Lets assume PI is rational.
Then PI = a/b, a and b are integers.
We also know pi is not an integer as its value lies between 3 and 4.
We know the equation:
e^(i PI) = -1
substituting its rational form:
e^(i * (a/b)) = -1
raising both sides to bth power.
e^(a i) = (-1) ^b
now there are two cases:
1. b is odd.
in this case e^(a i) = -1
this means cos a = -1
this implies a is an odd multiple of PI. That is PI, 3PI, 5PI etc. As cosx takes -1 only in these points.
but we know PI is not an integer. so a is not an integer either.
2. b is even
in this case e^(a i) = 1
this means cos a =1
this means a is an even multiple of PI. That is 0, 2 PI, 4 PI etc.
but again 2 PI etc. are not integers. So only option left is a = 0 which would mean PI = 0!
This proves by contradiction that PI is irrational.
