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| Acmlm's Board - I3 Archive - General Chat - (1 / 0) ^ (-1) |
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beneficii Broom Hatter                  Since: 11-18-05 Last post: 6300 days Last view: 6296 days |
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| Hmm, we all know that 1/0 is undefined, but what about (1/0)^(-1)? In other words, the inverse of (1/0)? Well, it would seem if you divided 1 by 1/0 the answer should be zero because it seems that:
1/(1/0) = 1 * (0/1) = 0 What do y'all think? I tried looking this up on Dr. Math, but I could not find anything that discussed this. (edited by beneficii on 10-28-06 11:25 PM) (edited by beneficii on 10-28-06 11:26 PM) (edited by beneficii on 10-29-06 12:00 AM) |
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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 Saying that is like saying that 0/0=1, and that it's wrong. What happens is: 1/(1/0) = 1/(undefined) So you can't continue. (That's just what I think though  ) |
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beneficii Broom Hatter Since: 11-18-05 Last post: 6300 days Last view: 6296 days |
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| Er, made an error I apologize. The final result was to be zero, not one. (Edited it.)
Uly, What I'm thinking is that, following from if you do a division with fractions like so: (a/b) / (c/d) then that is the same as: (a/b) * (d/c) where you take the reciprocal of the second number, dividing 1 by (1/0) like so: 1 / (1/0) should be 1 * (0/1) and then 0 because in the second number you do the reciprocal which comes out to being valid. |
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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 HyperHacker After some research I want to know what do others think too, because I think it makes sense, though it renders ∞, -∞, and -0 useless  |
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neotransotaku Sledge Brother Liberated from school...until MLK day    Since: 11-17-05 From: In Hearst Field Annex... Last post: 6299 days Last view: 6296 days |
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any coherent argument is my take on this. Division should not be allowed as an operation.  |
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NSNick Gohma IF ALL ELSE FAILS USE BOOZE  Since: 11-17-05 From: Last post: 6298 days Last view: 6298 days |
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| Makes sense that (1/0)^(-1) just equals (0/1), and therefore equals zero.
Edit: In fact, going by order of operations, that should be correct. (edited by NSNick on 10-29-06 03:32 AM) |
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Alastor Fearless Moderator Hero  Since: 11-17-05 From: An apartment by DigiPen, Redmond, Washington Last post: 6296 days Last view: 6296 days |
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| No. I can see the logic you're going by, but we get to 1/0 and then system shut down. No inverse operation ever happens because system down. | |||
Hiryuu Sword Maiden Retired Admin  Since: 11-17-05 From: Nerima District - Tokyo, Japan Last post: 6296 days Last view: 6296 days |
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| How often have people tried to go...
I ZEROED INFINITY ...when they should've gone... I SCREWED AN UNDEFINED This probably isn't the first time I've seen that actually and people thought they could do that but I'd have to agree with Alastor. Unless you define dividing by something normally impossible then it won't work. You bend that rule then sure, why not? You could have a bunch of other additions to that math but just like code in a series...if one thing goes wrong then the whole thing dies. Division by zero would be one such. Kaboom. |
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MathOnNapkins  1100  In SPC700 HELL Since: 11-18-05 Last post: 6296 days Last view: 6296 days |
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| 1/0 is the same as saying 1 * the inverse of zero. In rings and fields, there is no divide operation. In the formal theory, there really is no division operator at all, just multiplication by an inverse or addition by an inverse. (This means subtraction is also not a formal operator). So yeah... bad Beneficii, bad! | |||
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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 MathOnNapkins Yes, I'll be captain obvious now and add (I'm writing all of this, not copying from a site ):
Multiplication is just queued sum, 5*10 just means: 5+5+5+5+5+5+5+5+5+5 = 50 Division is just the question: "How many times should I sum y+y to get x?", as in: 50/5 = 5+5+5+5+5+5+5+5+5+5 = 10 times So then, 1/0 is "How many times should I sum 0+0 to get 1?", as in: 1/0 = 0+0+0+0+0+0+0+0+0+0+0... And so, not even ∞ would work, because you can do 0+0 forever, but you will never get 1. On the other hand, 0/0 would be "How many times should I sum 0+0 to get 0?" as in 0/0 = 0+0 = 1 time or 0/0 = 0+0+0+0+0+0+0+0+0+0 = 10 times That means 0/0 can be anything, since we are searching for a number that multiplied by 0 gives 0, and all of them do! That's why 0/0 is undefined and 1/0 gives an error. However, since we can change 0 for a x, then everything should be possible, as in: (1/x) / (1/x) = (1/x) * (x/1) = x/x So we have that: (1/0) / (1/0) = 0/0 XD |
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spel werdz rite Since: 11-19-05 Last post: 6297 days Last view: 6296 days |
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| You wouldn't get a defined answer, but you could use Calculus and say the
limit as x->0 of (1/x)^(-1) is 0. (edited by spel werdz rite on 10-29-06 05:20 PM) |
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Tommathy Since: 11-17-05 From: Cloud Nine, Turn Left and I'm There~ Last post: 6296 days Last view: 6296 days |
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| Damn it, someone beat me to limits. | |||
Black Lord + Flurry Since: 11-17-05 From: Where indians still roam... Last post: 6298 days Last view: 6299 days |
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| I'm not quite sold on the (1/0) ^ (-1) being undefined... when you go by order of operations, you take the power or inverse first... like was said and then you get 0/1, which is in fact zero...
