(New page: I got (1 - 2x) as the multiplicitave inverse, since that multiplied by (2x + 1) is 1. --~~~~) |
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| − | I got (1 - 2x) as the multiplicitave inverse, since that multiplied by (2x + 1) is 1. | + | I got (1 - 2x) as the multiplicitave inverse, since that multiplied by (2x + 1) is 1. ... which i guess is the same as the original function. |
--[[User:Jcromer|Jcromer]] 20:50, 1 April 2009 (UTC) | --[[User:Jcromer|Jcromer]] 20:50, 1 April 2009 (UTC) | ||
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| + | The unity of <math>\scriptstyle Z_4[x]</math> is 1, so we're looking for some polynomial <math>\scriptstyle g(x)</math> such that <math>\scriptstyle(2x+1)\cdot g(x)\ =\ 1</math>. So, just divide 1 by <math>\scriptstyle(2x+1)</math> modulo 4: | ||
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| + | <math>\scriptstyle1+2x</math> | ||
| + | <math>\scriptstyle1+2x)\overline{1+0x+0x^2}</math> | ||
| + | <math>\scriptstyle-\underline{1+2x}</math> | ||
| + | <math>\scriptstyle2x+0x^2</math> | ||
| + | <math>\scriptstyle-\underline{2x+0x^2}</math> | ||
| + | <math>\scriptstyle0</math> | ||
| + | |||
| + | From this it's clear that <math>\scriptstyle1\ =\ (2x+1)\cdot(2x+1)</math>, so the multiplicative inverse of <math>\scriptstyle(2x+1)</math> is itself. | ||
| + | :--[[User:Narupley|Nick Rupley]] 01:44, 2 April 2009 (UTC) | ||
Latest revision as of 20:44, 1 April 2009
I got (1 - 2x) as the multiplicitave inverse, since that multiplied by (2x + 1) is 1. ... which i guess is the same as the original function. --Jcromer 20:50, 1 April 2009 (UTC)
The unity of $ \scriptstyle Z_4[x] $ is 1, so we're looking for some polynomial $ \scriptstyle g(x) $ such that $ \scriptstyle(2x+1)\cdot g(x)\ =\ 1 $. So, just divide 1 by $ \scriptstyle(2x+1) $ modulo 4:
$ \scriptstyle1+2x $
$ \scriptstyle1+2x)\overline{1+0x+0x^2} $
$ \scriptstyle-\underline{1+2x} $ $ \scriptstyle2x+0x^2 $ $ \scriptstyle-\underline{2x+0x^2} $ $ \scriptstyle0 $
From this it's clear that $ \scriptstyle1\ =\ (2x+1)\cdot(2x+1) $, so the multiplicative inverse of $ \scriptstyle(2x+1) $ is itself.
- --Nick Rupley 01:44, 2 April 2009 (UTC)
