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?
Best Answer
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 $x<\frac{m}{N}<y$.