Question

Since {1, 2, . . . , 6} is the set of all possible outcomes of a throw with a regular die, the set of all possible outcomes of a throw with two dice is Throws := {1, 2, . . . , 6} × {1, 2, . . . , 6}. We define eleven subsets P2, P3, . . . , P12 of Throws as follows: Pk := {<m, n>: m + n = k} for k ∈ {2, 3, . . . , 12}. For example, P3 is the set of all outcomes for which the sum of the two numbers of dots thrown is 3.

(a) Show that the sets P2, P3, . . . , P12 form a partition of the set Throws.

(b) Let R be the equivalence relation on Throws that has P2, P3, . . . , P12 as its equivalence classes. Give a definition of R by means of a description.

(c) Give a complete system of representatives for the equivalence relation R.

Answer 1

Given,

Let S be the set of all possible outcomes of two dices,therefore

S := {(1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6)

   (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6)

(3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6)

(4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6)

(5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6)

(6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6)}

According to Question,

Pk := {<m, n>: m + n = k}

P2 := {(1,1)}

P3 := {(1,2) , (2,1)}

P4 := {(1,3) , (3,1) , (2,2)}

P5 := {(1,4) , (4,1) , (2,3) , (3,2)}

P6 := {(1,5) , (5,1) , (2,4) , (4,2) , (3,3)}

P7 := {(1,6) , (6,1) , (2,5) , (5,2) , (3,4) , (4,3)}

P8 := {(2,6) , (6,2) , (3,5) , (5,3) , (4,4)}

P9 := {(3,6) , (6,3) , (4,5) , (5,4)}

P10 := {(4,6) , (6,4) , (5,5)}

P11 := {(5,6) , (6,5)}

P12 :={(6,6)}

Now,if we count the number of elements in each set from P2 to P12

We have the sum as

which is equal to the number of elements in set S

also,

No subset is empty

Every element is contained in only one subset

Hence Proved that, the sets P2, P3 ,..., P12 form a partition of set S

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In the sub section 'b',it is given that P2 ,P3 ,......., P12 are equivalence classes of Relation R that is equivalent for set Throws.

We define Relation R and the equivalence as a ~ b such that sum(a)=sum(b)

This relation is equivalent as

(i) :Reflexive:  Any sum is equal to itself For eg. (1,4) =(1,4) [sum is 5]

(ii) Symmetric : For eg. (1,3)=(2,2) [sum is 4]  then (2,2)=(1,3) [sum is 4]

(iii) Transitive :For eg (1,6)=(2,5) [sum is 7] and (2,5) =(3,4) [sum is 7] from the two equations we can also see that (1,6) =(3,4) [sum is 7]

Hence Relation R is equivalent

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The complete system of representatives of the Relation R will be P2 , P3 , P4 ,...., P12 as no other sum is possible and the equivalence relation is based on sums.