Math related logic problem (nested roots)

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[img]http://www.bgoncalves.com/cgi-bin/mimetex/mimetex.cgi?\mathrm%20Looking\%20for\%20all\%20integer\%20values\%20of\%20m\%20where\%20c\%20is\%20also\%20an\%20integer%20\mathcal%20\\c%20=%20\sqrt%20{m+sqrt{m+\sqrt%20{m+\sqrt{m+...}}}}%20\\c^2%20=%20m%20+%20sqrt{m+\sqrt%20{m+\sqrt{m+...}}}%20\\c^2%20=%20m%20+%20c%20\\c(c-1)%20=%20m%20\\m%20=%20\{0,%202,%206,%2012,%2020,%2030,%2042,\%20...%20\}[/img]

My assumption is that after squaring both sides the remaining infinite series of sqrt(m+...) is still equal to c. Do you agree with this solution?

to start with
for any integer c that's greater than 0:
m=c(c-1)
the numbers for m you have already are m for c= {1,2,3,4,5,6,7, ...}

edit maybe pending because I can't instantly see how the roots also work yet.

edit1:I agree with this but I don't know how it works yet:

Yes.

Last edited by ValMikeSmith on 29 Nov 2008, 2:23 am, edited 1 time in total.

Its just another aspect of binomial expansion. And its not for any integer c, its for any positive integer c.

The thing that bugs me about this is that if m=0 then c=0 or c=1. This would mean that...

[img]http://www.bgoncalves.com/cgi-bin/mimetex/mimetex.cgi?%20\Large%201%20=%20\sqrt%20{0%20+%20\sqrt%20{0%20+%20\sqrt%20{0%20+%20...%20}}}%20\\and\\0%20=%20\sqrt%20{0%20+%20\sqrt%20{0%20+%20\sqrt%20{0%20+%20...%20}}}[/img]

Also since we know that if my logic in the first post is correct then when c=3 then m=6. If we square it again we get...

[img]http://www.bgoncalves.com/cgi-bin/mimetex/mimetex.cgi?%20\Large%20(c^2)^2%20=%20(m+c)^2\\c^4=m^2+2mc+c^2\\c^4=m^2+2mc+c^2\\\mathrm%20\%20\%20\%20%20Substitute\%20\mathcal%20m=6,%20c=3\\%203^4=6^2+2*6*3+3^2\\81=36+36+9\\81=81[/img]

EDIT: Which seems totally reasonable now that I didnt make a stupid mistake.

Last edited by Death_of_Pathos on 29 Nov 2008, 2:53 am, edited 2 times in total.

You're Right.
This is one of those funny maths that have an exception.
I once saw one that proofed that 2=3, and it was because 2 caused a division by zero
and all other numbers divided by 1. Here's one like it:
http://www.jimloy.com/algebra/two.htm

Edited my second post.

Yes but that is really just saying that 2*0 = 1*0, which I have no problem with. Their problem is that you CANT divide both sides by 0. That does NOT prove that 2=1 because the last step is impossible. Does not comply with the transformation rules of this grammar.

- - -

Should we agree that c must be a whole number then? Because when c=0 you get the statement 1=0. So c is defined as all whole numbers.

EDIT: Removed some stuff that was based on me pulling a number out of my ass.

Maybe. I'm not sure it really says 0=1. Does it?
I just assume that c can't be 0 because the m set doesn't have two 0's.
I don't really get a 1 from adding square roots of 0 like you did.
There are many other math functions that give the same answer for different numbers.

edit:
It seems that 0 is ok and 1 is not . so c = {0,2,3,4,5,6,7...}
1 does seem to fail because you can't get it from square roots of 0.

If c=1, then c(c-1)=0=m. There is nothing wrong with that, but the worrisome part comes when you make c=0 because 0(0-1)=1(1-1)=0=m.

Heh, I just realized I was falling victim to the same fallacy. 0(0-1)=1(1-1) is true, but that does not necessitate 1=0 because there is no way to get c=c without diving by zero. I wonder what happens when we use a limit to try to find c for m=0.

