Omnimaga
General Discussion => Other Discussions => Math and Science => Topic started by: Munchor on February 09, 2011, 04:22:26 pm
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No, I'm not asking you to tell me how they work and general stuff.
But the other day my teacher said:
x²>0
And I thought... Hm, any number ² will be bigger than 0, since all squared numbers are positive. However, I just checked:
>>> 0**2
0
>>>
And it seems, 0² will be 0, so x²>=0.
What I want to know is if any imaginary number when squared will be minor than 0. Thanks.
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Any purely imaginary number squared will be a negative number. X2>0 is only true for numbers along the real line. Along the imaginary line, numbers like i2 (which equals -1) are not necessarily greater than 0.
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Well, the definition of i is "the square root of negative one", so if you did i2, you'd get -1. Another example: 3i2=-3
You can check this on your calculator by hitting [mode] and changing to to a+bi. :)
And it seems, 0² will be 0, so x²>=0.
Correct. I believe what your teacher meant to say was "any non-zero number squared will be greater than zero." :D
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Any purely imaginary number squared will be a negative number. X2>0 is only true for numbers along the real line. Along the imaginary line, numbers like i2 (which equals -1) are not necessarily greater than 0.
Yes!
/me wins.
My teacher wrote this:
x²>0
I immediately shouted "x²>=0". And he said, "Oh thanks *myname*".
Then I said:
"x²>0,x (http://www.somatematica.com.br/figuras/simbolos/pertence.gif) IR"
And then he said "Well, that doesn't really matter".
But it seems it does. Thanks.
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It does matter, but probably not as far as your class is concerned. ;-)
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Yes, usually, teachers don't really want to mess with imaginary numbers unless they are actually teaching you imaginary numbers, so they would always say something like "x² ≥ 0 ∈ ℝ".
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Yes, usually, teachers don't really want to mess with imaginary numbers unless they are actually teaching you imaginary numbers, so they would always say something like "x² ≥ 0 ∈ ℝ".
Yeah, they just say that, but nobody really asks why it has to be in IR, at least in my class, they only care about maths for tests and grades, not maths as Maths :)
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I immediately shouted "x²>=0". And he said, "Oh thanks *scout".
we all know you're scout...:P/me wins
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I immediately shouted "x²>=0". And he said, "Oh thanks *scout".
we all know you're scout...:P/me wins
But nobody here knows my real name xD
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But nobody here knows my real name xD
Silly David...
* ZTrumpet can haz wins!!! O0
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But nobody here knows my real name xD
Silly David...
* ZTrumpet can haz wins!!! O0
?? Sorry, I don't understand you. My real name isn't David Gomes xD
Or is it?
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I have one question i figured I should just add in here.
Can imaginary numbers be prime? I don't think negative numbers could be prime but I don't know about imaginaries. any help would be appreciated.
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I think prime numbers are described as natural numbers, so they must be integers >0
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OK thanks for the help I thought that would be the answer
(I'm so happy I'm not completely stupid :D)
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I feel happy that I actually understand what I just typed.
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I'm happy i understood what you typed and what I am writing about in my book(mostly ;))
Thanks again for the help
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I'm happy i understood what you typed and what I am writing about in my book(mostly ;))
Thanks again for the help
Imaginary numbers of the form a+bi where b=0 can be prime. ;-) Technically, a+bi would have to only have multiples of 1 and itself. I'm sure you could generalize the multiplication of to complex numbers to get another. For example with P+Qi and S+Ti:
(P+Qi)(S+Ti)=PS+QSi+PTi+QTi^2=PS+QSi+PTi-QT=PS+PTi-QT+QSi=P(S+Ti)-Q(T+Si)
So, I guess as long as (PS-QT)=a and (PT-QS)=b, and as long as there were no P, Q, T, and S that satisfied that, it would be prime.
This is only in theory. If the idea of being prime isn't extended to complex numbers, then... yeah.
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Now I'm kondof confused. Must think about this...
Are prime numbers only the natural numbers?
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Now I'm kondof confused. Must think about this...
Are prime numbers only the natural numbers?
Well, technically, it is defined as a natural number, so yes. However, I believe you could extend the pretext of it to other numbers. For example, above, if you had 0 for Q and T, you would get:
P(S+Ti)-Q(T-Si)=PS=A+Bi=A. So, the factors of A would be P and S. So it makes sense. Take Q=2 and T=5:
P(S+Ti)-Q(T-Si)=P(S+5i)-2(5-Si)=PS+5i-10+2Si=S(P+2i)+5i-10
and P=9, S=3:
3(9+2i)+5i-10=27+6i+5i-10=17+11i
So the factors of 17+11i are (9+2i) and (3+5i) so 17+11i wouldn't be prime by this definition.
But yeah, it's defined as natural (real) numbers only.
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A complex extension to prime numbers are the Gaussian primes. Here, normal primes like 2 are no longer primes (2 = (1+i)(1-i)), though 3 is still a prime, for instance.
[link: http://en.wikipedia.org/wiki/Gaussian_prime#As_a_unique_factorization_domain]