Author |
Topic: matrix (Read 1929 times) |
|
temporary
Full Member
Posts: 255
|
Construct a matrix where all numbers in it add up to 0, but with no negatives, no imaginary part greater than 0, and no 0s. This is not impossible, nor does it use infinity or something like that. Hint:there are infinite solutions
|
| « Last Edit: Apr 16th, 2008, 8:23pm by temporary » |
 IP Logged |
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.
|
|
|
Obob
Senior Riddler
Gender: 
Posts: 489
|
 |
Re: matrix
« Reply #1 on: Apr 9th, 2008, 9:50am » |
 Quote  Modify
|
Yeah, no...
You are asking for a series of positive numbers summing to zero. Even in nonstandard interpretations of "number," this is not possible, because those interpretations do not also have an ordering.
|
|
IP Logged |
|
|
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Re: matrix
« Reply #2 on: Apr 9th, 2008, 12:55pm » |
Quote Modify
|
[ 2 1 ]
[ 1 2 ]
mod 3 works.
|
|
IP Logged |
Regards,
Michael Dagg
|
|
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Re: matrix
« Reply #3 on: Apr 9th, 2008, 1:19pm » |
Quote Modify
|
Since you might have also meant the sums along
the diagnoals then
[3 1 2]
[1 2 3]
[2 3 1]
mod 6 works.
Anyone got a general form?
|
|
IP Logged |
Regards,
Michael Dagg
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: matrix
« Reply #4 on: Apr 9th, 2008, 2:47pm » |
Quote Modify
|
But is 2 > 0 mod 3?
|
|
IP Logged |
|
|
|
temporary
Full Member
Posts: 255
|
|
Re: matrix
« Reply #5 on: Apr 11th, 2008, 11:18pm » |
Quote Modify
|
I have no idea what all this "mod" stuff is, but it is not part of the answer.
|
|
IP Logged |
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: matrix
« Reply #6 on: Apr 11th, 2008, 11:25pm » |
Quote Modify
|
Is it not part of the answer because there is no mod stuff in the answer, or is it not part of the answer because there is no answer?
I suspect both are true...
|
|
IP Logged |
|
|
|
TJMann
Newbie
Posts: 18
|
|
Re: matrix
« Reply #7 on: Apr 12th, 2008, 7:52am » |
Quote Modify
|
I get zero with both matrices. What makes this possible is the integers mod 3 and mod 6 wrap around. In mod 3, 2+1 = 3 = 0, so 2+1 + 1+2 = 6 = 0. In mod 6, 3+1+2 = 6, so (3+1+2) + (1+2+3) + (2+3+1) = 18 = 0.
|
| « Last Edit: Apr 12th, 2008, 7:57am by TJMann » |
IP Logged |
|
|
|
pex
Uberpuzzler
Gender:
Posts: 880
|
|
Re: matrix
« Reply #8 on: Apr 12th, 2008, 8:14am » |
Quote Modify
|
It's a bit of a stretch, but the following 2x1 matrix seems to fit the criteria:
[ 1-i ]
[ i-1 ]
- All numbers add up to zero: obvious.
- No negatives: nonreal complex numbers are neither "positive" nor "negative".
- No imaginary numbers: the numbers in this matrix are complex, but not imaginary, because the real parts are nonzero.
- No zeros: obvious.
- No "infinity or something like that": obvious.
|
|
IP Logged |
|
|
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Re: matrix
« Reply #9 on: Apr 15th, 2008, 10:42pm » |
Quote Modify
|
> But is 2 > 0 mod 3?
If you are asking if there is order in the
ring Z_3 then the answer is no. This in no
way invalidates addition modulo 3, including
multiplication; in fact, it is the other way around.
|
| « Last Edit: Apr 15th, 2008, 10:44pm by Michael Dagg » |
IP Logged |
Regards,
Michael Dagg
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: matrix
« Reply #10 on: Apr 16th, 2008, 8:10am » |
Quote Modify
|
Yes, I slightly misread the original post. By reading "no negatives, no imaginary numbers, and no 0s" I automatically concluded that meant "positive numbers." You could either interpret imaginary numbers as "purely imaginary" numbers, as pex did, or do some modular arithmetic example and just say that the elements are not negative or imaginary, since those terms don't even make sense.
At any rate, though, this riddle is mostly a wad of nonsense. At least the intended solution is.
|
|
IP Logged |
|
|
|
temporary
Full Member
Posts: 255
|
|
Re: matrix
« Reply #11 on: Apr 16th, 2008, 8:17pm » |
Quote Modify
|
on Apr 12th, 2008, 8:14am, pex wrote:It's a bit of a stretch, but the following 2x1 matrix seems to fit the criteria:
[ 1-i ]
[ i-1 ]
- All numbers add up to zero: obvious.
- No negatives: nonreal complex numbers are neither "positive" nor "negative".
- No imaginary numbers: the numbers in this matrix are complex, but not imaginary, because the real parts are nonzero.
- No zeros: obvious.
- No "infinity or something like that": obvious. |
|
I might have misphrased it, but you did kind of use negative numbers anyway. I'll fix the misphrase to avoid confusing anyone.
|
| « Last Edit: Apr 16th, 2008, 8:29pm by temporary » |
IP Logged |
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.
|
|
|
pex
Uberpuzzler
Gender:
Posts: 880
|
|
Re: matrix
« Reply #12 on: Apr 17th, 2008, 1:27am » |
Quote Modify
|
Okay... so we either need a nonstandard definition of addition, or a different set of numbers than real or complex numbers.
