【Basic Abstract Algebra】Exercises for Section 1.3 — Equivalence relation and equivalence classes
1.【Basic Abstract Algebra】Exercises of Section 1.1 — Sets2.【Basic Abstract Algebra】Exercises for Section 1.2 — Maps
3.【Basic Abstract Algebra】Exercises for Section 1.3 — Equivalence relation and equivalence classes
4.【Basic Abstract Algebra】Exercises for Section 1.4 — Factor Sets5.【Basic Abstract Algebra】Exercises for Section 1.5 — Number Theory6.【Basic Abstract Algebra】Exercises for Section 1.6 — The Chinese Remainder Theorem7.【Basic Abstract Algebra】Exercises for Section 2.1 — Definitions and examples8.【Basic Abstract Algebra】Exercises for Section 2.2 — Subgroups9.【Basic Abstract Algebra】Exercises for Section 2.3 — Cyclic groups10.【Basic Abstract Algebra】Exercises for Section 2.4 — Permutation groups11.【Basic Abstract Algebra】Exercises for Section 2.5 — Dihedral groups12.【Basic Abstract Algebra】Exercises for Section 3.1 — Cosets and Lagrange's Theorem13.【Basic Abstract Algebra】Exercises for Section 3.2 — Normal subgroups and factor groups14.【Basic Abstract Algebra】Exercises for Section 3.3 — Homomorphism of groups15.【Basic Abstract Algebra】Exercises for Section 3.5 — Fundamental Isomorphism theorem of group- Define a relation R on R2 by stating that (a,b)∼(c,d) if and only if a2+b2≤c2+d2. Show that ∼ is reflexive and transitive, but itis not symmetric.
Solution: (1) Obviously, (a,b)∼(a,b), so ∼ is reflexive.
(2) If (a,b)∼(c,d), (c,d)∼(e,f), then we have a2+b2≤c2+d2, c2+d2≤e2+f2. So a2+b2≤e2+f2. Thus (a,b)∼(e,f), so ∼ is transitive.
(3) Suppose (a,b), (c,d)∈R2 and a2+b2<c2+d2. Thus (a,b)∼(c,d). Since c2+d2>a2+b2, (c,d)\not\sim(a,b). Therefore, \sim is not symmetric. #
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