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Search - "formula 1"
Question - is this meaningful or is this retarded?
2*3 = 6
2*2 = 4
2*1 = 2
2*0 = 0
2*-1 = -2
then why doesnt this work?
6/3 = 2
6/2 = 3
6/1 = 6
6/0 = 0
6/-1 = -6
if n/0 is forbidden and 1/n returns the inverse of n, why shouldn't zero be its own inverse?
If we're talking "0" as in an infinitely precise definition of zero, then 1/n (where n is arbitrarily close to 0), then the result is an arbitrarily large answer, close to infinite, because any floating point number beneath zero (like an infinitely precise approximation of zero) when inverted, produces a number equal to or greater than 1.
If the multiplicative identity, 1, covers the entire set of integers, then why shouldn't division by zero be the inverse of the multiplicative identity, excluding the entire set? It ONLY returns 0, while anything n*1 ONLY returns n.
This puts even the multiplicative identity in the set covered by its inverse.
Ergo, division by zero produces either 0 or infinity. When theres an infinity in an formula, it sometimes indicates theres been
some misunderstanding or the system isn't fully understood. The simpler approach here would be to say therefore the answer is
not infinity, but zero. Now 'simpler' doesn't always mean "correct", only more elegant.
But if we represent the result of a division as BOTH an integer and mantissa
1.234567 or 0.1234567,
i.e. a float, we can say the integer component is the quotient, and the mantissa
is the remainder.
Logically it makes sense then that division by zero is equivalent to taking the numerator, and leaving it "undistributed".
I.e. shunting it to the remainder, and leaving the quotient as zero.
If we treat this as equivalent of an inversion, we can effectively represent the quotient from denominators of n/0 as 1/n
Meaning even 1/0 has a representation, it just happens to be 0.000...
(n * (n/0)) = 1
the multiplicative identity
(n* (n/0)) == (n * ( 1/n ))
People who math. Is this a yea or nay in your book?25