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Posted: 9/15/2005 1:14 AM
Post subject: some math challenges.
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Vatchern
Player (86)
Joined: 3/8/2005
Posts: 973
Location: Newfoundland, Canada
Anyone want to try and Prove why ( 1 / 3) = 0.33333333(etc..)? I wanna see what people come up with. good luck.
http://tasvideos.org/Vatchern.html
Posted: 9/15/2005 1:24 AM
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Bob_Whoops
Former player
Joined: 3/19/2004
Posts: 710
Location: USA
x = .3... 10x = 3.3... 10x - x = 3.3... - .3... 9x = 3 x = 1/3
Posted: 9/15/2005 1:24 AM
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DeHackEd
Emulator Coder, Site Developer, Former player
Joined: 11/6/2004
Posts: 833
For the other direction, you can carry out the long division algorithm. 1/3 = 1.00000 divided by 3 = 0.3 + 0.1/3, and you'll find yourself repeating yourself pretty fast.
Posted: 9/15/2005 3:00 AM
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Gigafrost
Former player
Joined: 5/3/2004
Posts: 366
A more formal way to prove it that way would be to use induction, but my first few rough attempts at it weren't very elegant looking, so... bleh.
07-24-2009: 2595 anime DVDs and counting...
Posted: 9/15/2005 5:06 AM
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Warp
Banned User
Joined: 3/10/2004
Posts: 7698
Location: Finland
A funnier thing to prove is that 0.99999... = 1.
Posted: 9/15/2005 5:19 AM
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Michael_Fried
Former player
Joined: 3/8/2004
Posts: 1107
A funnier thing to prove is that 0.99999... = 1. *is too lazy to think of something more original so copies Bob Whoops* x = .9... 10x = 9.9... 10x - x = 9.9... - .9... 9x = 9 x = 1 Edit: Now let's see who can prove that 1 = 2! :P
Posted: 9/15/2005 5:44 AM
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todd
Joined: 4/4/2004
Posts: 66
Here's a more formal proof which requires some familiarity with some concepts from calculus (infinite sums, limits, and geometric series): Let Rx = Sum(n=0...x, 0.9 / 10^n) In other words, R0= 0.9, R1= 0.99, R2= 0.999, R3= 0.9999, etc. Rx = 0.9 * Sum(n=0...x, 1 / 10^n) Rx = 0.9 * ( [ (1/10^n) - 1 ] / [ (1/10) - 1 ] ) lim(x->inf) Rx = 0.9 * (-1 / -0.9) = 1
Posted: 9/15/2005 6:21 AM
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nifboy
Former player
Joined: 5/31/2004
Posts: 375
The above topic (.99... =1) is the subject of the stupidest argument I ever got in over the internet. Naturally, it was on GameFAQs.
Posted: 9/15/2005 7:10 AM
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Randil
Skilled player (1827)
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
Edit: Now let's see who can prove that 1 = 2! :P
Allright! I'v shown this before, but it deserves a recap :) Here it is: 1 = √1 = √-1)*(-1 = √(-1) * √(-1) = i * i = -1 This means 1 = -1. Now, just add 1 to both lines. That means 2 = 0. And if 2 = 0, then all integers are zero. And if all integers are zero, then all integers are all integers (if you see what I mean). For example, 14=0 and 20=0 means that 14=20. So with this it's very easy to prove that 1 = 2 :) Oh, by the way, the 0.999999999.... = 1 only works if 0.99999 have unlimited decimals. If the number would be exactly something, like 0.9999999 then the equation won't work. If you'd like I could show you how. I have a wonderful proof here, but it just didn't fit the post :P
My own page on this site.
Posted: 9/15/2005 1:25 PM
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Xaphan
Former player
Joined: 5/24/2005
Posts: 405
Location: France
I have a funny one (and easy to find) : assume that a = b and a > 0 a = b >> a² = ab >> a² - b² = ab - b² >> (a - b)(a + b) = b(a - b) >> a + b = b so, if a = b = 1 then 1 = 2 :)
Not dead yet... still very busy... damn...
