Yes, of course. I think that things like this should be made clear, however, for the benefit of thick people like me. We have rights too, you know. I have a very good excuse for being superdim tonight though so you'll have to let me off.(Original post by joostan)
The point is that the Riemann sum converges to the integral in the limit as , so the limit exists iff the integral exists.

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 03042016 21:32

Smaug123
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 03042016 23:00
(Original post by atsruser)
Yes, of course. I think that things like this should be made clear, however, for the benefit of thick people like me. We have rights too, you know. I have a very good excuse for being superdim tonight though so you'll have to let me off. 
Lord of the Flies
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 04042016 03:32
C'mon people, what has happened to this thread. Problem 570 from earlier requires two skills: knowing what a rectangle is, and knowing what an integer is.
Anyway, more!
Problem 577
Prove/disprove that there is a family of disjoint subsets of the unit circle such that one can get from any member to any other through a rotation, and such that is the unit circle.
(this however is not trivial) 
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 04042016 11:57
(Original post by Lord of the Flies)
C'mon people, what has happened to this thread. 
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 04042016 12:24
(Original post by Lord of the Flies)
The harder questions aren't gathering very much attention it seems, so here's a nice dinnertable problem!
Problem 570
If a rectangle can be subdivided into smaller rectangles, each having at least one side of integer length, must the initial rectangle also have at least one side of integer length?
Spoiler:Show
It can't be done. Say a rectangle has a *good* subdivision if it has the subdivision you require. Let R be a rectangle with noninteger sides, which has a good subdivision; let R be minimal (in perimeter) with this property.
I'll do this in large stages, but the first claim is not necessary.
Claim: if the topleft subrectangle in a good subdivision is integerbyinteger, then we can find a smaller mainrectangle with noninteger side lengths which has a good subdivision. (And hence a contradiction.)
Proof: slice the main rectangle completely along the two internal edges of the topleft subrectangle, keeping only the bottomright chunk (the "complement". The complement has noninteger side lengths, and the subdivision from R inherits to a good subdivision on the complement, since we only ever chop off integer lengths from any subrectangle.
So suppose the topleft subrectangle S has integer height only. Chop in the widthdirection along the bottom edge of S, making the "heightcomplement". We've lost an integer height from every subrectangle, so the complement now has a good (inherited) subdivision but is smaller: a contradiction.
If S has integer width only, then we can chop in the heightdirection instead for the same effect.
Last edited by Smaug123; 04042016 at 22:08. 
Lord of the Flies
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 04042016 14:02
(Original post by Smaug123)
...
An alternative is the following:
Eliminate superfluous rectangles. Now colour in blue those with horizontal integer length and in red those with vertical integer length. Define a blue path as a chain of blue rectangles where the left edge of the next rectangle is aligned with right edge of the current one. Sim. for vertical path with top/bottom edges. If there is no blue path connecting the left & right sides of the big rectangle there must be a red path connecting the top & bottom.Last edited by Lord of the Flies; 04042016 at 21:29. 
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 04042016 15:28
(Original post by Lord of the Flies)
Interesting approach
An alternative is the following:
Eliminate superfluous rectangles. Now colour in blue those with horizontal integer length and in red those with vertical integer length. Define a path as a chain of adjacent rectangles of the same colour. If there is no red path connecting the top & bottom sides, there must be a blue path connecting the left & right sides, and vice versa. 
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 04042016 17:17
(Original post by Lord of the Flies)
C'mon people, what has happened to this thread. Problem 570 from earlier requires two skills: knowing what a rectangle is, and knowing what an integer is.
Anyway, more!
Problem 577
Prove/disprove that there is a family of disjoint subsets of the unit circle such that one can get from any member to any other through a rotation, and such that is the unit circle.
(this however is not trivial)
Spoiler:Show
Also, could you expand upon your solution to 570? I am not seeing why a chain of a single colour solves the problem. 
physicsmaths
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 04042016 18:40
Problem 578
Show that a graph is bipartite if and only if it contains no odd cycles.
A very nice problem.
Posted from TSR Mobile 
Renzhi10122
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 04042016 18:52
(Original post by physicsmaths)
Problem 578
Show that a graph is bipartite if and only if it contains no odd cycles.
A very nice problem.
Posted from TSR Mobile 
physicsmaths
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 04042016 18:56
(Original post by Renzhi10122)
Hmm, I wonder where you got that fromLast edited by physicsmaths; 04042016 at 18:58. 
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 04042016 21:13
(Original post by physicsmaths)
Problem 578
Show that a graph is bipartite if and only if it contains no odd cycles.
A very nice problem.
Posted from TSR Mobile
: Suppose G is bipartite then there is two independent vertex sets A and B.
Now any edge in a walk in G must go from A to B (or B to A) hence for any cycle we will take an even number of steps so there can be no odd cycles in G.
: Let G be a graph with no odd cycles WLOG we may assume G is connected (since if G is not connected we can show that each connected part of G is bipartite and then the union of each part is also bipartite).
Pick a vertex . Define and .
We claim that and hence we have that G is in fact bipartite.
WLOG suppose there exists an edge so that (of course we could suppose that they both are in B). Then are both even. is the shortest path length between the two vertices.
Consider the shortest path from u to a: and similarly then consider the cycle which has length which is odd therefore we have a contradiction and no such edge may exist.
Hence G is bipartite.
Problem 579: Let be a graph with at least one edge show that has at least edges.Last edited by poorform; 04042016 at 21:20. 
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 04042016 21:33
(Original post by DJMayes)
Assuming I am reading this correctly, this question feels like the same construction regarding the existence of a set which is not Lebesgue measurable.
Also, could you expand upon your solution to 570? I am not seeing why a chain of a single colour solves the problem. 
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 04042016 23:19
(Original post by DJMayes)
Assuming I am reading this correctly, this question feels like the same construction regarding the existence of a set which is not Lebesgue measurable.

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 04042016 23:36
(Original post by Smaug123)
That was indeed what I intended 
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 05042016 01:55
(Original post by Lord of the Flies)
Yep bang on.
Bad use of the word adjacent, should be clearer now.
(Original post by atsruser)
This is the construction of the Vitali sets, isn't it? Or is there some difference? 
EnglishMuon
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 08042016 18:47
Probably a well known one but quite nice
Problem 580
Prove that the product of four consecutive integers cannot equal the product of two consecutive integers (Referring to only). 
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 08042016 18:52
(Original post by EnglishMuon)
Probably a well known one but quite nice
Problem 580
Prove that the product of four consecutive integers cannot equal the product of two consecutive integers (Referring to only). 
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 08042016 19:09
Problem 581*
You select a real number entirely at random from the interval . What is the probability that is greater than 1 in magnitude?
Disclaimer: I made this up myself. I think my logic is correct, but my probability is **** so if this is an impossible question (or significantly harder than the * I gave it) then my apologies. 
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 08042016 19:12
(Original post by EnglishMuon)
Probably a well known one but quite nice
Problem 580
Prove that the product of four consecutive integers cannot equal the product of two consecutive integers (Referring to only).
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