3
$\begingroup$

I am trying to comprehend a proof for a theorem that states between any $x, y \subset{\Bbb{R}}$, there exists $q\subset{\Bbb{Q}}$ such that $x<q<y$.

The proof in question is as follows:

Choose n such that $\frac{1}{n}< y-x$. Consider multiples of $\frac{1}{n}$ which are unbounded. Choose the first multiple such that $\frac{m}{n}>x.$ We claim that $\frac{m}{n}<y.$

If that is not the case, we would have $\frac{m-1}{n}<x$ and $\frac{m}{n}>y$. but these imply that $\frac{1}{n}>y-x,$ which is a contradiction.

I understand that the first statement follows from the Archimedean principle, but I cannot grasp the intuition for the rest of the proof.

Most of my confusion arises at the last line, starting from how we arrived to the conclusion that $\frac{m-1}{n}<x.$

$\endgroup$
4
  • 1
    $\begingroup$ This is a case where a picture of the number line would help. If $\frac{m}{n}$ were not in $(x,y)$ it would have to be greater than $y$ (and $x$). But since the step of $\frac{1}{n}$ just before, i.e., $\frac{m-1}{n}$, would have been less than $x$, we have a step of size $\frac{1}{n}$ larger than $y-x$. But we defined $\frac{1}{n}$ to be less than that interval... $\endgroup$ Commented Jan 3 at 2:04
  • $\begingroup$ Which form of the Archimedean axiom do you use? $\endgroup$ Commented Jan 3 at 7:21
  • $\begingroup$ if x,y in R in x>0, the there exists a natural number N such that nx>y. Also (The one used here, substituting x with y-x): if x>0, the there exists a natural number n such that 1/n < x. $\endgroup$ Commented Jan 3 at 20:00
  • $\begingroup$ Then user2661923's answer is the best and you should accept it. $\endgroup$ Commented Jan 4 at 0:11

4 Answers 4

Reset to default
4
$\begingroup$

Personally, I think that the concept behind the official proof is sound, but that the proof was written in a confusing manner.

Choose $~n \in \Bbb{Z^+}~$ such that $~n > \dfrac{1}{y-x} \implies$

$y - x > \dfrac{1}{n} \implies y > x + \dfrac{1}{n}.$

Choose $~\displaystyle k \in \Bbb{Z}~$ such that $~k \leq nx < k+1 \implies $

$\displaystyle \frac{k}{n} \leq x < \frac{k+1}{n} = \frac{k}{n} + \frac{1}{n} \leq x + \frac{1}{n} < y.$

Actually, this is exercise 3.12.6 from chapter I (the introductory chapter) in "Calculus" 2nd Ed., volume 1, 1966 (Tom Apostol). Preceding this problem is

  • Theorem I.29 which proves that for any (finite) real number $~r,~$ there exists $~n \in \Bbb{Z^+},~$ such that $~n > r.$

  • Exercise 3.12.4 from chapter I that asks the math student to prove that for any real number $~r~$ there exists (unique) $~n \in \Bbb{Z}~$ such that $~n \leq r < n+1.$

$\endgroup$
2
$\begingroup$

The latter part of the proof is a proof by contradiction where the claim they've made is that $\frac{m}{n}>x$. Note that it says to choose the first multiple $m$ such that the claim holds; then if thats the case any multiple $k$ smaller than $m$ should result in $\frac{k}{n}<x$. In this proof, they've just used $k=m-1$. Then using this together with $\frac{m}{n}>y$, we can finish the proof algebraically.

$\endgroup$
1
$\begingroup$

To give you an example, think of the integers number line. Positive direction is to the right.

Suppose I claim that the multiple of 3 nearest in the positive direction of 4 is 6. That is, between 4 and 6 , there is no multiple of 3. Then it follows that the multiple of 3 just less than 6 has to be smaller than 4. That is the exact conclusion your proof arrived at to showcase the arising contradiction.

$\endgroup$
0
$\begingroup$

Once $\frac{m}{n}$ has been chosen as the first multiple of $\frac{1}{n}$ that is greater than $x$ (here "first" means "least"), you then use the fact that $\frac{m-1}{n}$ is a smaller multiple of $\frac{1}{n}$ to conclude that $\frac{m-1}{n}$ is not greater than $x$: if it were, then $\frac{m}{n}$ would not have been the least.

In other words, $\frac{m-1}{n} \le x$ (your post is somewhat sloppy about the difference between strict and nonstrict inequality).

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.