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| Acmlm's Board - I3 Archive - General Chat - (1 / 0) ^ (-1) |
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rubixcuber Mole                    Since: 09-08-06 From: St. Louis, MO Last post: 6406 days Last view: 6406 days |
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| What do you mean by clearly undefined? I don't see how (1/0)^(-1) is any less clearly undefined than (1/0)^(1). They are both powers of an undefined number. Thus to assume that one of them is defined is to assume all of them are defined. And if that is the case, you end up with nonsense results such as 1 = 0. That's all I was trying to say.
EDIT: And Gamma is just an extension of the factorial to non integers. It is still the same values at integers (i.e. -1). And the factorial is undefined for all negative integers. (edited by rubixcuber on 10-30-06 12:33 PM) (edited by rubixcuber on 10-31-06 10:16 AM) |
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beneficii Broom Hatter  Since: 11-18-05 Last post: 6300 days Last view: 6297 days |
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| rubixcuber,
Because, at the time (1/0) ^ (-1) was not clearly undefined, but you then multiplied both sides by (1/0), which we all know was undefined. It was the step should not have even been there. Still, it is clear now that (1/0) ^ (-1) is undefined. 
Re EDIT: That's what I was worried about. Still, from what I understand, for combinations (which are also called binomial coefficients), 0 <= k <= n. But it causes us to wonder what's outside of those ones in Pascal's triangle. In that case also, what would be: SIGMA [j = 0 to k - 1] C(j, k) As you know: 
Referencing article: http://en.wikipedia.org/wiki/Binomial_coefficient But, if you started j from 0 instead, and went to k - 1, what should the sum be? For one particular problem I'm working on to work out (which I'll try to get uploaded here), it would seem that such sum from 0 to k - 1 should be 0. Which would mean, that each case to the right of the triangle should be 0. Question: What program would be good for generating images containing mathematical notation? EDIT: Added article link. (edited by beneficii on 10-30-06 02:35 PM) |
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Deleted User Banned    Since: 05-08-06 Last post: None Last view: 6297 days |
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Originally posted by beneficii Well, when you have a /0 somewhere, "it is possible to disguise a special case of division by zero in an algebraic argument, leading to spurious proof: 1) For any real number x: x2 − x2 = x2 − x2 2) Factoring both sides in two different ways: (x − x)(x + x) = x(x − x) 3) Dividing both sides by x − x, simplified, yields: (1)(x + x) = x(1) 4) Which is: 2x = x 5) Since this is valid for any value of x, we can plug in x = 1. 2 = 1" (This time I copy pasted, I hope I got it right  ). |
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Guy Perfect Since: 11-18-05 Last post: 6299 days Last view: 6297 days |
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| Summary time!
It's easy to get confused with algorithmic management, which often leads to some people making blatanly incorrect statements like "1+ 1 might not always be 2!" Thing is, you can count to 2 with your eyes, so there's another aspect to the whole deal. Numbers that in and of themselves evaluate to certain values, such as 0 or (Undefined), can make correct, logical progression of exp And the original example, (1 / 0) ^ (-1)... (Undefined) ^ (-1). Can't do it. Like was mentioned, you can only consider the idea if you understand that the idea is impossible. It's a perfectly reasonable way to do mathematics, much like the imaginary unit, but you're still throwing in non-real numbers to get the job done. As was also mentioned, an important thing to remember is that order of operations must be defined prior to the mangling of operators. (1 / 0) ^ (-1) sounds good to make 0 / 1, but it's that 1 / 0 in the first place that makes it not work. You can't transform it into a defined value, because the exponentiation happens afterwards. |
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Cirvania Cyball I guess this is as close as Xkeeper will get to spell it right. :< Since: 11-17-05 From: The Island of Puerto Rico. Last post: 6299 days Last view: 6297 days |
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Let's consult the almighty tunabot, shall we?
Originally posted by tunabot on IRC (edited by Cirvante on 10-30-06 09:03 PM) |
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Guy Perfect Since: 11-18-05 Last post: 6299 days Last view: 6297 days |
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| We may now go away from this a little wiser and more informed. No post that any of us can make will be more insightful than that. | |||
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Deleted User Banned Since: 05-08-06 Last post: None Last view: 6297 days |
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| Indeed, knowing that the meaning of life is (undefined) means I can finally die =) | |||
MathOnNapkins  1100  In SPC700 HELL Since: 11-18-05 Last post: 6297 days Last view: 6297 days |
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| I already told y'all what 1/0 was and y'all didn't believe me. It was a thread on the old board. Consisted of constructing a continuous vector space of the real numbers based on multiplication and division by powers of zero. | |||
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Deleted User Banned Since: 05-08-06 Last post: None Last view: 6297 days |
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| Would you give us the link? | |||
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MathOnNapkins 1100 In SPC700 HELL Since: 11-18-05 Last post: 6297 days Last view: 6297 days |
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| Would if I could find it. Last I checked it was in Lost Threads and Lost Threads was locked. Forgot which forum it got moved to (98?) but there's only 20 threads in there. | |||
HyperHacker Star Mario Finally being paid to code in VB! If only I still enjoyed that. <_< Wii #7182 6487 4198 1828 Since: 11-18-05 From: Canada, w00t! My computer's specs, if anyone gives a damn. STOP TRUNCATING THIS >8^( Last post: 6297 days Last view: 6297 days |
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Well, Windows Calculator agrees with Kyouji. I put in "(1/0)" and  . |
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Omega45889 Shyguy Since: 11-18-05 Last post: 6330 days Last view: 6345 days |
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| MoN is clearly right.
(1/0) = undefined |
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