Mathematics, 24.01.2020 22:31 mikeysoulemison

Suppose m is a positive integer. is the set consisting of 0 and all polynomials with coefﬁcients in f and with degree equal to m a subspace ofp(f)?

Answers: 3

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Answers

For this case we don't have any problems for the conditions 1) and 3), but we have a problem with condition 2) since is not satisfied.

We just need to find a counterexample to show that the statement is False. If we find two elements in the subset provided S, and we show that the sum is not in S, then we have the counter example.

Let's say that we have two elements $a^m +1 , -a^m +1 \in S$ so both elements are in S, and if we apply the condition for the addition closed we got:

(a^m +1) +(-a^m +1) = a^m -a^m +1+1 = 2

And by definition of S, 2 is not in S so then since we can't satisfy the closed addition property then S can't be a subsapce

Step-by-step explanation:

For this case the answer would be no.

It can't be a subspace because we not satisfy the condition of closed under addition.

We need to remember that a subset U of V is a subspace of V if and only if U satisfies the following 3 conditions

1) Additive identity $0 \in U$

2) Closed under addition $u,v \in U \Rightarrow u+x \in U$

3) Cloases under scalar multiplication $a \in F, u \in U \Rightarrow au \in U$

Proof

For this case we don't have any problems for the conditions 1) and 3), but we have a problem with condition 2) since is not satisfied.

We just need to find a counterexample to show that the statement is False. If we find two elements in the subset provided S, and we show that the sum is not in S, then we have the counter example.

Let's say that we have two elements $a^m +1 , -a^m +1 \in S$ so both elements are in S, and if we apply the condition for the addition closed we got:

(a^m +1) +(-a^m +1) = a^m -a^m +1+1 = 2

And by definition of S, 2 is not in S so then since we can't satisfy the closed addition property then S can't be a subsapce

Answer from: Quest

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that $\frac{ab}{de}=\frac{bc}{ef}$ .

also, it is given that $\frac{ab}{de}=\frac{3}{4}$

so, from the options, we need that $\frac{bc}{ef}=\frac{3}{4}$ .

a. $\frac{bc}{ef}=\frac{6}{9}=\frac{2}{3}$

b. $\frac{bc}{ef}=\frac{4}{6}=\frac{2}{3}$

c. $\frac{bc}{ef}=\frac{8}{12}=\frac{2}{3}$

d. $\frac{bc}{ef}=\frac{12}{16}=\frac{3}{4}$

as, we can see that only option d gives the same ratio as $\frac{ab}{de}=\frac{3}{4}$ i.e. $\frac{bc}{ef}=\frac{3}{4}$ .

hence, the possible lengths are bc = 12 and ef = 16.

Answer from: Quest

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2 polynomial long division

dividing : x4+5x3-4x2-8x+6

("dividend")

by : x-1 ("divisor")

dividend x4 + 5x3 - 4x2 - 8x + 6

- divisor * x3 x4 - x3

remainder 6x3 - 4x2 - 8x + 6

- divisor * 6x2 6x3 - 6x2

remainder 2x2 - 8x + 6

- divisor * 2x1 2x2 - 2x

remainder - 6x + 6

- divisor * -6x0 - 6x + 6

remainder 0

quotient : x3+6x2+2x-6 remainder: 0

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Suppose m is a positive integer. is the set consisting of 0 and all polynomials with coefﬁcients in...