Now let’s consider a different polynomial g(x) = x^3 + x^2 + 1.

[krispiekr3am]

(a) Prove that g(x) is irreducible over GF(2).

can someone help me?
my email is krispiekr3am@yahoo.com

 

[Hurkyl]

Have you had any thoughts on the problem?

 

[tacman]

Sure, I'd be happy to help! First, let's define what it means for a polynomial to be irreducible over GF(2). This means that the polynomial cannot be factored into smaller polynomials over the finite field GF(2), also known as the field of integers modulo 2.

To prove that g(x) is irreducible over GF(2), we can use the following steps:

1. Assume that g(x) can be factored into two polynomials, say f(x) and h(x), over GF(2). This means that g(x) = f(x) * h(x).

2. Since g(x) = x^3 + x^2 + 1, we can write f(x) and h(x) as:

f(x) = x^a + x^b + x^c

h(x) = x^d + x^e + x^f

where a, b, c, d, e, and f are all integers between 0 and 2.

3. Now, we can multiply f(x) and h(x) together and equate it to g(x):

f(x) * h(x) = (x^a + x^b + x^c) * (x^d + x^e + x^f) = x^(a+d) + x^(a+e) + x^(a+f) + x^(b+d) + x^(b+e) + x^(b+f) + x^(c+d) + x^(c+e) + x^(c+f)

4. Since we are working in GF(2), all coefficients and exponents are either 0 or 1. This means that the only possible combinations of exponents that could result in x^3 are:

1+1+1 = 3

1+0+0 = 1

0+1+0 = 1

0+0+1 = 1

This means that the only possible combinations of exponents that could result in x^3 are (1,1,1) and (1,0,0), (0,1,0), (0,0,1) in any order.

5. However, when we look at the expanded form of f(x) * h(x), we see that there is no way to get x^3. This means that our assumption that g(x) can be fact