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08/13/2008 11:47:45 AM · #26 |
According to Richard Bach, in The Bridge Across Forever: the sum of one and one, if they're the right ones, can be infinity!
Message edited by author 2008-08-13 11:48:09.
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08/13/2008 12:15:29 PM · #27 |
Originally posted by violinist123:
Originally posted by Hot_Pixel:
Originally posted by FocusPoint:
Originally posted by bnilesh:
Originally posted by Judi:
Errr...what does 1 + 1 = ? |
so simple 1 + 1 = 1 |
That's the problem with DPC voters... can't get clearer than this :/ |
Well in Binary it would be:
1+1= 10 :) |
In binary 1+1=1 again, 1+0=0 as does 0+0. Binary addition is very pessimistic. |
Actually it really depends on what logic gate you are doing your addition thru :) |
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08/13/2008 12:30:15 PM · #28 |
And politics:
Fuzzy Math vid
I think this might be slander. But it's interesting if you like this sorta thing. |
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08/13/2008 01:28:16 PM · #29 |
Here's a mathematics challenge for you:
Let p > 0 be congruent to 4n or 4n + 1 (where n is an integer greater than zero). Show that p is the order of some self-complementary graph.
It's easy to show that if G is a self-complementary graph of order p, that p is congruent to 4n or 4n + 1. It's the converse I'm struggling with.
Any ideas? Am I the only mathematician here? |
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08/13/2008 01:41:38 PM · #30 |
Originally posted by bvy:
Here's a mathematics challenge for you:
Let p > 0 be congruent to 4n or 4n + 1 (where n is an integer greater than zero). Show that p is the order of some self-complementary graph.
It's easy to show that if G is a self-complementary graph of order p, that p is congruent to 4n or 4n + 1. It's the converse I'm struggling with.
Any ideas? Am I the only mathematician here? |
I dunno if you're the only one, but I do know a few mathematician jokes. Here's one:
An engineer and a mathematician check into different hotels. There's a fire in the hall outside their doors that night.
The engineer wakes up smelling smoke, feels the doorknob, opens the door, locates the fire extinguisher, puts out the fire and goes back to bed.
The mathematician wakes up smelling smoke, feels the doorknob, opens the door, locates the fire extinguisher, exclaims, "A solution exists!" and goes back to bed.
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08/13/2008 03:00:34 PM · #31 |
Don't foget fractals, and geometry. Lots of things "multiply" lol.
Would color addition count? as in red + blue = an open challenge |
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08/13/2008 03:28:32 PM · #32 |
Originally posted by Donna21:
Don't foget fractals, and geometry. Lots of things "multiply" lol.
Would color addition count? as in red + blue = an open challenge |
Sure if you could show their RGB numbers ;p
Message edited by author 2008-08-13 15:28:40. |
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08/14/2008 01:21:43 AM · #33 |
| My favourite idea is just not going to work :( Must move on... Must move on... Can't wait to share the out-takes post challenge though :) |
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08/14/2008 11:42:16 AM · #34 |
12 hours to go and only 5 pictures
if it stays like that I have a good chance of winning a ribbon (then give it for charity) |
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08/14/2008 11:43:29 AM · #35 |
Originally posted by Oded:
12 hours to go and only 5 pictures
if it stays like that I have a good chance of winning a ribbon (then give it for charity) |
It closes on the 17th, not the 14th... |
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08/14/2008 01:28:08 PM · #36 |
Originally posted by jeger:
Originally posted by Oded:
12 hours to go and only 5 pictures
if it stays like that I have a good chance of winning a ribbon (then give it for charity) |
It closes on the 17th, not the 14th... |
Maybe math isn't his challenge (or maybe it is!). |
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08/14/2008 02:03:04 PM · #37 |
Originally posted by bvy:
Originally posted by jeger:
Originally posted by Oded:
12 hours to go and only 5 pictures
if it stays like that I have a good chance of winning a ribbon (then give it for charity) |
It closes on the 17th, not the 14th... |
Maybe math isn't his challenge (or maybe it is!). |
oooppsss
to many beers (but I have to admit that something felt rather strange) |
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08/16/2008 02:07:57 AM · #38 |
| Just submitted - yay! Finally got over my obsession with the first idea, and I think I ended up with something better (well at least more appealing to the majority). |
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08/16/2008 02:36:15 AM · #39 |
Originally posted by bvy:
Originally posted by scrybz:
Okay, here's my original idea, and I don't want anyone copying this novel idea:
"I see somebody standing in front of a chalkboard full of complicated looking algebra." |
Before you fill that chalkboard with "complicated looking algebra" make sure it's something substantial -- or at least correct! I see a lot of bad math going up just for the visual, and I suspect I won't be the only one put off by it. I'll be grading -- er, I mean, voting in this one. |
This lot's correct, isn't it?
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08/16/2008 02:53:29 AM · #40 |
Well, I can't see any problem with it. People should certainly divide minus infinity by infinity a lot more often in my opinion - give them some idea where they stand.
But are you sure you want to reveal so much of the formula
for the vanishing cream? |
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08/16/2008 04:53:51 AM · #41 |
"Here's a mathematics challenge for you:
Let p > 0 be congruent to 4n or 4n + 1 (where n is an integer greater than zero). Show that p is the order of some self-complementary graph.
It's easy to show that if G is a self-complementary graph of order p, that p is congruent to 4n or 4n + 1. It's the converse I'm struggling with.
Any ideas? Am I the only mathematician here?"