For example, in the 10-adic numbers, ...9999999 + 1 = 0. (The representation ...9999999 is infinite, but the number it represents isn't - does that count?)
|
|
IP Logged |
|
|
|
pex
Uberpuzzler
Gender:
Posts: 880
|
|
Re: matrix
« Reply #13 on: Apr 17th, 2008, 4:34am » |
Quote Modify
|
Just a thought that crossed my mind: is there a sensible way to define matrices of dimensions m x n, with m or n equal to zero? If yes, one might argue that the sum of its elements (an empty set) equals zero, and matrices of dimensions 0x0, 0x1, 1x0, 0x2, 2x0, 0x3, 3x0, ... would work.
|
|
IP Logged |
|
|
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Re: matrix
« Reply #14 on: Apr 17th, 2008, 3:24pm » |
Quote Modify
|
Of course I would have hardly guessed that
those kind of matrices would be involved in
solution a here, but some matrices like these can
pop up when finite groups are wrote using Smith
Normal Form, in particular, subgroups of
G = (Z/p^2 Z) x (Z/pZ) for example.
|
|
IP Logged |
Regards,
Michael Dagg
|
|
|
TJMann
Newbie
Posts: 18
|
|
Re: matrix
« Reply #15 on: Apr 17th, 2008, 5:37pm » |
Quote Modify
|
Being curious but knowing a just little about groups, how do you make the product (Z/p^2Z) X (Z/pZ)? I looked my algebra book about it but I did not see an explicit way to actually do it or should I think that it is just a "product" through the use of notation?
|
|
IP Logged |
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: matrix
« Reply #16 on: Apr 17th, 2008, 7:58pm » |
Quote Modify
|
Given two groups G and H, you form the product G X H as the set of all pairs (g,h), where g is in G and h is in H. You multiply pairs (g1,h1) and (g2,h2) by the rule (g1,h1) * (g2,h2) = (g1 * g2,h1 * h2). So this is totally explicit, but it is also essentially just taking the product through the use of notation.
|
|
IP Logged |
|
|
|
TJMann
Newbie
Posts: 18
|
|
Re: matrix
« Reply #17 on: Apr 18th, 2008, 5:05pm » |
Quote Modify
|
Thx. I did know that and it seems to fit the idea that I thinking because if we have p = 2 then we have the product (Z/4Z) X (Z/2Z) which is more specific that the general form wrote but aren't both Z/4Z and Z/2Z finite fields? One is of length 4 and the other is of length 2 and now that means the product would effectively be a 4 x 2 - tuple? Groups that I know about don't go that far as to require what I suppose are called lists.
|
|
IP Logged |
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: matrix
« Reply #18 on: Apr 19th, 2008, 7:46pm » |
Quote Modify
|
Z/4Z is not a field. 2 has no multiplicative inverse in Z/4Z. But you can view Z/4Z x Z/2Z as a two-tuple of numbers, one coming from the set {0,1,2,3} and the other coming from {0,1}. The group Z/4Z x Z/2Z does in fact have 8 elements.
|
|
IP Logged |
|
|
|
temporary
Full Member
Posts: 255
|
|
Re: matrix
« Reply #19 on: Apr 22nd, 2008, 5:11pm » |
Quote Modify
|
on Apr 17th, 2008, 4:34am, pex wrote:| Just a thought that crossed my mind: is there a sensible way to define matrices of dimensions m x n, with m or n equal to zero? If yes, one might argue that the sum of its elements (an empty set) equals zero, and matrices of dimensions 0x0, 0x1, 1x0, 0x2, 2x0, 0x3, 3x0, ... would work. |
|
Yep. That pretty much hits the nail on the head.
|
|
IP Logged |
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.
|
|
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Re: matrix
« Reply #20 on: Apr 22nd, 2008, 7:02pm » |
Quote Modify
|
> Yep. That pretty much hits the nail on the head.
That's stated lightly -- got a proof?
|
|
IP Logged |
Regards,
Michael Dagg
|
|
|
temporary
Full Member
Posts: 255
|
|
Re: matrix
« Reply #21 on: Apr 24th, 2008, 6:53pm » |
Quote Modify
|
Isn't it obvious and self-proven. A matrix with no numbers has to add up to 0. And I am prepared to bet 1ud8f7g6g6v7v7v7 dollars that it does. And the only time I bet something I don't have is when it doesn't matter because I win.
|
|
IP Logged |
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.
|
|
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Re: matrix
« Reply #22 on: Apr 24th, 2008, 7:11pm » |
Quote Modify
|
That is not a solution to the problem you posted.
It is technically clear, do you know why?
|
|
IP Logged |
Regards,
Michael Dagg
|
|
|
temporary
Full Member
Posts: 255
|
|
Re: matrix
« Reply #23 on: Apr 24th, 2008, 7:26pm » |
Quote Modify
|
There is an answer. A matrix with no numbers(like 0x3, 1x0, 9x0, 0x(pi+e), etc) Where have you been for the last several posts.
|
|
IP Logged |
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.
|
|
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Re: matrix
« Reply #24 on: Apr 24th, 2008, 7:42pm » |
Quote Modify
|
A matrix with no numbers is not a solution to the
problem you posted. It is true that a sum on an
empty set is zero but it does not satisfy the
hypothesis of your problem.
|
|
IP Logged |
Regards,
Michael Dagg
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print