Suppose that
r1={(24)},
r2={(14)},
r3={(2,,2)} ,
r4={(1,,,4)},
r5={(14)},
r6={(14)},
determine which of these statements are correct.
type all correct answers below.
a. r1 is not symmetric.
b. r3 is symmetric.
c. r3 is transitive.
d. r2 is reflexive.
e. r4 is symmetric.
f. r1 is reflexive.
g. r5 is not reflexive.
h. r5 is transitive.
i. r3 is reflexive.
j. r2 is not transitive.
k. r4 is transitive.
l. r4 is antisymmetric.
m. r6 is symmetric.

A, B, D, H are the correct statements

Step-by-step explanation:

A) If R1 were symmetric, since we have the relation (2,4), we should have for symmetry the relation (4,2), which is not the case. Therefore R1 is not symmetric and A is True.

B) There are only 2 relations in R3, (2,4) and (4,2). One relation is the symmetric relation of the other, therefore R3 is symmetric and B is True.

C) If R3 were transitive it should satysfy the transitive rule. Since 2 is related with 4 and 4 is related with 2, then for transitivity 2 should be related with itself, which is not true. As a consecuence, R3 is not transitive and C is False.

D) The elements that appear in R2 are 1, 2, 3 and 4. R2 is reflexive because we have all the relations (1,1), (2,2), (3,3) and (4,4) that relate an element with itself. We conclude that D is True.

E) R4 is not symmetric because we have the relation (1,2) but we dont have the symmetric relation (2,1) in R4. Therefore E is False.

F) The element 4 appears in R1 but we dont have the relation (4,4) in R1. This means that R1 in not reflexive, so F is False.

G) Similar to what happened with R2, the elements 1,2,3 and 4 appear in R5 and all the relations (1,1), (2,2), (3,3) and (4,4) are present in R5. As a consecuence, every element is related to itself, which means that R5 IS reflexive, and because of that, G is False.

H) All the realtions in R5 are of the form '(a,a)' for a in {1,2,3,4}. Suppose we have a chain of relations (a,b), and (b,c) in R5, and we want to know if (a,c) is in R5. Necessarily, a = b, and b = c. Then, a and c are equal, which means that the relation (a,c) is the relation (a,a), which is in R5 for any a in {1,2,3,4}. This proves the transitivity of R5 and therefore, H is True.

I) R3 is not reflexive because 2 is an element of the set and the relation (2,2) is not in R3. I is False

15 batches are the most you can make

step-by-step explanation:

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !