Matthew F. answered • 05/13/22
Graduate Student in the PhD Program at UNC Greensboro
Recall that sets X and Y are equivalent if and only if X ⊆ Y AND Y ⊆ X. Further, sets in general don't include duplicates nor are they ordered (unless otherwise stated).
Thus, we can apply this to the given pairs above!
a. Notice that W has the element a in it twice. This means that we can write W as {a,b,c} which is the same as Q. Therefore, W is equivalent to Q!
b. Z contains -3 which is not en element of E, so Z ⊄ E. Also, E contains 1 which is not an element of Z, so E ⊄ Z. Therefore, Z is not equivalent to E.
c. We'll first show that K ⊆ U. If x is in K, then x is an integer that is strictly less than zero. Since this means x cannot equal to zero, we get that x must be a negative integer, so x is in U; that is, K ⊆ U.
Next, let x be an element of U, meaning that X is a negative integer. This implies that x is in integer and x < 0, meaning that x is in K; that is U ⊆ K.
Therefore, K is equivalent to U.
d. N appears to be all integer multiples of 7. So let x be an arbitrary element of N. Then x = a*7 for some integer a. By definition, this means that x is divisible by 7 by definition. Hence x is in H, so N ⊆ H.
Now let's go the other direction. Suppose that x is an element of H. Then x is divisible by 7, which means that there exists an integer a such that x = 7*a; i.e., x is a multiple of a. Therefore, x must be an element of N, so H ⊆ N.
Therefore, N is equivalent to H.
e. Notice that L only contains 5 elements; -6, -3, 0. 3, and 6. So it's clear that L ⊆ J. However, J is written in what is called roster notation; that is, the elements are listed. But J is written with ellipses, implying that the elements of the set continue on. So we may assert that J is the set of all integer multiples of 3. So J contains 12, for instance, and 12 is not an element of L. In other words, J ⊄ L.
Therefore J is not equivalent to L.
