Chapter 3, Problem 3.1E

Program Plan Intro

Hexadecimal numbers:

In mathematics, hexadecimal numbers are positional numbers with a base or a radix of “16” which uses symbols “0–9” to represent values zero to nine, and “A–F” (or alternatively “a–f”) to represent values ten to fifteen.

Given Instruction: $\text{5ED4-07A4}$

Explanation of Solution

Hexadecimal numbers Operations:

Step-1:

Converting “$5ED4$” hexadecimal number to decimal format.

Hexadecimal numbers are written with the base of “16”, which means that each digit takes 4 bits as “$16=2^4$”.

Base 16 uses symbols “0–9” to represent values zero to nine, and “A–F” (or alternatively “a–f”) to represent values ten to fifteen.

The 16 digits of hexadecimal representation are: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

“$\text{5ED4}$”

$\begin{array}{l}=5*{16}^3+E*{16}^2+D*{16}^1+4*{16}^0\\ =5*{16}^3+14*{16}^2+13*{16}^1+4*{16}^0\\ =5*4096+14*256+13*16+4*1\\ =20480+3584+208+4\end{array}$

 = $24276indecimalbase$

Hence, the decimal value of “$5ED4$” is “24276”.

Step-2:

Converting “$07A4$” hexadecimal number to decimal format.

$\begin{array}{l}\text{07A4}\\ =0*{16}^3+7*{16}^2+A*{16}^1+4*{16}^{\begin{array}{l}0\\ \end{array}}\\ =0*{16}^3+7*{16}^2+10*{16}^1+4*{16}^{\begin{array}{l}0\\ \end{array}}\\ =0*4096+7*256+10*16+4*1\end{array}$

$\begin{array}{l}=0+1792+160+4\\ =1956indecimalbase\end{array}$

Hence, the decimal value of “$\text{07A4}$” is “1956”.

Step-3:

Performing “$\text{5ED4-07A4}$”

As the numbers are represented in unsigned magnitude format, a simple subtraction operation occur which results in:

$\begin{array}{l}\text{=5ED4 –}\text{07A4}\left(\text{hex}\right)\\ \text{=24276 - 1956 }\left(\text{decimal}\right)\\ \text{=22320}\end{array}$

Step-4:

The above value is in decimal format and hence one needs to convert the result into hexadecimal form.

In order to convert the value into hexadecimal, one needs to find the biggest power of 16.

“${16}^3=4096$” and hence the biggest power is “3”.

Now, one needs to divide the number by the highest power to see how many times it fits, which results as:

$\begin{array}{l}22320/4096\\ =\mathrm{5.45.}\\ \text{= 5}\end{array}$

Now “$5*4096=20480$”, and one needs to subtract that from the number which results in:

“$22320-20480=1840$”.

Now, the next power that fits in this number is “${16}^2=256$”, and it fits:

$1840/256=\mathrm{7.18..}times$.

Also, “$256*7=1792$”.

Next, one needs to subtract the above result:

“$1840-1792=48$”.

The next power that fits is “${16}^1=16$”, and it fits “$48/16=3$” times exactly.

Hence, “$\text{5ED4-07A4}$”

$\begin{array}{l}=22320(decimal)\\ =5*{16}^3+7*{16}^2+3*{16}^1+0*{16}^0\\ =5730(hexadecimal)\end{array}$

Hence, the unsigned operation “$5ED4-07A4$” results into “5730” as hexadecimal.