Question About Floating Point System I have begun reading Numerical Analysis by Walter Gautschi. On page $3$, the author introduces the floating point number system as follows: a floating point number is a number representible as 
$$
\begin{equation}
\pm (.b_{-1}b_{-2}...b_{-t})_2 \cdot 2^{\pm(c_{s-1}c_{s-2}...c_0.)_2} \tag {1}
\end{equation}
$$
where $s, t$ are fixed. We denote this number system by $\mathbb{R}(t, s)$. 

Then, Walter says that a number $x \in \mathbb{R}(t, s)$ is normalized if $b_{-1}=1$, and that we $"$assume all numbers in $\mathbb{R}(t, s)$ are normalized $"$.

My Question: Consider consider the number 
$$x = +(.00100) \cdot 2^{-(111.)}$$
Is this number an element of $\mathbb{R}(5, 3)$? According to the definition, I want to say yes. But when the author says $"$we assume all numbers in $\mathbb{R}(t, s)$ are normalized,$"$ I don't know if he is implicitly redefining $\mathbb{R}(t, s)$ to be numbers of the form $(1)$ with the additional requirement that $b_{-1}=1$. If we try to impose this requirement on $x$, I believe $x$ would come out to 
$$x=+(.10000) \cdot 2^{-1001}$$
which shouldn't be part of $\mathbb{R}(5, 3)$

Background
I am a math major taking my first numerical analysis. I do not have any prior experience with computer science or numerical mathematics.
After reading further, I found the answer. The answer is that the $x$ given in the question is not in $\mathbb{R}(5, 3)$. The way I figured this out is as follows:

On the next page, the author states that 
$$\min_{x \in \mathbb{R}(t, s)} |x| = 2^{-2^s}$$
This is consistent with $(.100 \cdots 0) \cdot2^{-(11\cdots 1)}$ being the least positive number in $\mathbb{R}(t, s)$, and not $(.000 \cdots 01) \cdot2^{-(11\cdots 1)}.$