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 |
T.R | Title | User | Personal Name | Date | Lines |
---|---|---|---|---|---|
1546.1 | SRQ(-1) = i | PLAYER::BRH932::VERHOEVEN | Fri Jan 24 1992 10:17 | 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 11:55 | 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 12:02 | 8 |
1546.4 | why did he say that? | BUZON::BELDIN_R | Pull us together, not apart | Fri Jan 24 1992 12:14 | 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 |