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Let X={1,2,3,4,5,6,7,8,9}.R1 be a relation on X given by R1={(x,y):x-y is divisible by 3}and R2 be another relation X given by R2={(x,y):{x,y}belongs to {1,4,7} or {x,y} belongs to {2,5,8} or {x,y} belongs to {3,6,9}.Show that R1=R2?
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Let X={1,2,3,4,5,6,7,8,9}.R1 be a relation on X given by R1={(x,y):x-y...
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Let X={1,2,3,4,5,6,7,8,9}.R1 be a relation on X given by R1={(x,y):x-y...
Introduction:
In this problem, we are given two relations R1 and R2 on a set X={1,2,3,4,5,6,7,8,9}. We need to prove that R1=R2 by showing that they have the same set of ordered pairs.

Definition of R1:
R1={(x,y):x-y is divisible by 3}

Definition of R2:
R2={(x,y):{x,y} belongs to {1,4,7} or {x,y} belongs to {2,5,8} or {x,y} belongs to {3,6,9}}

Proof:
To prove that R1=R2, we need to show that every ordered pair in R1 is also in R2, and vice versa.

Proof of R1 ⊆ R2:
Let (x,y) be an arbitrary ordered pair in R1. This means that x-y is divisible by 3. We need to show that (x,y) is also in R2.

Since x-y is divisible by 3, there are three possibilities for the values of x and y:

1. If x and y both belong to {1,4,7}, then (x,y) belongs to R2.
2. If x and y both belong to {2,5,8}, then (x,y) belongs to R2.
3. If x and y both belong to {3,6,9}, then (x,y) belongs to R2.

In all three cases, (x,y) is in R2. Therefore, R1 ⊆ R2.

Proof of R2 ⊆ R1:
Let (x,y) be an arbitrary ordered pair in R2. We need to show that (x,y) is also in R1.

Since (x,y) belongs to R2, there are three possibilities for the values of x and y:

1. If {x,y} belongs to {1,4,7}, then x and y differ by a multiple of 3, so (x,y) belongs to R1.
2. If {x,y} belongs to {2,5,8}, then x and y differ by a multiple of 3, so (x,y) belongs to R1.
3. If {x,y} belongs to {3,6,9}, then x and y differ by a multiple of 3, so (x,y) belongs to R1.

In all three cases, (x,y) is in R1. Therefore, R2 ⊆ R1.

Conclusion:
Since R1 ⊆ R2 and R2 ⊆ R1, we can conclude that R1=R2. Thus, the relations R1 and R2 are equal.

Summary:
- We have shown that every ordered pair in R1 is also in R2, and vice versa.
- This proves that R1=R2.
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Let X={1,2,3,4,5,6,7,8,9}.R1 be a relation on X given by R1={(x,y):x-y is divisible by 3}and R2 be another relation X given by R2={(x,y):{x,y}belongs to {1,4,7} or {x,y} belongs to {2,5,8} or {x,y} belongs to {3,6,9}.Show that R1=R2?
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