Greatest Power of Two not Divisor

Lemma
Let $S = \{ {2, 3, 4, \ldots, n}\}$ be a set of natural numbers from $2$ to $n$.

Let $t$ be the greatest power of two in $S$.

Then $t$ does not divide any other number in $S$.

That is, no other member of $S$ is a multiple of $t$.

Proof
Seeking a contradiction, assume otherwise.

That is, assume that there is some multiple of $t$ in $S$ other than itself. Call it $k$.

As $k$ cannot be smaller than $t$, $k$ must be in $\left \{ {t+1, t+2, \ldots, n}\right\}$.

By the definition of multiple, $k$ is of the form:


 * $k = t \times m$

for some $m \in \Z$.

Suppose $m$ is even.

Then:

But if $2 \times t \times j \in S$, then so too must $2 \times t$ be.

This contradicts $t$ being the greatest power of $2$ in $S$.

So $k$ must be odd.

But $k$ is a multiple of $t$, which is a multiple of $2$, so cannot be odd.

Thus $k$ is neither even nor odd, a contradiction.

Also see

 * Harmonic Numbers not Integers