zero by zero

[papernuke]

Any number divided by itself is one correct?
Then what would zero divided by zero be?

Idea derived from reply from wendy from "Points in a Line" (geometry section)

[Keiji]

If you take the limit as x->0 of x/x, it suggests 0/0 = 1.
If you take the limit as x->0 of 0/x, it suggests 0/0 = 0.
If you take the limit as x->0 of a/x where a is a nonzero constant, it suggests 0/0 = infinity.

Things such as 0/0 can't really be explained in the general sense from arithmetic laws - where needed, such singularities are defined to take the most useful or appropriate value. For example, the number circle defines a/0 to be infinity for any real a (and the number sphere for any complex a).

[wendy]

You can't actually do something like "0/0", becasue you can never guarrantee that the known bits of 1/0 align.

What you do is operations that assume that 0/0 or 1/0 has a fixed orientation. Euclidean geometry depends on it.

[Alex S]

Theoretically, I see that 0/0 is either 0 or 1, depending on whether you think that there is 1 zero in zero, or zero zeros in zero. Anything else divided by 0 would by infinity.

[wendy]

It is relatively easy to show that 0^0 = 1, but this can be done without using division. [hint: 0 can be the result of a count].

For division, one should read 0 as 1/u, for a variously large and indefinite u. Some information exists for u in particular instances, so 0/0, as v/u, comes to be definite when there is a definite relation between u and v. Most of the time this is not the case, but there are particular instances where v/u has a definite value. It is this particular instance, that allows Euclidean geometry (where R ~ u), to work, and the relation v/u allows Euclidean geometry to resize without error (ie from R ~ u, to R ~ v).

[sher77]

If you take the limit as x->0 of 0/x, it suggests 0/0 = 0.
If you take the limit as x->0 of a/x where a is a nonzero constant, it suggests 0/0 = infinity.

(Spambot post)

[student91]

When u take lim x->0 ((x^2+ax)/x), this becomes 0/0, but when u evaluate the limit, you get a

[Prashantkrishnan]

0/0 can assume any value...
Verify it by multiplying zero on both sides.

If this explanation is unsatisfactory, assuming that we are not allowed to multiply or divide zero on both sides of an equation, just take the limits. Eg:- lim x -> a (x² - a²)/x - a = 0/0 = x + a or 2a

Since a can assume any value, so can x + a or 2a.

[ICN5D]

Ohh no. Not another bot. Uggggh, we've been plagued with them lately. Username "sher77" is not a real person, it's a bot program that registers and unearths ancient threads that are 7 years old. The signature "NOOR" will become an ad for websites in about a couple of weeks. Bots typically repeat a small snippet of a previous comment, usually word for word. Very few of them change it a little, but the pattern is always the same. They only make one post. Then silence. Then the sig change.

[anderscolingustafson]

0/0 can be anything depending on what equation is used. If you use x/x then it's 1. If you use 2x/x then it's 2. If you use x/0 then it's positive or negative infinity. If you use 0/x it's 0.

[ICN5D]

Although ........ since it seems to have sparked an intelligent conversation, maybe it wasn't such a bad thing, after all.