Timeline for How do we find the numbers?
Current License: CC BY-SA 4.0
28 events
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| Sep 24 at 22:31 | history | edited | Nautilus | CC BY-SA 4.0 |
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| Sep 24 at 19:28 | comment | added | Nautilus | Had to give up on my initial answer. I tried hard with it because I wanted to generalize it for another question here. This one is far simpler. Maybe even too simple. | |
| Sep 24 at 19:27 | history | undeleted | Nautilus | ||
| Sep 24 at 19:27 | history | edited | Nautilus | CC BY-SA 4.0 |
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| Sep 24 at 17:50 | history | deleted | Nautilus | via Vote | |
| Sep 24 at 17:49 | comment | added | Nautilus | I understand I'm wrong, but I thought I was so close. This proof will take a long time to complete, so I have to delete it. | |
| Sep 24 at 17:33 | comment | added | tehtmi | I also don't get it. Specific objection: it seems with regard to the colorings you only use "any coloring of a grid width>4 has a monocolor rectangle" -- the argument doesn't seem to care about the number of total colors as long as this property holds. We can satisfy a stronger property: If we color e.g. 1, 3, and for k>4 k*10^k+1 and k*10^k+3 red, and every other number gets its own unique color -- there is a monocolor (red) rectangle for every choice of width>4 -- but the monocolor a,b,a+b clearly does not hold (every sum of reds has an even one's digit). Why doesn't your argument work here? | |
| Sep 24 at 14:39 | history | edited | Nautilus | CC BY-SA 4.0 |
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| Sep 24 at 13:12 | comment | added | Tim Seifert | I don't understand this, maybe I'm very dense. The number $a$ found in the topleft corner of a rectangle certainly depends on the given colouring, does it not? But later on, in the attempted proof by contradiction you seem to treat it as an independent parameter to shift the colouring by. So what is $a$? (And what is $n$, for that matter?) | |
| Sep 24 at 12:47 | comment | added | Nautilus | Because if there was at least one coloring (independently from the one on our grid) where all of $x, y$ and $x+y$ must have two colors between them, we could take a possible initial coloring, add the $a$ to all the numbers matched with a color, and put them on our grid. Any distinct number over $a$ would be addends that satisfy this condition on our grid. But we can have certain $n^2$ different numbers of the same color that are orthogonally aligned, so the colors would be the same, eliminating the possibility of never having $x, y$ and $x+y$ be of the same color. | |
| Sep 24 at 12:32 | comment | added | Nautilus | "There are plenty colourings of such a grid in which the top vertex is not part of a monochromatic rectangle." True, but let the top left vertex of a same colored vertex be a certain integer ($a$). It can be any positive integer, or even zero. Even if this constant isn't zero, it can be proven. | |
| Sep 24 at 12:09 | comment | added | Tim Seifert | Tbh I don't really follow the argument. Why should the top vertex being zero have no influence on the outcome? There are plenty colourings of such a grid in which the top vertex is not part of a monochromatic rectangle. I don't see how "shifting the colouring" could adress that, since you need to show the claim for every colouring, not just everything up to translation | |
| Sep 24 at 11:09 | history | edited | Nautilus | CC BY-SA 4.0 |
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| Sep 24 at 10:52 | comment | added | Nautilus | Hope now things are cleared up. | |
| Sep 24 at 10:51 | history | edited | Nautilus | CC BY-SA 4.0 |
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| Sep 20 at 10:08 | history | edited | Nautilus | CC BY-SA 4.0 |
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| Sep 20 at 9:53 | comment | added | Nautilus | The grid has a finite length of $n>4$ but an infinite height. | |
| Sep 20 at 9:51 | comment | added | Tim Seifert | Huh? But how, then, does the next row "continue from there"? | |
| Sep 20 at 9:49 | comment | added | Nautilus | The whole grid is meant to be colored though. The $0$ is just t the beginning or somewhere in the middle (that can correspond to the top left corner of any rectangle formed in it), and everything that comes before (if any) is a negative integer. | |
| Sep 20 at 9:43 | history | edited | Nautilus | CC BY-SA 4.0 |
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| Sep 20 at 9:42 | comment | added | Tim Seifert | Ah, I thought you meant to colour the whole grid. But then, how do you decide on an adquate width of the rectangle? As stated, it now sounds like this should work for any width, but that is clearly false (for just length 2 rows, this would amount to always being able to find an integer $x$ such that $1,x$ and $1+x$ are always the same colour, which is not necessarily possible). | |
| Sep 20 at 9:23 | comment | added | Nautilus | It doesn't matter where the $0$ is as long as it's at the top left corner of a rectangle., because $0$ or negative integers don't change the $x, y, x+y$ relationship. | |
| Sep 20 at 9:16 | comment | added | Nautilus | Each row is a sequence of consecutive integer, and the next row continues from there, so for a rectangle, the difference between two vertices on the same row or column is equal to the one between the remaining two. If the top left vertex equals $0$, the rest will be $x, y, x+y$. | |
| Sep 20 at 8:55 | comment | added | Tim Seifert | Why should the vertices of such a rectangle (containg 0, say) necessarily be of the form $x,y,x+y$? | |
| Sep 19 at 10:41 | history | edited | Nautilus | CC BY-SA 4.0 |
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| Sep 18 at 19:27 | history | edited | Nautilus | CC BY-SA 4.0 |
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| Sep 18 at 17:09 | history | edited | Nautilus | CC BY-SA 4.0 |
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| Sep 18 at 15:56 | history | answered | Nautilus | CC BY-SA 4.0 |