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castortr0y [4]
3 years ago
7

Let H be the set of all polynomials having degree at most 4 and rational coefficients. Determine whether H is a vector space. If

it is not a vector space, determine which of the following properties it fails to satisfy. A: Contains zero vector B: Closed under vector addition C: Closed under multiplication by scalars
Mathematics
1 answer:
Verizon [17]3 years ago
7 0

Answer:

Yes. It is a vector space over the field of rational numbers \mathbb{Q}

Step-by-step explanation:

An element p of the set H has the form

p(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}

where a_{0},a_{1},a_{2},a_{3},a_{4} are rational coefficients.

The operations of addition and scalar multiplication are defined as follows:

p(x)+q(x)=(a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+x_{4}x^4)+(b_{0}+b_{1}x+b_{2}x^{2}+b_{3}x^{3}+b_{4}x^{4})=(a_{0}+b_{0})+(a_{1}+b_{1})x+(a_{2}+b_{2})x^{2}+(a_{3}+b_{3})x^{3}+(a_{4}+b_{4})x^{4}

\lambda p(x)=\lambda (a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4})=\lambda a_{0}+\lambda a_{1}x+\lambda a_{2}x^{2}+\lambda a_{3}x^{3}+\lambda a_{4}x^{4}

The properties that H, together the operations of vector addition and scalar multiplication,  must satisfy are:

  1. Conmutativity
  2. Associativity of addition and scalar multiplication
  3. Additive Identity
  4. Additive inverse
  5. Multiplicative Identity
  6. Distributive properties.

This is not difficult with the definitions given. The most important part is to show that H has a additive identity, which is the zero polynomial, that is closed under vector addition and scalar multiplication. This last properties comes from the fact that \mathbb{Q} is a field, then it is closed under sum and multiplication.

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Answer:

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From the information given, we have:

Two angles (<L and <M) in ∆LMN that are congruent to two corresponding angles (<O and <P) in ∆OPQ.

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1.) Determine the type of solutions for the function (Picture 1)
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Answer:

1) 2 nonreal complex roots

2) 1 Real Solution

3) 16

4) Reflected, narrower by a factor of 2/5, slides right 4 units and slides up 6 (units)

Step-by-step explanation:

1) The graph does not intercept the x-axis, therefore, there are no real solutions at the point y = 0

We get;

y = a·x² + b·x + c

At y = 6, x = -2

Therefore;

6 = a·(-2)² - 2·b + c = 4·a - 2·b + c

6 = 4·a - 2·b + c...(1)

At y = 8, x = 0

8 = a·(0)² + b·0 + c

∴ c = 8...(2)

Similarly, we have;

At y = 8, x = -4

8 = a·(-4)² - 4·b + c = 16·a - 4·b + 8

16·a - 4·b = 0

∴ b = 16·a/4 = 4·a

b = 4·a...(3)

From equation (1), (2) and (3), we have;

6 = 4·a - 2·b + c

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6 - 8 = -b

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b = 2

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The equation is therefor;

y = (1/2)·x² + 2·x + 8

Solving we get;

x = (-2 ± √(2² - 4 × (1/2) × 8))/(2 × (1/2))

x =( -2 ± √(-12))/1 = -2 ± √(-12)

Therefore, we have;

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The general form of the quadratic function is f(x) = a·x² + b·x + c

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The discriminant of the function, D = b² - 4·a·c, therefore, for the function, we have;

D = 8² - 4 × 2 × 6 = 16

The discriminant of the function, D = 16

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