# Normalised binary exponential form

I am doing some mathematics, and I am currently stuck on something. I do not understand this part at all, expressing the result in normalized binary exponential form, the book also mentions characteristics and exponent bias. An example from the book:

Find the 32-bit computer representation of $$-1873.42$$, where 8 bits are used for the characteristics, and the exponent bias is $$2^7 - 1$$.

Solution:

$$-1873.42_{10} = -1101010001.011010111000_2$$

Express the result in normalized binary exponential form:

$$-11101010001.011010111000_2 = -0.11101010001011010111000 \times 2^{11}$$

• Sign bit: 1
• Characteristic: $$10001010$$
• Computer representation:

$$11000101\enspace01110101\enspace00010110\enspace10111000_2$$

Does someone here possibly know how these steps are performed to arrive at this? I am unable to find this in the book and I also Googled and went to YouTube. Finally, I decided to post here in the hopes of some light.

Reference:

• Discrete Mathematics for Computing, 3rd Edition by Peter Grossman

A binary representation of $$1873_{10}$$ is \begin{align*} 1873_{10}&=\color{blue}{111\,0101\,0101_{2}}\\ \\ 1873&=2^{10}+2^9+2^8+2^6+2^4+2^2+2^0\tag{1}\\ &=1024+512+256+64+16+4+1 \end{align*} A binary representation of $$0.42_{10}$$ using $$12$$ binary digits is, using a truncated representation \begin{align*} 0.42_{10}&\doteq \color{blue}{0.0110\,1011\,1000_2}\\ &=\frac{1}{4}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\tag{2}\\ &=0.419\,921\,875 \end{align*} Note that $$0.0110\,1011\,100\color{blue}{1}_2>0.42_{10}$$.

From (1) and (2) we obtain the representation \begin{align*} \color{blue}{-1873.42_{10}=-111\,0101\,0101.0110\,1011\,1000_2}\tag{3} \end{align*} We want a normalized representation such that the significand is a fractional part. From (3) we obtain the normalised representation \begin{align*} \color{blue}{-1873.42_{10}=-0.111\,0101\,0101\,0110\,1011\,1000_2\times 2^{11}}\tag{4} \end{align*} Here we multiply the right-hand side by $$2^{11}$$ to compensate for the shift of the significand by $$11$$ digits.

We have now in (4) for the fractional part $$23$$ bits out of $$32$$ in use. One sign bit $$b_0$$ and eight bits $$b_1\ldots b_8$$ for the exponent are left.

• We use the most significant bit $$b_0$$ as sign bit and set it to $$1$$ to indicate the minus sign of $$-1873.42_{10}$$.

• We use bits $$b_1b_2\ldots b_8$$ as exponent. The exponent is biased by $$2^7-1=127$$. We therefore use due to the exponent $$11$$ from (4) \begin{align*} (11+127)_{10}=138_{10}=128_{10}+8_{10}+2_{10}\color{blue}{=1000\,1010_2}\tag{5} \end{align*}

Putting finally sign bit $$b_0$$, the exponent from (5) and the fractional part (4) together we obtain the $$32$$ bit representation \begin{align*} \color{blue}{1100\,0101\,0111\,0101\,0101\,0110\,1011\,1000} \end{align*}

• How do you know what exponent to use? Commented Jul 24, 2023 at 17:30
• @AlixBlaine: If the number is not equal to zero, the significand contains at least one digit $1$. We shift the number such that the fractional binary part starts with $1$, in this case we shift the number by $11$ digits. Commented Jul 24, 2023 at 17:52
• Aha, so the 11 in the exponent is actually the length of the shift. How do you determine this? Commented Jul 24, 2023 at 19:50
• @AlixBlaine: This shift is performed until the first bit to the right of the decimal point is $1$ and there are no $1$ left of the decimal point. Commented Jul 24, 2023 at 19:56
• I have seen shifts by 10 for example. Commented Jul 24, 2023 at 20:20