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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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Answer:

m = -1

Step-by-step explanation:

Solve for m:

5 (m - 7) = -8 (3 m + 8)

Hint: | Write the linear polynomial on the left hand side in standard form.

Expand out terms of the left hand side:

5 m - 35 = -8 (3 m + 8)

Hint: | Write the linear polynomial on the left hand side in standard form.

Expand out terms of the right hand side:

5 m - 35 = -24 m - 64

Hint: | Move terms with m to the left hand side.

Add 24 m to both sides:

24 m + 5 m - 35 = (24 m - 24 m) - 64

Hint: | Look for the difference of two identical terms.

24 m - 24 m = 0:

24 m + 5 m - 35 = -64

Hint: | Group like terms in 24 m + 5 m - 35.

Grouping like terms, 24 m + 5 m - 35 = (5 m + 24 m) - 35:

(5 m + 24 m) - 35 = -64

Hint: | Add like terms in 5 m + 24 m.

5 m + 24 m = 29 m:

29 m - 35 = -64

Hint: | Isolate terms with m to the left hand side.

Add 35 to both sides:

29 m + (35 - 35) = 35 - 64

Hint: | Look for the difference of two identical terms.

35 - 35 = 0:

29 m = 35 - 64

Hint: | Evaluate 35 - 64.

35 - 64 = -29:

29 m = -29

Hint: | Divide both sides by a constant to simplify the equation.

Divide both sides of 29 m = -29 by 29:

(29 m)/29 = (-29)/29

Hint: | Any nonzero number divided by itself is one.

29/29 = 1:

m = (-29)/29

Hint: | Reduce (-29)/29 to lowest terms. Start by finding the GCD of -29 and 29.

The gcd of -29 and 29 is 29, so (-29)/29 = (29 (-1))/(29×1) = 29/29×-1 = -1:

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