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

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).... Nov 14, 2019 at 8:59
• ....(continued).... but in any such extension there will be non-empty subsets with upper bounds but no least upper bounds. Nov 14, 2019 at 9:02

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 $$Nx and finally that $$x<\frac{m}{N}.

• +1....I posted a proof by contradiction, just to show how. It's not my preferred method either. Nov 13, 2019 at 23:29

If there was no $$q\in\mathbb{Q}$$ such that $$x then there are no $$m\in\mathbb{Z} , n\in\mathbb{Z}^*$$ such that $$x<\frac{m}{n} $$\Rightarrow\quad \forall n\in\mathbb{N}^*\quad nx\geq \lfloor ny \rfloor$$ $$\Rightarrow\quad \forall n\in\mathbb{N}^*\quad ny-nx\leq ny-\lfloor ny \rfloor<1$$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.

• Why does the absence of integers that render $x < m/n < y$ imply that $ny - nx \le 1$, for all $n$? Nov 13, 2019 at 21:39
• I edited it for you, let me know if you still have troubles understanding. Nov 13, 2019 at 21:55
• Thank you for adding more details! What does $E$ mean? Nov 14, 2019 at 9:24
• It is the floor function check : 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. Nov 14, 2019 at 10:58
• 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 $nx < \lfloor ny \rfloor < ny$. Nov 14, 2019 at 17:56

(i).Suppose $$0\le x and $$\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

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

(iii).Finally, if $$x<0 then $$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 $$(x,y).$$

• 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 $q$. Nov 14, 2019 at 10:50
• I don't follow you............. Nov 15, 2019 at 2:46
• 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 $(x, y)$. Nov 15, 2019 at 6:45
• How can a human being possibly conceive of something like this. Impressive. Nov 15, 2019 at 7:20