The number of binary digits is less than the value

Question

I want a proof that the number of binary digits is less than the value. There is a log relationship $$base_{digits} = \log_2(value) + 1.$$ I can't seem to prove $base_{digits}<value$. Or perhaps using the definition $$n=a_m2^m +\dots+a_02^0.$$ Where $m+1$ is the number of binary digits. We must also recognize that for this inequality to hold $value>2$ where the $value\in\mathbb{W}$.

Answer 1

As commenters have pointed out, we need to assume we are talking about integers here, and we must assume $n\geq 3$.

After this, we need not use any facts about logarithms or exponents, we can simply observe that incrementation of a number by one can increase its number of binary digits (or for that matter, its digits in any base) by at most one. Therefore if the result is true for $n=k$, it must hold for $n=k+1$ as well. Since the result is true for $n=3$, it follows by induction that it is true for all $n\geq 3$.