MHB Are These Polynomial Factorizations Correct in Z7?

ertagon2
Messages
36
Reaction score
0

Attachments

  • Maths61-2.png
    Maths61-2.png
    20.8 KB · Views: 150
  • Maths61-3.png
    Maths61-3.png
    15.7 KB · Views: 150
  • Maths61-4.png
    Maths61-4.png
    16.2 KB · Views: 148
Physics news on Phys.org
Oh, dear. You have left almost all of them blank and those you have tried to answer, the "multiple choice" questions, are mostly wrong.

The first question asks you to find the Greatest Common Divisor of $$f(x)= x^4+ x^3+ x+ 1$$ and [math]g(x)= x^5- x^3+ x^2- 1[/math]. We need to factor those to see what possible divisors there are. The first thing I notice is that [math]f(-1)= 1- 1- 1+ 1= 0[/math] which means f has x+ 1 as a factor. Dividing f by x+ 1 gives [math]x^3+ 1[/math]. But [math](-1)^3+ 1= -1+ 1= 0[/math] also so there is another factor of x+ 1. Dividing [math]x^3+ 1[/math] by x+ 1 we get [math]x^2- x+ 1[/math]. That does not factor (in the real numbers) since the quadratic formula give complex roots. [math]f(x)= x^4+ x^3+ x+ 1= (x+ 1)^2(x^2- x+ 1)[/math].

And I see that [math]g(1)= 1- 1+ 1- 1= 0[/math] so g(x) has x- 1 as a factor. In fact [math]g(x)= (x- 1)(x^4+ x^3+ x+ 1)[/math]. And (-1)^4+ (-1)^3+ (-1)+ 1= 1- 1- 1+ 1= 0 so we have an additional factor of x+ 1: [math]g(x)= (x- 1)(x+ 1)(x^3+ 1)[/math]. Again [math](-1)^3+ 1= -1+ 1= 0[/math] so there is another factor of x+ 1: [math]g(x)= (x- 1)(x+ 1)^2(X^2- x+ 1)[/math]. The quadratic formula again shows that [math]x^2- x+ 1[/math] does not factor. [math]g(x)= x^5- x^3+ x^2- 1= (x- 1)(x+ 1)^2(x^2- x+ 1)[/math].<br /> <br /> So what factors do f(x) and g(x) have in common? <br /> <br /> If you expect help with the others, show what you have tried so we will understand what you <b>do</b> know about them!
 
For number 9, "word 5" is "every".

Let P(n) b statement about the natural numbers. If
1. P(1) is true[/b] and
2. P(k) implies P(k+1) for every integer k> 0

then we can conclude that P(n) is true for every $n\in N$.
 
For 10, you have the statement "Every proof by mathematical induction requires at least two base cases to be checked" marked "true". That is incorrect. Proof by induction requires only one "base case" be checked.

And you have the statement "Mathematical induction differs from the kind of induction used in the experimental sciences because it is actually a form of deductive reasoning" marked "false". That is incorrect. "Mathematical Induction" is "deductive reasoning" while the type of "induction" used in the experimental sciences is "inductive reasoning".
 
In 7, $x^3- 1= (x- 1)(x^2- x+ 1)$ and $x^4- x^3+ x^2- 1= (x- 1)(x^3+ x+ 1 )$.
The two polynomials, $x^2- x+ 1$ and $x^3+ x+ 1$ are "irreducible" over the natural numbers. But working in $Z_7$, we need to check their values for x= 0, 1, 2, 3, 4, 5, and 6 "mod 7". $3^2- 3+ 1= 7$ and $5^2- 5+ 1= 25- 5+ 1= 21$, a multiple of 7 so $x^2- x+ 1= (x- 5)(x- 3)$ (mod 7) so we can write $x^3- 1= (x- 1)(x- 3)(x- 5)$ mod 7. No value from 0 to 6 makes $x^3+ x+ 1$ a multiple of 7 so it is irreducible even in $Z_7$.

In $Z_7$ $x^3- 1= (x- 1)(x- 3)(x- 5)$ and $x^4- x^3+ x^2- 1= (x- 1)(x^3+ x+ 1)$. What is the greatest common divisor of those?
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top