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I tried a proof using mainly the Archimedean property of $\Bbb R$:

I am to show that for any $x\in \Bbb R$, every neighborhood of $x$ contains a rational number. I shall first prove it for non-negative real numbers. Suppose $x$ is a non-negative real number and let $\epsilon >0$ be arbitrary. By the Archimedean property, there is an $n\in \Bbb N$ such that $n\epsilon >1$ i.e. $\frac 1n<\epsilon$. Again, using the Archimedean property, we can find an $m\in \Bbb N$ such that $m\frac 1n >x$.

Let $S=${$m\in \Bbb N : \frac mn>x$}. Since $S$ is a nonempty subset of $\Bbb N$, it has a least element, say, $m'$. It means one of the following:

A) $m'-1$ is a natural number with the property that $\frac {m'-1}n≤x$; or

B) $m'-1=0$ whence $m'=1$.

A) If possible assume that neither $\frac {m'-1}n$ nor $\frac {m'}n$ belong in $(x-\epsilon,x+\epsilon)$. Which means :

$\frac{m'-1}n<x-\epsilon<x+\epsilon<\frac{m'}n \implies (x+\epsilon) - (x-\epsilon)<\frac{m'}n - \frac{m'-1}n \implies 2\epsilon<\frac 1n<\epsilon$ which is a contradiction. Thus one of them must be in $(x-\epsilon,x+\epsilon)$.

B) $m'=1$ means that $\frac 1n>x$. But since $x$ is non-negative we also have $\frac 1n<\epsilon<x+\epsilon$. Hence, $\frac 1n\in (x-\epsilon,x+\epsilon)$.

Thus every neighborhood $(x-\epsilon,x+\epsilon)$ of $x$ contains a rational number $q$. But then, the neighborhood $(-x-\epsilon,-x+\epsilon)$ of $-x$ contains the rational number $-q$. This completes my proof.

Is the proof okay?

Not Euler
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  • It seems to be correct. You can probably try to make it simpler. – fleablood Jun 29 '18 at 22:44
  • @fleablood Okay. I will try. – Not Euler Jun 29 '18 at 22:45
  • Just a comment: This seems to be a strange problem, because $\mathbb{R}$ is the completion of $\mathbb{Q}$ with respect to the normal absolute value. What I mean is, $\mathbb{Q}$ is dense in $\mathbb{R}$ by definition. – Dzoooks Jun 29 '18 at 22:45
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    @Dzoooks there are many definitions of $\mathbb{R}$. For example, $\mathbb{R}$ can be defined as the only complete ordered field (up to isomorphism), by Dedekind cuts, by completing $\mathbb{Q}$ as a metric space, etc. – qualcuno Jun 29 '18 at 22:47
  • @Dzoooks I don't know the definition of density you used. What I understand by $\Bbb Q$ being dense in $\Bbb R$ is that every point of $\Bbb R$ is a closure point of $\Bbb Q$ – Not Euler Jun 29 '18 at 22:49
  • @Dzoooks to say the R is defined as the completion of Q is to assume Q has a completion. – fleablood Jun 29 '18 at 22:50
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    @HritRoy Dzoooks' point is that R by definition is Q and its closure points. But an excercise equivalent to yours must, IMO, be done eventually to prove Q can be completed with closure points. – fleablood Jun 29 '18 at 22:54
  • @fleablood I don't know what is meant by the "completion" of $\Bbb Q$ yet. However, if $\Bbb Q$ was dense in $\Bbb R$ axiomatically or something, wouldn't I, at some point, have to invoke that axiom (directly or indirectly) in order to prove the above? – Not Euler Jun 29 '18 at 23:07
  • @HritRoy It depends on your definition of $\mathbb{R}$. If your definition is anything other than the completion wrt to usual $|\cdot |$ of $\mathbb{Q}$, then proving $\mathbb{Q}$ is dense in $\mathbb{R}$ is an exercise worth doing. If, as is probably the case, your definition of $\mathbb{R}$ is the completion of $\mathbb{Q}$, then this exercise is true by definition. – Dzoooks Jun 29 '18 at 23:32
  • @HritRoy Definitions and particulars aside though, the easiest way to see that $\mathbb{Q}$ is dense in "$\mathbb{R}$", imo is to look at decimal expansions. Take $\pi$, for example. It's approximated arbitrarily well by the sequence of rational numbers $3, 3.1, 3.14, $ etc. – Dzoooks Jun 29 '18 at 23:37
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    With all due respect to Dzoooks, he is assuming there is an intuitive sense of the meaning of the reals as being "the rationals with all the holes filled in". The decimmal expansions assume that the reals are well-defined and that every real number has an infinite expansion and every infinite decimal is a real number. IMO that is not an acceptable assumption. However, if I were you, Hrit Roy, I wouldn't worry too much. Eventually, we must prove that rational numbers are dense ... within themselves (that is within any interval there is an rational number). So I'm glad you did it. – fleablood Jun 30 '18 at 00:04
  • @Dzoooks - The essential property of the real numbers is that it is - up to isomorphism - the unique topologically complete ordered field. Building from sets $\to \Bbb N \to \Bbb Z \to \Bbb Q \to \Bbb R$ is simply a way of showing such a field exists. But $\Bbb R$ itself should be considered an otherwise undistinguished complete ordered field. $\Bbb Q$ is then the smallest subfield, $\Bbb Z$ is the smallest unital subring, and $\Bbb N$ the smallest set containing $1$ (and $0$, depending on your choice of definition) which is closed under addition. – Paul Sinclair Jun 30 '18 at 16:09
  • @PaulSinclair Please direct helpful responses at the OP. It's pedantic to assume that I didn't know this stuff.. – Dzoooks Jun 30 '18 at 22:14
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    @Dzoooks - why then did you proclaim to the OP that $\Bbb Q$ is dense in $\Bbb R$ "by definition" if you knew differently? – Paul Sinclair Jul 01 '18 at 03:28

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