Difference between revisions of "Talk:Modular equivalence"

From Why start at x, y, z
m (Alternate Mod Notation)
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Examples:  
Examples:  
In lieu of $x\equiv y \mod{z}$, we have $x\bm{z}y$
In lieu of $x\equiv y \mod{z}$, we have $x\bm{z}y$
Rather than take both sides $\mod{m}$, we would take both sides $\bm{m}$
Rather than take both sides $\mod{m}$, we would take both sides $\bm{m}$
Taking all $m\bm{7}1\in \mathbb{Z}_{+}$ gives $\{1,8,15,\dots\}$
Taking all $m\bm{7}1\in \mathbb{Z}_{+}$ gives $\{1,8,15,\dots\}$
A non-equivalence can be written $9\bmn{4}3$
A non-equivalence can be written $9\bmn{4}3$
It can be chained ala $27\bm{11}5\bm{3}2$
It can be chained ala $27\bm{11}5\bm{3}2$
and even be adapted to carry quotient information as in $77\bmq{13}{5}12$.  
and even be adapted to carry quotient information as in $77\bmq{13}{5}12$.  



Revision as of 18:28, 2 July 2021


Mod notation can be bulky and counter-intuitive. Here is an alternate mod notation that simply removes the middle line from the equivalence sign, places the modulus inside and drops the now moot 'mod'.

The following macros are defined for the Latex that follows and the second graphic shows the result:

Alternate mod Latex macros

Examples: In lieu of $x\equiv y \mod{z}$, we have $x\bm{z}y$

Rather than take both sides $\mod{m}$, we would take both sides $\bm{m}$

Taking all $m\bm{7}1\in \mathbb{Z}_{+}$ gives $\{1,8,15,\dots\}$

A non-equivalence can be written $9\bmn{4}3$

It can be chained ala $27\bm{11}5\bm{3}2$

and even be adapted to carry quotient information as in $77\bmq{13}{5}12$.

Here's how that looks when rendered:

Alternate mod notation as rendered in Latex