This question came up to my mind while dealing with complexity of binary operations, speciffically with multiplication.
The question. Let's say we have two binary numbers $n$ and $m$, such that $n \geqslant m.$ Can we guarantee (only with this information) that the number of digits of $n$ (say $k$) is greater or equal than the number of digits of $m$ (say $l$), i.e, can we guarantee that $k \geqslant l?$
My thoughts. This obviously should be a trivial question but I was unable to find an answer for it with a quick search on the internet. Therefore, I started thinking: Firstly, by contraposition, it suffices to show that if $l > k$, then $m > n.$ Therefore, let us fix arbitrary binary numbers $m$ and $n$ such that $l>k.$ The worst case scenario (where our thesis is most likely to fail) is if $m$ is just a sequence of zeros that ends in a $1$, i.e., if $m = 10\dots 0$ and $n$ is a sequence of ones with exactly one less digit than $m$ (here, we are just picking the smallest possible value for $m$ and the highest possible value for $n$). Therefore, transforming into decimal we can say that $(k = l-1)$
$$ m = 2^{k} \quad \text{ and } \quad n = \sum_{i=0}^{k-1} 2^i. $$
Now, it suffices to see that
$$ n = \sum_{i=0}^{k-1} 2^i = 1 + \sum_{i=1}^{k-1} 2^i = 1 + \frac{2(1-2^{k-1})}{1-2} = 1 - 2 + 2^k = 2^k - 1 < 2^k = m, $$ proving what's wanted.
Is this result true and are my thoughts behind it acceptable?