problem I found myself with...

Post by A Random Player »

robly18 wrote:No, I don't think that's what I'm doing. I'm taking numbers... Wait... This means.. i IS in fact 1! Holy dip! Did I revolutionize mathemathics?
(
Anyway, what I'm doing is taking into account that 1 = 4/4 and x^(4/4) equals √(√(1^4)). This equals √(√1) which equals √1 and √-1, which equals 1, -1, i and -1
I think I've found something similar to your problem: Look at the bottom of this section. Or:
Wikipedia wrote:Alternatively, imaginary roots are obfuscated in the following:
sqrt(-1) = (-1)^(2/4) = ((-1)^2)^(1/4) = 1^(1/4) = 1
The error here lies in the last equality, where we are ignoring the other fourth roots of 1, which are −1, i and −i (where i is the imaginary unit). Seeing as we have squared our figure and then taken roots, we cannot always assume that all the roots will be correct. So the correct fourth are i and −i, which are the imaginary numbers defined to be sqrt(-1).
(Expressions converted to plaintext to make them easier to read)
$1 = 100¢ = (10¢)^2 = ($0.10)^2 = $0.01 = 1¢ [1]
Always check your units or you will have no money!
User avatar
testtubegames
Site Admin
Posts: 1148
Joined: Mon Nov 19, 2012 7:54 pm

Re: A small math problem I found myself with...

Post by testtubegames »

Yeah, I think ARP has found a good explanation of this one. It looks like it's one of those cases of needing to be extra careful with particular operations.

(Kind of how like you can't just square both sides of an equation and then act like you didn't... since then x = 1 becomes x^2 = 1, and if we solve we find x = +-1... which is clearly not where we started)
User avatar
robly18
Posts: 413
Joined: Tue Jun 04, 2013 2:03 pm

Re: A small math problem I found myself with...

Post by robly18 »

Hm. Things like these make me feel like there should be a sign for regular square roots and then a sign for only the positive result and another for only the negative, because using only context to figure which one we need to use sounds a bit... Meh.
Convincing people that 0.9999... = 1 since 2012
exfret
Posts: 585
Joined: Sun Jul 28, 2013 8:40 pm

Re: A small math problem I found myself with...

Post by exfret »

Well, you found out that (1^4)^(1/4)={1,-1,i,-i}, which IS true. {1,-1,i,-1}=1 as well. The problem is when you just take out one of those solutions without testing if they are extraneous. When you raise to the power of 4, you lose information, because the result can be traced back to 4 numbers, even though you started with one. The power of a fourth you placed in there is trying to restore that information by giving you the four possible numbers that the number you gave it could have once been. You can't just assume that each of those numbers was the starting number, so don't blame it on the fourth root, it has no way of knowing that you raised 1 and not some other number to the power of four! You have to be the one who figures out which of those solutions you need. Also, using the radical only returns positive roots. If you want both positive and negative, use +/-sqrt or raise it to the power of 1/2.
Nobody ever notices my signature. ):
Post Reply