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Author Topic: Math wizards look here.  (Read 13681 times)

Storyyeller

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Re: Math wizards look here.
« Reply #105 on: February 19, 2011, 04:42:47 pm »

What's wrong with my answer?
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Kwiftee

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Re: Math wizards look here.
« Reply #106 on: February 19, 2011, 05:53:24 pm »

Because you don't have zero, you have an undefined value.
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zyxfrlpso
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Kwiftee

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Re: Math wizards look here.
« Reply #107 on: February 21, 2011, 02:02:29 am »

Double post! Can anyone explain to me why Cantor's diagonal argument can't also be used to prove that the natural numbers are uncountable? I mean, they're obviously not uncountable, but the proof seems to work on them just as well as real numbers.

It's pretty likely that I just don't understand how the argument works at all. <_<
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zyxfrlpso
Quote from: Yuletideriz
My god, if I could shit gold and I give it to you, you would still bitch because it was technically crap.
how about you GO ON IRC RIGHT NOW PLEASE

Venser

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Re: Math wizards look here.
« Reply #108 on: February 21, 2011, 02:10:16 am »

"A set X is uncountable if and only if any of the following conditions holds:
There is no injective function from X to the set of natural numbers.
X is nonempty and any ω-sequence of elements of X fails to include at least one element of X. That is, X is nonempty and there is no surjective function from the natural numbers to X.
The cardinality of X is neither finite nor equal to aleph-null.
The set X has cardinality strictly greater than aleph-null."

The argument doesn't prove any of that for the set of natural numbers.
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Kwiftee

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Re: Math wizards look here.
« Reply #109 on: February 21, 2011, 02:52:23 am »

I don't know what a ω-sequence is. Is it just any list, finite or infinite, of elements of X?
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zyxfrlpso
Quote from: Yuletideriz
My god, if I could shit gold and I give it to you, you would still bitch because it was technically crap.
how about you GO ON IRC RIGHT NOW PLEASE

Venser

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Re: Math wizards look here.
« Reply #110 on: February 21, 2011, 01:05:30 pm »

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Neko Desu

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Re: Math wizards look here.
« Reply #111 on: February 24, 2011, 06:51:02 pm »

I'm back from doing CEMC's Grade 10 Cayley math contest:

http://img196.imageshack.us/img196/2798/2011cayley.png

I only did up to question 21 before I ran out of time. Can you guys tell me what you got for questions 17 and 20, and hopefully  questions 22 - 25?.
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Kwiftee

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Re: Math wizards look here.
« Reply #112 on: February 24, 2011, 08:54:10 pm »

: question 17! • show
400cm2 (C). You know that 2w+2h=40, so w+h=20, and the area is equal to (w+h)2.


: question 20! • show
Pretty sure this should be 8 (B). Proof is kind of confusing, and I haven't given it TOO much thought, but it's definitely either 8, or less than 8.


I'll do the rest later.
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zyxfrlpso
Quote from: Yuletideriz
My god, if I could shit gold and I give it to you, you would still bitch because it was technically crap.
how about you GO ON IRC RIGHT NOW PLEASE

Neko Desu

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Re: Math wizards look here.
« Reply #113 on: February 24, 2011, 09:14:27 pm »

Thanks, Kwiftee. I got #17 right, but didn't get 8 for #20, unfortunately, and I saw the error in my calculations.
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Kwiftee

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Re: Math wizards look here.
« Reply #114 on: February 24, 2011, 09:22:42 pm »

Oh, and according to my good pal Rohan, question 22 is 7.5. I can't prove it myself, but he did.
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zyxfrlpso
Quote from: Yuletideriz
My god, if I could shit gold and I give it to you, you would still bitch because it was technically crap.
how about you GO ON IRC RIGHT NOW PLEASE

Neko Desu

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Re: Math wizards look here.
« Reply #115 on: March 24, 2011, 07:07:48 pm »

Two numbers 'a' and 'b' with 0 <= a <= 1 and 0 <= b <= 1 are chosen at random. The number 'c' is defined by c = 2a + 2b. The numbers 'a', 'b', and 'c' are each rounded to the nearest integer to give 'A', 'B', and 'C', respectively (for example, if a = 0.432 and b = 0.5, then A = 0, B = 1, and C = 2). What is the probability that 2A + 2B = C?
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LudorExperiens

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Re: Math wizards look here.
« Reply #116 on: April 04, 2011, 12:08:46 pm »

That's one of my favourites:

Given a parallelogram of area S, draw a parallelogram of area S/6, using only a straightedge.
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Kwiftee

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Re: Math wizards look here.
« Reply #117 on: April 10, 2011, 09:55:52 am »

Alright, here's one for you Mark.

Let a, b be the two solutions of the second-degree equation in z:
x2 - 2px + p2 - 2p - 1 = 0
Find a real value p such that (0.5*[a-b]2-2) / ([a+b]2+2) = n, where n is an integer.
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zyxfrlpso
Quote from: Yuletideriz
My god, if I could shit gold and I give it to you, you would still bitch because it was technically crap.
how about you GO ON IRC RIGHT NOW PLEASE

ybbald

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Re: Math wizards look here.
« Reply #118 on: April 11, 2011, 09:44:09 am »

That's one of my favourites:

Given a parallelogram of area S, draw a parallelogram of area S/6, using only a straightedge.
how is that hard? just draw a parallelogram and mark the sides w/e you want
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LudorExperiens

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Re: Math wizards look here.
« Reply #119 on: April 12, 2011, 02:54:24 pm »

That's one of my favourites:

Given a parallelogram of area S, draw a parallelogram of area S/6, using only a straightedge.
how is that hard? just draw a parallelogram and mark the sides w/e you want
Are you sure you understood the problem? In mathematics a straightedge means you can only draw straight lines because straightedges aren't marked. So you have to find a way to draw a parallelogram of area S/6 using only straight lines and that's not as easy as it seems to be.
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