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docker41 [41]
3 years ago
13

Which of these relations on the set {0, 1, 2, 3} are equivalence relations? If not, please give reasons why. (In other words, if

a relation is not an equivalence relation, please list each property that is missing and the reason why it is missing.) (1) {(0,0), (1,1), (2,2), (3,3)} (2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)} (3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)} (4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}
Mathematics
1 answer:
Delicious77 [7]3 years ago
5 0

Answer:

(1)Equivalence Relation

(2)Not Transitive, (0,3) is missing

(3)Equivalence Relation

(4)Not symmetric and Not Transitive, (2,1) is not in the set

Step-by-step explanation:

A set is said to be an equivalence relation if it satisfies the following conditions:

  • Reflexivity: If \forall x \in A, x \rightarrow x
  • Symmetry: \forall x,y \in A, $if x \rightarrow y,$ then y \rightarrow x
  • Transitivity: \forall x,y,z \in A, $if x \rightarrow y,$ and y \rightarrow z, $ then x \rightarrow z

(1) {(0,0), (1,1), (2,2), (3,3)}

(3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)}

The relations in 1 and 3 are Reflexive, Symmetric and Transitive. Therefore (1) and (3) are equivalence relation.

(2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)}

In (2), (0,2) and (2,3) are in the set but (0,3) is not in the set.

Therefore, It is not transitive.

As a result, the set (2) is not an equivalence relation.

(4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}

(1,2) is in the set but (2,1) is not in the set, therefore it is not symmetric

Also, (2,0) and (0,1) is in the set, but (2,1) is not, rendering the condition for transitivity invalid.

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raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the ci
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Answer:

Equation:

{x}^{2}   +  {y}^{2} +  2x  - 2y   -  35= 0

The point (0,-5), (0,7), (5,0) and (-7,0)also lie on this circle.

Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

The center of this circle is the midpoint of (-3, 4) and (5, -2).

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r =  \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }

r =  \sqrt{ {(5 -  - 1)}^{2} + {( - 2- - 1)}^{2} }

r =  \sqrt{ {(6)}^{2} + {( - 1)}^{2} }

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The equation of the circle is given by:

(x-h)^2 + (y-k)^2 =  {r}^{2}

Where (h,k)=(-1,1) and r=√37 is the radius

We plug in the values to get:

(x- - 1)^2 + (y-1)^2 =  {( \sqrt{37}) }^{2}

(x + 1)^2 + (y - 1)^2 = 37

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{x}^{2}  + 2x  + 1 +  {y}^{2}  - 2y + 1 = 37

{x}^{2}   +  {y}^{2} +  2x  - 2y +2 - 37= 0

{x}^{2}   +  {y}^{2} +  2x  - 2y   -  35= 0

We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

{x}^{2}   +  {0}^{2} +  2x  - 2(0)  -   35= 0

{x}^{2}   +2x   -   35= 0

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x = 5 \: or \: x =  - 7

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When x=0

{0}^{2}   +  {y}^{2} +  2(0)  - 2y   -  35= 0

{y}^{2} - 2y   -  35= 0

(y - 7)(y + 5) = 0

y = 7 \: or \: y =  - 5

The point (0,-5) and (0,7) lie on this circle.

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