I am taking an introduction class on set theory. We have formally constructed the natural numbers, integers, rationals and reals. I am now trying to think of how to define the complex numbers inside of set theory. My idea is to follow the idea of how the reals were constructed from Dedekind cuts. The complex numbers would be Dedekind cuts of the real numbers that allow for the imaginary numbers to be defined. I am not sure exactly how to formally explain this or if this would be the correct way of going about defining the complex numbers. Any answers and comments would be greatly appreciated. Thank you in advance.
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$\begingroup$ You cannot make a proper Dedekind cut on the reals. $\endgroup$– DanielWainfleetNov 15, 2015 at 21:24
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2 Answers
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The simplest thing is just to define a complex number to be an ordered pair $(x,y)$ of reals. Define the sum in the obvious way, define the product $$(x,y)(a,b)=(ax-by,ay+bx),$$and show you have a field.
Then $x\mapsto(x,0)$ is an embedding of $\Bbb R$ into $\Bbb C$. If you define $i=(0,1)$ you get $i^2=(-1,0)$. Complex numbers.
answered Nov 15, 2015 at 19:29
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$\begingroup$ Not sure "isomorphism into" is the best term. Embedding of $\mathbb R$, maybe. $\endgroup$ Nov 15, 2015 at 19:39
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$\begingroup$ Yeah - I never got those thisamorphism and thatamorphism things straight. Embedding is better. $\endgroup$ Nov 15, 2015 at 20:17
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Your proposed construction will not work. In order to construct Dedekind cut, you must have an ordered field. The completion of the space will also be an ordered field. However, the is no order one can put on $\Bbb C$ which would make $\Bbb C$ an ordered field.
answered Nov 15, 2015 at 19:32
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$\begingroup$ In the def'n of an ordered field ,if a<b then a+c<b+c ; Also if a<b and 0<c then ac<bc. These won't work when you try to decide whether -i<0<i or i<0<-i. $\endgroup$ Nov 15, 2015 at 21:27