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I am asked to prove the density of irrationals in $\mathbb{R}$. I understand how to do this by proving the density of $\mathbb{Q}$ first, namely, adding a known irrational number such as $\sqrt{2}$ to $x,y \in \mathbb{R}$ ($x<y$), then there exists $r_0 \in \mathbb{Q}$ such that $x+\sqrt{2}<r_0<y+\sqrt{2}$, then substract $\sqrt{2}$ from all sides of the inequality to yield $x<r_0-\sqrt{2}<y$, and $r_0-\sqrt{2}$ is irrational ($r_0$ is rational). However, my professor has said that I can prove density of $\mathbb{R} \setminus \mathbb{Q}$ without even using the density of $\mathbb{Q}$ and it is a simple proof. I have puzzled over this for quite some time. I appreciate any help provided on this question in advance.

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  • $\begingroup$ do you know that $\mathbb{R}$ is uncountable and $\mathbb{Q}$ is countable? $\endgroup$ – clark Jan 7 '18 at 8:17
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Fix $x\in \mathbb{R}$.

Case 1: If $x\in\mathbb{Q}$, then the irrational sequence $x+\sqrt{2}/n$ converges to $x$.

Case 2: If $x\notin\mathbb{Q}$, then the irrational sequence $x+1/n$ converges to $x$.

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Even the set $\mathbb{A}=\{m+n\sqrt{2}: m,n\in\mathbb{Z}\}$ is dense in $\mathbb{R}$, and that's easy to see: for every $\epsilon>0$, there is a $\delta\in\mathbb{A}$ with $0<\delta<\epsilon$, and we can construct it as $(\sqrt{2}-1)^k$ with sufficiently big $k$, because $0<\sqrt{2}-1<1/2$, and as you can see using the binomial theorem, $(\sqrt{2}-1)^k\in\mathbb{A}$. Now, if you want to approximate $x\in\mathbb{R}$ with an error $<\epsilon$, you just take $$\left\lfloor\frac{x}{\delta}\right\rfloor\delta\in\mathbb{A}.$$ It's easy to divide by a $\delta$ of the above form, since $\displaystyle\frac1{(\sqrt{2}-1)^k}=(\sqrt{2}+1)^k$.
This is entirely constructive: if you want to approximate $\displaystyle x=\frac23$ with an error $10^{-8}$, you observe that $\delta=(\sqrt{2}-1)^{21}=-54608393 + 38613965\sqrt{2}<10^{-8}$, $\displaystyle \left\lfloor\frac{x}{\delta}\right\rfloor=72811190$, so our approximation in $\mathbb{A}$ is $-3976102078317670 + 2811528742268350\sqrt{2}$.

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Let us take any $\varepsilon$-neighborhood, for $\varepsilon > 0$, of any number $a \in \mathbb{R}$.

It suffices to show that $A = (a - \varepsilon, a + \varepsilon)$ must contain an irrational number. Let us show this to be the case with a proof by contradiction: Suppose our neighborhood does not contain any irrational numbers, so that every number in there is rational; i.e., that $A \subset \mathbb{Q}$.

Note that $a \in A \subset \mathbb{Q}$, and $a + \varepsilon/2 \in A \subset \mathbb{Q}$; by closure of $\mathbb{Q}$ under subtraction, $\varepsilon/2 \in \mathbb{Q}$. By closure of $\mathbb{Q}$ under multiplication/division, this means that $\varepsilon \in \mathbb{Q}$.

This is all looking a bit nonsensical, but let us persist and write down an explicit contradiction:

Observe that $\sqrt{5} > \sqrt{4} = 2$, and $\sqrt{5} \not\in \mathbb{Q}$; so the positive irrational number $\varepsilon/\sqrt{5} < \varepsilon/2$, which means that the irrational number $a + \varepsilon/\sqrt{5} \in A \subset \mathbb{Q}$. Contradiction!

Therefore, we must have been incorrect in our supposition, so that any neighborhood of any real number contains at least one irrational number.

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  • $\begingroup$ [This could really have been done with $\varepsilon/\sqrt{2}$; such is life.] $\endgroup$ – Benjamin Dickman Jan 7 '18 at 8:34
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Let $\Bbb F$ be an ordered field with $\Bbb Q \subsetneqq \Bbb F.$ Then $\Bbb Q^+\subsetneqq \Bbb F^+.$

(1)... $\forall q\in Q^+\;\exists f\in \Bbb F^+$ \ $\Bbb Q^+\;(f<q).$ .. Proof: Take $x\in F^+$ \ $\Bbb Q^+.$ We have:

(1A). If there exists $q'\in \Bbb Q^+$ such that $x<q'$ then let $f=xq/q'.$

(1B). If $\forall q'\in \Bbb Q^+\;(x>q')$ then let $f=1/x.$

(2). For $x,y\in \Bbb F$ with $x<y$ we wish to show $\exists g\in \Bbb F$ \ $\Bbb Q\; (x<g<y).$ We have $x<a<b<y$ where $a=(2x+y)/3$ and $b=(x+2y)/3.$ We have:

(2A). If either $a$ or $b$ belongs to $\Bbb F$ \ $\Bbb Q$ then we're done.

(2B). If $a,b\in \Bbb Q$ then $x=2a-b\in \Bbb Q$ and $y=2b-a\in \Bbb Q,$ so $y-x\in \Bbb Q^+. $ By (1) take $f\in \Bbb F^+ \setminus \Bbb Q^+$ with $f<y-x.$ Let $g=x+f.$

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