math problem for you all
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Date: May 24th, 2015 12:16 PM Author: lime rehab stain
Consider the usual x-y coordinate plane. Assign every point (ordered pair of real numbers) the letter A,B, or C. Prove there exist two points with the same letter that are 1 unit apart.
This problem has the advantage of being both easier and more interesting than the dumb combinatorics problems that one dude was posting.
(http://www.autoadmit.com/thread.php?thread_id=2888104&forum_id=2#27954385) |
Date: May 24th, 2015 7:07 PM Author: Diverse federal indian lodge useless brakes
1. What does "every point" mean? Do you mean "every coordinate pair with integral x and y", or do you mean "every coordinate pair with real x and y"? If the latter, LJL at this joke of a question. If the former, read on for a disproof.
2. Assuming you mean that, then the statement is false. If you define the set of A points as {(x,y)|(x+y)mod2=0} and B points as {(x,y)|(x+y)mod2=1} where x and y are integers, then there is no A point within 1 unit of another A, and there is no B unit within another B, and there is no C unit within another C since there are no C units.
Proof that there is no A point within 1 unit of another A:
Choose two points in A, n and m. n = (x1,y1), m=(x2,y2).
The distance between m and n = sqrt((x1-x2)^2+(y1-y2)^2)).
Assume these two points are within 1 unit of each other. Because by postulation x1,y1,x2,y2 are all integers, in order to be within 1 unit of each other, either x1=x2 and |y1-y2|=1 or y1=y2 and |x1-x2|=1. The proof of this is left as an exercise for the reader, but it's pretty fucking obvious, merely tedious.
Now we defined A such that (x+y)mod2=0, so let's look at the first point, W/OLOG:
(x1+y1)mod2=0, and now do some substitution
(x2+y1)mod2=0, and by a theorem of absolute value addition I don't wnat to prove here,
(x2+(y2+1))mod2=0
x2mod2 + y2mod2 + 1mod2 = 0
By the same definition of A, we know
(x2+y2)mod2 = 0
x2mod2 + y2mod2 = 0
So combine those,
x2mod2 + y2mod2 + 1mod2 = x2mod2 + y2mod2
simplify
1mod2 = 0
1 = 0
This is a contradiction. The proof regarding two points in B is almost exactly the same and is left as an exercise for the reader.
Accordingly, the theorem has been disproved.
In fact, you can generalize the opposite statement as a theorem in N dimensions:
For any set of N-dimensional points with integral components, you can assign these points with N different labels such that no two points with the same label have a distance <= 1
(http://www.autoadmit.com/thread.php?thread_id=2888104&forum_id=2#27956314) |
Date: May 24th, 2015 11:30 PM Author: contagious hideous area factory reset button
am i a faggot if i say the regular name of this problem?
(http://www.autoadmit.com/thread.php?thread_id=2888104&forum_id=2#27957679)
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Date: May 25th, 2015 2:11 AM Author: Hairless Cruise Ship Dragon
They can touch diagonally (sqrt 2???)? This is so stupid
Put A at 0,0. Theres 8 spots away that 0,0 can touch. Theres no way you can B and C in those spots without any touching eachother
Did i fuck up
(http://www.autoadmit.com/thread.php?thread_id=2888104&forum_id=2#27958522)
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