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This helped a lot! Thanks...Neely | This helped a lot! Thanks...Neely | ||
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| + | I believe that (-2x+1) is the inverse, if we first assume (2x-1) is the inverse, we'll get | ||
| + | <math> (2x-1)(2x+1) = 4x^2 - 1 </math> | ||
| + | |||
| + | and since 4 = 0, we'll get the above equation to equal -1 mod 4. | ||
| + | So, -(2x-1) = (-2x+1) is the inverse. | ||
Latest revision as of 21:38, 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
Yea, he or she did mean that. Look at the line:
$ 4x^2+2x+2x+1=1 $ from that you can see he or she multiplied $ (2x+1)(2x+1)=1 $
or it would look like $ 4x^2-2x-2x+1=1 $
This helped a lot! Thanks...Neely
I believe that (-2x+1) is the inverse, if we first assume (2x-1) is the inverse, we'll get $ (2x-1)(2x+1) = 4x^2 - 1 $
and since 4 = 0, we'll get the above equation to equal -1 mod 4. So, -(2x-1) = (-2x+1) is the inverse.
