Let E be the congruence modulo 5 relation on Z For all m n e

Let E be the congruence modulo 5 relation on Z. For all m, n element of Z, m E n if and only if 5| (m - n). Show that E is an equivalence relation. computer programming team has 13 people. How many ways can a group of 7 be chosen to work on the team? Suppose 7 team members women and 6 are men. (a) How many groups of 7 can be chosen that contain 4 women and 3 men? (b) How many groups of 7 can be chosen that contain at least 3 men? (c) How many groups of 7 can be chosen that contain at most 4 women? Ten points labeled A, B, C, D, E, F, G, H, I, J are arranged in a plane in such a way that no 3 lie on the same straight line. (a) How many straight lines are determined by the ten points? (b) How many straight lines do not pass through points A? (c) How many triangles have three of the ten points as vertices? (d) How many triangles do not have A as a vertex? Let A = R. For all x, y element of A, define x R y if and only if |x| = |y|. Determine if R is an

Solution

If E is an equivalence relation then surely it would be reflexive, symmetric and transitive. Now by defintion congrunce modulo 5 means a=b(mod 5) i.e. difference a-b is completely divisible by a-b.

Now for E to be reflexive, mEm must be true. Here 5(m-m) = 0 that is surely congruenc modulo 5. Thus E is reflexive.

Again E is symmetric if mEn, then nEm,

So if 5(m-n) is congruent modulo 5, then surely 5(n-m)=-5(m-n) is also congruent modulo 5 ,being same factors.

So it is symmetric also.

Further E is transitive, if mEn and nEr, then m E r

Now if 5(m-n)= 5d and 5(n-r)= 5e, then 5(m-n)+5(n-r) = 5(m-r) = 5(d+e) that is also congruence modulo 5.

Thus E is transitive also.

Or E is equivalence.

This is the answer of question (1)