Math & Science Home  Proficiency Tests  Mathematical Thinking in Physics  Aeronauts 2000 

The Irrationality ofProblem: Solution:
for a and b = any two integers. We must then show that no two such integers can be found. We begin by squaring both sides of eq. 1: If b is odd, then b^{2} is odd; in this case, a^{2} and a are also odd. Similarly, if b is even, then b^{2}, a^{2}, and a are even. Since any choice of even values of a and b leads to a ratio a/b that can be reduced by canceling a common factor of 2, we must assume that a and b are odd, and that the ratio a/b is already reduced to smallest possible terms. With a and b both odd, we may write where we require m and n to be integers (to ensure integer values of a and b). When these expressions are substituted into eq. 2a, we obtain Upon performing some algebra, we acquire the further expression The Left Hand Side of eq. 6 is an odd integer. The Right Hand Side, on the other hand, is an even integer. There are no solutions for eq. 6. Therefore, integer values of a and b which satisfy the relationship = a/b cannot be found. We are forced to conclude that is irrational.


Please send suggestions/corrections to: Web Related: David.Mazza@grc.nasa.gov Technology Related: Joseph.C.Kolecki@grc.nasa.gov Responsible NASA Official: Theresa.M.Scott (Acting) 