(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

On the set of Natural Numbers from 1 to 10000 are given the following identity relations.

R1 ; n R1 m where m and n have the same remainder by division by 24, that is mod n 24 == mod m 24.

R2 ; n R2 m where n and m have in decimal notation the same number of 2s

R3; n R3 m where n and m have in decimal notation the same number of 4s.

The questions.

1. Show that R = R1 intersection R2 intersection R3 is an indentity relation

2. Show that no identity class has more than 653 elements

3. Find identity class with the least number of elements

4. Can there be an equivalence class with exactly 2 elements? Why.

2. Relevant equations

3. The attempt at a solution

1. Is easy. Since ab definitio R1, R2, R3 are reflexive, symmetric and transitive by the theorem of relation properties, R is aswell reflexive, symmetric and transitive since this properies are preserved by the intersection operation.

2. I have no idea. Please give me some pointers

3. Same as above.

4. Easy. e.g. let R be relation on the set of integers where nRm ; n^2 == m^2, then n-th identity class would be

[n] = {n,-n} with the exception of [0] = {0}, therefore I have shown that there can be a identity class with exactly 2 elements.

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# [Identity relations] Need help at some odd identity relation problem

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