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Why can't you divide by 0?
Why is 0/0 "indeterminate" and 1/0 "undefined"?
For one thing, when you divide one number by another, you expect the result to be another number. Look at the sequence of numbers 1/(1/2), 1/(1/3), 1/(1/4), ... . Notice that the bottoms of the fractions are 1/2, 1/3, 1/4, ..., and that they're going to zero. If there's a limit to this sequence, we would take that number and call it 1/0, so let's see if there is.
Well, the sequence turns out to be 2, 3, 4, ..., and that goes to infinity. Since infinity isn't a real number, we don't assign any value to 1/0. We just say it's undefined.
But let's say we did assign a value. Let's say that infinity is a real number, and 1/0 is infinity. Then look at the sequence 1/(-1/2), 1/(-1/3), 1/(-1/4), ..., and notice again that the denominators -1/2, -1/3, -1/4, ..., are going to zero. So again, we would want the limit of this sequence to be 1/0. But looking at the sequence, it simplifies to -2, -3, -4, ..., and it goes to negative infinity. So which would we assign to 1/0? Negative infinity or positive infinity? Instead of just assigning one willy nilly, we say that infinity isn't a number, and that 1/0 is undefined.
The reason is related to the associated multiplication question. If you divide 6 by 3 the answer is 2 because 2 times 3 IS 6. If you divide 6 by zero, then you are asking the question, "What number times zero gives 6?" The answer to that one, of course, is no number, for we know that zero times any real number is zero not 6. So we say that division by zero is undefined, for it is not consistent with division by other numbers.
- Dr. Robert
Because there's just no sensible way to define it.
For example, we could say that 1/0 = 5. But there's a rule in arithmetic that a(b/a) = b, and if 1/0 = 5, 0(1/0) = 0*5 = 0 doesn't work, so you could never use the rule. If you changed every rule to specifically say that it doesn't work for zero in the denominator, what's the point of making 1/0 = 5 in the first place? You can't use any rules on it.
But maybe you're thinking of saying that 1/0 = infinity. Well then, what's "infinity"? How does it work in all the other equations?
Does infinity - infinity = 0?
Does 1 + infinity = infinity?
If so, the associative rule doesn't work, since (a+b)+c = a+(b+c) will not always work:
1 + (infinity - infinity) = 1 + 0 = 1, but
(1 + infinity) - infinity = infinity - infinity = 0.
You can try to make up a good set of rules, but it always leads to nonsense, so to avoid all the trouble we just say that it doesn't make sense to divide by zero.
What happens if you add apples to oranges? It just doesn't make sense, so the easiest thing is just to say that it doesn't make sense, or, as a mathematician would say, "it is undefined."
Maybe that's the best way to look at it. When, in mathematics, you see a statement like "operation XYZ is undefined", you should translate it in your head to "operation XYZ doesn't make sense."
There's a special word for stuff like this, where you could conceivably give it any number of values. That word is "indeterminate." It's not the same as undefined. It essentially means that if it pops up somewhere, you don't know what its value will be in your case. For instance, if you have the limit as x->0 of x/x and of 7x/x, the expression will have a value of 1 in the first case and 7 in the second case. Indeterminate.
>1/ Is 0,0,0,... a geometric sequence ?
>Well, sort of. You could say that it's a geometric sequence with common ratio 4, or whatever, but I wouldn't. The reason is that you can't find the common ratio by looking at the sequence, dividing one term by the previous term. So I guess I'd say no. What you might say is that this is a degenerate case of a geometric sequence.
>2/ What is the value of 0/0 ? (is it really undefined or are there an infinite number of values)
>There's a special word for stuff like this, where you could conceivably give it any number of values. That word is "indeterminate." It's not the same as undefined. It essentially means that if it pops up somewhere, you don't know what its value will be in your case. For instance, if you have the limit as x->0 of x/x and of 7x/x, the expression will have a value of 1 in the first case and 7 in the second case. Indeterminate.
>3/ What is the value of 0^0 (^ means exponent)? Can it be considered to be the limit of a^0 as a approaches 0 ?
>0^0 is indeed indeterminate. It turns out that you could make it have any value between 0 and 1, inclusive. You could have 0 if it's the limit as a->0 of 0^a, you could have 1 if it's the limit as a->0 of a^0, and for x in between 0 and 1, (and this is the neat part from Dr. Shimimoto) look at the expression (x^n)^(1/n). This just equals x for all positive values of n. As n->Infinity, this fraction goes to 0^0, but if it's just x the whole time, the limit of the expression as it goes to 0^0 is x. So we could make it anything in between 0 and 1, so it's got to be indeterminate.
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