Re: (tye)Re: What is zero divided by zero anyway?

[jordanh]

I think the reason is 0/0 is not defined as 1 is that division by 0 is undefined, in all cases. Division by zero is simply undefined, it makes no sense to divide by zero. Put another way, division is an operation on real numbers defined when the numerator is a real number and the denominator is a real number != 0.

The following definitions hold:

x/x = 1         ( x != 0 )
0/x = 0         ( x != 0 )
x/y = x * (1/y) ( y != 0 )

That's all there is to it. We disallow division by zero and dispense with long winded explanations.

Consider the following formulation that proves 2 = 1.

(1)              a = b            ( a != 0, b != 0 )
(2)             aa = ab
(3)        aa - bb = ab - bb
(4) (a - b)(a + b) = b(a - b)
(5)          a + b = b
(6)          b + b = b            (from a = b, (1))
(7)             2b = b
(8)              2 = b/b
(9)              2 = 1

This flaw in this "proof"? To get to step 5 we divide both sides by (a - b) (which, because a=b, (a-b) = 0).

You could formulate step 5 as:

(5)   ( ( a - b ) / ( a - b ) )( a + b ) =  ( ( a - b ) / ( a - b ) )b

In which case, you'd be multiplying both sides by 1 if you allowed that 0/0 = 1. But, 0/0 doesn't = 1, x/0 is simply disallowed, always, even when x = 0.

[tye]

It doesn't matter. Sure, it doesn't "make sense" to divide by 0. But there are several cases where you can decide to define that some specific expression that doesn't really "make sense" should be considered to have a specific value because doing so would simplify having to deal with some edge cases.

For example, 0**0 is generally defined to be 1 even though 0**X is 0 for all values of X > 0. Also, 0! (factorial) is defined to be 1. Because these "conventions allow us to extend definitions in different areas of mathematics that would otherwise require treating 0 as a special case."

I wish I could remember cases where continuous functions were the complete justification for such conventions (I know there is at least one such, but I can't think of it).

        - tye (but my friends call me... um... something)

[Zaxo]

An example where 0/0 really is defined and finite: $y = sin($c * $x) / $x; , considered continuous with $c constant, should give $y == $c where $x is zero.

After Compline,
Zaxo

[jordanh]

For example, 0**0 is generally defined to be 1 even though 0**X is 0 for all values of X > 0. Also, 0! (factorial) is defined to be 1. Because these "conventions allow us to extend definitions in different areas of mathematics that would otherwise require treating 0 as a special case."

Seems to me that this is why we don't allow 0/0 = 1, or in fact, n/0 ever. Because, it doesn't work without making a lot of special cases. If 0/0 = 1, then the proof I gave before would prove that 2=1.

Here's a definition that we'd have to apply a special case to if 0/0 = 1 were allowed:

   0 * x = 0  (except when x = n/0)

If we simply say that n/0 is not a number, all these special cases go away.

I think the examples of continuous functions that seem to work when the denominator goes to 0 is conflating division by 0 with taking the limit as the denominator goes to 0. I'm not sure, as I'm no expert in math, but I believe that these are all examples of functions whose values are approximations represented by infinite series. In such cases, you would have to examine the infinite series to understand what's really going on.

[Abigail-II]

The fact that 0! == 1 is a logical extension of the definition of factorial: n! == n * (n - 1)!.

    1 == 1! == 1 * (1 - 1)! == 1 * 0! == 0!

It's not an arbitrary value, it's a logical one.

But we can't do so for 0/0. Consider:

        0 
    lim - == 0
   x->0 x 
 
but,

        x
    lim - == 1
   x->0 x
   

It would also mean the function x / 0 is undefined, except for the case x == 0.

Abigail