Posted: 9/15/2005 1:58 PM
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Baxter
Skilled player (1410)
Joined: 5/31/2004
Posts: 1821
TAS of 2013 SNES TAS of 2013 Speedy TAS of 2007 Innovative TAS of 2007
Xaphan: you can't divide by zero (but you probably knew that). If that "problem" was written as: 1*0 = 2*0 you'd see it right away.
http://tasvideos.org/Baxter.html
Posted: 9/15/2005 2:08 PM
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feitclub
Former player
Joined: 3/13/2004
Posts: 1118
Location: Kansai, JAPAN
Xaphan wrote:
I have a funny one (and easy to find) : assume that a = b and a > 0 a = b >> a² = ab >> a² - b² = ab - b² >> (a - b)(a + b) = b(a - b) >> a + b = b so, if a = b = 1 then 1 = 2 :)
This problem was actually part of Randil's signature until recently.
Do Not Talk About Feitclub http://www.feitclub.com
Posted: 9/15/2005 2:34 PM
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MahaTmA
Joined: 3/8/2004
Posts: 185
Location: Denmark
Here's a funny wee thing, try to find the fault: Let f(x) be continuos in [a;b] and differentiable in ]a;b[ Let us assume that f(x) isn't constant in the interval [a;b], this means that there exist numbers x1 and x2 within [a;b] , so that: x1 < x2 => f(x1) < f(x2) or x1 < x2 => f(x1) > f(x2) Assume the first is true, and: x1 < x2 => f(x1) < f(x2) (1) We only look at this, since the proof for the other sentence follows the same form. It is then known, that there exists a number c in [x1,x2] so that: f'(c) = (f(x2)-f(x1))/(x2-x1) As a cause of (1), the numerator and denominator are both positive, so the fraction is also positive. Ergo: f'(c) > 0 (2) Define the function g as: g(x)=(f(x)-f(c))^(1/3) , a=<x=<b Ergo: g(c) = 0 (3) From the definition of g comes: g³(x) = f(x)-f(c) <=> f(x) = g³(x)+f(c) We differentiate the last expression, and get (since f(c) is a constant): f'(x)= 3*g²(x)*g'(x) + 0 = 3*g²(x)*g'(x) (4) since g(c)=0 according to (3), (4) gives that f'(c)=0, which is inconsistent with (2). This means that our assumption (1) is wrong, and thus: If f(x) is continuos in [a;b] and differentiable in ]a;b[, f(x) is constant in [a;b] Sure does make the life of a lot of people easier.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
Posted: 9/15/2005 4:09 PM
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Randil
Skilled player (1827)
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
It is then known, that there exists a number c in [x1,x2] so that: f'(c) = (f(x2)-f(x1))/(x2-x1)
If I undestand you correct, since c is a number, the coordinates for c would be (c, f(c)). f'(c) = (f(x2)-f(x1))/(x2-x1) only derives the line that is drawn from (X1, f(X1)) and (X2, f(X2)). I can agree that f'(c) = (f(x2)-f(x1))/(x2-x1) if f(x) is a linear function, but if it's not, then I'm not sure I follow. Perhaps I'm all wrong about this, MahaTmA, but it would be nice if you'd explain all of this for me.
Sure does make the life of a lot of people easier.
No, but it was fun reading :)
My own page on this site.
