# Subfields of complex numbers and the inclusion of rational#s

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1. Dec 6, 2016

### VrhoZna

1. The problem statement, all variables and given/known data
Prove that each subfield of the field of complex numbers contains every rational number. '

From Hoffman and Kunze's Linear Algebra Chapter 1 Section 2

2. Relevant equations

3. The attempt at a solution
Suppose there was a subfield of the complex numbers that did not contain every rational number (from now on referred to as F), that is there is a rational number p/q, where p and q denote integers, that is not an element of F. Then it follows that either p ∉ F or 1/q ∉ F (as their product is not an element.) We consider each case separately.

Suppose p ∉ F, then (p - 1) ∉ F as (p - 1 + 1 = p) and similarly (p - 2) ∉ F, we proceed stepwise and find that p - (p - 1) ∉ F but of course (p - (p - 1)) = 1 contradicting our assumption that F is a subfield of the complex numbers.

Now suppose 1/q ∉ F, then q ∉ F (as there would be no element x such that x*q = 1) and a similar argument as above finds 1 ∉ F contradicting our assumption that F is a subfield of the complex numbers. Thus every subfield of the complex numbers contains as elements every rational number.

I feel my reasoning is correct but given that my knowledge of fields is limited to that narrow introduction in the section I'm not sure if any misunderstandings on my part have cropped up.

2. Dec 6, 2016

### Staff: Mentor

Looks o.k., although the many indirect arguments are a bit confusing.
Why don't you reason positively? $1 \in F$ and $char F = 0$, which means $1+\ldots +1 \neq 0$ no matter how many $1$ are added. This automatically results in $\mathbb{Q}$: first $\mathbb{N}$, then $\mathbb{Z}$ and finally $\mathbb{Q}$. It's basically the same argument as yours, only without negations.

3. Dec 7, 2016

### VrhoZna

I suppose I hadn't realized the full implications of said subfields having characteristic zero at the time of writing the proof, but I understand a bit better now. Thank you for your answer.