I know that you usually can't divide by zero, but under certain circumstances, is it possible?
Let me give you 3 examples:
1. 0 / 0
2. (1 / 0) ^ 0
3. (a / 0) * 0
Please explain to me which of these cases (if any) can equal a number or pronumeral that is not "undefined". Thanks!
Let me give you 3 examples:
1. 0 / 0
2. (1 / 0) ^ 0
3. (a / 0) * 0
Please explain to me which of these cases (if any) can equal a number or pronumeral that is not "undefined". Thanks!
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Usually the symbols you've written are interpreted as "indeterminate forms" (or perhaps they're not indeterminate). In that case, the various values are interpreted as limits of functions, where each function approaches the given constant as some variable goes to zero (or some constant, or infinity; it doesn't matter). If you haven't encountered limits, the rest of this post probably won't make much sense; sorry. Here are some examples:
0/1: the limit of x/1 as x goes to 0 is 0, and in fact for any function f(x) where the limit of f(x) as x goes to 0 is 0, we have that f(x)/1 goes to 0 as x goes to 0. This is not an indeterminate form.
0/0: the limit of x/x as x goes to 0 is 1, since x/x is 1 everywhere except at 0. However, the limit of (2x)/x is 2, even though the limit of both 2x and x is 0 as x goes to 0. Similarly, you can make this limit be any constant you wish--which is why this is called an indeterminate form.
(1/0)^0: this is essentially the indeterminate form (infinity)^0, as listed on the Wikipedia page I've linked.
(a/0)*0: when a is not 0, this is essentially the indeterminate form (infinity)*0, as listed on the Wikipedia page I've linked. When a is 0, the whole thing must be 0.
0/1: the limit of x/1 as x goes to 0 is 0, and in fact for any function f(x) where the limit of f(x) as x goes to 0 is 0, we have that f(x)/1 goes to 0 as x goes to 0. This is not an indeterminate form.
0/0: the limit of x/x as x goes to 0 is 1, since x/x is 1 everywhere except at 0. However, the limit of (2x)/x is 2, even though the limit of both 2x and x is 0 as x goes to 0. Similarly, you can make this limit be any constant you wish--which is why this is called an indeterminate form.
(1/0)^0: this is essentially the indeterminate form (infinity)^0, as listed on the Wikipedia page I've linked.
(a/0)*0: when a is not 0, this is essentially the indeterminate form (infinity)*0, as listed on the Wikipedia page I've linked. When a is 0, the whole thing must be 0.
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Sometimes, when the nature of zero that you are dividing with is known, division is permitted. I can sight a simple example.
We know that
n=infinity
Sigma 1/n = 0
n=1
We also know that
n=infinity
Sigma 5/n = 0
n=1
You can divide the second limit by the first. Though they are 0's, the 0/0 here is actually 5.
Many such examples do exist.
:-)
We know that
n=infinity
Sigma 1/n = 0
n=1
We also know that
n=infinity
Sigma 5/n = 0
n=1
You can divide the second limit by the first. Though they are 0's, the 0/0 here is actually 5.
Many such examples do exist.
:-)
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Actually, it is for you, and only you, Mr. Norris.
For anyone else, all 3 of those cases are defined in calculus (usually in examples revolving around L'Hospitals rule) to be of indeterminate form, not necessarily undefined.
For anyone else, all 3 of those cases are defined in calculus (usually in examples revolving around L'Hospitals rule) to be of indeterminate form, not necessarily undefined.
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no it is not possible to divide a number by 0. division by 0 is not defined in any case. 0/0, (1/0)^0, (a/0)*0 is undefined in these cases.unde any circumstances, it is not possible.
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You can divide by a function that approaches the limit 0, which will generally give you an infinite answer. But you can never divide by zero, it is undefined in mathematics.
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Its like you asking to divide something amongst no people. Answer : Not possible.