Posted: 9/15/2005 4:41 PM
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MahaTmA
Joined: 3/8/2004
Posts: 185
Location: Denmark
f(x) doesn't have to be a linear function. f'(c) is a number which corresponds to the inclination constant of the tangent in (c,f(c)). If the definitions for f are true, you can pick a c so that those two are equal. A rough translation of what it's called in Danish (Since I'm not really too stiff in academical English) would be the "mean value sentence." It's a proven mathematical sentence, so that step is valid. To clarify, think of a "curved" function. Pick two points on that function's graph and put a line between them. Now raise/lower the line until it coincides with a tangent for f. There's your proof in informal form.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
Posted: 9/15/2005 4:50 PM
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Randil
Skilled player (1827)
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
Oh, allright, then I'm following you. It's a pretty interesting problem, although I'm not sure I know what you mean with ]a;b[. Do you mean outside the interval (I think that's the english word for it) outside of the a-b interval? Or do you mean the interval that is inside a-b but never actually IS a or b? I hope you understand what I'm talking about, mathematical terms in English isn't really my strong side :)
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Posted: 9/15/2005 4:53 PM
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MahaTmA
Joined: 3/8/2004
Posts: 185
Location: Denmark
Feel the same way. ]a;b[ is the interval which is between a and b, but is never actually a or b, like you said :)
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
Posted: 9/15/2005 4:56 PM
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Cazlab
Joined: 7/5/2004
Posts: 551
Location: Karlstad, Sweden
lol, are u actually doing these stuff for fun? SICK! :D
Posted: 9/15/2005 5:00 PM
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Randil
Skilled player (1827)
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
Cazlab wrote:
lol, are u actually doing these stuff for fun? SICK! :D
Are you actually NOT doing this for fun??? :) There's not much that's funnier than math. Well, some stuff, but not much :)
My own page on this site.
Posted: 9/15/2005 5:02 PM
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MahaTmA
Joined: 3/8/2004
Posts: 185
Location: Denmark
Math gets your brain a-swinging, and I like that.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
Posted: 9/15/2005 8:02 PM
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Walker_Boh
Former player
Joined: 4/16/2004
Posts: 1276
Location: Uppsala, Sweden
Knapsack problem. You plan to go hiking, and you want to pack your stuff in a knapsack. Your knapsack has a total colume of V cubic inches, and you have n different items of volumes v1, v2, ....., vn that you want to take with you. Unfortunately, you can't take them all, since their total volume exceeds the volume of your knapsack. So you need to decide which items to take with you and which items to leave at home. Suppose that you want to make your decision so as to utilize the volume of the knapsack as much as possible. That is, you want to find a subset of the items so as to maximize the sum of their volumes subject to the constraint that this sum is no more than V. We call the subset which maximizes this sum the optimum packing of v1, v2, .... , vn in volume V. 1) Fix a list of items v1, v2, .... vn. For any i <= n and any U <= V donate by OPT i(U) the volume of the optimum packing of v1,v2, .... , vn in volume U. Prove that for all i and U, we have: OPT i(U) = max { OPT i-1(U), vi + OPT i-1(U- vi) }. ------ i here is just a constant, and as an index (lowered that is, so for example OPT(lowered i)(U), if that makes sense ;P).
/Walker Boh
Posted: 9/15/2005 8:13 PM
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MahaTmA
Joined: 3/8/2004
Posts: 185
Location: Denmark
Post the answer to that in binary code, and I'll set you up for a Nobel prize.
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
Posted: 9/15/2005 8:13 PM
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Walker_Boh
Former player
Joined: 4/16/2004
Posts: 1276
Location: Uppsala, Sweden
Hehe. I don't know the answer unfortunately =/
/Walker Boh
Posted: 9/15/2005 8:18 PM
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MahaTmA
Joined: 3/8/2004
Posts: 185
Location: Denmark
That's because the problem has, up until now anyhow, been unsolveable. Mathematician's are getting close, but if they can't come up with some clever trick such as the one for Fermat's last Theorem, they will never solve it. Exhaustive methods just aint doing the trick here. For the record, Fermat's last theorem: Prove, that X^n + Y^n = Z^n has no solutions for n>2
"We observe the behaviour of simple folk, and derive pleasure from their defects." -Aristotle - Book of Humour
Posted: 9/15/2005 8:21 PM
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Walker_Boh
Former player
Joined: 4/16/2004
Posts: 1276
Location: Uppsala, Sweden
Oh. So it's insolvable? I really thought it was solvable. Crap.
/Walker Boh
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