Lecture 16 Enoch Cheung October 22 2013 1 Let A n Z n 100 Consider the relation R on A defined by a b R iff a and b agree in their first 3 decimal digits For example 999 9990 R Clearly R is an equivalence relation because R just looks at equality of the first three digits so reflexivity symmetry and transitivity follows from the properties of equality The elements in the equivalence class of 2013 has the form 201 In general if x y z are decimal digits x 0 then xyz R n Z n xyz Therefore each equivalence class can be identified with the first three digits xyz There are 9 choices for x 10 choices for y and 10 choices for z Therefore A R 9 10 10 900 there are 900 equivalence classes 2 Let X the set of triangles in the cartesian plane For x y X define x y x and y are similar Is an equivalence relation Why or why not If it is an equivalence relation describe X The answer to this question depends on whether or not we consider a degenerate triangle to be a triangle i e a triangle with 0 area with vertices 0 0 0 0 0 0 If we do consider degenerate triangles then depending on our notion of similarity the degenerate triangle of a single point could be similar to every triangle which would mean that would not be an equivalence relation Therefore we will consider X to contain only triangles with area 0 In which case would be an equivalence relation One presentation of similarity of triangles is to look at their angles two triangles are similar if and only if their angles are the same up to permutation so 30 60 90 would be the same as 60 30 90 We also know that for any triple x y z R 0 x y z 180 and x y z 180 there is a triangle with angles x y z Therefore let T be a triangle with angles x y z then T R the set of triangles with angles x y z 3 Define a relation R on Z by x y R 4 x2 y 2 Prove that R is an equivalence relation Describe the equivalence class 0 R How many equivalence classes are in Z R Note that 4 x y x y x2 y 2 x y both even or both odd because x y both even or both odd x y x y both even 4 x2 y 2 On the other hand x y not both even or both odd x y x y both odd 4 6 x2 y 2 Obviously x x so x x are either both even or both odd Suppose x y then x y has the same parity so y x If x y y z then x y z all have the same parity so x z The equivalence class 0 R is the set of integers with the same parity as 0 so 0 R x Z x is even There are exactly two equivalence classes 0 R the set of even integers and 1 R the set of odd integers so Z R 2 4 Define a relation R on R by x y R k l N xk y l R is an equivalence relation Reflexivity clearly x1 x1 so x x R Symmetry If x y R then k l N xk y l so y l xk so y x R Transitivity if x y y z R then k l m n N xk y l and y m z n so xkm xk m y l m y lm y m l z n l z nl 1 where km nl N so x z R 0 R 0 1 R 1 2 R 2r r Q r 0 2r r Q r 0 where in general x R xr r Q r 0 xr r Q r 0 because for any x R x2 x 2 so x R x R Now for any y R since y x R y x R we can consider the case where y is positive For any k l N k xk y l x l y and we know that kl must be positive because k l N and given any positive r Q r positive integer k l 2 k l for some