How do you prove #log_2 3# is irrational number ?

1 Answer
Cesareo R.
Oct 27, 2017

See below.

Explanation:

#y = log_2 3 hArr 2^y = 3# If #y# is rational then #y = n/m# with #{n,m} in ZZ# and # m ne 0# or

#2^(n/m) = 3 rArr 2^n = 3^m#

This last equality is an absurd because #3^m# is odd and #2^n# is even.

Concluding #y = log_2 3# is not rational, being then irrational.

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