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docker41 [41]
3 years ago
13

given examples of relations that have the following properties 1) relexive in some set A and symmetric but not transitive 2) equ

ivalence relation in some set A 3) serial in some set A but not transitive
Mathematics
1 answer:
rodikova [14]3 years ago
6 0

Answer: 1) R = {(a, a), (а,b), (b, a), (b, b), (с, с), (b, с), (с, b)}.

It is clearly not transitive since (a, b) ∈ R and (b, c) ∈ R whilst (a, c) ¢ R. On the other hand, it is reflexive since (x, x) ∈ R for all cases of x: x = a, x = b, and x = c. Likewise, it is symmetric since (а, b) ∈ R and (b, а) ∈ R and (b, с) ∈ R and (c, b) ∈ R.

2) Let S=Z and define R = {(x,y) |x and y have the same parity}

i.e., x and y are either both even or both odd.

The parity relation is an equivalence relation.

a. For any x ∈ Z, x has the same parity as itself, so (x,x) ∈ R.

b. If (x,y) ∈ R, x and y have the same parity, so (y,x) ∈ R.

c. If (x.y) ∈ R, and (y,z) ∈ R, then x and z have the same parity as y, so they have the same parity as each other (if y is odd, both x and z are odd; if y is even, both x and z are even), thus (x,z)∈ R.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial but not transitive, so the relation provided in (1) satisfies this condition.

Step-by-step explanation:

1) By definition,

a) R, a relation in a set X, is reflexive if and only if ∀x∈X, xRx ---> xRx.

That is, x works at the same place of x.

b) R is symmetric if and only if ∀x,y ∈ X, xRy ---> yRx

That is if x works at the same place y, then y works at the same place for x.

c) R is transitive if and only if ∀x,y,z ∈ X, xRy∧yRz ---> xRz

That is, if x works at the same place for y and y works at the same place for z, then x works at the same place for z.

2) An equivalence relation on a set S, is a relation on S which is reflexive, symmetric and transitive.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial and not transitive.

QED!

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The following system has one solution: x = 2, y = −2, and z = 3. 4x − 2y + 5z = 27 Equation 1 x + y = 0 Equation 2 −x − 3y + 2z
I am Lyosha [343]

Answer:

x = t

y = -t

z = (17-6t)/5

Step-by-step explanation:

First we will write down Equation 1 and Equation 2

4x-2y+5z = 17------(1)

x + y = 0--------(2)

Since we are asked to express our answer in terms of the parameter t so lets suppose x = t and  y = -t

Now substitute the value of x and y in equation (1) we get

4(t) -2(-t) +5z = 17

6t + 5z = 17

5z = 17 -6t

z = (17-6t)/5

hence we get the following answer

x = t

y = -t

z= (17-6t)/5

we can also verify our answer by substituting the values of x ,y and z in any of the two equations above, for example lets substitute these values in equation 1

4(t) -2(-t) + 5((17-6t)/5)) = 17

6t + 17 -6t = 17

6t -6t = 17 -17

0 = 0

6 0
3 years ago
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jonny [76]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

5400 ÷ 37000

Cancel out extra zeros

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Answer:

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1 ÷ 1/10

Steps

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1/1 ÷1/10

Flip the second fraction in the equation.

1/1 ÷ 10/1

Then change the division sign to a multiplication sign. So were multiply instead.

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Multiply the fractions and your answer should be 10/1 .

This fraction is also changed to a whole number.

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