# Thread: Proof involving rational numbers

1. ## Proof involving rational numbers

a) Prove that between any two rational numbers there is another rational number; that is, if a, b are in Q and a < b, then there exists z in Q such that a < z < b.

b) Prove that between any two rational numbers there are infinitely many rational numbers.

2. ## Proof that there is a rational between any two real numbers

Since $\displaystyle a<b$ then

$\displaystyle 0< \frac{1}{b-a}$ by the archimedian principle we can pick an Integer N such that

$\displaystyle 0< \frac{1}{b-a}<N$ This implies that

$\displaystyle 1 < N(b-a)$ we can now pick another integer n such that

$\displaystyle n < N(a) < n+1$ adding 1 we get

$\displaystyle n+1 < N(a)+1$ usin g the fact that $\displaystyle 1 < N(b-a)$ we get

$\displaystyle n+1 <N(a) +1 < N(a) +N(b-a)=N(b)$ this implies that

$\displaystyle N(a) < n+1 < N(b) \implies a < \frac{n+1}{N}< b$

3. You don't have to be that complicated! If a and b are rational numbers, then $\displaystyle c= \frac{a+ b}{2}$ is a rational number between a and b.

Suppose there were only a finite number of rational numbers between a and b. Then there must be largest such number: c. But (b+c)/2 is a rational number between c and b and so between a and b and is larger than c: contradiction.