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| User | Post |
| Omega45889 Posts: 88/92 |
MoN is clearly right.
(1/0) = undefined |
| HyperHacker Posts: 3821/5072 |
Well, Windows Calculator agrees with Kyouji. I put in "(1/0)" and  . |
| MathOnNapkins Posts: 907/1106 |
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. |
| Deleted User Posts: 797/-7750 |
Would you give us the link? |
| MathOnNapkins Posts: 901/1106 |
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. |
| Deleted User Posts: 791/-7750 |
Indeed, knowing that the meaning of life is (undefined) means I can finally die =) |
| Guy Perfect Posts: 403/451 |
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. |
| Cirvania Posts: 960/1181 |
Let's consult the almighty tunabot, shall we?
Originally posted by tunabot on IRC |
| Guy Perfect Posts: 402/451 |
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 expressions become incorrect at the end. In Uly's example, x(x - x) / (x - x) would theoretically yield x, but the fact that (x - x) equals 0 means, essentially x(0) / 0... 0 / 0... It's (Undefined), not x. 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. |
| Deleted User Posts: 771/-7750 |
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  ). |
| beneficii Posts: 276/310 |
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. |
| rubixcuber Posts: 322/356 |
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. |
| MathOnNapkins Posts: 896/1106 |
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... |
| beneficii Posts: 275/310 |
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)!) |
| Young Guru Posts: 199/279 |
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. |
| MathOnNapkins Posts: 895/1106 |
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 expression is undefined because of a domino effect. |
| rubixcuber Posts: 311/356 |
(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. |
| Black Lord + Posts: 218/273 |
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... |
| Deleted User Posts: 755/-7750 |
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'?  |
| Black Lord + Posts: 216/273 |
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. |
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