# Difference between revisions of "HowTo:Use Modular Numbers"

From CoCoAWiki

m |
(Strange behaviour of modular numbers) |
||

Line 1: | Line 1: | ||

− | + | == Question == | |

− | |||

What's wrong in this? | What's wrong in this? | ||

Use R::=Z/(5)[x]; | Use R::=Z/(5)[x]; | ||

Line 11: | Line 10: | ||

Moreover this command throws an error | Moreover this command throws an error | ||

Use R::=Z/(5); | Use R::=Z/(5); | ||

− | + | ||

+ | == Answer == | ||

Type(5); | Type(5); | ||

INT | INT | ||

------------------------------- | ------------------------------- | ||

5 is an integer, and integers do not depend on the current ring: could you imagine what would happen to a For cycle over Z/(2)? ;-) | 5 is an integer, and integers do not depend on the current ring: could you imagine what would happen to a For cycle over Z/(2)? ;-) | ||

+ | |||

If you want to use 5 as a modular number you should either use this syntax (similar to C/C++) | If you want to use 5 as a modular number you should either use this syntax (similar to C/C++) | ||

5 % 5; | 5 % 5; |

## Revision as of 16:50, 29 November 2005

## Question

What's wrong in this?

Use R::=Z/(5)[x]; 5x+2y; 2y ------------------------------- 5=0; FALSE -------------------------------

Moreover this command throws an error

Use R::=Z/(5);

## Answer

Type(5); INT -------------------------------

5 is an integer, and integers do not depend on the current ring: could you imagine what would happen to a For cycle over Z/(2)? ;-)

If you want to use 5 as a modular number you should either use this syntax (similar to C/C++)

5 % 5; 0 % 5 -------------------------------

or embed your integer into the polynomial ring

Use R::=Z/(5)[x]; Poly(5); 0 -------------------------------

The creation of a polynomial ring with no indeterminated has been disabled to highlight this (unexpected?) behaviour.