Are These Polynomial Factorizations Correct in Z7?

In summary, the two polynomials $x^2- x+ 1$ and $x^3+ x+ 1$ are irreducible over the natural numbers, but they have a greatest common divisor in $Z_7$.
  • #1
ertagon2
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  • #2
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 \(\displaystyle f(x)= x^4+ x^3+ x+ 1\) and \(\displaystyle g(x)= x^5- x^3+ x^2- 1\). We need to factor those to see what possible divisors there are. The first thing I notice is that \(\displaystyle f(-1)= 1- 1- 1+ 1= 0\) which means f has x+ 1 as a factor. Dividing f by x+ 1 gives \(\displaystyle x^3+ 1\). But \(\displaystyle (-1)^3+ 1= -1+ 1= 0\) also so there is another factor of x+ 1. Dividing \(\displaystyle x^3+ 1\) by x+ 1 we get \(\displaystyle x^2- x+ 1\). That does not factor (in the real numbers) since the quadratic formula give complex roots. \(\displaystyle f(x)= x^4+ x^3+ x+ 1= (x+ 1)^2(x^2- x+ 1)\).

And I see that \(\displaystyle g(1)= 1- 1+ 1- 1= 0\) so g(x) has x- 1 as a factor. In fact \(\displaystyle g(x)= (x- 1)(x^4+ x^3+ x+ 1)\). And [tex](-1)^4+ (-1)^3+ (-1)+ 1= 1- 1- 1+ 1= 0 so we have an additional factor of x+ 1: \(\displaystyle g(x)= (x- 1)(x+ 1)(x^3+ 1)\). Again \(\displaystyle (-1)^3+ 1= -1+ 1= 0\) so there is another factor of x+ 1: \(\displaystyle g(x)= (x- 1)(x+ 1)^2(X^2- x+ 1)\). The quadratic formula again shows that \(\displaystyle x^2- x+ 1\) does not factor. \(\displaystyle g(x)= x^5- x^3+ x^2- 1= (x- 1)(x+ 1)^2(x^2- x+ 1)\).

So what factors do f(x) and g(x) have in common?

If you expect help with the others, show what you have tried so we will understand what you do know about them!
 
  • #3
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 $n\in N$.
 
  • #4
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".
 
  • #5
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?
 

FAQ: Are These Polynomial Factorizations Correct in Z7?

What is induction and how is it used in mathematics?

Induction is a mathematical proof technique used to prove that a statement holds true for all values in a given set. It involves proving a base case and then showing that if the statement holds true for a particular value, it also holds true for the next value in the set.

How is induction used to prove properties of polynomials?

Induction can be used to prove properties of polynomials by showing that the statement holds true for a base case, such as a polynomial of degree 1, and then using the inductive step to show that if the statement holds true for a polynomial of degree n, it also holds true for a polynomial of degree n+1.

Can induction be used to prove all properties of polynomials?

No, induction can only be used to prove properties that follow a particular pattern or sequence. If the property does not have a predictable pattern, then induction cannot be used to prove it.

How can induction be used to show the binomial theorem for polynomials?

The binomial theorem for polynomials can be proved using induction by showing that it holds true for a base case, such as (x+y)^2, and then using the inductive step to show that if it holds true for (x+y)^n, it also holds true for (x+y)^(n+1).

Are there other proof techniques besides induction that can be used for polynomials?

Yes, there are other proof techniques such as direct proof, proof by contradiction, and proof by contrapositive that can also be used to prove properties of polynomials. The choice of proof technique depends on the specific property being proven.

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