23 - 28 or as much as you can. Thanks! For the integers Z we take nRm iff n mod_

Question

23 - 28 or as much as you can. Thanks!

For the integers Z we take nRm iff n mod_5 = m mod_5 Show R is an equivalence relation. Take a mod_5 = b iff b elementof N union {0} where b lessthanorequalto 5 and 5 divides a - b. More generally for w elementof N, a mod_w = b if and only if b elementof N union {0} where b lessthanorequalto w and w divides a - b. Show that for the R in problem 22 we have Z/R = {[0], [1], [2], [3], [4]}. The standard notation for Z/R is Z_5. Show that if (nRm iff n mod_5 = m mod_5), then R is an equivalence relation on Z. Show that if [a], [b] elementof Z_m, then [a + b] = {x + y: x elementof [a] and y elementof [b]}. To show this you will need to show if x elementof [a + b], then there exists r elementof [A] and t elementof [b] where r + t = x, and you will have to show that if r elementof [a] and t elementof [b], then r + t elementof [a + b]. Because [a + b] = {x + y: x elementof [a] and y elementof [b]} we define an operator + on Z_m to be [a] + [b] = [a + b] where + in [a + b] is the standard + on Z and + in [a] [b] is our new operator. Give a first order set of axioms for the binary predicate ~ to be an equivalence relation Note that at the top of the page the axioms from 1 to 7 are in a first order format. Show that given (X ~ Y iff there exists a bijection f: X rightarrow Y), we then have ~ is an equivalences relation. Given ~ is from problem 27 would there be a problem with us claiming that there is a set U where ~ subsetofequalto (U times U) If so what is the problem? We now denote the equivalence classes for ~ as |X|.

Explanation / Answer

24) 1) since n mod5=n mod5 we get nRn that is reflexivity

2) n mod5=m mod5 implies m mod5=n mod5 that is symmetry.

3) If n mod5=m mod5 and m mod5= r mod5 then n mod5=r mod5 that is transitivity

This shows that R is equivalence relation.

27) we know that every map from X to X is bijective that is reflexivity

If F from X to Y is bijective then it's inverse is also bijective that is symmetry

If F from X to Y and G from Y to Z are bijective then composite G•F from X to Z is bijective that is transitivity

This shows equivalence.

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