Deal with these relations on the set of real numbers: r₁ = {(a, b) ∈ r² | a > b}, the "greater than" relation, r₂ = {(a, b) ∈ r² | a ≥ b}, the "greater than or equal to" relation, r₃ = {(a, b) ∈ r² | a < b}, the "less than" relation, r₄ = {(a, b) ∈ r² | a ≤ b}, the "less than or equal to" relation, r₅ = {(a, b) ∈ r² | a = b}, the "equal to" relation, r₆ = {(a, b) ∈ r² | a ≠ b}, the "unequal to" relation. find
a)r1◦r1.b)r1◦r2.c)r1◦r3.d)r1◦r4.e)r 1◦r5.f)r1◦r6.g)r2◦r3.h)r3◦r3.

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 =

d) R1ºR4 =

e) R1ºR5 = R1

f) R1ºR6 =

g) R2ºR3 =

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(a,c) is in R1ºR1 if there exists b such that (a,b) is in R1 and (b,c) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2  

This case is similar to the previous one. (a,c) is in R1ºR2 if there exists b such that (a,b) is in R2 and (b,c) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a < b and b > c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that a<b and b >c. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~

i think it would be c

mark brainliest if im right

step-by-step explanation: