Here is quiz 5 and here is quiz 6.
Problem 1 asks to show that the relation defined as follows is antisymmetric: Given a set the relation is defined on the subsets of by setting for iff where denotes set-theoretic difference of sets.
To show that is antisymmetric, we need to show that whenever are such that and then Suppose satisfy these assumptions. We need to show that they are equal as sets, i.e., that they have the same elements.
By definition, holds iff and holds iff Recall that if are sets, then It follows that iff every element of is also an element of and that iff every element of is also an element of But these two facts together mean precisely that and have the same elements, i.e., that as we needed to show.
Problem 2(a) of quiz 6 asks to consider the relation defined on by setting for iff and to show that is an equivalence relation. This means that is reflexive, symmetric, and transitive. Problem 2(a) of quiz 5 asks to show one of these properties.
To show that is reflexive, we need to show that for any we have that i.e., that But is certainly divisible by 5.
To show that is symmetric, we need to show that for any if it is the case that then it is also the case that Suppose then that This means that i.e., there is an integer such that But then showing that also i.e.,
To show that is transitive, we need to show that if and both and hold, then also holds. But if and then and But then it is certainly the case that Since this proves that, indeed or as we needed to show.
(If one feels the need to be somewhat more strict: That means that there is an integer such that Similarly, means that there is an integer such that But then showing that there is an integer such that namely, we can take )
Problem 2(b) asks to find all natural numbers such that where is as defined for problem 2(a).
That means exactly the same that Since and an integer is a multiple of 5 iff it ends in 5 or 0, we need to find all natural numbers such that ends in or Since and is even for all naturals we actually need to find all natural numbers such that ends in 8. For this, we only need to examine the last digit of the numbers and we find that these last digits form the sequence which is periodic, repeating itself each 4. This means that the numbers we are looking for are precisely i.e., the natural numbers of the form with
I was working with one of students(Josue Gomez) in your class.
From problem 2 (b,)
I guess 5 divides 0 since 5*0 = 0. so need to say n is the form 4k + 3 where k is greater than or equal to 0.
I understand 0 is not natural number. I guess k can be 0 also.
Thanks for your lecture!
The natural numbers begin with 0. That is the definition we are using. The integers that are larger than 0 are, naturally, the positive integers.
Oh thank you!