Elena & Fabrice's Web

m (Created page with "= Multiplication = The action of repeated additions: $a\times b=\underbrace{$b$ times}a+a+\cdots+a$") |
m (→Multiplication) |
||

(6 intermediate revisions by one user not shown) | |||

Line 1: | Line 1: | ||

= Multiplication = | = Multiplication = | ||

− | The action of repeated [[addition]]s: $a\times b=\underbrace{$b$ times}a | + | The action of repeated [[addition]]s: $a\times b=\underbrace{a+a+\cdots+a}_{\hbox{$b$ times}}$. |

+ | |||

+ | By looking at it from a geometric point of view, it is not difficult—although not trivial either—to see that $a\times b=b\times a$. In typesetting, we do not use $.$ or $\cdot$ and usually omit the multiplication sign $\times$ ($\mathrm{\LaTeX}$ <tt>times</tt>, not x (ex)), so that we write $ab$. The most important rule of multiplication is how it combines with [[addition]], with the property of ''distributivity'': | ||

+ | |||

+ | $$(a+b)(c+d)=ac+ad+bc+bd\,.$$ | ||

+ | |||

+ | An interesting problem regards the complexity, or cost of multiplication. In 1960, Karatsuba improved on the "school method" by finding a clever way to write $(a+b)(c+d)$ as a sum of three products, rather than four. | ||

+ | |||

+ | == Links == | ||

+ | |||

+ | * [http://www.ccas.ru/personal/karatsuba/divcen.pdf Karatsuba recollections and comments on his methods]. |

The action of repeated additions: $a\times b=\underbrace{a+a+\cdots+a}_{\hbox{$b$ times}}$.

By looking at it from a geometric point of view, it is not difficult—although not trivial either—to see that $a\times b=b\times a$. In typesetting, we do not use $.$ or $\cdot$ and usually omit the multiplication sign $\times$ ($\mathrm{\LaTeX}$ `times`, not x (ex)), so that we write $ab$. The most important rule of multiplication is how it combines with addition, with the property of *distributivity*:

$$(a+b)(c+d)=ac+ad+bc+bd\,.$$

An interesting problem regards the complexity, or cost of multiplication. In 1960, Karatsuba improved on the "school method" by finding a clever way to write $(a+b)(c+d)$ as a sum of three products, rather than four.