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: 156
  • Maths61-3.png
    Maths61-3.png
    15.7 KB · Views: 158
  • Maths61-4.png
    Maths61-4.png
    16.2 KB · Views: 153
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?
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 24 ·
Replies
24
Views
695
  • · Replies 1 ·
Replies
1
Views
2K
Replies
0
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K