Respuesta :
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)
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Let say that g(x) = [tex]b_qx^q +b_{q-1}^{q-1}+...+b_1x +b_0[/tex] with deg(g(x)) = q
f(x) = [tex]a_nx^n +a_{n-1}^{n-1} +...+a_1x +a_0[/tex] with its deg(f(x)) =n
What this means is that [tex]a_n[/tex] and [tex]b_q[/tex] are not non-zero coefficients
That to say [tex]a_n \neq 0[/tex] and [tex]b_q \neq 0[/tex]
Looking at the product of the two function
f(x)g(x) =[tex](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)[/tex]
=[tex]a_nb_qx^{n+q}+...+a_0b_0[/tex]
Looking at the above equation in terms degree
de(f(x)g(x)) = [tex]deg(a_nb_qx^{n+q}+...+a_0b_0)[/tex]
= n+q
= [tex]deg(f(x))+deg(g(x))[/tex]
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) = [tex]3x^2[/tex] ∈ [tex]R_2[x][/tex] and g(x) = [tex]3x^2[/tex] ∈ [tex]R_2[x][/tex]
Notice that f(x) = g(x) [tex]\neq[/tex] 0
Let take a closer look at their product
f(x) g(x) = ([tex]4x^2[/tex] ).([tex]4x^2[/tex] )
= [tex]16x^4[/tex]
To obtain their degree we input mod (power in this case 4)
= [tex]0x^4[/tex]
= 0
So the degree of the polynomial is 0
Since the both polynomial are the same then
deg(f(x)) = deg(g(x))
= deg ([tex]2x^2[/tex])
= 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
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) = with deg(g(x)) = q
f(x) = with its deg(f(x)) =n
What this means is that and are not non-zero coefficients
That to say and
Looking at the product of the two function
f(x)g(x) =
=
Looking at the above equation in terms degree
de(f(x)g(x)) =
= n+q
=
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) = ∈ and g(x) = ∈
Notice that f(x) = g(x) 0
Let take a closer look at their product
f(x) g(x) = ( ).( )
=
To obtain their degree we input mod (power in this case 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 ()
= 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 Step-by-step explanation:
