| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I was reading the STAR_TREK conference recently about warp speeds etc.
This made me think about a question which has bothered me for some
time been seems to have crept into general parlance without anyone
noticing.
The discussion also included a transgression into why objects with a
mass cannot attain the speed of light (c). This was pointed out as
being the result of an equation in general or special relativity (I
forgot which). To paraphrase:
"... the central part in this equation is:
---------
| /
| / v�
| / 1 - --
|/ c�
This means that when v approaches c this equation goes to 0. This means
the mass goes to infinity. And when v goes beyond c this is
the square root of a negative number, which means the mass becomes
imaginary. ...."
^^^^^^^^^
My question concerns the last statement. I feel is sloppy in the sense
that I think it is incorrect.
I was always under the impression that taking the square root of a
negative number was not allowed. There seems however to be a trend to
say that the square root of -1 is the imaginary number i. So that:
---- ---
|/ -n = i*|/ n
Does the definition of the square root include this?
I have always thought that i was defined as that number who's square
was -1, i.e. i� = -1, and never as the square root of -1.
Thanks, Wildrik
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1546.1 | SRQ(-1) = i | PLAYER::BRH932::VERHOEVEN | Fri Jan 24 1992 07:17 | 10 | |
SQR(-1) = i. This is tru for imaginary numbers For real numbers SQR(-1) does not exists Somebody else can probaly explay you why in good English, I studied math in Dutch an French and don't know the porper English terms. Johnny | |||||
| 1546.2 | WONDER::COYLE | Fri Jan 24 1992 08:55 | 10 | ||
The trouble with the number whose square is -1 being the definition
of <i> is that would include two numbers: +i and -i.
_____
With i = \/ -1 we assume the positive root by conventional
use of the square root operator. If we want to use both roots
we use: _______
+ \ /
- \/
-Joe
| |||||
| 1546.3 | and "imaginary" is not a good term, either | VMSDEV::HALLYB | Fish have no concept of fire | Fri Jan 24 1992 09:02 | 8 |
> With i = \/ -1 we assume the positive root by conventional
Hmmm, just what do you mean by "positive" here? (A trick question).
Actually there are 2 complex number systems that have i� = -1, and we
don't really care "which" i is used, since the two systems are isomorphic.
John
| |||||
| 1546.4 | why did he say that? | BUZON::BELDIN_R | Pull us together, not apart | Fri Jan 24 1992 09:14 | 15 |
To be precise, one would have to define what s/he includes as a "number" in the particular discussion. So, if one only counts the "reals", then sqrt(-1) does not describe any number. If you admit complex numbers, then there is an operation which, when restricted to the reals, yields square roots, and which, when applied to -1, yields +i and -i. You could make a case that this is a reasonable definition for the square root of -1, but very pedantic people would require the kind of qualifications I made here. As asked in .0, is it sloppy? Yes, I guess so. But then, what is the purpose the author is trying to accomplish. As I read .0, the author (or speaker) was trying to help students understand why velocities beyond the speed of light were "peculiar" in some way, and he used this example. I guess he was successful. Dick | |||||