Is #pi^2# rational or irrational?

1 Answer
Oct 29, 2016

#pi^2# is irrational

Explanation:

#pi# is transcendental, meaning that it is not the root of any polynomial equation with integer coefficients.

Hence #pi^2# is transcendental and irrational too.

If #pi^2# were rational, then it would be the root of an equation of the form:

#ax+b = 0#

for some integers #a# and #b#

Then #pi# would be a root of the equation:

#ax^2+b = 0#

Since #pi# is not the root of any polynomial with integer coefficients, let alone a quadratic, this is not possible.

Further, if #pi^2# was the root of any polynomial equation with integer coefficients then #pi# would be the root of the same equation with each #x# replaced by #x^2#. So since #pi# is transcendental, so is #pi^2#.