1) For any two subsets and of a universal set , prove that: a) b) For part a), we must show that and that First, let . This means that is in the complement of , so therefore . This means that and (draw a Venn diagram to verify this for yourself if you don’t see it – if is not in the union of and then cannot be in and also cannot be in ). Since and , and , so . Therefore, Next, let means . This means . and , so therefore . But this means that cannot be in either . Thus or , which , and so . Part b) is similar. 3) Show that a) b) For a), you can work with the left hand side as follows: Part b) is very similar. 7) Let , and let Determine if the relation is a partial order on set . If it is a partial order, then construct a Hasse diagram. is reflexive since every element of is related to itself (that is, is antisymmetric since the only way that both is transitive since if and then since and , we need to also see The Hasse diagram looks like this: and ). is if . . Note that here we only need to check this once – (and we do since ). 4 3 5 8) Let and means divides . Determine if the relation partial order on set . If it is a partial order, then construct a Hasse diagram. is reflexive since Also, is antisymmetric since if and divides is if . , i.e. divides and for all is a . , this implies , i.e. the only way that divides Further, is transitive since if and , then . This means if and , then . To show this, consider what it means for to divide (or ). It means that such that . Similarly, means such that . Using these two facts, we can show that , i.e. that such that . Since and , we can replace with in the first equation. This gives . Since , this means that (i.e. we have found our such that , namely ). Since is reflexive, antisymmetric and transitive, it is a partial order on the set . The Hasse diagram is: 72 9) Is the relation Explain why or why not. The relation paired with is not a function with domain . 3 2 6 4 24 20 120 a function with domain ? since there is no element from the codomain that is 10) Prove the results in the following exercises by mathematical induction. a) For all positive integers is the statement “For all positive integers , show that is true: ”. First we is the statement , which is true since both sides of the equation equal . Next we assume is true for some We would now like to show that would like to show that implies . is the statement is true, based on the assumption that . is the statement is true; that is, we or Notice that, because of our assumption, we can replace the first terms on the left-hand side with , which gives the equation All that is left now is to show that this equality is true, so we first work with the left-hand side and show that it is equal to the right-hand side. Find a common denominator on the left-hand side and add the two terms together, which gives us Expand this to get Expanding the right-hand side gives the same thing, so the equality is true. Thus by the principle of mathematical induction, is true for all . is true, and so b) For all positive integers is the statement “For all positive integers that is true: is the statement equation are equal to Next we assume ”. First we show , which is true since both sides of this . is true for some We would now like to show that would like to show that implies . is the statement is true, based on the assumption that . is the statement Notice that, because of our assumption, we can replace the first terms with is true; that is, we , which gives us the equation All that is left now is to show that this equality is true, so we work with the left-hand side and show that it is equal to the right-hand side. Find a common denominator on the left-hand side and add the two terms together, which gives us Simplifying this gives us Thus is indeed true, and so is true for all by mathematical induction. 11) How many subsets of Here we use the formula where contain exactly 5 elements? (since there are elements in the given set), and Thus there are 56 subsets of . This gives us which contain exactly 5 elements. 12) A basketball coach must choose a 5-person starting team from a roster of 12 players. In how many ways is this possible? Again we use the formula for , where in this case and , so we get Thus there are 792 possible starting teams of 5 from a roster of 12 players. Note that on the actual exam, you will be required to show all work for problems like #11 and #12 (so just writing a calculator answer will not be enough).
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