Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Why is $\mathbf{Q}$ dense in $\mathbf{R}$?

Does this imply that (Edit: the field of algebraic numbers) $\overline{\mathbf{Q}}$ is dense in $\mathbf{C}$?

For the first question, I guess one has to show that for every $x$ in $\mathbf{R}$ and every $\epsilon >0$, there exists a rational number $q$ such that $\vert x- q \vert < \epsilon$. Does the proof of this rely upon methods from Diophantine Approximation or is it easier than this?

share|improve this question
10  
$\overline{\mathbb Q}\supset\mathbb Q+i\mathbb Q$ –  Pierre-Yves Gaillard Sep 28 '11 at 16:02

5 Answers 5

up vote 5 down vote accepted

The density of the rational numbers inside the real numbers follows from the very definition of real numbers as (classes of equivalence of) Cauchy sequences of rational numbers. By definition, every real number is the limit of some sequence of rational numbers.

As for your second question: which is the meaning of your $\overline{\mathbf{Q}}$?

Added: If $\overline{\mathbb Q}\supset\mathbb Q+i\mathbb Q$, you can just prove density separately in the real and imaginary axes

share|improve this answer
8  
I guess $\overline{\mathbf{Q}}$ should mean the algebraic closure of $\mathbf{Q}$ –  Julian Kuelshammer Sep 28 '11 at 15:51
    
@AgustiRoig: I put the Added into your answer instead of mine by accident. Feel free to delete it if you want. –  Ross Millikan Sep 28 '11 at 18:47
    
Don't worry, Ross. –  a.r. Sep 29 '11 at 2:09

As I said in my comment, the algebraic closure $\overline{\mathbb Q}$ of $\mathbb Q$ in $\mathbb C$ is dense because it contains $\mathbb Q+i\mathbb Q$, which, as others explained, is dense.

share|improve this answer

For the first, just take a rational with large enough denominator ($2/\epsilon$ will work)

For the second, what do you mean by $\overline{\mathbf{Q}}$? If it is the closure of $\mathbf{Q}$, that is $\mathbf{R}$, but you don't get very close to $i$ with any real numbers. Maybe you want $\mathbf{Q}\times\mathbf{Q}$?

Added: If $\overline{\mathbb Q}\supset\mathbb Q+i\mathbb Q$, you can just prove density separately in the real and imaginary axes

share|improve this answer

If q is a rational number, then the sequence {q,q,..,q,...,q} converges to q.

If $q$ is irrational, then, by the LUB property, it has a (non-periodic) decimal

expansion $a.q_oq_1...q_n.....$ for $a, q_0,q_1,...,q_n,... $integers . Then the sequence : {a,$a+\frac {q_0}{10}, a+\frac{q_0}{10}+\frac{q_1}{10^2},......., a+\frac{q_0}{10}+.......+\frac{q_n}{10^{n+1}},....$ converges to q.

share|improve this answer

Well, if $\mathbb R$ is defined as some complete ordered field, then first prove the archimedean property, so that there is no positive number $< 1/n$ for all natural numbers $n$, and then the "rational with large enough denominator" trick will work.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.