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 )

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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

[download]

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

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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.

In reply to Re: (tye)Re: What is zero divided by zero anyway? by jordanh
in thread What is zero divided by zero anyway? by BrowserUk

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