24 Jul 1998 12:26:53 -0400

Related articles |
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[12 earlier articles] |

Re: Is infinity equal to infinity? larry.jones@sdrc.com (Larry Jones) (1998-07-20) |

Re: Is infinity equal to infinity? darcy@usul.CS.Berkeley.EDU (1998-07-20) |

Re: Is infinity equal to infinity? darcy@usul.CS.Berkeley.EDU (1998-07-20) |

Re: Is infinity equal to infinity? darcy@usul.CS.Berkeley.EDU (1998-07-20) |

Re: Is infinity equal to infinity? joachim.durchholz@munich.netsurf.de (Joachim Durchholz) (1998-07-20) |

Re: Is infinity equal to infinity? miker3@ix.netcom.com (1998-07-21) |

Re: Is infinity equal to infinity? dwcantrell@aol.com (1998-07-24) |

From: | dwcantrell@aol.com (DWCantrell) |

Newsgroups: | comp.compilers |

Date: | 24 Jul 1998 12:26:53 -0400 |

Organization: | AOL http://www.aol.com |

References: | 98-07-159 |

Keywords: | arithmetic |

Michael Rubenstein writes:

*>>However, if the numerator converges*

*>>on some non-zero value c, the ratio is +/-infinity. Therefore,*

*>>in IEEE arithmetic x/0 is +/-infinity for any finite, non-zero x.*

*>*

*>let x(i) = 1; y(i) = (-1)^i/i*

*>*

*>then lim(i->inf) x(i) = 1*

*> lim(i->inf) y(i) = 0*

*> lim(i->inf) x(i) / y(i) = ?*

*>*

*>[I think the word "continuous" got lost somewhere in there. -John]*

Actually it was not the idea of continuity that got lost. A similar

example could be given taking x(t) = 1 and y(t) = sin(t)/t. Note that

y(t) is continuous for all nonzero real t, and lim(t->inf) y(t) = 0 from

a standard mathematical viewpoint, in which 0 is *unsigned*.

What actually got lost was the fact that in IEEE 754 arithmetic, all

zeros are *signed*. Therefore in both the quoted example and in my

example, if zeros must be signed, then lim y does not exist! Of

course lim x/y does not exist either, but that fact then ceases to be

problematic.

This all goes to show that the projective mode, having one unsigned 0

and one unsigned infinity, does have its uses. In that mode (which

unfortunately does not appear in IEEE 754 itself, although it did appear

in some drafts of the standard), all of the limits discussed above exist.

David Cantrell

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