Do you know Square Root
Leibniz , a German mathematician stated that
SQRT(1 + SQRT(-3)) + SQRT(1 - SQRT(-3)) = SQRT(6)
What do you think ?
SQRT(1 + SQRT(-3)) + SQRT(1 - SQRT(-3)) = SQRT(6)
What do you think ?
Labels: mathschallenge
posted by ToM at
9:13 AM
13 Comments:
As long as it's understood that the principal values of sqrt are implied, then Leibinz was right.
That the principal value is being used is implicit in the expression.
super duper!
square that whole thing and do binomial playaround =)
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You should have actually tried it - you wouldn't have found anything to apply the binomial theorem to (except for the pure heck of it).
Also squaring is dodgy, you turn -'s into +'s.
e.g. 1 = -1 because (1)^2 = (-1)^2, I don't think so.
well, i DID try, thought it wasnt worth to post after what we discussed here... also the only thing with squaring is your double solution... 1^2=(-1)^2 => 1=+-(-1)
and thats correct
sqrt(x)=s(x)
s(1-s(-3))+s(1+s(-3)) |square
(1-i*s(3))+2*s(1-i*s(3))*s(1+i*s(3))+(1+i*s(3)) |i*s(3) cancel once
2+2*s[(1-i*s(3))*(1+i*s(3))] |(a-b)(a+b)=a^2-b^2
2+2*s(1^2-3*i^2) | i^2=-1
2+2*s(1+3)=2+2*s(4)=2+4=6
take root -> =+-s(6)
You didn't use the binomial theorem, you used the "difference of two squares (identity)".
I didn't check it by squaring, but I accept that your method is valid despite the general warning. I can be quite a pedant (as you've probably noticed).
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Because Leibniz explicitly distinguish between +/- Sqrt(3) and
because I suspect that he would have used the square root symbol, I
took it that principal values were to be assumed throughout.
Although implicit, this is the standard convention and a
mathematician would be expected to assume it.
Using s(x) to mean the principal value => s(-3) = i*s(3) etc.
s(1 + i*s(3)) = s(e^(i*pi/3)) = 2*e^(i*pi/6)
Similarly get 2*e^(-i*pi/6) for the other term.
NB in above only principal values are taken. I've omitted the other values.
Using Euler's identity, e^(i*x) = cos(x) + i*sin(x) we obtain
the whole sum = 4*cos(pi/6) = s(6)
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If I had actually being processing a similar expression for real, I would have said that Sqrt(6) was ONE of the values of the original expression.
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