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kherson [118]
3 years ago
10

HELLLLLP ESGESGKSEGKESKGKSEKGESKKKS KKSGEKSGKSKEK GKSEGKKSEGKK what is 1+!

Mathematics
2 answers:
aliya0001 [1]3 years ago
8 0
So hard I want to know but it’s so hard
uranmaximum [27]3 years ago
7 0
1 Kakfngn s s c xbsnanndnsjs
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(Degree Rule) LetDbe an integral domain andf(x), g(x)∈D[x]. Prove that deg(f(x)·g(x)) = degf(x) + degg(x). Show, by example, tha
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Complete Question

The complete question is shown on the first uploaded image

Answer:

From the question we are told that

Let D be an integral domain  and f(x),g(x)∈ D[x]

We are to prove that

if f(x),g(x)∈ D[x] then deg(f(x)g(x)) = deg f(x) + deg  g (x)

to do this we need to show that the commutative ring R is possible. That  f(x) and g(x) are non-zero elements in R[x], deg f(x)g(x) < g=deg f(x) + deg g(x)  

\

Let say that g(x) = b_qx^q +b_{q-1}^{q-1}+...+b_1x +b_0 with deg(g(x)) = q

f(x) = a_nx^n +a_{n-1}^{n-1} +...+a_1x +a_0 with its deg(f(x)) =n

What this means is that a_n and b_q are not non-zero coefficients

That to say  a_n \neq 0 and b_q \neq  0

Looking at the product of the two function

f(x)g(x) =(a_nx^n+a_{n-1}x^{n-1}+...+a_1x +a_0)(b_qx^q+b_{q-1}x^{q-1}+...+b_1x+b_0)

           =a_nb_qx^{n+q}+...+a_0b_0

Looking at the above equation in terms degree

  de(f(x)g(x)) = deg(a_nb_qx^{n+q}+...+a_0b_0)

                    = n+q

                   = deg(f(x))+deg(g(x))

Looking at the above equation we have proven that

                      deg(f(x)g(x)) = deg f(x) + deg  g (x)

Now considering this example

 f(x) = 3x^2  ∈ R_2[x]  and g(x) = 3x^2 ∈  R_2[x]  

Notice that f(x) = g(x) \neq 0

Let take a closer look at their product

   f(x) g(x) = (4x^2 ).(4x^2 )

                = 16x^4

    To obtain their degree we input mod (power in this case 4)

               = 0x^4

               = 0

So the degree of the polynomial is 0

Since the both polynomial are the same then

      deg(f(x)) =  deg(g(x))

                     = deg (2x^2)

                     = 2

and we know from our calculation that the

       deg(f(x)g(x)) = deg (0)

                           = 0

     and looking at this we can see that it is not equal to the individual degrees of the polynomial added together i.e 2+2 = 4

     Thus  deg(f(x)g(x))  < degf(x) + deg g(x)

Note   This only true when the ring is a cumulative ring

Step-by-step explanation:

In order to get a better understanding of the solution above Let explain so terms

RING

  In mathematics we can defined a  ring a a collection R of item that has the ability to perform binary operations that define generally addition and multiplication

Now when talk about commutative ring it mean that the operation that the collection is equipped to perform is commutative in nature  

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