I'm assuming that the procedure falls apart because its inverse isn't a function.

Thanks for your input, reminding me that dividing by quantity k, if k=0, does not work. It should be obvious, but I think the fact that I treat numbers as a special case single element group, instead of groups as a special case representation of single numbers caused me to overlook it. This was probably caused by me working with groups more often then individual numbers.

I am now confident in my solution.

What interests me still is the curiosity of {root 0 + root 0 + ...} equaling 1 and 0. Is it ever explicitly defined as strictly 0, or are we allowed to express 1 in the same manner, without saying that 1=0?

I think that since square root of 0 can't equal 1, we can't use 1.
(I edited my last post.)

I see your point, but I don't think that that necessarily excludes it. Every rule we used to get there holds true.

I say that, counter intuitive as it may seem, unless we define [img]http://www.bgoncalves.com/cgi-bin/mimetex/mimetex.cgi?\Large0=\sum%20_{k=1}^{\infty}%200[/img] it is still true. (and defining the sum of infinitely many 0's to 0 seems tempting enough)

Ive not had formal schooling at this level of math. Can you provide any insight?

EDIT:

Ahh. The sum of n many k is n*k. In this case since n is infinity we have infinity*0. Looks like [img]http://www.bgoncalves.com/cgi-bin/mimetex/mimetex.cgi?%20\Large%200%20\neq%20%20\sum%20_{k=1}^{\infty}%200%20=%20\lim_{x\to0^+}%20x[/img] because of the proof here.

So you were right. c={0,2,3,4,5,6,7, ... }

EDIT2: Heh, apparently when c=1 m= the golden ratio.

Deleted

Last edited by mystyc on 29 Nov 2008, 2:24 pm, edited 1 time in total.

The problem, as posed to me, restricted m (and thus c) to /positive/ integers. I did neglect that part.

[img]http://www.bgoncalves.com/cgi-bin/mimetex/mimetex.cgi?\mathrm%20Looking\%20for\%20all\%20positive\%20integer\%20values\%20of\%20m\%20where\%20c\%20is\%20also\%20an\%20integer%20\mathcal%20\\c%20=%20\sqrt%20{m+sqrt{m+\sqrt%20{m+\sqrt{m+...}}}}%20\\c^2%20=%20m%20+%20sqrt{m+\sqrt%20{m+\sqrt{m+...}}}%20\\c^2%20=%20m%20+%20c%20\\c(c-1)%20=%20m%20\\m%20=%20\{0,%202,%206,%2012,%2020,%2030,%2042,\%20...%20\}[/img]

After you square both sides you are not left with

[img]http://www.bgoncalves.com/cgi-bin/mimetex/mimetex.cgi?%20\Large%20c^2=m\%20\pm\%20\sqrt{m+\sqrt{m+...}}[/img]

because that behavior only happens when you take the square root. Consider the quantity j which is equal to the square root of 3. If you were to square j your answer is not +/- 3, but simply 3. If you were to take the square root of k, a quantity which is equal to j squared, then you would have +/- j (assuming both are within the domain of j).

How do you claim c=(-1),m=0 ? Because c(c-1) of -1 is not 0 but instead (-1)(-2)=2.

There are no +/- signs in the original problem. When you square the right hand side of the equation you get m + (series). You in no way get m +/- (series).

Consider this:

[img]http://www.bgoncalves.com/cgi-bin/mimetex/mimetex.cgi?%20\Large%20x=\sqrt{7+\sqrt{4}}\\x^2=7+\sqrt{4}\\x^2=7+2\\x^2=9[/img]

[img]http://www.bgoncalves.com/cgi-bin/mimetex/mimetex.cgi?%20\Large%20x=\sqrt{7+\sqrt{4}}\\x^2=7-\sqrt{4}\\x^2=7-2\\x^2=5[/img]

[img]http://www.bgoncalves.com/cgi-bin/mimetex/mimetex.cgi?%20\Large%20\sqrt{7+\sqrt{4}}\\\sqrt{7+2}\\\sqrt{9}\\3[/img]

No not really... since we are looking for all m where both m and c are positive integers, and since c increases less rapidly then m, we can simply define m to be the result every non 1 positive integer value of c in the equation c(c-1), which is exactly what I have shown here.

