Foundations of algebra

Transcription

1 Foundations of algebra Equivalence relations - suggested problems - solutions P1: There are several relations that you are familiar with: Relations on R (or any of its subsets): Equality. Symbol: x = y. Less than. Symbol: x < y. Less than or equal to. Symbol: x y. Greater than. Symbol: x > y. Greater than or equal to. Symbol: x y. Relations on the set of possible sets: Subset. Symbol: A B. Proper subset. Symbol: A B. Set equality. Symbol: A = B. Logical relations: Logical equivalence. Symbol: p q. For each of these, you should know which of the four properties the relation has [reflexive, symmetric, transitive, anti-symmetric], and be able to identify if it s an equivalence relation or a partial ordering. You can use your intutive understanding of those relations on the reals - we haven t looked at the truly formal definition of less than, for example - but should know the formal definitions for those set relations and the logical equivalence relation. Some of these were covered in the lecture. I ll run through a few here (in the solutions to the suggested problems), and leave a couple for the graded assignment. Equality: Reflexive: yes. x = x for all x R. Symmetric: yes. If x = y, then y = x for all x, y R. Transitive: yes. If x = y and y = z, then x = z for all x, y, z R. Anti-symmetric: yes. If x = y and y = x, then x = y for all x, y R. Trivially true, of course, but true nonetheless! Equivalence relation: yes, since reflexive, symmetric, transitive. Partial order: yes, since reflexive, antisymmetric, transitive.

2 Note that the idea with these is simply to be sure you can translate the xry notation into the notation associated with each relation. There s no real proof that these hold...because equality is defined as the relation on the reals for which these hold. Fortunately, this fits with our intuitive idea of how we d expect equality to work. Less than: Reflexive: no. x < x for all x R is a false statement. Symmetric: no. If x < y, then y < x for all x, y R is a false statement. Transitive: yes. If x < y and y < z, then x < z for all x, y, z R is true. Anti-symmetric: yes. If x < y and y < x, then x = y for all x, y R is (vacuously) true since x < y and y < x is always false, and F anything is true. This was noted in the lecture. Equivalence relation: no. Partial order: no. Less than or equal to: Reflexive: yes. x x for all x R is a true statement. Symmetric: no. If x y, then y x for all x, y R is a false statement (it s not impossible, but it s not true for all x, y R). Transitive: yes. If x y and y z, then x z for all x, y, z R is true. Anti-symmetric: yes. If x y and y x, then x = y for all x, y R is true. Equivalence relation: no. Partial order: yes. Reflexive, anti-symmetric, transitive. Greater than: Greater than or equal to: Are analogous to their counterparts. They re also redundant - if we have a relation less than, we really don t need a greater than relation - just flip the numbers you re comparing around.

3 Subset: Is similar to less than or equal to, but in the context of sets. You should know the formal definition for the subset relation: (Every element of A is also an element of B.) A B iff x A implies x B. Reflexive: yes. A A for all A K is a true statement. Symmetric: no. If A B, then B A for all A, B K is a false statement (it s not impossible, but it s not true for all A, B K). Transitive: yes. If A B and B C, then A C for all A, B, C K is true. Anti-symmetric: yes. If A B and B A, then A = B for all A, B K is true. By definition - that is how we define equality for sets. Equivalence relation: no. Partial order: yes. Reflexive, anti-symmetric, transitive. Proper subset: Is similar to less than. Formal definition is A B iff x A implies x B, and A B. Neither an equivalence relation nor a partial order, since not reflexive. Equality (of sets): A = B iff A B and B A Both an equivalence relation and a partial order, as you d expect from the use of equals. Logical equivalence: Two logical expressions are logically equivalent if they have the same truth table; formally p q iff p q is a tautology For example, you should recall that (p q) p q (DeMorgan s Law). So, Reflexive: yes. p p is a tautology, so p p. Symmetric: yes. If p q is a tautology, then so is q p (the operator is commutative). p q implies q p. Transitive: yes. If p q and q r, then p r (they would all have the same pattern of trues and falses). Anti-symmetric: doesn t make sense in this context. We don t have an idea of equality for logical expressions in the abstract. Equivalence relation: yes. Partial order: no (non-sensical).

