a*0 = 0 isn’t a problem in the one-element ring because there’s no failure of cancellation, I think.
Doesn’t that mean 0/0=0 because 0 is nothing. what’s all this ‘undefined’ nonsense?
a/b is defined to be c iff a = bc and there’s no other number d such that a = bd. There’s no unique c if a = b = 0, so it’s undefined.
I concede. Maths is like Klingon to me. I don’t even know why I posted in this thread (Yes I do. I saw an analogy that I misinterpreted as being able to produce an answer of zero)
It’s funny that. Maths is alien to me, yet I took to programming like a duck to water.
You’d think a good coder would be a good mathmatician.
The reason for calling it “undefined” is that if you define it to be anything (even zero), then you can come up with some absolutely ridiculous mathematical proofs. Here’s a favorite of mine, which “proves” that 1=2:
1: Let a=b
2: Then a[sup]2[/sup] = ab
3: a[sup]2[/sup] + a[sup]2[/sup] = a[sup]2[/sup] + ab
4: 2a[sup]2[/sup] = a[sup]2[/sup] + ab
5: 2a[sup]2[/sup] - 2ab = a[sup]2[/sup] + ab - 2ab
6: 2a[sup]2[/sup] - 2ab = a[sup]2[/sup] - ab
7: 2(a[sup]2[/sup] - a b) = 1(a[sup]2[/sup] - a b)
8: cancel the (a[sup]2[/sup] - ab) from both sides to show 1=2
There are much simpler “proofs” like this, but the more complex versions tend to be more fun, because it takes a bit longer to find the problem.
The answer depends on what zero is. If it is a function of x then you may use l’Hopital’s rule to find the answer. The limit as g(x) goes to zero of f(x)/g(x) =f’(x)/g’(x). For example, limit as x goes to zero of sin (x) / x = limit as x goes to zero cos (x) / 1 = 1.
See posts #5 and #7.
Huh? Who else mentioned l’Hopital’s rule or derivatives?
Imagine if 0/0 were defined as 1, as has been proposed. Then we could do the following:
0 = 0
Anything times zero is still zero, so we can do this:
20 = 10
This is still 0=0, so we’re fine. Now let’s divide both sides by zero:
2*(0/0) = 1*(0/0)
And, since 0/0=1, it follows that:
2 = 1
This is the sort of nonsense we can “prove” if we allow division by zero to be defined.
This is probably one of the simpler proofs InvisibleWombat refers to. I like it because the use of actual numbers rather than variables sometimes makes the point more clear to the uninitiated.
They mentioned limits. L’Hopital’s rule is but one way of evaluating a limit–there are others.
It is indeterminate mathematically speaking.
It is a DEAD HORSE!
Stop beating it to death!
It’s A Black Hole!!! I Swear It!
But 0! (zero factorial) is defined as 1. :eek:
This is getting really odd, if 0! is equal to 1…found this great website:
If n! is defined as the product of all positive integers from 1 to n, then:
1! = 11 = 1
2! = 12 = 2
3! = 123 = 6
4! = 1234 = 24
…
n! = 123…(n-2)(n-1)n
and so on.
Logically, n! can also be expressed n(n-1)! .
Therefore, at n=1, using n! = n*(n-1)!
1! = 1*0!
which simplifies to 1 = 0!
n: Number of Permutations (n!): Visual example:
1 1 {1}
2 2 {1,2}, {2,1}
3 6 {1,2,3}, {1,3,2}, {2,1,3}, {2,3,1}, {3,1,2}, {3,2,1}
10 3,628,800 ummm, you get the idea…
Therefore,
0 1 { }
It can be said that an empty set can only be ordered one way, so 0! = 1
obviously if 0 can be arranged in the permutation of the empyt set, then why can’t one permutation over another be equal to 1?
Huh? This does not parse.
Because a permutation is not a real number. How can you divide a permutation by another? You have define what you mean by division over permutations.
I think you are a bit confused. n! is the number of permutations of a set with n members. For instance, it is what you get when you try to count the number of ways to arrange n different balls from left to right so that each ball appears exactly once. The fact that 0! = 1 reflects that the number that there is one way to arrange 0 different balls from left to right so that each balls appears exactly once.
Saying you want to divide permutations is like saying you want to divide the things you are trying to count. If you are counting pencils, do you think you can divide a pencil by another pencil?
Be careful. When rigorously working out the model of number theory afforded by set theory, certain numbers are defined as sets of functions. In particular, given a set A in the equivalence class of sets labelled “n”, n! is represented by the set of set automorphisms (permutations) on A. When you want to calculate something like factorials of transfinite numbers this is pretty much necessary.
Admittedly, your point that within the context of this discussion such a concept as dividing sets hasn’t been well-defined still holds.
Well, to even talk about invertibility we have to assume we’re talking about rings with unit. In this case a0 = a(1-1) = a-a = 0.
Now, assume there exists an element i (left-)inverse to zero. We have 1 = i*0 = 0, which implies the additive and multiplicative identities are, well, identical.
Now, given any a, a = a1 = a0 = 0, so the ring has only one element. In this case, the ring is actually a (rather degenerate) field, and 0/0 = 0.
It’s not indeterminate, it’s undefined. The whole statement that “0/0 is indeterminate” is waving the question away. It stems from the question:
If lim_{x->x_0} f(x) = 0 and lim_{x->x_0} g(x) = 0, then what is lim_{x->x_0} f(x)/g(x)?
This question is indeterminate, since to answer it you need to know more information – the particular forms of f and g. Once you have those forms the answer is very well determined.
However, the question purely in terms of 0/0 (as in the OP) cannot even reasonably be interpreted in this way. It mentions nothing but 0 and division, which are both algebraic concepts. There is no assumption that this is the zero of R (as opposed to zero in Q or zero in Z). To even set this up as a question about limits inherently assumes a rich topological structure to the concept under study which simply is not afforded in the OP as it stands. The only tenable position is that it is inherently a ring-theoretic question, where the answer is that division by zero is undefined except in the degenerate case of a ring with a single element.
I suppose somebody has got to ask. What is this “ring” business?
A ring is a set of numbers which is closed under two binary opertaions (+, *) and satisfies the following conditions:
Additive identity: there is an element (0) such that a + 0 = a for all a.
Additve inverse: For all a there is an elemnt -a such that a + (-a) = 0
Associtvity: for all a, b and c (a + b) + c = a + (b + c) and (ab)c = a(bc)
Additve commutativity: for all a and b a + b = b + a
Distributivity: For all a, b and c a*(b + c) = (ab) + (ac) and (a + b)c = (ac) + (b*c)