1) Determine whether the relation R on the set of all real numbers is reflexive,

Question

1) Determine whether the relation R on the set of all real numbers is reflexive, symmetric, antisymmetric, and/or transitive, where (x, y) e R if and only if:

A) x + y = 0

B) X = Y v X = -Y

C) x = 2y

D) xy >= 0

2) Let R be the relation on the set {0, 1, 2, 3} containing the ordered pairs (0,1), (1, 1), (1, 2), (2, 0), (2, 2), and (3, 0).

Find the

reflexive closure of R

symmetric closure of R

3) Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) e R if and only if ad = bc. Show that R is an equivalence relation.

4) Which of these collections of subsets are partitions of the set {1, 2, 3, 4, 5, 6}?

A. {1, 2}, {2, 3, 4}, {4, 5, 6}

B. {1}, {2, 3, 6}, {4}, {5}

C. {2, 4, 6}, {1, 3, 5}

D. {1, 4, 5}, {2, 6}

Explanation / Answer

1.

A. R is not reflexive, as (a, a) is not in R unless a = 0. In particular, (1, 1) is not in R since 1 + 1 = 2 =/=0

R is symmetric since addition is commutative. (a, b) in R implies a + b = 0, which means b + a = 0, which implies (b,a) is in R.

R is not antisymmetric, since we have for instance (2, -2) in R and (-2, 2) in R, but -2 =/= 2

R is not transitive, since for instance we have (2, -2) and (-2, 2) in R, but (2, 2) is not in R.

B

R is reflexive since a = a for all a, so (a, a) is in R by the first part of the definition of R.

R is symmetric since if (x, y) is in R then either (i) x = y, in which case y = x and we have (y, x) is in R. or (ii) x = -y, in which case y = -x, so (y, x) is in R.

R is not antisymmetric since (2, -2) is in R and (-2, 2) is in R, but 2 =/= -2

If (x, y), (y,z) is in R, then we have several cases

(i) x = y, y = z. Then x = z and (x,z) is in R.

(ii) x= y, y = -z. Then x = -z and (x, z) is in R.

(iii) x = -y, y = z. Then x = -z and (x, z) is in R.

(iv) x = -y, y = -z. Then x = z and (x, z) is in R.

In all cases we have (x, z) in R, so R is transitive.

C. x = 2y.

R is not reflexive since (a, a) is not in R unless a = 0. In particular (1,1) is not in R.

R is not symmetric since (2,1) is in R, but (1, 2) is not.

R is antisymmetric. If (a, b), (b, a) are in R, then we have that a = 2b and b = 2a. Thus 2a = 4b, and we have b = 4b, or 3b = 0. Thus b = 0 = a. So in the only case where (a, b), (b, a) are in both in R, we have a=b.

R is not transitive. We have (4, 2) and (2, 1) in R but (4, 1) is not in R.

D. R is reflexive since x^2 >= 0 for all real x, so (x,x) is in R.

R is symmetric due to the commutivity of multiplication. (x, y) in R implies xy >=0 implies yx >= 0 implies (y, x) is in R.

R is not antisymmetric since (1, 2), (2, 1) are in R, but 1 =/= 2.

R is not transitive, since (-1, 0), (0, 1) are in R, but (-1, 1) is not in R.

2.

(a) R = {(0,0), (0,1), (1, 1), (1, 2), (2, 0), (2, 2), (3, 0), (3,3)}

(b) R = {(0,1), (0,2), (0,3), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1) (2, 2), (3, 0)}

3. We must show R is reflexive, symmetric and transitive

Note that ((a, b), (a, b)) is in R, since ab = ba, so R is reflexive.

Also if ((a, b), (c,d)) is in R, then ad = bc, which implies cb = da, which implies ((c, d), (a, b)) is in R. So R is symmetric.

Now, if ((a, b), (c, d)), ((c, d), (e, f)) is in R, then we have ad = bc, and cf = de.

Now all of our numbers are positive integers, so we can write d=cf/e

Then ad = a(cf)/e = bc. Canceling the c's and multiplying by e, we have af = be. Thus ((a, b), (e, f)) is in R and thus R is transitive. So R is an equivalence relation.

4. Only B and C are partitions. A has sets that aren't disjoint, and the union of the sets of D is not the whole set {1, 2, 3, 4, 5, 6}

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