# Re: comparison operators and *typos

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``` | Date: Mon, 20 Jun 2005 12:48:21 +0800
| From: "Chongkai Zhu" <mathematica@xxxxxxxxx>
|
| ======= At 2005-06-20, 10:06:21 Aubrey Jaffer wrote: =======
| >...
|
| (= -0 0) should be #f.

Then (eqv? -0 0) ==> #f; which will break much existing Scheme code
which tests for 0.  The ZERO? procedure expects a number; one can test
whether an arbitrary object is 0 with (eqv? obj 0).  Because of -0,
this test must be replaced by (and (number? obj) (<= -0 obj 0)).

Another example where -0 breaks existing code is:

(case val
((0) ...)
((1) ...)
(else ...))

will not match when VAL is -0.

(exact->string (my* -5 0)) ==> "-0".  So -0 will occur often.

| > | procedure: numerator q
| > | procedure: denominator q
| > |     These procedures return the numerator or denominator of their
| > |     argument; the result is computed as if the argument was
| > |     represented as a fraction in lowest terms.  The denominator is
| > |     always positive or zero.  The denominator of 0 is defined to be
| > |     1.

(my-numerator |-0|)	==> 0
(my-denominator |-0|)	==> 1
(numerator 0)		==> 0
(denominator 0)	==> 1

If the NUMERATORs and DENOMINATORs of -0 and 0 are equal, then -0 and
0 must be the same number.  But as you wrote, "(= -0 0) should be #f."

If (< -0 0), then -0 must be negative; but:

(my-negative? |-0|) ==> #f

If (< -0 0), then (- 0 -0) must be nonzero; but

(exact->string (my- 0 |-0|)) ==> "0"

```