Terence Tao, Analysis I, Ex. 5.4.5: There is a rational between any two reals

Question

Terence Tao, Analysis I, 3e, Exercise 5.4.5:

Prove Proposition 5.4.14. (Hint: use Exercise 5.4.4. You may also need to argue by contradiction.)

Proposition 5.4.14:

Given any two real numbers $x < y$, we can find a rational number q such that $x < q < y$.

Exercise 5.4.4:

Show that for any positive real number $x > 0$ there exists a positive integer $N$ such that $x > 1/N > 0$.

What I've found so far (with the help of this answer, and Pratik Apshinge's comment):

From Exercise 5.4.4, there is a positive real number

$$y - x > 1/N > 0, $$

$$yN - xN > 1, $$

$$yN - 1 > xN.$$

Since there is an integer $m$ between $yN$ and $yN - 1$, we have that

$$yN > m \ge yN - 1 > xN$$

$$yN > m > xN$$

$$y > m/N > x$$

Since $m$ and $N$ are integers, there exists a rational between $y$ and $x$.

But how could a proof by contradiction help in this case?

I think the proof in your Q is as simple as possible. A proof by contradiction in this case is basically the same thing made complicated.... Much of the logical foundation of $\Bbb R$ depends on the Archimedean Property: If $x>0$ then for any $y$ there exists $n\in \Bbb N$ with $nx>y$.... We can extend $\Bbb R$ to a larger ordered field in which the basic rules of $+,-,\times, /,$ and $<$ still apply, but which includes objects larger than any $n\in \Bbb N;$ and their reciprocals are positive but less than any member of $\Bbb Q^+$..... (continued).... — DanielWainfleet, Nov 14 '19 at 08:59
....(continued).... but in any such extension there will be non-empty subsets with upper bounds but no least upper bounds. — DanielWainfleet, Nov 14 '19 at 09:02

score 5 · Answer 1 · answered Nov 13 '19 at 22:04

5

Why would you want to do it by contradiction? It's easier to do it directly. If $y-x>0$ then there is an integer $N$ such that $y-x>1/N>0$, which means that $Ny-Nx>1$ and this implies that there is an integer $m$ such that $Nx<m<Ny$ and finally that $x<\frac{m}{N}<y$.

answered Nov 13 '19 at 22:04

Matematleta

29,139

1

+1....I posted a proof by contradiction, just to show how. It's not my preferred method either. – DanielWainfleet Nov 13 '19 at 23:29

Laassila souhayl · Accepted Answer · 2019-11-14T17:57:27.967

3

If there was no $q\in\mathbb{Q}$ such that $x<q<y$ then there are no $m\in\mathbb{Z} , n\in\mathbb{Z}^*$ such that $$x<\frac{m}{n}<y$$ $$\Rightarrow\quad \forall n\in\mathbb{N}^*\quad nx\geq \lfloor ny \rfloor$$ $$\Rightarrow\quad \forall n\in\mathbb{N}^*\quad ny-nx\leq ny-\lfloor ny \rfloor<1$$Thus for all $n\in\mathbb{N}^*$ $ny-nx\leq1$ this implies that $\forall n\in\mathbb{N}^*$ $y-x\leq \frac{1}{n}$ contradiction with exercise 5.4.4 since $y-x$ is a positive number.

edited Nov 14 '19 at 17:57

answered Nov 13 '19 at 21:29

Laassila souhayl

730

Why does the absence of integers that render $x < m/n < y$ imply that $ny - nx \le 1$, for all $n$? – Max Herrmann Nov 13 '19 at 21:39
I edited it for you, let me know if you still have troubles understanding. – Laassila souhayl Nov 13 '19 at 21:55
Thank you for adding more details! What does $E$ mean? – Max Herrmann Nov 14 '19 at 09:24
1

It is the floor function check : https://en.wikipedia.org/wiki/Floor_and_ceiling_functions for more detail , sorry if I used the french notation for it I didn't know that this notation doesn't exist in english. – Laassila souhayl Nov 14 '19 at 10:58
Your first implication claims that $x < m/n < y \Rightarrow nx \ge \lfloor ny \rfloor$. Choose, e.g. $x = 0$ and $y = 1$. Then $n \cdot 0 = 0 < n \cdot 1 = \lfloor n \rfloor$ for $n > 0$. – Max Herrmann Nov 14 '19 at 15:51
1

I meant the negation of it , if there were no m ,n such that $x<\frac{m}{n}<y$ then if $m=\lfloor ny \rfloor$ there is no n such that $nx < \lfloor ny \rfloor < ny $. – Laassila souhayl Nov 14 '19 at 17:56
In your last conclusion, shouldn't be $y - x$ be strictly less than $1/n$? – Max Herrmann Nov 15 '19 at 06:06
Well both give a contradiction, right? – Laassila souhayl Nov 15 '19 at 12:02
Yes. Absolutely. – Max Herrmann Nov 15 '19 at 12:14

DanielWainfleet · Answer 3 · 2019-11-13T23:19:57.513

2

(i).Suppose $0\le x<y$ and $\Bbb Q\cap (x,y)=\emptyset.$

There exists $N\in \Bbb N$ with $1/N<(y-x)/2.$

There exists $n\in \Bbb N$ with $n(1/N)\ge x.$ This is obvious if $x=0.$ And if it were false with $x>0$ then $1/n> x'=1/(Nx)>0$ for all $n\in \Bbb N,$ contrary to 5.4.4.

Let $n_0=\min \{n\in \Bbb N: n(1/N)\ge x\},$ so $(n_0-1)/N<x\le n_0/N.$

Then $x<(n_0+1)/N,$ so $y\le(n_0+1)/N,$ otherwise $(n_0+1)/N\in \Bbb Q\cap (x,y).$ But then $$2/N=(n_0+1)/N-(n_0-1)/N \ge$$ $$\ge y-(n_0-1)/N>$$ $$> y-x>2/N,$$ a contradiction.

(ii).If $x<y\le 0$ then by (i) there exists $q\in \Bbb Q\cap (-y,-x),$ so $\;-q\in \Bbb Q\cap (x,y).$

(iii).Finally, if $x<0<y$ then $0\in \Bbb Q\cap (x,y).$

The idea behind (i) is that if $N$ is large enough then consecutive values of $1/N,2/N,3/N,...$ cannot "skip over" the interval $(x,y).$

edited Nov 13 '19 at 23:19

answered Nov 13 '19 at 23:13

DanielWainfleet

57,985

Thank you for your elaborate answer! I'm surprised that you seem to negate $y > x > 0$. From my understanding, the negation of Proposition 5.4.14 states: There are two real numbers $x < y$ s.t. $\neg(x < q < y)$, for all rational numbers $q$. – Max Herrmann Nov 14 '19 at 10:50
I don't follow you............. – DanielWainfleet Nov 15 '19 at 02:46
I got you wrong, sorry! Your strategy is to investigate into the three possible cases when $x < y$. In all of these three cases you are assuming that there is no rational in between $x$ and $y$. Since all of them lead to contradiction, we have that there is a rational in $(x, y)$. – Max Herrmann Nov 15 '19 at 06:45
How can a human being possibly conceive of something like this. Impressive. – Max Herrmann Nov 15 '19 at 07:20

Terence Tao, Analysis I, Ex. 5.4.5: There is a rational between any two reals

3 Answers3