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WITCHER [35]
3 years ago
8

3. (6 points) Determine whether the relation R on the set A is an equivalence relation a. (3 pts) A = {1,2,3,4, 5) R={(1,1), (1,

2), (1,3), (2,2), (2,3), (3,1), (3,2), (3,3), (4,4), (5,5) OPC b. (3 pts) A = {a, b, c R={(a, a), (a, c), (b, b), (c, a), (c, c)}
Mathematics
1 answer:
dezoksy [38]3 years ago
4 0

Answer:

a is not an equivalence relation.

b is an equivalence relation.

Step-by-step explanation:

a.

A = {1,2,3,4, 5) R={(1,1), (1,2), (1,3), (2,2), (2,3), (3,1), (3,2), (3,3), (4,4), (5,5)

To see if is an equivalence relation you need to see if you have these 3 things:

Part 1: xRx for all x in A. This is the reflexive property.

Do we? Yes we have all these points in R: (1,1), (2,2) ,(3,3) ,(4,4), and (5,5).

Part 2: If xRy then yRx. This is the symmetic property.

Do we? We have (1,2) but not (2,1). So it isn't symmetric.

Part 3: If xRy and yRz then xRz.

Do we? We are not going to check this because there is no point. We have to have all 3 parts fot it be an equivalence relation.

b.

A = {a, b, c} R={(a, a), (a, c), (b, b), (c, a), (c, c)}

To see if is an equivalence relation you need to see if you have these 3 things:

Part 1: xRx for all x in A. This is the reflexive property.

Do we? Yes we have all these points in R: (a,a),(b,b), and (c,c).

Part 2: If xRy then yRx. This is the symmetric property.

Do we? We have (a,c) and (c,a). We don't need to worry about any other (x,y) since there are no more with x and y being different. This is symmetric.

Part 3: If xRy and yRz then xRz.

Do we? We do have (a,c), (c,a), and (a,a).

We do have (c,a), (a,c), and (c,c).

So it is transitive.

Question b has all 3 parts so it is an equivalence relation.

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Answer: (6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0) are the required points.

or  (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

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Using quadratic formula : x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}

so, (6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0) are the required points.

since \sqrt{3}=1.732

so, (6+7(1.732),0)\text{ and }(6-7(1.732),0) are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

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