Discrete Mathematics · Sets, functions, relations, and equivalence classes
Let R be a relation on a set A
Problem
Let \( R \) be a relation on a set \( A \). Prove that if \( R \) is both an equivalence relation and a partial order, then \( R \) is the equality relation on \( A \) (that is, \( R = \{(a,a) : a \in A\} \)).
Hint
The two extra axioms you have, beyond reflexivity, point in opposite directions: one produces the reverse pair, the other forbids distinct reverse pairs.
Check your work
Work the problem yourself first. Then open it in Training to check your answer and read the full worked solution.
The answer check and full solution for this problem come with ProofAnvil Practice membership ($19 USD monthly). See membership. Or start with the free Discrete Mathematics sample problem: Try the free sample problem.
More Discrete Mathematics practice problems
- Let p, q, and r be propositionsPropositional and predicate logic with proof methods
- Let A={1,2,3,4} and B={3,4,5,6}Propositional and predicate logic with proof methods
- Use strong induction to prove that every positive integer n can be written as n=2^k m…Sets, functions, relations, and equivalence classes
- (Honors) Let u_n be defined by u_0=0, u_1=1, and u_n=4u_(n-1)-4u_(n-2)+n for n≥ 2Induction, recursion, and invariants
- [Honors] Using inclusion-exclusion, find the number of permutations π of {1,2,…,9} such…Induction, recursion, and invariants
- Let n be a positive integerRecurrences, generating functions, and discrete asymptotics
- RSA decryption needs Euler, and Euler needs a coprime messageDivisibility, modular arithmetic, and elementary number theory
- Four labs sharing ten hours cannot exceed a per-lab capCombinatorics and inclusion-exclusion
- BFS two-coloring is the certificate that no odd cycle existsGraphs, trees, connectivity, and matchings