Relations (only do 4 -8) 1. (Relations.) (a) Here\'s an exercise we started in c
ID: 3111901 • Letter: R
Question
Relations (only do 4 -8)
1. (Relations.) (a) Here's an exercise we started in class in week 5: consider the set C of people in this class You can define lots of relations on C: things like "has the same hair color as," or "is taller than," or "went to the same high school as." A table is presented below, with an entry for each combination of truth values for the properties reflexive, symmetric, and transitive. Fill in these entries! That is, find a relation on C for each of these combinations. Make sure to explain why each of your claimed relations has the desired property. [16 marks] trans? T FTFTFTFExplanation / Answer
.T T T
{(1, 1),(1, 2),(2, 1),(2, 2),(3, 3),(4, 4)}
Reflexive because (a, a) is in the relation for all a = 1, 2, 3, 4.
Symmetric because for every (a, b), we have (b, a).
. Transitive because while we have (1, 2) and (2, 1), we also have (1, 1) and (2, 2) in the relation
T T F
R={(a,a),(a,b),(b,a),(b,b),(c,c),(b,c),(c,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 (a,b)Rand (b,a)R and (b,c)R and (c,b)R
T F T
Define a relation R in R as:
R = {a, b): a3 b3}
Clearly (a, a) R as a3 = a3.
R is reflexive.
Now,
(2, 1) R (as 23 13)
But,
(1, 2) R (as 13 < 23)
R is not symmetric.
Now,
Let (a, b), (b, c) R.
a3 b3 and b3 c3
a3 c3
(a, c) R
R is transitive.
T F F
not possible
because every reflexive set is symmetrical with respect to its own samples.
F T T
xy 1. Not reflexive because we can’t have (0, 0).
Is symmetric because we have xy = yx.
Is transitive because if we have (a, b) R and that (b, c) R, it follows that (a, c) R.
Note that in order for the relation to be true, a, b, and c will have to be all positive or all negative.
F T F
Let A = {5, 6, 7}.
Define a relation R on A as R = {(5, 6), (6, 5)}.
Relation R is not reflexive as (5, 5), (6, 6), (7, 7) R.
Now, as (5, 6) R and also (6, 5) R, R is symmetric.
=> (5, 6), (6, 5) R, but (5, 5) R
R is not transitive.
Hence, relation R is symmetric but not reflexive or transitive.
F F T
Consider a relation R in R defined as:
R = {(a, b): a < b}
For any a R, we have (a, a) R since a cannot be strictly less than a itself. In fact, a = a.
R is not reflexive.
Now,
(1, 2) R (as 1 < 2)
But, 2 is not less than 1.
(2, 1) R
R is not symmetric.
Now, let (a, b), (b, c) R.
a < b and b < c
a < c
(a, c) R
R is transitive.
Hence, relation R is transitive but not reflexive and symmetric.
F F F
{(1, 3),(1, 4),(2, 3),(2, 4),(3, 1),(3, 4)}
Not reflexive because we do not have (a, a) for all a = 1, 2, 3, 4.
Not symmetric because the relation does not contain (4, 1), (3, 2), (4, 2), and (4, 3).
. Not transitive because we do not have (2, 1) for (2, 3) and (3, 1).