What is root of minus four multiplied by root of minus 9?
There is this faggot teacher at my uni that keeps bugging us with this question.
From my point, in elementary school we would say, it does not exists as number, in high school it would be something like 2i * 3i = -6 and at uni it is still unidentified since we dont define "i" as square root of minus 1 but as i^2 = -1.
So what would be the most logical and simplified explanation of this?
OP
...I am doing math degree and I begin to feel embarrassed about this and similar problem..
Also, getting disappointed at narrow education and interest my teachers have. fucking hell.
Begin, thread about usage and examples in science mathematics has.
Fucking english
>>7752011
That's how most complex analysis texts treat it.
Just like how you can't make zero a denominator because it has no multiplicative inverse in the field axioms.
So we make square root of minus four a set {-2i,2i} in most of text books. And still, when we try to find some roots of polynomial equations we use the absolute value (2i) (like in dif.equations when locating fundamental set) as solution to this problem.
So square root of minus four multiplied by square root of minus nine should be set {-6,6} ?
[math]2i * 3i = 2*3*i*i=2*3*i^2=6*(-1)=-6
>>7752815
Sqrt(-4) = + or - 2i, and similarly -9 = +-3i
So you get 6 or -6, depending on which roots you multiply.
[math]
\displaystyle \sqrt{(-4) \cdot \sqrt{-9}} = -2 (-1)^{\frac{3}{4}} \sqrt{3} = (1-i) \sqrt{6}
[/math]
>>7751979
This has been freaking me out as well, ever since Deligne wrote in one of his three papers on Hodge theory that it didn't matter for his purposes which root of -1 we pick i to be. This seems entirely tautological to me, and it makes me feel like I'm making some kind of incredibly dumb Gal(C/R)-related mistake in daily life.
>>7751979
x = sqrt(-4)*sqrt(-9)
x = 6*sqrt(-1)*sqrt(-1)
x = -6
sqrt(-1) = +/- i
Also, sqrt(x)*sqrt(y)=sqrt(x*y). Thus, sqrt(-4)*sqrt(-9)=sqrt(-4*-9)=sqrt(36)=+/-6
(−4)⋅−9−−−√−−−−−−−−−√=−2(−1)343√=(1−i)6√
>>7752823
No, you should always work in the same branch., so in either case it's -6.
>>7751979
How is this different from asking what is sqrt(4)*sqrt(9)? Both these also have positive and negative solutions. We just say that it is sqrt(4*9) = +/- 6
>>7751979
sqrt(-4)*sqrt(-9)
i*sqrt(4)*i*sqrt(9)
i*2*i*3
please note that i*i=-1
so, -6
By what I assume are valid root rules, we get sqrt(-4*-9)=sqrt(36)=+-6
>>7752011
But that’s true. Try to construct the complex number field. There are several ways to do it but all of them define [math]\mathrm i[/math] as [math]\mathrm i^2\,=\,1[/math] rather than [math]\mathrm i\,=\, \sqrt{-1}[/math].
OP your teacher probably alludes to this:
[eqn]6 \,=\, \sqrt{36} \,=\, \sqrt{(-4)\,(-9)} \,=\, \sqrt{-4}\,\times\sqrt{-9} \,=\, 2\mathrm i \, 3\mathrm i \,=\, -6[/eqn]
which shows how the square root notation is misleading when using it with negative numbers.
>>7753186
Not true. It only works for x,y > 0 by definition of sqrt
>>7753308
It is different. If you are working in set R, that expresion has one answer, 36. Look at the graphic of sqrt function.
>>7753308
x = y
x^2 = y^2
are not equivalent.
Can't you just multiply the insides and get root 36
>>7751979
>There is this faggot teacher at my uni that keeps bugging us with this question.
This is year 12 content, what the hell kind of shit-tier uni are you at?
[math]sqrt{-4} sqrt{-9}[/math]
[math](\pm 2i )(\pm 3i)[/math]
[math]\pm 6[/math]
>>7754721
actually, its not shitty. This guy teaches some computer development subject, and dropped this question in the middle of lectures. nothing to do with complex plane
>>7751979
Just straight up define i as Sqrt[-1] and run with it to get -6.
If some asshat wants to play with definitions further by saying -i = Sqrt[-1] must also be included in the definition then you say the answer is ±6.
To say the answer is undefined is flat out wrong though. The only thing that is in limbo is whether the final value is + or - and that is only because asshat can't decide how he wants to define i.
I think I'm taking crazy pills oO
The result of the sqrt of any number is ALWAYS positive! The function is defined this way!
Explaination: The square-function from R to R isn't bijective, therefore it has no inverse. But you can construct a well defined inverse ( the sqrt) to it by splitting the square function into two, where one goes from R+ to R+ and the other from R- to R+! Now we would also get two inversefunctions, but we CHOOSE the R+ to R+ to be the sqrt! You could also choose the other one without trouble as long as you don't switch between them within a calculation ... but we won't!
Thing is within a quadratic equation it is not possible to distinguish between the two definitions of the square, and therefore two solutions are possible. But only in this case!
ALSO: I'm typing this from my Phone and am german, Happy new year /sci/fags.
>>7752011
he's right, you fuck off
>>7755099
It depends. If we take square root of positive real number, we get a positive real number. If we take squareroot of complex number, we get two numbers. But real numbers are complex numbers aswell. This causes problems.