As of right now, my TI-89 Platinum is saying that is true also, and I don't think Texas Instruments would allow an error like this to fall through.... TI also says that 1/(1/0) = 1 * (0/1) = 0 is true... which would make sense. Edit: Maple 10 does not agree with my Calculator... granted I don't rely on either in my ventures through math, it is interesting they don't agree. (edited by Black Lord + on 10-29-06 08:39 PM) |
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Deleted User Banned Since: 05-08-06 Last post: None Last view: 6297 days |
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| Oh, there are persons that like math because "It's a science engraved in stone, no matter what happens, 2+2 will always be 4"...
But what happens when you tell them one calculator says 'yes', and the other says 'no'?  |
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Black Lord + Flurry Since: 11-17-05 From: Where indians still roam... Last post: 6298 days Last view: 6299 days |
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Originally posted by Uly Well, I hate to admit it, but I am a Math major... and this is kinda throwing me for a loop... although I'm only a soph. I mean, both seem plausible, the whole order of operations seems plausible to me, but then again, the whole divide by zero = no, no matter what also makes sense... I almost wonder if I missed a bug fix in Maple because I never download them... |
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rubixcuber Mole  Since: 09-08-06 From: St. Louis, MO Last post: 6405 days Last view: 6405 days |
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| (1/0)^(-1) is undefined.
The TI-89 will probably say 0 but give a warning down at the bottom. I'm pretty sure that's what the 92 does. Here's why (1/0)^(-1) can't be 0: Assume (1/0)^(-1) = 0: (1/0)^(-1) * (1/0)^(1) = 0 * (1/0)^(1) (1/0)^0 = 0 1 = 0 And that's clearly not true. |
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MathOnNapkins 1100 In SPC700 HELL Since: 11-18-05 Last post: 6296 days Last view: 6296 days |
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| Black Lord: by order of operations, whatever is in parentheses is 'executed' first, and that is why 1 divided by zero would execute prior to the power outside of it. Hence the whole exp |
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Young Guru Snifit  Since: 11-18-05 From: Notre Dame, IN Last post: 6303 days Last view: 6296 days |
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| Rubixcuber: 0*(1/0)^1 != 0. Zero multiplied by undifined does not equal zero.
As for limits, they don't tell you the value of 1/0, or 0/0, or any other ambiguous fraction. Say you take the limit as x --> 0 of sin(x)/x (aka 0/0 at x=0) The result of the limit will give you 1, but sin(x)/x at x=0 will still be undefined. You use the limit to make a point substitution for f(x) at x=0 to be f(0)=1 and then sin(x)/x will be continuous if you define f(x) to be sin(x)/x for all x != 0 and f(0) to be 1 As for the arguement at hand (1/0)^-1 is undefined. If you do order of operations you get 1/0 first as others have stated. If you chose to rewrite it you get (1^-1)/(0^-1) which gives you 1/undefined. And you shouldn't be using calculators to prove mathmatical ideas, use theorems and proofs, because calculators take shortcuts that don't use mathmatical principles all the time. They often use approximations and physical relations to come up with methods that give the correct solution to a good degree of accuracy but that fail under certain situations. (edited by Young Guru on 10-30-06 11:51 AM) |
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beneficii Broom Hatter Since: 11-18-05 Last post: 6300 days Last view: 6296 days |
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Originally posted by rubixcuber But that doesn't prove it wrong. Your error was that you multiplied both sides by what is clearly an undefined number (1/0)^(1) which clearly equals just (1/0), which is clearly undefined. MON, Your answer in that last one probably makes the most sense. If you look at the problem itself, and you distribute the exponent into the parantheses, you will have: 1 ^ -1 / 0 ^ -1 and within you're going to have 0^-1 power, which is undefined. Thank you for helping. 
What I'm trying to do is find the value of 1 / (-k)!, where k is an integer > 0. Basically, I'm working on the inverse of a negative factorial. The reason goes to the binomial coefficients (Pascal's triangle), where I was trying to find the value of C(0, 1)*. My best guess was to take C(1, 1) - C(0, 0), which is 1 - 1, and 0. I noticed if you simplified the problem, it would come out to 1 / (-1)!. I wondered if then the inverse of a factorial of a negative integer would be 0. *C(n, k) = n! / (k! (n - k)!) |
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MathOnNapkins 1100 In SPC700 HELL Since: 11-18-05 Last post: 6296 days Last view: 6296 days |
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| You know... the factorial is just the Gamma function, and most calculators nowadays have it built in. You can also set it up as an integral. As I recall it doesn't work for certain negative values but I don't remember the specifics. Maybe it was negative odd integers... |
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