Written multiplication

Written multiplication is a computation process ( algorithm ) that can be used to multiply two multi-digit numbers using a written representation. The following describes the procedure for natural numbers. The extension to real numbers with a finite number of decimal places then takes place.

Procedure

The common practice is to split the multiplication of a multi-digit number (multiplicand) by a second multi-digit number (multiplier) into several multiplications of the first number by a single-digit number by breaking the second number into its digits. Then you have to multiply these results with the place value of the respective digit of the multiplier by adding the required number of zeros and finally add everything. The notation used is shown in the example below.

Math background

A natural -digit number with the sequence of digits ${\displaystyle n}$ ${\displaystyle a}$

${\displaystyle a_{n-1}a_{n-2}\dotsb a_{1}a_{0}}$

can be represented as the sum of single-digit multiples of powers of ten:

${\displaystyle a_{n-1}a_{n-2}a_{n-3}\dotsb a_{1}a_{0}=a_{n-1}\cdot 10^{n-1}+a_{n-2}\cdot 10^{n-2}+\dotsb +a_{1}\cdot 10^{1}+a_{0}\cdot 10^{0}}$

${\displaystyle a=\sum _{i=0}^{n-1}a_{i}\cdot 10^{i}}$

The multiplication of a -digit number with a -digit number corresponds to the multiplication ${\displaystyle n}$ ${\displaystyle a}$ ${\displaystyle m}$ ${\displaystyle b}$

${\displaystyle (a_{n-1}\cdot 10^{n-1}+a_{n-2}\cdot 10^{n-2}+\dotsb +a_{1}\cdot 10^{1}+a_{0}\cdot 10^{0})\cdot (b_{m-1}\cdot 10^{m-1}+b_{m-2}\cdot 10^{m-2}+\dotsb +b_{1}\cdot 10^{1}+b_{0}\cdot 10^{0})}$

${\displaystyle a\cdot b=\left(\sum _{i=0}^{n-1}a_{i}\cdot 10^{i}\right)\cdot \left(\sum _{k=0}^{m-1}b_{k}\cdot 10^{k}\right)}$

If the products of the digits and their place value are understood as elements of two vectors, then the multiplication can be understood as the sum of the elements of the dyadic product of the vectors to form a matrix:

${\displaystyle {\begin{pmatrix}a_{n-1}\cdot 10^{n-1}\\a_{n-2}\cdot 10^{n-2}\\\vdots \\a_{1}\cdot 10^{1}\\a_{0}\cdot 10^{0}\end{pmatrix}}\otimes {\begin{pmatrix}b_{m-1}\cdot 10^{m-1}\\b_{m-2}\cdot 10^{m-2}\\\vdots \\b_{1}\cdot 10^{1}\\b_{0}\cdot 10^{0}\end{pmatrix}}={\begin{pmatrix}{a_{n-1}b_{m-1}\cdot 10^{n+m-2}}&{a_{n-1}b_{m-2}\cdot 10^{n+m-3}}&\dotsb &{a_{n-1}b_{1}\cdot 10^{n}}&{a_{n-1}b_{0}\cdot 10^{n-1}}\\a_{n-2}b_{m-1}\cdot 10^{n+m-3}&a_{n-2}b_{m-2}\cdot 10^{n+m-4}&\dotsb &a_{n-2}b_{1}\cdot 10^{n-1}&a_{n-2}b_{0}\cdot 10^{n-2}\\\vdots &\vdots &\dotsb &\vdots &\vdots \\a_{1}b_{m-1}\cdot 10^{m}&a_{1}b_{m-2}\cdot 10^{m-1}&\dotsb &a_{1}b_{1}\cdot 10^{2}&a_{1}b_{0}\cdot 10^{1}\\a_{0}b_{m-1}\cdot 10^{m-1}&a_{0}b_{m-2}\cdot 10^{m-2}&\dotsb &a_{0}b_{1}\cdot 10^{1}&a_{0}b_{0}\cdot 10^{0}\end{pmatrix}}}$

At the above Procedure, all matrix elements are calculated and added column by column. These column sums are noted and then added up in writing so that the overall result is obtained.

The results of the columns are:

${\displaystyle (a_{n-1}\cdot 10^{n-1}+a_{n-2}\cdot 10^{n-2}+\dotsb +a_{1}\cdot 10^{1}+a_{0}\cdot 10^{0})\cdot (b_{m-1}\cdot 10^{m-1})}$

${\displaystyle (a_{n-1}\cdot 10^{n-1}+a_{n-2}\cdot 10^{n-2}+\dotsb +a_{1}\cdot 10^{1}+a_{0}\cdot 10^{0})\cdot (b_{m-2}\cdot 10^{m-2})}$

${\displaystyle \dotsb }$

${\displaystyle (a_{n-1}\cdot 10^{n-1}+a_{n-2}\cdot 10^{n-2}+\dotsb +a_{1}\cdot 10^{1}+a_{0}\cdot 10^{0})\cdot (b_{1}\cdot 10^{1})}$

${\displaystyle (a_{n-1}\cdot 10^{n-1}+a_{n-2}\cdot 10^{n-2}+\dotsb +a_{1}\cdot 10^{1}+a_{0}\cdot 10^{0})\cdot (b_{0}\cdot 10^{0})}$

example

As an example we take the numbers and . Then the sub-steps result ${\displaystyle a=8642}$ ${\displaystyle b=9731}$

${\displaystyle (8\cdot 10^{3}+6\cdot 10^{2}+4\cdot 10^{1}+2\cdot 10^{0})\cdot (9\cdot 10^{3})}$

${\displaystyle (8\cdot 10^{3}+6\cdot 10^{2}+4\cdot 10^{1}+2\cdot 10^{0})\cdot (7\cdot 10^{2})}$

${\displaystyle (8\cdot 10^{3}+6\cdot 10^{2}+4\cdot 10^{1}+2\cdot 10^{0})\cdot (3\cdot 10^{1})}$

${\displaystyle (8\cdot 10^{3}+6\cdot 10^{2}+4\cdot 10^{1}+2\cdot 10^{0})\cdot (1\cdot 10^{0})}$

so

${\displaystyle (8000+600+40+2)\cdot 9000}$

${\displaystyle (8000+600+40+2)\cdot 700}$

${\displaystyle (8000+600+40+2)\cdot 30}$

${\displaystyle (8000+600+40+2)\cdot 1}$

With the help of a staggered placement of the values ​​on preferably squared paper , you can save having to write down the powers of ten (shown in red in the graphics). Using the multiplication table and adding them, one gets for the lines:

The whole scheme with the shortened notation of the lines is then:

The multiplication is now complete.

Decimal places and signs

If at least one factor has decimal places, the multiplication is initially carried out as if they were whole numbers. Then you have to set the comma so that the number of decimal places of the result corresponds to the sum of the number of decimal places of the factors.

If at least one factor has a negative sign, the amounts are first multiplied and then the sign is determined using the sign rules .

literature

• Friedhelm Padberg, Andreas Büchter: Introduction to Mathematics Primary Level - Arithmetic . 2nd Edition. Springer, 2015, ISBN 978-3-662-43449-9 , pp. 50-55.
• Petra Knöß: Fundamental ideas of computer science in mathematics lessons: Basic considerations and examples for the primary level . Springer, 1989, pp. 189-201.
• Schülerduden - Mathematics I . 8th edition. Duden-Verlag, 2008, ISBN 978-3-411-04208-1 , pp. 198, 202, 412-414.

Web links

• Instructions for Written Multiplication (Teacher Training Baden-Württemberg)