The fact that c=(-|k|) and c=|k|+1 have the same m value is cause to discard negative values of -|k|, not to add in a plus or minus after the actual squaring has been done.

- - -

The slick math pictures are from a free LaTeX rendering service that I found online. You just type in what you want, using LaTeX math notation, and bam, you get this cool image you can link to. Its pretty easy once you get the hang of it.

- - -

PS: Not intending to be rude but simply to alert you, adding QED on inaccurate forum posts comes of as a bit pretentiously moot. I understand that it has no such negative intrinsic qualities, and I have always used it as a declaration of relief after a tedious exercise, but adding it to the end of a forum post attempting to disprove someone could be easily viewed as less then respectful, regardless of whether or not your post was based off of an honest mistake.

As I said before I am not trying to be rude, but to help you in good faith, as you have done by trying to add insight into this problem.

I'm afraid you have missed the point, mystyc. The square root notation is usually taken to mean the positive square root. Adding in an infinite number of undecided signs isn't really going to make the problem very tractable.

==========

The root of the problem, Death_of_Pathos, as I see it, is that you have made no attempt to define define what convergence, on the infinitely nested set of square root signs, might mean to you. In fact, that's the end of it, really.

The case with c=1 and m=0 seems to be a perfectly good solution.



I think you can easily see that, if m=0 and n=1, there is no great doubt that c(m,n,j)=1 for all j, so c(m,n)=1, presumable.

However, we are at liberty to choose "n" as any value, I guess, and its pretty clear that m=0, n=0 leads to c(m,n,j)=0 for all j, so c(m,n)=0.

In fact, I'd suggest that all positive, non-zero values of "n", tend, in the limit, to result in "c=1". Only choosing "n=0" makes the limit zero.

Basically, the function "c(m)" can't be particularly well defined, as it is susceptible to knowing "n".

In parting, the way I have chosen to "define" c(m,n) is still a little suspect. I can't say I like this sort of limit definition, as it relies on the function concerned being "nice", in some obscure way. I'm sure I could toy with alternate ways of (loosely) defining the game.

=============
Note that the image above was produced using Laeqed - LaTeX equation editor, so you can edit it.

Cool. Ill check out that tool in a sec.

c(m,n) only converges to 1 if m is 0 and n is non-zero.

But if n=m, then c(m,n) = c(m,m). And since, the addition of n was spontaneous (there is no mention of it in the problem as I presented here or as I was given) we can only assume that n must equal m.

I wholly agree with your interpretation that if this problem was derived from some other source so that we could know that n != 0. But we don't, so we can't.

- - -

EDIT: Brings up a thought. What if c(0,-1,INF ) ? What happens when you take the square root of i?

Deleted.
You don't deserve to read it.
There's no point in communicating with you on this level.

Last edited by mystyc on 30 Nov 2008, 2:37 am, edited 1 time in total.

As I said when I adressed you adding QED to your post, it was done in good faith trying to let you see how I could have taken it wrong. I realize that you replied to a thread of mine trying to help. I appreciate that.

But this is one of those sublimely complicated problems. The math will become complicated. Could you please specifically state where you see a problem with our "unnecessarily complex" math?

I would again like to draw your attention to the fact that the below is clearly inaccurate because x=3:

[img]http://www.bgoncalves.com/cgi-bin/mimetex/mimetex.cgi?%20\Large%20x=\sqrt{7+\sqrt{4}}\\x^2=7-\sqrt{4}\\x^2=7-2\\x^2=5[/img]

Deleted.
You don't deserve to read it.
There's no point in communicating with you on this level.

Last edited by mystyc on 30 Nov 2008, 2:34 am, edited 2 times in total.