Congruence Class Question


by Brimley
Tags: class, congruence
hochs
hochs is offline
#19
Nov1-10, 07:26 PM
P: 39
Alright I'm wasting way too much time on this triviality. Here's the complete ****ing proof, leave me alone now:

if x = 1 (mod lambda) then x = 1 + a*(1 - w) + b*(1 - w)^2 for some integers a,b.

Then x-1 = a*(1 - w) + b*(1 - w)^2 is divisible by 1-w clearly.

x-w = 1 - w + a*(1 - w) + b*(1 - w)^2 = (1-w)(1 + a + b*(1 - w)) is divisible by 1-w clearly.

Finally,
x-w^2 = 1 - w^2 + a*(1 - w) + b*(1 - w)^2 = (1 - w)(1 + w) + a*(1 - w) + b*(1 - w)^2 = (1 - w)(1 + w + a + b(1 - w)) is divisible by 1-w.

So all of x-1, x-w, and x-w^2 are divisible by lambda. Do you not see that (x - 1)(x - w)(x - w^2) is now divisible by lambda^3 ?
Brimley
Brimley is offline
#20
Nov1-10, 07:26 PM
P: 77
Quote Quote by hochs View Post
Just wow... are you a native English speaker?

I told you, you need to show separately that x-1, x-w, and x-w^2 are all divisible by lambda. That's 3 statements to show.

my god..
I'm sorry about your frustration hochs! I'm really trying here! (Btw, it takes a ton of time to write it in LaTeX, but I think its easier on the eyes of those reading it herein). I'm putting in the time and effort to understand this, so I do appreciate the patience!

If you remember, I told you in several posts that I had trouble understanding your translation from using [itex]a,b,[/itex] and [itex]w[/itex] to just [itex]x[/itex] and [itex]w[/itex]. I'm strill trying to make that connection, then I could better attempt explaining why [itex]\lambda[/itex] divides [itex]x-1,x-w,[/itex] and [itex]x-w^2[/itex].
Brimley
Brimley is offline
#21
Nov1-10, 07:31 PM
P: 77
Quote Quote by hochs View Post
if x = 1 (mod lambda) then x = 1 + a*(1 - w) + b*(1 - w)^2 for some integers a,b.

Then x-1 = a*(1 - w) + b*(1 - w)^2 is divisible by 1-w clearly.

x-w = 1 - w + a*(1 - w) + b*(1 - w)^2 = (1-w)(1 + a + b*(1 - w)) is divisible by 1-w clearly.

Finally,
x-w^2 = 1 - w^2 + a*(1 - w) + b*(1 - w)^2 = (1 - w)(1 + w) + a*(1 - w) + b*(1 - w)^2 = (1 - w)(1 + w + a + b(1 - w)) is divisible by 1-w.

So all of x-1, x-w, and x-w^2 are divisible by lambda. Do you not see that (x - 1)(x - w)(x - w^2) is now divisible by lambda^3 ?
I do see it now. I really appreciate it, but please don't get mad at me. I'm just trying to understand what you were doing, and I had most of it but was confused in the translation. Seeing that performed makes sense of it. I do appreciate it and I understand your frustration, but please don't get mad at someone who is just trying to learn mathematics.

I take it to alter this for part B, all I would need to do is flip the + and - signs for part A's?
Is there a quick explanation on why part C's is trivial?
stoolie77
stoolie77 is offline
#22
Nov1-10, 11:57 PM
P: 7
Quote Quote by Brimley View Post

I take it to alter this for part B, all I would need to do is flip the + and - signs for part A's?
interesting thread. ive tried my best to follow along and i believe that this is all you need to do. hochs should confirm it first though as hes beek working with you this whole time.
Brimley
Brimley is offline
#23
Nov5-10, 11:54 PM
P: 77
Quote Quote by stoolie77 View Post
interesting thread. ive tried my best to follow along and i believe that this is all you need to do. hochs should confirm it first though as hes beek working with you this whole time.
Can anyone help me finish this? Stoolie I appreciate your input, but yes, I'd really appreciate if someone could confirm this or give me a hand finishing this.


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