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I can't understand the following proof I found in Rudin, "Principles of mathematical analysis".

This is the statement: if $x\in\mathbb{R}, y\in\mathbb{R}$ and $x<y$, then there exists $p\in\mathbb{Q}$ such that $x<p<y$.

Here is the proof I can't understand.

proof.

1) Since $x<y$, we have $y-x>0$ (ok, I got it)

2) Archimedean principle furnishes a positive integer $n$ such that $n(y-x)>1$ (ok)

3) Apply archimedean principle again to obtain positive integers $m_1,m_2$ such that $m_1>nx$ and $m_2>-nx$ (ok)

4) Then $-m_2<nx<m_1$ (ok)

5) Hence there is an integer $m$ (with $-m_2\leq m\leq m_1$) such that $m-1\leq nx< m$

Here is where I have some problems. Why do I need integers $-m_2$ and $m_1$? Couldn't just say: every real number lies between an integer and its successor? Moreover, even knowing the existence of integers $m_1, m_2$, how do I deduce the fact that $m$ exists with that property?

6) If we combine these inequalities, we obtain $nx<m\leq 1*nx<ny$ (ok)

7) Since $n>0$, it follows that $x<\frac{m}{n}<y$ (ok)

8) Take $p=\frac{m}{n}$. (ok)

Maryam
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  • How will you state 'For any real number $x$, there exists a integer $n$ such that $n\leq x\leq n+1$' from the definition of the real number (equivalent class of Cauchy sequences in rationals)? If this is not the definition that you are using, then which definition of the real number are you using? – Jingeon An-Lacroix Apr 28 '20 at 16:29
  • I am following Rudin, this proof precedes the construction of the reals and is supposed to be based only on axioms for an ordered field. With these axioms, I don't know how to prove what you are asking, but still don't understand point 5) of that proof – Maryam Apr 28 '20 at 16:31
  • It is easy question if it asks the truth value of the statement, but it gets hard when you ask it's intension.. I will say we just needed the interval that $nx$ is contained. – Jingeon An-Lacroix Apr 28 '20 at 16:35
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    You may find this answer to an earlier question about this proof helpful. – Brian M. Scott Apr 28 '20 at 16:36

3 Answers3

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"Couldn't just say: every real number lies between an integer and its successor?"

You could say that, but could you prove it? That's what Rudin is doing.

Moreover, even knowing the existence of integers $m_1,m_2$, how do I deduce the fact that $m$ exists with that property?

The set of integers $k$ such that $nx < k$ is nonempty (since it contains $m_1$) and bounded below (by $m_2$), so it has a minimal element: that is $m$.

Robert Israel
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Question $1$: in general, what you do in proofs is sufficient but not necessary.

Question $2$: yes, but you have to prove it.

Question $3$: suppose not, then $\forall m \in \mathbb{Z}.\ (nx \ge m) \lor (nx < m-1)$. In both cases you reach a contradiction with $4)$.

Riccardo
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I think this question had already been asked, but here we go.

I'm going to prove that between two real numbers there's always one rational and one irrational. Let $x,y \in \mathbb{R}$ with $x<y$. Using the fact that the natural numbers are not upperly bounded in the real numbers, we can choose $n \in \mathbb{N}$ such that $n\overset{(3)}>\frac{1}{y-x} \in \mathbb{R}$. Now, let $m \in \mathbb{N}$ such that $m-1\leq nx < m$.

Then, we divide by $n$, which i possible because $n>\frac{1}{y-x}>0$, we have,

$$\frac{m-1}{n}\leq x < \frac{m}{n}$$

$$\frac{m}{n}-\frac{1}{n}\leq x < \frac{m}{n}$$

and we can deduce that,

$$\frac{m}{n}-\frac{1}{n}\overset{(1)}\leq x \hspace{2mm} \text{and} \hspace{2mm} x\overset{(2)}<\frac{m}{n}$$

thus,

$$x\overset{(2)}<\frac{m}{n}\overset{(1)}\leq x+\frac{1}{n}\overset{(3)}<x+y-x=y$$

We get,

$$x<\frac{m}{n}<y$$

This proves the first part of the statement, which is the existance of a rational number between two real numbers.

Let's prove the existance of an irrational number between this two real numbers.

Let $z>0 \in \mathbb{R}\setminus\mathbb{Q}$ and we consider the interval $(xz,yz)$. From what we've seen before there's a rational $s \in (xz,yz)$. Therefore, $\frac{s}{z} \in (x,y)$ and $\frac{s}{z}$ is in fact irrational.

HornyPigeon54
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