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03-29-2010, 08:16 PM
| | Registered User | | Join Date: Jul 2008 Location: Somewhere in Canada | | | Mathematical Proofs Anyone?
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I think I've asked on here before about a Mathematical Foundations course and if it was common. Well now I'm having a bit of trouble and I'm wondering if any of you might be able to help me out.
I have to prove that (for all x in the set of the Natural Numbers excluding 1) if (there exists a natural number t) where 10 = xt and (there exists a natural number s) where 15 = xs then (there exists a natural number q) where 5 = xq.
Does anyone have any idea of how I could do this? If so, help would be GREATLY appreciated.
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Last edited by rarisgod : 03-29-2010 at 08:27 PM.
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03-29-2010, 08:27 PM
| | Registered User | | Join Date: Mar 2008 Location: Bridgewater, Virginia | | | Only x has to be a natural number im assuming.
x = 2
t = 5
s = 7.5
q = 2.5
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03-29-2010, 08:28 PM
| | Registered User | | Join Date: Jul 2008 Location: Somewhere in Canada | | Quote:
Originally Posted by Araillac Only x has to be a natural number im assuming.
x = 2
t = 5
s = 7.5
q = 2.5 | Ahh sorry, they all have to be natural numbers.
And it's more of a mathematical proof than finding a solution. Yeah "advanced mathematics"...
__________________ Quote:
Originally Posted by Thunderscreech Social Networking is a plague upon the face of the Earth. | Quote:
Originally Posted by Milky I'd get an Itouch myself | |
03-29-2010, 08:32 PM
| | Registered User | | Join Date: Mar 2008 Location: Bridgewater, Virginia | | Quote:
Originally Posted by rarisgod Ahh sorry, they all have to be natural numbers.
And it's more of a mathematical proof than finding a solution. Yeah "advanced mathematics"... | This is why I quit college and went to a trade school. When would I ever have to use this in a real world application? However, I would answer
x =/= 1 therefore xq =/= 5
and demand proof if that is not correct in a substantial mathematical equation.
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03-29-2010, 08:51 PM
| | Registered User | | Join Date: May 2009 Location: kansas city, mo | | | I'm not sure of the proofs, but the answers are q=1, t=2, s=3, and all of those are natural numbers if x = 5.
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03-29-2010, 09:15 PM
|  | Registered User | | Join Date: Feb 2001 Location: San Francisco, CA | | | |
03-29-2010, 09:16 PM
|  | Registered User | | Join Date: Jul 2004 Location: Fort Collins, Colorado | | | Sorry, I ran out of fingers and toes early on this one.
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03-29-2010, 09:22 PM
| | Registered User | | Join Date: Jul 2008 Location: Somewhere in Canada | | | Success!
Proof:
15 - 10 = xs - xt
5 = x(s - t) [I factored here]
Let q = s - t
5 = xq
QED
Now I'm trying to prove
if x = 10t and x = 15s then x = 30q
I've tried putting 10t = 15s, which is true, but now where to go from there to get 30 times something = x?
__________________ Quote:
Originally Posted by Thunderscreech Social Networking is a plague upon the face of the Earth. | Quote:
Originally Posted by Milky I'd get an Itouch myself | |
03-29-2010, 09:47 PM
| | Registered User | | Join Date: May 2009 Location: kansas city, mo | | Quote:
Originally Posted by rarisgod Success!
Proof:
15 - 10 = xs - xt
5 = x(s - t) [I factored here]
Let q = s - t
5 = xq
QED
Now I'm trying to prove
if x = 10t and x = 15s then x = 30q
I've tried putting 10t = 15s, which is true, but now where to go from there to get 30 times something = x? |
Are these natural numbers as well? If so, then let t =3, s = 2 and q =1, and x = 30 and for the proof 30q-X=10t-15s
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Last edited by blendermassacre : 03-29-2010 at 09:56 PM.
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03-29-2010, 09:47 PM
| | Registered User | | Join Date: Jul 2008 Location: Somewhere in Canada | | | yes, all
__________________ Quote:
Originally Posted by Thunderscreech Social Networking is a plague upon the face of the Earth. | Quote:
Originally Posted by Milky I'd get an Itouch myself | |
03-29-2010, 09:52 PM
|  | Me? Solecistic? That's unpossible! | | Join Date: Nov 2003 Location: Toronto, Ontario, Canada eh? | | | (see below)
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Last edited by wabbit : 03-29-2010 at 10:26 PM.
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03-29-2010, 10:00 PM
| | Registered User | | Join Date: May 2009 Location: kansas city, mo | | | please tell me if I'm right. It's bothering me.
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03-29-2010, 10:12 PM
| | Me? Solecistic? That's unpossible! | | Join Date: Nov 2003 Location: Toronto, Ontario, Canada eh? | | | I think I have it ....
"if x = 10t and x = 15s then x = 30q"
t = x/10 and s = x/15
t - s = x/10 - x/15
t - s = (3x - 2x)/30
30(t - s) = 3x - 2x
30(t - s) = x
let q = t - s
30q = x
Is that right?
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Did you learn to play through an instructor or on your own?
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Last edited by wabbit : 03-29-2010 at 11:03 PM.
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03-29-2010, 10:17 PM
| | Registered User | | Join Date: May 2009 Location: kansas city, mo | | Quote:
Originally Posted by wabbit I think I have it ....
if x = 10t and x = 15s then x = 30q
t = x/10 and s = x/15
t - s = x/10 - x/15
t - s = (3x - 2x)/30
30(t - s) = 3x - 2x
30(t - s) = x
let q = t - s
30q = x
Is that right? | You lost me at the bold portion. I figured it out. NVM.
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Last edited by blendermassacre : 03-29-2010 at 10:20 PM.
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03-29-2010, 10:40 PM
| | Me? Solecistic? That's unpossible! | | Join Date: Nov 2003 Location: Toronto, Ontario, Canada eh? | | Quote:
Originally Posted by blendermassacre please tell me if I'm right. It's bothering me. | blender, your numbers jibe with the proof that i came up with so i think we're both good to go 
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Did you learn to play through an instructor or on your own?
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Last edited by wabbit : 03-30-2010 at 09:07 AM.
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