# wath is apening here ? i = 1/i -) 1 = -1 ?

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18 years ago
#1
i = sqr(-1)

i = sqr ( 1 / -1 )

i = sqr 1 / sqr (-1) = +- (1 / i)

with - gives: i = i, ok

with + gives i = -i ?????????
0
18 years ago
#2
Ï "Sekhmet" <[email protected]> Ýãñáøå óôï ìÞíõìá news:[email protected]...
[q1]> i = sqr(-1)[/q1]
[q1]>[/q1]
[q1]> i = sqr ( 1 / -1 )[/q1]
[q1]>[/q1]
[q1]> i = sqr 1 / sqr (-1) = +- (1 / i)[/q1]
[q1]>[/q1]
[q1]> with - gives: i = i, ok[/q1]
[q1]>[/q1]
[q1]> with + gives i = -i ?????????[/q1]
[q1]>[/q1]
[q1]>[/q1]

You make error, because

i=sqr(1)/sqr(-1)=1/i ==> (sqr(1)=1 and sqr(-1)=i) i=1/i==>i^2=1

tilemachos
0
18 years ago
#3
"Sekhmet" <[email protected]> wrote:
[q1]> i = sqr(-1)[/q1]
[q1]>[/q1]
[q1]> i = sqr ( 1 / -1 )[/q1]
[q1]>[/q1]
[q1]> i = sqr 1 / sqr (-1) = +- (1 / i)[/q1]
[q1]>[/q1]
[q1]> with - gives: i = i, ok[/q1]
[q1]>[/q1]
[q1]> with + gives i = -i ?????????[/q1]

I'm slightly surprised that there hasn't yet been a response showing where the error lies, so I'll
try to do so now.

There are two different approaches.

(1) If your "sqr" is used to denote the _principal_ square root, as your first line seems
to indicate:

Let me now denote it as Sqrt instead. Then i = Sqrt(-1) = Sqrt(1/(-1)) is correct, and
Sqrt(1)/Sqrt(-1) = 1/i = -i is also correct [and, BTW, Sqrt(1)/Sqrt(-1) = -(1/i) = +i is
incorrect]. So the error lies in claiming that Sqrt(1/(-1)) = Sqrt(1)/Sqrt(-1). The "law" that
Sqrt(a/b) = Sqrt(a)/Sqrt(b) does not always hold! [Note, however, that it does hold if both a and
b are positive.]

(2) If your "sqr" is used to denote the _set_ of _all_ squares roots, as the end of your third line
seems to indicate:

Let me now denote it as sqrt instead. Then the left side of each of your lines should read +/-i,
as an abbreviation for {-i, +i}. Then the law sqrt(a/b) = sqrt(a)/sqrt(b) _does_ hold.
Specifically, we have +/-i = sqrt(-1) = sqrt(1/(-1)) = sqrt(1)/sqrt(-1) = (+/-1)/(+/-i) = +/-i. No

David Cantrell

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