person Posted July 6, 2008 Posted July 6, 2008 Again, in Stephen Hawking's "A Brief History of Time" he states that he created a mathematical theorem that incorperated "imaginary" numbers to descible the possible expansion of the universe. The differance between "real" and "imaginary" numbers is that when one squares a real number, the result will always be positive(2*2 = 4 and -2*-2 = 4) but with imaginary numbers; the square is always negative. Hawking uses a variable to explain this(eg. "i") (i*i = -1 and 2i * 2i = -4 and so on). Please will someone explain the mathematics behind this.
insane_alien Posted July 6, 2008 Posted July 6, 2008 i is not a variable, it is the base unit for imaginary numbers (the 1 of imaginary numbers). i is the square root of negative 1. this has no normal number counterpart. so, the mathematicians decided to say, well, what if we let this equal some imaginary number i, and see if we can do anything with it, as it turns out you can, you can do a lot with it. though the term now is complex numbers, not imaginary numbers. it can be difficult to get your head round at first but basically you are turning the numberline you learned in your first years of school into a number plane with two axes(like an x-y plot where x is a normal number and y is the imaginary number)
DJBruce Posted July 6, 2008 Posted July 6, 2008 (edited) An imaginary number is any number with [math]i[/math] in it. As said before [math]i[/math] is not a variable but rather somewhat of a constant. It is used to sole equations which have no real solutions. For example try and find a answer to [math]\sqrt{(-2)}[/math]: As you quickly see there is no answer because [math]\sqrt{(x)}=y[/math] where [math]y^{2}=x[/math] and if you square any number you get a positive because if for example [math]y=-2[/math] then [math](-2)^{2}=-2\cdot-2=4[/math]. But surely there must be an answer right. So a long time ago mathematicians like Euler and Gauss came up with the idea of creating a imaginary number. This number is [math]i[/math]. The entire idea of this was to use it to represent solutions to systems that otherwise could not be represented. Here is a little bit on how [math]i[/math] works. For starts [math]i[/math] is normally defined as having a value equal to [math]\sqrt{(-1)}[/math] So [math]i^{2}=-1[/math]. Knowing this you should be able to figure out an power of [math]i[/math]. But here is a table of some of them. [math]i^{-3}=i i^{-2}=-1 i^{-1}=-i i^{0}=1 i^{1}=i i^{2}=-1 i^{3}=-i i^{4}=1 i^{5}=i i^{6}=-1 [/math] As you can see there is a pattern and this pattern repeats. Another use for imaginary numbers is using the to write complex numbers. A complex number is a number written in [math]a+bi[/math] form where a and b are real numbers. Here are some links that I hope help you: http://www.math.toronto.edu/mathnet/answers/imaginary.html http://en.wikipedia.org/wiki/Complex_number http://en.wikipedia.org/wiki/Imaginary_number http://www.purplemath.com/modules/complex.htm I hope this answered your questions if not ask again and I will try to help you. PS: The answer to [math]\sqrt{(-2)}\approx1.41421i[/math]. Edited July 11, 2008 by Cap'n Refsmmat
the tree Posted July 7, 2008 Posted July 7, 2008 though the term now is complex numbers, not imaginary numbers.Just to be clear, a complex number is one with both real and imaginary parts.5 is real, 3i is imaginary, (5+3i) is complex.
DJBruce Posted July 7, 2008 Posted July 7, 2008 To continue on my little description imaginary numbers are very useful in many field. They are used in electronics, physics and where even behind Riemann's Hypothesis.
Anubisboy Posted July 11, 2008 Posted July 11, 2008 (edited) Are you certain this statement is correct? [math]i^{-3}=ii^{-2}=-1 i^{-1}=-i i^{0}=1 i^{1}=i i^{2}=-1 i^{3}=-i i^{4}=1 i^{5}=i i^{6}=-1 [/math] In the above expression, odd power of [math]i[/math] should be not real. Edited July 11, 2008 by Cap'n Refsmmat
DJBruce Posted July 12, 2008 Posted July 12, 2008 Yes you are right it should be [math]i[/math]. I apologize for my mistake.
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