You might look at ..
computation with the extended rational numbers...
www.cs.berkeley.edu/~fateman/papers/extrat.pdf?
(1994)
RJF
On 8/5/2013 6:19 AM, Henry Baker wrote:
> I've been playing with rational numbers which allow the denominator to be zero. While
> these don't solve the problem of continued fractions (discussed previously), they do
> allow for the representation of +oo ("inf") and -oo ("minf"), and comparisons with them
> work correctly.
>
> x=xn/xd < y=yn/yd iff xn*yd < yn*xd.
>
> If x=xn/xd is finite, then -1/0 < xn/xd and xn/xd < 1/0.
>
> Proof. -1*xd < 0*xn=0 and 0=xn*0 < 1*xd (xd>0 because x is finite).
>
> (It's a bad idea to allow denominators to be negative, because it does screw up comparisons.)
>
> The good news: k/0 = sign(k)/0, because gcd(k,0)=abs(k), so all versions of inf are = and all versions of minf are =.
>
> The bad news: -1/0 = 1/0 (-1*0=1*0), so we don't have minf<inf.
>
> We could cheat by using IEEE -0.0: -1*0.0 = -0.0 < 1*0.0 = +0.0.
>
> Or we simply recognize this case specially, and force -1/0 < 1/0.
>
> Perhaps Common Lisp err'ed in not allowing -1/0, 1/0.
>
> Common Lisp desperately needs identities for min & max, so that
> (min) = 1/0 = inf ('min' with 0 arguments) and
> (max) = -1/0 = minf ('max' with 0 arguments).
>
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