Re: A83: Re: Square root of 0, OT

To: assembly-83@lists.ticalc.org
Subject: Re: A83: Re: Square root of 0, OT
From: Linus Akesson <lairfight@softhome.net>
Date: Sat, 05 Sep 1998 23:39:26 +0100
Delivered-To: assembly-83-outgoing@towerguard.unix.edu.sollentuna.se
In-Reply-To: <a5241d1d.35f19edd@aol.com>
Reply-To: assembly-83@lists.ticalc.org


On 05-Sep-98, JUSTINJS@aol.com wrote:

>In a message dated 9/5/98 8:38:49 AM Pacific Daylight Time,
>lairfight@softhome.net writes:

>> Subj: Re: A83: Re: Square root of 0, OT
>>  Date:9/5/98 8:38:49 AM Pacific Daylight Time
>>  From:lairfight@softhome.net (Linus Akesson)
>>  Sender:owner-assembly-83@lists.ticalc.org
>>  Reply-to:assembly-83@lists.ticalc.org
>>  To:assembly-83@lists.ticalc.org
>>  
>>  
>>  On 05-Sep-98, JUSTINJS@aol.com wrote:
>>  
>>  >ok enuff about deviding by 0 and it square root, what would the square
>root
>>  of
>>  >Iota be, lodgicaly it should be imposible becouse Iota = the smalest
>amount
>>  >posible( sqr. root of Iota = ERROR)   what my calc says is
>.7071067812+.70..
>> .
>>  >??????? 
>>  
>>  Sigh. If your calculator says .7071067812+.70... then it means that if you
>>  press right, you'll see the rest of the answer. Probably ends with an i.
>ok, yes it did but...
>Iota is the smalest # corect? so how could it get smaller?

um.. where did you find iota on the calculator?
That's the imaginary i you're playing around with. 2nd-decimalpoint, right?
FYI (people on this list seem to love that tla): i = sqrt(-1), so it isn't a
real number.

Linus

References:

Re: A83: Re: Square root of 0, OT

From: JUSTINJS@aol.com