Prove that between every two rational numbers on the line th

Prove that between every two rational numbers on the line there is an irrational number.

Solution

First you must show that between any two real number a<b there exists a rational number c/d such that a < c/d < b
so lets prove that first.
Pf.
Given a,b are real numbers and a<b we know
b-a > 0
So there exists a natural number n such that
n(b-a) > 1
=>
nb - na > 1
=>
nb > 1 + na

Choose another natural number k such that
k < na < k+1
then
k+1 < nb as we know 1 + na < nb and k < na
so we have
na < k+1 < nb
so
a < (k+1)/n < b
and since k, and n are natural numbers then we know (k+1)/n is a rational number.

Now that we know between any two real numbers there is a rational number we can now prove that between any two real numbers there is an irrational number.

Pf. We know between any two real numbers a<b there exists a rational number.
Given a<b then we can divide both sides by sqrt(2) to get the two real numbers
a/sqrt(2) < b/sqrt(2)
But we know there exists a rational number c/d between any two real numbers.
so we can say
a/sqrt(2) < c/d < b/sqrt(2)
now multiply both everything by sqrt(2)
we get
a < c*sqrt(2)/d < b
and we know that a rational number times an irrational number is always irrational so we are done.