Division by 0

This is a discussion on Division by 0 within the A Brief History of Cprogramming.com forums, part of the Community Boards category; Let's assume such a number exists...call it a. How far can we get? What rules of division and multiplication do ...

1. Let's assume such a number exists...call it a. How far can we get? What rules of division and multiplication do we need to meet?

That would be quite impractical. If you're to work it out, changing the division in real numbers in such a way would create two absorbant elements in real mulitplication, which would then lead to having to reformulate many of the other division properties. That would also lead to restating what a positive and negative number would be, which would then make you...get my point ?

You can go ahead and define such a number "a", but unless you find practical use for it, and allow it to coincide with the other properties defined in regular arithmetic, I doubt it will be wide-spread and accepted.

Even at that, I doubt a ring will ever exist in which division by 0 is defined, simply due to the fact that no one needs to know what a number divided into 0 distinct entities is equal to. The same cannot be said about negative roots and logarithms, whose necessity were the prime motive for the creation of complex numbers.

When they find a motive for 3/0 to have a result, then the appropriate result will be given. Until then, it will be undefined.

2. not sure if anyone has pointed this out yet (damn thread is too long to read) but:

>>I was thinking about why you can't divide anything by 0 (except 0).

You can't divide 0 by 0 either. Its undefined as well.

3. Originally Posted by Happy_Reaper
Even at that, I doubt a ring will ever exist in which division by 0 is defined,
It doesn't. Proof:
z is the zero in the ring.
u is the one in the ring.
u/z = u*z', where z' is the multiplicative inverse to z.
z * z' = u
but by definition z*'z=z
There exist no z' so that zz'=u.

4. Makes sense. I guess I should've seen that. This very same proof should satisfy all the doubtful out there. You cannot divide anything by 0 because division as a binary relation has no way of defining it in the first place.

5. The person who invented zero probably had plenty of problems with it. Talking about zero it gets really philosophical. Some people would say 0 is the measure of nothing, but how can you measure nothing, so if nothing is being measured than it must be something. So people need to look at it as perceptual nothingness(what we percieve to be nothing). So lets say 0 is the measure of nothing or a value assigned to nothing(sounds pretty messed up) so 5/nothing would surely be 5(taking something and not splitting it then you have the same thing as before). As for 0/0 nothing cant go into nothing at all so I don't see how it could possibly be infinite. How many times does it go in? None at all, thus the answer is 0 or nothing. Your teachers may not even care what you think so if they act negatively when you bring it up it would be better to put up with their arrogancy rather than flunk. Mathematicians, teachers, and scholars have probably adressed this problem and decided to teach it one way everywhere, but questioning is human nature and I think its best you point out them being wrong.

6. that's what limits are for anyway, you can't calculate x/y when y = 0, so you calculate the limit of x as y approaches 0.

7. >>I was thinking about why you can't divide anything by 0 (except 0).

You can't divide 0 by 0 either. Its undefined as well.
Exactly. Here's why:

If 0/0 = x, then x*0 = 0. So x can be anything. 0/0 can be anything. Rather than let it be anything, it's undefined.

8. what I never understood is how x^0 = 1. I'm sure there's some proof for it out there, but it just doesn't make sense to me.

9. Again because its defined as such. (of course 0^0 is the exception)

What that defination does is allows for a smooth transition from postive exponents to negative exponents:

x^3 = x^3 / 1
x^2 = x^2 / 1
x^1 = x^1 / 1
x^0 = x / x
x^-1 = 1 / x^1
x^-2 = 1 / x^2
x^-3 = 1 / x^3

10. Actually, 0^0 is also equal to 1.

11. Originally Posted by Happy_Reaper
Actually, 0^0 is also equal to 1.
Actually no its not. Its indeterminate. Most calculators and even computer implmentation take a shortcut and if the exponent is 0 returns 1.

12. 0^0 is undefined, but
lim x->0 (x^x) exists and is equal to 1

Here's the graph of Re(x^x):

13. Originally Posted by Sang-drax
0^0 is undefined, but
lim x->0 (x^x) exists and is equal to 1

Here's the graph of Re(x^x):
that's good enough for me then

14. Originally Posted by Sang-drax
0^0 is undefined, but
lim x->0 (x^x) exists and is equal to 1

Here's the graph of Re(x^x):
That only approaches the point of discontinuity from a particular direction, though. Exponentiation is a multivariable function. Let f be a function defined by f(x,y) = x^y, where x and y are real numbers and x >= 0.

lim x->0 [f(x,0)] = 1.
lim y->0 [f(0,y)] = 0.

Hence lim (x,y)->(0,0) [f(x,y)] is undefined.

0^0 = 1 is usually more useful, though, especially with binomial expansion. To say (a + b)^n = (the sum for i = 0 to n of (n choose i) * a^i * b^(n - i)) is rather elegant, but it implies that 0^0 = 1.

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