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| CarpeGuitarrem |
 Posted: Tue Sep 25, 2007 11:07 pm |
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Joined: 07 Apr 2006
Posts: 3361
Location: Franciscan University of Steubenville
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| Zero does not equal infinity. I don’t get at all how your equation works. There’s no way you can logically step to the final step. So you’ve got X ^ Infinity = Infinity. You’ve got 0 ^ X = 0. Now what? You can’t substitute anything in. |
| "ok, change of plans. the Cobracards christmas party is coming to my house, and we’re gunna teach FM how to hunt." (mm) |
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| AnxiousChieftain |
| Posted: Wed Sep 26, 2007 6:40 am |
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Joined: 15 Jul 2005
Posts: 3947
Location: Maryland
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I think the error occurs here:
NBarden wrote: X to the infinity power = infinity (or X^∞ = ∞), while X is a number.
I think it should be X^∞ = ∞, while X is a non-zero number. Check out this article:
http://www.philforhumanity.com/Zero_Times_Infinity.html
- AC |
| MODS RULE. - lem0nhead |
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| CarpeGuitarrem |
| Posted: Wed Sep 26, 2007 10:10 am |
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Joined: 07 Apr 2006
Posts: 3361
Location: Franciscan University of Steubenville
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Yeah, AC’s right, I just realized that this morning. That’s an excellent example of how forgetting a little detail in math can lead to big problems. Like the proof of 4 = 5...
a = b + c
b + c = a
4a = 4b + 4c
5b + 5c = 5a
4a + 5b + 5c = 5a + 4b + 4c
[subtract 9a from both sides]
5b + 5c - 5a = 4b + 4c - 4a
5(b + c - a) = 4(b + c - a)
5 = 4 |
| "ok, change of plans. the Cobracards christmas party is coming to my house, and we’re gunna teach FM how to hunt." (mm) |
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| AnxiousChieftain |
| Posted: Wed Sep 26, 2007 12:41 pm |
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Joined: 15 Jul 2005
Posts: 3947
Location: Maryland
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Hehe, I like that proof... never seen it before, so it took me a minute of searching to find the error.
- AC |
| MODS RULE. - lem0nhead |
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| bobtheorc |
| Posted: Wed Sep 26, 2007 3:06 pm |
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Joined: 19 Mar 2006
Posts: 1218
Location: Wow, its hot in Iowa
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| Elessar's Socks |
| Posted: Wed Sep 26, 2007 3:32 pm |
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Joined: 29 Jul 2007
Posts: 349
Location:
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 That proof’s new to me, too. I think the error occurs in this step:
5(b + c - a) = 4(b + c - a)
5 = 4
because that involves division by zero, which is undefined. Or to be more straightforward:
5 * 0 = 4 * 0 --> 5 = 4 *t3h gaspery!!*
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Here’s one of my favorite "proofs":
sqrt[-1] = sqrt[-1]
sqrt[-1/1] = sqrt[1/-1]
sqrt[-1]/sqrt[1] = sqrt[1]/sqrt[-1]
sqrt[-1] * sqrt[-1] = sqrt[1] * sqrt[1]
-1 = 1
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| Merrick_Hale |
| Posted: Wed Sep 26, 2007 5:15 pm |
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Joined: 13 Apr 2006
Posts: 445
Location: Austin, Texas
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Elessar’s Socks wrote: That proof’s new to me, too. I think the error occurs in this step:
5(b + c - a) = 4(b + c - a)
5 = 4
because that involves division by zero, which is undefined. Or to be more straightforward:
5 * 0 = 4 * 0 --> 5 = 4 *t3h gaspery!!*
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Here’s one of my favorite "proofs":
sqrt[-1] = sqrt[-1]
sqrt[-1/1] = sqrt[1/-1]
sqrt[-1]/sqrt[1] = sqrt[1]/sqrt[-1]
sqrt[-1] * sqrt[-1] = sqrt[1] * sqrt[1]
-1 = 1
4A + 5B + 5C = 5A + 4B + 4C
which is correct. Substitute in any values such that A = B + C and you will find that the equation is still an equation.
The problem comes when trying to "subtract 9 A from both sides". You can add equal amounts to both sides as much as you want and still maintain the equality, and you can subtract as much as you want from both sides as long as one or both sides don’t have a factor that goes negative. The equality will be maintained if you subtract 4A from both sides:
5B + 5C = A + 4B + 4C
Substitute back in A = B + C and you wind up with 5B + 5C = 5B + 5C
You just can’t take a factor negative. |
Aaron Rhinehart
Rider of Rohan (retired)
DAgent #417 (retired)
This Incognito is a smooth operator - Lem0nhead |
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| Elessar's Socks |
| Posted: Wed Sep 26, 2007 6:01 pm |
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Joined: 29 Jul 2007
Posts: 349
Location:
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You can subtract as much as you want from both sides as long as you keep your signs straight. The result of subtracting 9a from both sides,
5b + 5c - 5a = 4b + 4c - 4a,
simplifies to
a = b + c,
which is the original equation.
However from a = b + c, we see that
b + c - a = 0.
Therefore you can’t cancel b + c - a from both sides of the equation in the final step, because you can’t divide by zero. |
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| NBarden |
| Posted: Thu Sep 27, 2007 12:31 pm |
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Joined: 28 Dec 2006
Posts: 5468
Location: I don't know...
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Its not zero times infinity, its zero times zero (which is defined), infinity times.
The point of the thread was to confuse some people who don’t like math. 
Oh, ES, when you go sqrt[-1], you are out of the realm of real numbers and thus real number proofs don’t work. Sorry.
SQRT[-1] is more commonly known as i. |
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 Add the popcorn smiley to your sig, help it achieve world domination.
What if the hokey pokey really IS what its all about? 
As I lay in bed staring at the stars last night, I thought to myself, "where the heck is the ceiling?"
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| Elessar's Socks |
| Posted: Thu Sep 27, 2007 12:59 pm |
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Joined: 29 Jul 2007
Posts: 349
Location:
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NBarden wrote: Oh, ES, when you go sqrt[-1], you are out of the realm of real numbers and thus real number proofs don’t work. Sorry.
SQRT[-1] is more commonly known as i.
Oh believe me, I know.
And you got it. Straight outta MathWorld: "The reason for the fallacy is that sqrt(-1) is not an ordinary (real) square root, hence the rule for computing the quotient of radicals does not apply to it." |
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