# Proof involving rational numbers

• Nov 18th 2008, 07:05 PM
Janu42
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.
• Nov 18th 2008, 07:24 PM
TheEmptySet
Proof that there is a rational between any two real numbers
Since $a then

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

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

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

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

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

$n+1 this implies that

$N(a) < n+1 < N(b) \implies a < \frac{n+1}{N}< b$
• Nov 19th 2008, 09:23 AM
HallsofIvy
You don't have to be that complicated! If a and b are rational numbers, then $c= \frac{a+ b}{2}$ 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.