(New page: Suppose the inverse of <math>2x-1</math> is <math>2x-1</math>, then <math>(2x-1)(2x-1)=1</math> <math>4x^2+2x+2x+1=1</math> <math>4x^2+4x+1=1</math>, but in <math>Z_4[x]</math>, 4=0. so...) |
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Therefore, <math>2x-1</math> has an inverse in <math>Z_4[x]</math> and specifically, that inverse is <math>2x-1</math> | Therefore, <math>2x-1</math> has an inverse in <math>Z_4[x]</math> and specifically, that inverse is <math>2x-1</math> | ||
| + | ------------------------------ | ||
| + | Did you mean to put 2x+1? | ||
| + | -Sarah | ||
Revision as of 11:39, 5 November 2008
Suppose the inverse of $ 2x-1 $ is $ 2x-1 $, then
$ (2x-1)(2x-1)=1 $
$ 4x^2+2x+2x+1=1 $
$ 4x^2+4x+1=1 $, but in $ Z_4[x] $, 4=0. so,
$ 0x^2+0x+1=1 $
$ 1=1 $
Therefore, $ 2x-1 $ has an inverse in $ Z_4[x] $ and specifically, that inverse is $ 2x-1 $
Did you mean to put 2x+1? -Sarah
