Your thoughts on Division by 0?
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I am mostly concerned with the operation 0/0. Would this not be the set of all numbers, real and imaginary? Surely it fits - all numbers multiplied by 0 are 0. But, why is it undefined?
I would assume because division was created to represent the concept of "split x things into y parts". And since you can't split anything into 0 parts, it is undefined.
You cannot divide by zero because it is a basic mathematical principle.
http://www.math.utah.edu/~pa/math/0by0.html
Hope this helps.
[edit] Sorry if it seems a little tautological... It just causes problems. Have fun with L'Hopital. :-)
http://www.math.utah.edu/~pa/math/0by0.html
Hope this helps.
[edit] Sorry if it seems a little tautological... It just causes problems. Have fun with L'Hopital. :-)
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So, um, isn't "the set of all numbers" not the same as "a number"?
I'm not assigning a number, I'm assigning a set...
| Again, we run into contradictions if we attempt to assign any number to 0/0. |
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| I'm not assigning a number, I'm assigning a set... |
But a set consists of numbers, and since no definite number can be assigned to the expression 0/0. The only set you could assign would be the empty set.
But even that can't be allowed, as the empty set can be an argument in many logical expressions, whereas 0/0 can occur only in certain limits.
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| all numbers multiplied by 0 are 0 |
Oh? What is 0*C? If you're assuming an equivalent definition as for mZ, it would be {0}, not 0.
Multiplication in (R/{0})^2 is a bijection. Multiplication in R^2 is not. There is nothing wrong with it. It's natural that it's inverse works with sets.
Your statement that 0/0 = R (and, using the same idea, x/0 = empty set) is a valid definition of division. It's a bit uncomfortable to work with anda bit useless though[edit] I'd rather say, that I don't see where it could be usefully applied. I don't see many things..[/edit]
@Athat, mathematics is not strongly typed.
Your statement that 0/0 = R (and, using the same idea, x/0 = empty set) is a valid definition of division. It's a bit uncomfortable to work with and
@Athat, mathematics is not strongly typed.
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There is no such thing as "set of all numbers". If you're working with N then you get N, if you're working with R, you get R. In some cases it's needed to extend R with infinities. In those cases, why not?
00 is a different thing. For all x, x0 = 1 and for all y, 0y = 0. I don't see a way to reason about this. I usually find it defined as 1.
00 is a different thing. For all x, x0 = 1 and for all y, 0y = 0. I don't see a way to reason about this. I usually find it defined as 1.
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| hamsterman wrote: |
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| There is no such thing as "set of all numbers". |
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| hamsterman wrote: |
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00 is a different thing. [...] I usually find it defined as 1. |
That's probably because
lim xx = 1 where x approaches zero
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@ascii - 0^0 = ?????
0^0 = 1. Anything raised to 0 is = 1.
0 / 0 = 0. If I have nothing and divide it into no groups, I still have nothing.
0^0 = 1. Anything raised to 0 is = 1.
0 / 0 = 0. If I have nothing and divide it into no groups, I still have nothing.
@LB, you think that you're always working with C. That's not really bad, you probably are. It's a bit silly to get complex results when you work in N, but that's probably the least of your problems in that case. It's a bit worse if you want to generalise this to some other ring.
@Cubbi, good idea. I would never have thought of it.
@Cubbi, good idea. I would never have thought of it.
On a not too related topic:
I called the math department at the Massachusetts Institute of Technology to find out the proper way to count and whether zero is a real number. Apparently, counting is not MIT's forte. I was told that no one in the math department would comment on that topic. As for zero, a department administrator said, "Our people are interested more in numbers invented after 1972." He told me I needed a number theorist. --Dick Teresi, writing in the Atlantic Monthly about his attempt to establish whether it is proper to count starting from zero |
@mats, yes seeing as I have passed 3rd grade I'm aware that "anything" to the power of 0 is 1. But 0 is interesting. 0^0 = 0^1 / 0 following basic rules of math (as in, x^y = x * x^y-1). However, since 0^1 = 0, this 0^0 = 0 / 0. As has been said, this is undefined. Not so simple.
x0
can be logically deduced to be 1 via the following math.
say we take xy where y can be any value.
Then xy / xy = 1
==> xy-y = 1
==> x0 = 1
However, in math, we usually define that x cannot be zero in order to do the division.
Otherwise this would lead to 0/0 = 1.
As one can see - this would lead to ambiguity - however I do agree with the human logic that if you divide nothing by nothing you would have nothing, however one could also ask the conceptual question - by how many times can you divide nothing amoung nothing and some philospher might argue 1 :)
can be logically deduced to be 1 via the following math.
say we take xy where y can be any value.
Then xy / xy = 1
==> xy-y = 1
==> x0 = 1
However, in math, we usually define that x cannot be zero in order to do the division.
Otherwise this would lead to 0/0 = 1.
As one can see - this would lead to ambiguity - however I do agree with the human logic that if you divide nothing by nothing you would have nothing, however one could also ask the conceptual question - by how many times can you divide nothing amoung nothing and some philospher might argue 1 :)
By the way, my http://cplusplus.com/forum/lounge/58607/#msg316089 was a silly post. How did nobody notice?
Multiplication in R2 is a surjection with or without 0. Also the definition of multiplication as a function R2->R is silly, because then its inverse should be R->R2, which division is not.
I'll try again. For all c in R/{0}, f(x) = c*x is a bijection. Then f-1(x) = x/c. When c = 0, f(x) = 0. Still, in that case, by the definition that f-1(x) = {y : f(y)=x}, f-1(0) = R and f-1(x) = {} when x!=0.
The thing to notice here is that division is not a group operation with or without 0. Really, a/b should be looked at as a*b-1. My the definitions of this post, 0-1 = 1/0 = {} and you can't multiply by an empty set.. (you could define that to be {} too; that would make sense)
@chrisname, no, those are from 16th century.
Multiplication in R2 is a surjection with or without 0. Also the definition of multiplication as a function R2->R is silly, because then its inverse should be R->R2, which division is not.
I'll try again. For all c in R/{0}, f(x) = c*x is a bijection. Then f-1(x) = x/c. When c = 0, f(x) = 0. Still, in that case, by the definition that f-1(x) = {y : f(y)=x}, f-1(0) = R and f-1(x) = {} when x!=0.
The thing to notice here is that division is not a group operation with or without 0. Really, a/b should be looked at as a*b-1. My the definitions of this post, 0-1 = 1/0 = {} and you can't multiply by an empty set.. (you could define that to be {} too; that would make sense)
@chrisname, no, those are from 16th century.
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