4 P2: A set A is given below: A = {a, b, c, d} Several relations are defined on set A. For each relation, determine which of the four properties it satisfies, and whether it is an equivalence relation, a partial order, or both. If it is an equivalence relation, give the equivalence class for each element of A under that relation. (a) R 1 = {(a, a), (b, b), (c, c), (d, d)} Reflexive: yes. For every x A, we have xrx (the ordered pair (x, x) appears for x = a, b, c, d - every element of A. Symmetric: yes. If xr 1 y, then yr 1 x. For every pair (x, y) in the relation, we also have the pair (y, x)...because every pair has x and y equal. Transitive: yes. If xr 1 y and yr 1 z, then xr 1 z. For this one, trivially true - we don t have any pairs that connect with each other. Anti-symmetric: yes. If xr 1 y and yr 1 x, then x = y. Note ar 1 a and ar 1 a, and a = a. br 1 b and br 1 b, and b = b, etc. Equivalence relation: yes. Equivalence class of each element: Partial order: yes. [a] = {a} [b] = {b} [c] = {c} [d] = {d} Note that this relation defines equality as a relation on A. (b) R 2 = {(a, a), (b, b), (c, c), (d, d), (a, b), (b,a), (a,c)} Reflexive: yes. For every x A, we have xr 2 x Symmetric: no. Since ar 2 c, we need cr 2 a for it to be symmetric, but the ordered pair (c, a) is not in the relation. Transitive: no. Since br 2 a and ar 2 c, we need br 2 c, but the ordered pair (b, c) is not in the relation. Anti-symmetric: no. ar 2 b and br 2 a, but b a. Equivalence relation: no. Equivalence class of each element: N/A. Partial order: no.

5 (c) R 3 = {(a, a), (b, b), (c, c), (d, d), (a, b), (b,a), (a,c), (c, a)} Reflexive: yes. For every x A, we have xr 3 x Symmetric: yes. If xr 3 y, then yr 3 x. Every pair also has its mirror image in the relation. Transitive: no. Since br 3 a and ar 3 c, we need br 3 c, but the ordered pair (b, c) is not in the relation. Anti-symmetric: no. ar 3 b and br 3 a, but b a. Equivalence relation: no. Equivalence class of each element: N/A. Partial order: no. (d) R 4 = {(a, a), (b, b), (c, c), (d, d), (a, b), (b,a), (a,c), (c, a),(b, c), (c,b)} Reflexive: yes. For every x A, we have xr 4 x Symmetric: yes. If xr 4 y, then yr 4 x. Every pair also has its mirror image in the relation. Transitive: yes. You need to consider all possible pairings; for example: Since ar 4 c and cr 4 b, we must have ar 4 b. Check. Since cr 4 b and br 4 c, we must have cr 4 c. Check. And so on. Too many to list them all out; the thing to observe is that you can t find a counterexample (I hope). Anti-symmetric: no. ar 4 b and br 4 a, but b a. Equivalence relation: yes. Equivalence class of each element: [a] = [b] = [c] = {a, b, c} [d] = {d} Partial order: no.

6 P3: Let A be the set of integers Z and let R = {(x, y) x y = 1}. Is R an equivalence relation on A? State explicitly which properties it passes and which it fails. Reflexive: no. x x 1 for any x A, so x Rx. Symmetric: no. If x y = 1, then y x = 1, not 1. xry does not imply yrx. Transitive: no. If x y = 1 and y z = 1, then (x y)+(y z) = 2, and therefore x z = 2. xry and yrz does not imply xrz. R is not an equivalence relation. P4: Let A be the set of integers Z and let R = {(x, y) x y = 0}. Is R an equivalence relation on A? State explicitly which properties it passes and which it fails. Reflexive: yes. x x = 0 for all x A, so xrx. Symmetric: yes. If x y = 0, then y x = 0, not 1. xry implies yrx. Transitive: yes. If x y = 0 and y z = 0, then (x y) + (y z) = 0, and therefore x z = 0. xry and yrz implies xrz. R is an equivalence relation.

7 P5: [#7 on p. 7] Define the relation R on the set of all ordered pairs of real numbers as follows: (x, y)r(s, t) if and only if 2(x s) = (y t). Prove that R is an equivalence relation. Find the equivalence class of the point (1, 1). Reflexive: We need to see if (x, y)r(x, y) - let s = x and y = t in the defining formula. It is true that 2(x x) = (y y), as both sides are 0, so the R is reflexive. Symmetric: We need to see if (x, y)r(s, t) implies (s, t)r(x, y). Suppose that (x, y)r(s, t) and see if the algebra takes you where it needs to go: (x, y)r(s, t) 2(x s) = (y t) ( 1)2(x s) = ( 1)(y t) 2( x + s) = ( y + t) 2(s x) = (t y) (s, t)r(x, y) Transitive: We need to see if (x, y)r(s, t) and (s, t)r(m, n) imply (x, y)r(m, n). Suppose (x, y)r(s, t) and (s, t)r(m, n). Then 2(x s) = (y t) and 2(s m) = (t n). Adding the equations gives 2(x s) + 2(s m) = (y t) + (t n) and simplifying gives and therefore (x, y)r(m, n). 2x 2s + 2s 2m = y t + t n 2x 2m = y n 2(x m) = y n Since R is reflexive, symmetric, and transitive, it is an equivalence relation. The equivalence class of the point (1, 1) is all pairs (x, y) such that (x, y)r(1, 1); all pairs such that 2(x 1) = (y 1) This can be expressed as all points on the line y = 2x 1.