Question

Prove that the grade-school multiplication algorithm (page 24), when applied to binary numbers, always gives the right answer.

Short answer

Grade school multiplication can be defined as finding the multiplication by multiplying the elements in parts.

Step-by-step solution

Introduction

An algorithm (or method) for multiplying two integers is known as a multiplication algorithm. Many alternative algorithms are employed, depending on the magnitude of the numbers. Since the introduction of the decimal system, efficient multiplication algorithms have been available.

Calculation

Let’s take an example of binary of $143$.

$\begin{array}{l}1101\\ \underset{}{\times 1011}\\ 1101\left(1101\mathrm{times}1\right)\\ 1101(1101\mathrm{times}1,\mathrm{shifted}\mathrm{once})\\ 0000(1101\mathrm{times}0,\mathrm{shifted}\mathrm{twice})\\ \underset{}{+1101}(1101\mathrm{times}1,\mathrm{shifted}\mathrm{thrice})\\ 10001111\end{array}$

Final Answer

So, the multiplication of the binary number $1101\mathrm{and}1011\mathrm{is}10001111\text{i}.\mathrm{e}.143$.