Colouring Numbers - Numberphile
video description
But how big is F(m, n? It looks like one of those combinatorical/Ramsey theoretic functions which grow insanely fast, and are the basis for Graham's number and TREE(3. Already, F(3, 2) looked pretty decently big in terms of the number of dots he was drawing out. My guess is that F(10, 10) must be unimaginably huge. Bigger than a googolplex?
Date: 2022-04-09
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Comments and reviews: 9
87vortex87
Ok, I love these video's but I dislike this one. Because every number already has a colour for me, also days of the week, months of the year, decades, centuries, letter of the alphabet, words, etc. Etc. Only thing I think is: -no, zero is gray, one is white, two is light pink, three normal light blue, four is yellow, five is orange, six is bordeaux, seven is dark green, eight is dark blue, nine orange-ish towards red. Etc. Never bothered me except in primary school where we did multiplying based on coloured sheets, where the colours made no sense with the numbers what so ever. That was very very frustrating.
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Ok, I love these video's but I dislike this one. Because every number already has a colour for me, also days of the week, months of the year, decades, centuries, letter of the alphabet, words, etc. Etc. Only thing I think is: -no, zero is gray, one is white, two is light pink, three normal light blue, four is yellow, five is orange, six is bordeaux, seven is dark green, eight is dark blue, nine orange-ish towards red. Etc. Never bothered me except in primary school where we did multiplying based on coloured sheets, where the colours made no sense with the numbers what so ever. That was very very frustrating.
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Rosie
5: 56 -these 6 configurations- where a configuration mapped to 5 integers in -ascending- order, and the same configuration mapped to 5 integers in -descending- order, count as 1. If you counted them as different, you have 12 configs. Another point; your 5th configuration is the same as your 1st, but shifted rightwards by 1. If they count as different, why not also consider the same, but shifted rightwards by another 1?
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5: 56 -these 6 configurations- where a configuration mapped to 5 integers in -ascending- order, and the same configuration mapped to 5 integers in -descending- order, count as 1. If you counted them as different, you have 12 configs. Another point; your 5th configuration is the same as your 1st, but shifted rightwards by 1. If they count as different, why not also consider the same, but shifted rightwards by another 1?
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Noah
Honestly, I don't think this is all that surprising. On an infinite line with finite colours, any and all sets would necessarily eventually exist. I think the generations are gaining a better understanding of infinity as time goes on, so while the proof may be interesting in of itself, it isn't necessary for the average viewer to comprehend, and may even confuse them more than aid in their understanding
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Honestly, I don't think this is all that surprising. On an infinite line with finite colours, any and all sets would necessarily eventually exist. I think the generations are gaining a better understanding of infinity as time goes on, so while the proof may be interesting in of itself, it isn't necessary for the average viewer to comprehend, and may even confuse them more than aid in their understanding
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education
So for any amount of colors (c, and for any sequence length (l, you can have a limit to the number of integers you can color before you run into a sequence of the correct length (x. This proof provides an upper bound for x, for any c and l. Has any progress been made on reducing that upper bound, or on adding a lower bound?
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So for any amount of colors (c, and for any sequence length (l, you can have a limit to the number of integers you can color before you run into a sequence of the correct length (x. This proof provides an upper bound for x, for any c and l. Has any progress been made on reducing that upper bound, or on adding a lower bound?
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Mike
Maybe I am missing something but it seems a rather intuitive conclusion. You have a finite number of colors which to color an infinite number of positions. Eventually you will cover all the possible permutations of colorings and eventually you will repeat all those permutations any number of times.
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Maybe I am missing something but it seems a rather intuitive conclusion. You have a finite number of colors which to color an infinite number of positions. Eventually you will cover all the possible permutations of colorings and eventually you will repeat all those permutations any number of times.
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Gregory
If there is ever practical application for anything being discussed in videos can you please make note of it. The more I watch the videos the more I start to think that mathematicians run out of practical solutions to solve and just make up problems so they can solve it just for the sake of it.
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If there is ever practical application for anything being discussed in videos can you please make note of it. The more I watch the videos the more I start to think that mathematicians run out of practical solutions to solve and just make up problems so they can solve it just for the sake of it.
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nivolord
Correction for 2: 54 for the video editor.
The 76th does not have to be a repeat. We only know that there must be at least two numbers smaller or equal to 76 which are similar. Take for example the first 75 numbers to be red, then the 76th number to be blue.
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Correction for 2: 54 for the video editor.
The 76th does not have to be a repeat. We only know that there must be at least two numbers smaller or equal to 76 which are similar. Take for example the first 75 numbers to be red, then the 76th number to be blue.
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JNCressey
0: 36 he didn't mention colours in the constraints for the progression.
-. with alternating colours-?
-. with the same colour-?
-. which touches every colour at least once-?
-. which touches every colour infinitely many times-?
Who knows.
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0: 36 he didn't mention colours in the constraints for the progression.
-. with alternating colours-?
-. with the same colour-?
-. which touches every colour at least once-?
-. which touches every colour infinitely many times-?
Who knows.
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FlutterBug
my understanding of this is, since we are using all of the integers, all of the multiples of those integers are in there too, so no matter what multiple you are trying to use, it will always come up, because it's just. inherently there somewhere
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my understanding of this is, since we are using all of the integers, all of the multiples of those integers are in there too, so no matter what multiple you are trying to use, it will always come up, because it's just. inherently there somewhere
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