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#1
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10-02-2003, 06:21 AM
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[Algebra] (What is the purpose of i?)
What the heck is 'i'. People just nod at me and say "'i' is an imaginary number." SO what? What is the point of "i"?
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 I tell you this, for when my days have come to an end, you shall be King. - Terenas Menethil II Last edited by Cody; 03-25-2005 at 11:07 PM. 
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#2
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10-02-2003, 06:47 AM
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Re: (maths)i?
i is the square root of -1, and is there so we can accomplish the idea of squaring a negative number. It's basically there to get around the fact that numbers under a square root can be negative... They also make very nice graphs called fractals that are pretty cool!
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#3
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10-05-2003, 09:16 PM
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Re: (maths)i?
yah, but i can also be any variable......
usually variables are things like x, y, and z though...
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serious quote: imagination is more important than knowledge- Albert Einstein funny quote: I woke up one morning and all my stuff had been stolen and replaced with perfect duplicates- Steven Wright -thanks to SearanoX for the great avy and siggy! -read the Rules, they Rule you -if you are ever in the chat and kedisar and pk-kitty are also in the chat, don't worry, they always do that..... -I want Andi! -I am really small for my age -currently attempting to learn japanese
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#4
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10-05-2003, 09:42 PM
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Re: (maths)i?
my geometry teacher says that Geo. is easy cause we make it up as we go along. I infers that it is imagenary. you can never reach the number it self.
(radical) -9 will get us 3i.
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#5
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10-15-2003, 12:38 AM
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Re: (maths)i?
Quote:
In electrical engienieering , "j" is used insted of "i" , but this is just to not get it confused with the current intensity. like alredy was explained, the imaginary numbers are the roots of negative numbers. A complex number is a number that has a real part and an imaginary part., they usually look like this: "2 + i2" or "2 - i8" or "-8 + 9i" and so on.
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#6
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10-17-2003, 05:09 AM
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Re: (maths)i?
The purpose of complex mathematics is (among other things) to validate certain theorems.
For instance, take the fundamental theorem of calculus. This theorem states that for any equation a1z^n+a2z^(n-1)+...+an=0 has exactly n roots. (if a1,...,an are constants) So how can that be? If we take the function x^2+9=0 we see that this isn't possible with our standard real numbers. (x1,2=(-9)^0.5) So therefor we expand our number system to include the number i=(-1)^0.5. It is now possible to state the above theorem as x1,2 in the example would be +-3i. This proves useful in many other cases such as the calculation of integrals, finding the number of zeros in one quadrant and so on.
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#7
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10-17-2003, 04:41 PM
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Re: (maths)i?
My friend mentioned something about Stephen Hawking using 'i' when talking about time, and that imaginary time could very well add another dimension to our linear concept of time. Think of a horizontal line that represent time, then a perpendicular axis known as 'i' with -i, 0, i, 2i, etc labelled on this vertical axis. Cool, eh? Although science has uncovered so many secrets, most of the things still undiscovered require knowledge of things that cannot be comprehended by man.
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