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I want to show that any subfield of the complex field must contain every rational number.

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    @chang Could you add what you have tried? It helps if you "ask" a question: "I want to show that ..." really isn't a question, and it also helps us to know what you've tried, or any ideas you have about how to approach it, etc. I see you have a suggestion in the comment above...can you see where you can get with that?2012-08-26

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Every subfield must contains the multiplicative identity $1$ and additive identity $0$. Now, fields are closed under addition and additive inverses, so $1+1+1+...+1 =n$ and $-n$ exists in the subfield for any $n\in\mathbb{N}$. Now, fields are also closed under multiplicative inverses, so $\frac{1}{n}$ is in the subfield for any $n\in\mathbb{Z}$. Subfields are closed under multiplication, so for any $m,n\in\mathbb{Z}$, we have $\frac{m}{n}$ is in the subfield. Hence, $\mathbb{Q}$ is contained in the subfield.

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    I think this is the most straightforward answer, but I was going to drop an alternative way to complete the second step (if you are comfortable with fields of fractions). Once you are comfortable that $\mathbb{Z}$ is contained in your subfield, you can appeal to the fact that the inclusion $\mathbb{Z}\rightarrow \mathbb{F}$ must extend to $\mathbb{Q}\rightarrow\mathbb{F}$, using the universal property of the field of fractions.2012-08-26