Math 4200
Theorem. Between any two real numbers there exists an irrational number.
Proof:
Case 1. Suppose B is positive and B  C. There exists a natural number 5 such that
È#
È
 C  B . Let A œ 5# . Notice that A is an irrational number. Starting at
5
B œ !, we will walk along the positive real line with a step size equal to A Þ We will
prove that we must step into the interval ÐBß CÑ since its width is greater than A Þ
Let W œ Ö4 À 4A
C×. It follows from the Archimedean property that W is not empty.
Let 7 be the least element of WÞ Then, 7A C and Ð7  "ÑA  C .
We must show that B  Ð7  "ÑAÞ Suppose Ð7  "ÑA Ÿ B . Then 7A  A Ÿ Bß
and 7A Ÿ B  A  B  C  B œ CÞ This implies that 7A  Cß a contradiction. Thus,
B  Ð7  "ÑA  CÞ Since A is irrational, Ð7  "ÑA is also an irrational number.
Case 2. Suppose B Ÿ ! Þ Choose a positive integer n such that B  8  !Þ Consider
the positive numbers B  8 and C  8Þ From case 1, there exists an irrational number ?
such that B  8  ?  C  8Þ Subtracting 8 from both sides of this inequality gives
B  ?  8  CÞ Now, ?  8 is irrational since ? is irrational, and 8 us rational.
Theorem. Between any two real numbers there exists a rational number.
Proof:
Modify the proof above by replacing
È#
5
with 51 .