CSCI-3310 Homework 1 Problem 1. Suppose we have two sets, A...

CSCI-3310 Homework 1 Problem 1. Suppose we have two sets, A = {10, 20, 30}, B = {20,b,c}. Find 1. A?B 2. A?B 3. B-A 4. B - A?B Problem 2. Write the following descriptions in set in the format of {element | condition}. 1. the set containing all positive integer less than 10. 2. the set containing all even negative integers but greater than -500. (hint: if x is an integer, you can use 2x to get even integers.) (Z is all integer set and N is the set of all natural numbers). Problem 3. Suppose we have some water Pokemons with the movement in a set. Samurott Empoleon Gyarados aqua tail flash cannon splash hydro cannon hydro cannon tackle water gun water gun Aqua Tail megahorn waterfall waterfall crunch Table 1: Movesets of pokemons 1. Write down the set of three pokemons using the initial letter S, E, M respectively. (e.g. you should use S denote the set of Samurott's moves). 2. Write down the set operation for the following question first, then give the set satisfies the question. (answer format: if your answer is S?E, then you write S?E = {x|x ? S and x ? E} = { hydro cannon }.) (a) what are the move(s) belongs to Samurott but not Gyrados? (b) what are the move(s) belongs to Empoleon, but not the other two?

Transcript

00:01 Okay, so let's get started with a.

00:04 Okay, well, the set of all odd integers, well, clearly this guy is gonna be the complement of b, because b is the set of all even integers.

00:18 Now let's take a look at b.

00:21 Well, here we have even integers such that the square of these integers is less than or equal to 100 so we are going to have b intersection c easy now see okay here we have six and such that n is an integer and n is greater than or equal to two well okay so in this case what do we have well we have that this guy here six and such that n is an integer and n is greater than or equal to two well this one let me call this set x can be written as can be written as three multiplied by two n like this such that n belongs to z and than or equal to two perfect now this one can be written as three multiplied by m such that okay what we have m is gonna be m is gonna be even because m is equal to to n so am so m belonging to z even and moreover m is greater than or equal to four perfect so as we can see this guy looks like our set a which is the subset consisting of integers of the form 3 n such that n is an additional condition here is that m is even so what are we gonna have well this guy x is gonna be a intersection what well this guy is gonna be a intersection b perfect now at this point we are left with d only okay so the only so this guy is very easy to write as is very easy to write as an intersection of our sets complement of b the subset consisting of all odd integers and what and see indeed we have that negative 9 negative 7 and so on up to 9 well this guy is just the set consisting of all odd integers such that day squares are less than 100 so this one as i already said can be written as the complement of b odd integers intersection c perfect and so we are done with exercise six now okay let's take a look at exercise number seven.

04:08 Well, let's get started with a times b.

04:14 Okay, well, this one is going to be 0 ,2, comma, the other element is going to be 03.

04:24 Then we're going to have 2, 2, 2, 3.

04:32 Finally we are gonna have three comma two three comma three perfect and the cardinality of this guy is clearly six perfect now this one was part a of our exercise part b of our exercise well b times a b times a is very easy b times a is very easy b times a is okay 2 .0 3 .0 and just switching the coordinates of these elements so 2 .2 3 .2 2 .3 and 3 .3 again perfect and clearly the cardinality is 6 the same as a times b.

05:35 Now let's take a look at c okay so here we have b squared which is b times b this one is 2 .2 2 .3 3 .2 and 3 .3 and clearly the cardinality of b squared is 4 perfect now let's take a look at d well here we have omega times the empty set so omega times the empty set is the empty set and the cardinality the cardinality of this guy is clearly zero perfect now e the power set of b okay so here the power set of b is going to be the set consisting of all this all subsets of b.

06:44 So we are going to have the empty set, b itself, then we are going to have the subset consisting of two only, the subset consisting of three only.

06:56 And the cardinality of the power set is clearly four.

07:01 Perfect.

07:02 Now finally, f, okay.

07:07 So f is the power set of c.

07:14 Okay, well the power set of c basically is the same thing as the power set of b.

07:19 That is a c is a subset consisting of two elements only.

07:25 So we are going to have the empty set c itself, the subset one, the subset four...

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