- Proposition 3.7.3 from text

- Theorem 3.7.4 from text

- Prove that log_3 2 is irrational.

 Proof: By way of contradiction, assume that
  this is rational, i.e. there are integers a, b such that 
  log_3 2 = (a/b); 
  (Note that b<>0 and also a<>0 because 3^0=1 <> 2)

  Therefore,  3^{a/b}=2;  
  Raising both sides to the power of b yields:  3^a = 2^b
  
  Now, by Fundamental Theorem of Arithmetic, there should be
  the same number of each prime number on the LHS and RHS.
  But here, we only have powers of 3 in the LHS and only powers
  of 2 on the RHS, so they cannot be equal.

  Alternatively, we can say that 3^a is an odd number for
  any value of a > 0 while 2^b is an even number so they cannot
  be equal.