Holocene
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Actually, can you take a root of any negative number?
The discussion revolves around the question of whether one can take an even root of a negative number, exploring the implications of complex numbers in this context. Participants delve into the definitions, existence, and interpretations of complex numbers, as well as their applications in various fields.
Participants do not reach a consensus on the existence of complex numbers or their interpretation in physical terms. Multiple competing views remain regarding the nature and application of complex numbers.
Limitations include unresolved definitions of existence for numbers, the ambiguity in mathematical definitions, and the varying interpretations of complex numbers in different contexts.
Holocene said:Actually, can you take a root of any negative number?
ehj said:how do you know that these complex number exists :P?
ehj said:how do you know that these complex number exists :P?
ehj said:I just thought there was a "proof" of its existence? Like nearly everything else in mathematics, shouldn't there be an argument for it? Or maybe that's quite advanced?
There are technical difficulties with that. Since complex numbers, just like real numbers, have two square roots, defining i as [itex]\sqrt{-1}[/itex], or as "the number such that i2= -1", is ambiguous.Tricore said:All complex numbers can be written on the form [tex]a + ib[/tex] where [tex]a[/tex] and [tex]b[/tex] are real numbers, and [tex]i[/tex] is the imaginary unit, DEFINED as [tex]i=\sqrt{-1}[/tex]. This is just definitions.
Are you asserting that real number "exist in a physical sense"? Can you draw a line of length [itex]\pi[/itex]? For that matter, can you draw a line of length 1?So complex numbers exist because we created them. They don't exist in a physical sense - you can't draw a line on paper with a complex/imaginary length, but perfect circles don't exist in a physical sense either.
I'm not complaining (too much). I agree with most of what you say.because of that, it doesn't make sense to ask for a proof that they exist. What we CAN prove however, is that there exists a solution to [tex]\sqrt[n]{-x}[/tex] within the complex numbers, and also that there is no solution within the real numbers.
HallsofIvy said:There are technical difficulties with that. Since complex numbers, just like real numbers, have two square roots, defining i as [itex]\sqrt{-1}[/itex], or as "the number such that i2= -1", is ambiguous.
What we can do is define the complex numbers as pairs of real number, (a, b), with addtion and multiplication defined by (a, b)+ (c, d)= (a+ c, b+ d), (a, b)*(c, d)= (ac- bd, ad+ bc). Of course one must then prove that this is "well defined" and forms a field (but it does NOT form an "ordered" field).
HallsofIvy said:Are you asserting that real number "exist in a physical sense"? Can you draw a line of length [itex]\pi[/itex]? For that matter, can you draw a line of length 1?
Can't I say that a line, of length [itex]\sqrt{5}[/itex], parallel to the imaginary (i.e. y-axis in an xy-coordinate system) is such a thing?Tricore said:Thanks for correcting me
No, I'm not.
What I was trying to say is that you can draw a line that's (almost) say 5cm, but it doesn't make sense to draw a line that's almost [tex]\sqrt{-5}[/tex]cm.