Proof
=> Every self complimentary graph has p(p-1)/4 edges since it must contain exactly half the edges in a completed graph of p nodes. therefore p must be congruent to either 0 or 1 mod 4.
<= By Induction
case 1 (p = 0 mod )
self complimentary graphs are isomorphisms.
i)We can construct a self complimentary graph of size 4 (it's shaped like an N)
ii) Suppose we could construct a self complimentary graph of size p-4, called R. Look at the isomorphism F between R and R' (the compliment). F maps verticies to verticies (for the size four example, if the verticies of N are labeled 1,2,3,4 from the top left clockwise, then N becomes Z via F(1) = 4, F(2)= 1, F(3) = 2 F(4) = 3 )
Consider how F permutes the verticies of R. Label the verticies of R a1, a2, a3, a4,... We construct a new graph of size p by drawing edges between R and the Graph N in the following way. a1 connects to vertice 1 and 3, F(a1) connects to vertices 2 and 4, F(F(a1)) connects to 1 and 3, F(F(F(a1))) connects to 2, 4. and so on.
That should get you started.
Prove F hits all the verticies in R (or, if not, how can you change it so that it does?)
Prove that this graph is self complimentary.
This is probably not the most efficient way, I havent taken graph theory in quite some time.
Message edited by author 2008-08-16 11:59:49. |
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08/16/2008 08:29:14 AM · #42 |
Originally posted by tntncsu:
"Here's a mathematics challenge for you:
Let p > 0 be congruent to 4n or 4n + 1 (where n is an integer greater than zero). Show that p is the order of some self-complementary graph.
It's easy to show that if G is a self-complementary graph of order p, that p is congruent to 4n or 4n + 1. It's the converse I'm struggling with.
Any ideas? Am I the only mathematician here?"
Proof
=> Every self complimentary graph has p(p-1)/4 edges since it must contain exactly the edges in a completed graph of p nodes. therefore p must be congruent to either 0 or 1 mod 4.
<= By Induction
case 1 (p = 0 mod )
self complimentary graphs are isomorphisms.
i)We can construct a self complimentary graph of size 4 (it's shaped like an N)
ii) Suppose we could construct a self complimentary graph of size p-4, called R. Look at the isomorphism F between R and R' (the compliment). F maps verticies to verticies (for the size four example, if the verticies of N are labeled 1,2,3,4 from the top left clockwise, then N becomes Z via F(1) = 4, F(2)= 1, F(3) = 2 F(4) = 3 )
Consider how F permutes the verticies of R. Label the verticies of R a1, a2, a3, a4,... We construct a new graph of size p by drawing edges between R and the Graph N in the following way. a1 connects to vertice 1 and 3, F(a1) connects to vertices 2 and 4, F(F(a1)) connects to 1 and 3, F(F(F(a1))) connects to 2, 4. and so on.
That should get you started.
Prove F hits all the verticies in R (or, if not, how can you change it so that it does?)
Prove that this graph is self complimentary.
This is probably not the most efficient way, I havent taken graph theory in quite some time. |
Thread killer! ;]
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08/16/2008 08:50:43 AM · #43 |
Originally posted by tntncsu:
Here's a mathematics challenge for you:
[...]
Proof
=> Every self complimentary graph has p(p-1)/4 edges...
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This is cool. Thanks for taking the time to post it! I'm actually about to leave for vacation, so I'm printing it out so I can look more closely at it. And, alas, this will be as close to the math challenge as I get.
Now back to your regularly scheduled banter.
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08/17/2008 03:15:33 PM · #44 |
Originally posted by Oded:
Originally posted by bvy:
Originally posted by jeger:
Originally posted by Oded:
12 hours to go and only 5 pictures
if it stays like that I have a good chance of winning a ribbon (then give it for charity) |
It closes on the 17th, not the 14th... |
Maybe math isn't his challenge (or maybe it is!). |
oooppsss
to many beers (but I have to admit that something felt rather strange) |
Now less than 9 hours to go and still only 44 entries. I like that. Everyone else, go do something else for the next 9 hours!
Message edited by author 2008-08-17 15:16:13.
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08/17/2008 03:18:00 PM · #45 |
Originally posted by Hot_Pixel:
Originally posted by Judi:
Errr...what does 1 + 1 = ? |
1+1=11 yea definately 11.... :) |
I'm thinking a picture of mom and dad and about 8 kids.
1+1 = 10
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08/17/2008 03:18:17 PM · #46 |
I just entered mine... << got all my fingers crossed for this one >>
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08/17/2008 04:01:19 PM · #47 |
Originally posted by DarkRider:
I just entered mine... << got all my fingers crossed for this one >> |
How will you count? |
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08/17/2008 04:03:03 PM · #48 |
Originally posted by raish:
Originally posted by DarkRider:
I just entered mine... << got all my fingers crossed for this one >> |
How will you count? |
I don't need to count...I plan on using my nose to hit the update button ;P |
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08/17/2008 04:04:42 PM · #49 |
I've unsubmitted my Steak and Kidney PIE shot as I feel there will be too many entered ;)
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08/21/2008 12:30:52 AM · #50 |
| I had an idea for this one, but didnt have the time nor the matierials needed since i didnt get a chance to look until monday....I wanted to take a abacus (sp?) a colorful one....and cut it in half with a saw or similar...then arrange teh 2 peices as if it were split by a samauri sword which would be in the picture and the bead peices spread somewhat all over....a offsetting colorful background. I was going to title it..."Simple Division" or " Quick Division". I have all these great ideas....and they come to me last minute....I need a hobby store closer than they are and one open 24/7 :) |
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