Is

log2 rational or irrational? Justify your answer.

Assume that log 2 is rational, that is,

…(1)

where p, q are integers.

Since, and therefore, p<q

From Eq. 1,

2^q = (2×5)^p

2^(q-p) = 5^p

Where p-q is an integer greater than 0.

Now, it can be seen that the L.H.S is even and the R.H.S is odd.

Hence, there is contradiction and log 2 is irrational.