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tristanreid
Topic Author
Posts: 441
Joined: May 12th, 2004, 6:58 pm

### 5 equivalence classes

In the natural numbers (1,2,3...) define 5 nonempty pairwise disjoint equivalence classes without using the number 5.In other words, if we were allowed to use the number 5, a simple way to define a class would be all numbers that are a multiple of 5 away from each other are in the same class, and we could just say that the class that a number n belongs to is:[n%5]but since we can't use the number 5, we can't use that one.Any ideas? I don't think this has any practical significance, that's why it's a brainteaser.-t.

Marsden
Posts: 3829
Joined: August 20th, 2001, 5:42 pm

### 5 equivalence classes

Couldn't you just use n<x; x<=n<y; y<=n<z; z<=n<u; u<=n?

mattcushman
Posts: 100
Joined: July 14th, 2002, 3:00 am

### 5 equivalence classes

Use the fact that, mod 13, the integers are a field with the multiplicative group being cyclic of order 12. Thus, the cubes yield 4 distinct elements of Z/(13). Add in 0, and you get 5 distinct powers of 3 in Z/(13). So, defined x~y to mean X^3 = Y^3 (mod 13).Explicitly, these equivalence classes are given by 0, 1, 5, 8 and 12.

tristanreid
Topic Author
Posts: 441
Joined: May 12th, 2004, 6:58 pm

### 5 equivalence classes

QuoteOriginally posted by: mattcushmanUse the fact that, mod 13, the integers are a field with the multiplicative group being cyclic of order 12. Thus, the cubes yield 4 distinct elements of Z/(13). Add in 0, and you get 5 distinct powers of 3 in Z/(13). So, defined x~y to mean X^3 = Y^3 (mod 13).Explicitly, these equivalence classes are given by 0, 1, 5, 8 and 12.Perfect! And even though the name of one of the equivalence classes is [5], I don't think that breaks the rule of no 5s since your relation itself doesn't contain 5. -t.

Aaron
Posts: 6433
Joined: July 23rd, 2001, 3:46 pm

### 5 equivalence classes

Here's another one. The group for N is (N mod 6)!The five possible values are 1, 2, 6, 24 and 120, nary a 5 in sight.