Quandaries and Queries Hi, I am taking a course in ring theory and i have problems with the following question: a) How many different equivalence relations can be defined on the set X={a,b,c,d}? i know a~a, b~b,c~c,d~d does that mean there are only 4 different equivalence relations? b)Show that 6 divides the product of any 3 consecutive integers. I know it is true that 6 divides the product of any 3 consecutive integers. However, i have problem showing the proof. Thank you and i do appreciate your help! Eva Hi Eva, For your first question think in terms of partitions. An equivalence relation on a set generates a partition of the set and a partitiion of a set generates an equivalence relation. Thus you can ask "how many partitions are there of the set {a,b,c,d}?" For a set with one element, {a}, there is one partitiion {{a}}. For a set with two elements {a,b} there are two partitions {{a,b}} and {{a},{b}}. For a set with three elements {a,b,c} we need to be more systematic. If there is a member of the partition with three elements then there is only 1 such partition, {{a,b,c}}. If the largest memeber of the partition has two elements then there are 3 such partitions, {{a,b},{c}}, {{a,c},{b}} and {{b,c},{a}} If the largerst element of the partition has only one element then there is only 1 such partition {{a},{b},{c}. Hence there are 1 + 3 + 1 = 5 partitions of a set with 3 elements. Now try a set with 4 elements. With respect to your second question, consider that in order to be divisible by 6, an integer must be divisible by both 2 and 3. If you have 3 consecutive integers you have at least one that is even so the product must have a factor of 2 because of that integer. Also, for every 3 consecutive integers, one must be divisible by 3. A quick proof of this using modulus 3: Given n, n+1 and n+2 consecutive integers. If n is divisible by 3, then the product n(n+1)(n+2) has a factor of 3. Suppose n is not divisible by 3. Then n is congruent to 1 modulo 3 or n is congruent to 2 modulo 3. Case 1: If n is congruent to 1 modulo 3, then n+2 is congruent to 0 modulo 3 Thus n+2 is divisible by 3. Case 2: If n is congruent to 2 modulo 3, then n+1 is congruent to 0 modulo 3 Thus n+1 is divisible by 3. Therefore, either n, n+1 or n+2 is divisible by 3 for all integers n. Thus the product n(n+1)(n+2) has a factor of 2 and a factor of 3 and so is always divisible by 6. Hope this helps, Leeanne and Penny Go to Math Central