Floating point bitwise comparator. If f1 and f2 are floating point numbers with the following properties can we always say f1 > f2?
If we study the expression for a floating point number, $$(-1)^{s}\cdot2^{e-127}(1 + \sum\limits_{k=1}^{23} b_{23-k}\cdot 2^{-k}),$$
and pay attention to the sum $\sum\limits_{k=1}^{23} b_{23-k}\cdot 2^{-k}$, we note that
$$0 \leq \sum\limits_{k=1}^{23} b_{23-k}\cdot 2^{-k} \leq \sum\limits_{k=1}^{23} 2^{-k} &lt; \sum\limits_{k=1}^\infty 2^{-k} = 1,$$
where the last equality is formed by evaluating the geometric series. Thus we find that
$$2^{e-127} \leq2^{e-127}(1 + \sum\limits_{k=1}^{23} b_{23-k}\cdot 2^{-k}) &lt; 2^{e-127}\cdot 2$$
Thus if $s = 0$ and $e_1 &gt; e_2$ we find that $f_2 &lt; 2^{e_2 - 127} \cdot 2 \leq 2^{e_1 - 127} \leq f_1$.
Note that if $s = 1$ and $e_1 &gt; e_2$, the signs are reversed and we actually have $f_2 &gt; f_1$. But always $|f_1| &gt; |f_2|$.

The extended result of the above is that if we ignore the effects of denormal numbers, NaNs, infinities and say that $-0 &lt; 0$, we have
$$f_1 &lt; f_2 \iff n_1 &lt; n_2,$$
where $n_1, n_2$ are the bit patterns of $f_1, f_2$ interpreted as $w$-bit two's complement integers.