help with all of them please Find the union and intersection of each of the foll
ID: 2969671 • Letter: H
Question
help with all of them please
Find the union and intersection of each of the following families or indexed collections. For each natural number n, let Bn = N - {1,2, 3,...,n } and let = {Bn:n N}. For each n N, let Mn = {..., -3n, -Zn, -n, 0, n, 2n, 3n,...}, and let M = {Mn: n N}. For each n Z, let An = {n, n + 1) and A = {An: n Z}. For each n Z, let Dn = (-n, 1/n) and D = {Dn: n N}. For each prime number p, let pN = {np: n N} and A be the family {pN: n N and p is prime}. Which families in Exercise 1 are pairwise disjoint? Give an example of an indexed collection of sets {An: n N} such that each . Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A. Claim. If for all , then . "proof." Suppose . Then, since for all , .Therefore, Claim. For every indexed family . "Proof." Let . Use the PMI to prove the following for all natural numbers n. 3 + 11 + 19 ++ (8n - 5) = 4n2-n. 13 + 23 ++ n3 + [n(n + 1/2]2 Use the PMI to prove the following for all natural numbers:Explanation / Answer
Finally, I have the time to answer this.
d) Each Bn = N - {1, 2, 3, ..., n)
Note how B1 has all N except 1, B2 has all N except 1 and 2, ...
Clearly, for n > m, Bn is a proper subset of Bm as all integers greater than n are in both sets, integers m+1, m+2, ..., n are only in Bm, and all integers <= m are in neither set.
Then, all Bi are subsets of B1, so the union is B1 = N - (1)
As Bn is a subset of B1, B2, ..., Bn-1, then the intersection for i = 1 to n of Bn is Bn. Then, the lim of the intersection of Bi is the null set, as only positive integers > 1 are in any of the sets, and for any positive integer n, Bn does not include n, so the intersection does not.
j) Note that each set Mi is a subset of M1, as M1 includes all integers. Thus, the union is M1, or all integers.
Note that, other than 0, the smallest positive integer is n and the largest negative number is -n. Then, any positive integer n is not in Mn+1 and any negative integer -n is not in Mn+1.
Yet, 0 is in each Mi. Thus, 0 is in the intersection.
m) Any real non-integer x is in (int(x), int(x) + 1), so it is in the union.
However, any integer s is not in the union, as (s-1, s) and (s,s+1) are adjacent sets, neither including S
Thus, the union is all real non-integers.
As the intersection of M1 and M2 is (1, 2) intersect (2, 3) is the null set, so the intersection of all sets is the null set.
n) Note that Dn = (-n, 0] U (0, 1/n)
Thus, if we consider the union of Di, this the union of (-n,0] for all n U the union of (0, 1/n) for all n
The union of (-n,0] for all n, as (-1, 0], (-2, 0], ... (-n+1, 0] is a subset of (-n, 0], is the limit of (-n,0] as n goes to infinity is (-infinity, 0]
However, the union of (0, 1/n), as each (0, 1/n) is a subset of (0, 1), is (0, 1)
Thus, the union is (-infinity, 0] U (0, 1) = (-infinity, 1)
Working similarly, the intersection of all (-n,0] is (-1, 0], and the intersection of all (0, 1/n) is the null set (0 and all numbers less than 0 or not in the intersection and for any x > 0, for n > int(1/x), (0, 1/n) does not include x.); thus, the intersection is
(-1, 0]
o) For the union, note that any integer > 1 has prime factors, so the set with this prime includes this integer, so the union includes this integer. However, as 1 is not divisible by any prime, it is not in any of these sets.
Thus, the union is all integers > 1 (Z > 1)
However, for any integer n, there exists a prime larger than it, so the set of this prime does not include n. Thus, the intersection does not include n.
Thus, the intersection is the null set.
2. The sets in m), (n, n+1), are pairwise disjoint.
All others are not.
To show, d) the intersection of i and j is B max(i,j), which is not null
j) All sets include 0, so none are disjoint
n) Again, 0 is in all sets, so none are disjoint
o) Consider the sets A2 and A3. 2 divides 6, and so does 3, so 6 is in both sets.
12. Exercise 1 n) was a good hint for this problem.
Let set Ai = (0, 1/i) Then, Ai intersect Aj = (0, 1/max(i,j))
However, A1 does not include any reals >= 1 or <= 0, and for any x in (0, 1), there exist n = 1 + int(1/x), and x is not in (0,1/(1+int(1/x)). Thus, the intersection is the null set.
18. b) C (partially correct) what is left out is saying that x is in A alpha for some alpha since it is in the union.
c) incorrect. Showing this is true for four sets has nothing to do with showing this is true.
Note: the proof is trivial. The first set A1 is a subset of the union, and the intersection is a subset of A1, so the intersection is a subset of the union.
(If you wish, consider any x in the intersection, then it has to be in A1; if it is in A1, then it is in the union; thus, every x in the intersection is in the union, so the intersection is a subset of the union)
6. b) Show 3 + 11 + ... + 8n - 5 = 4n^2 - n
For n = 1, the left hand side is 3. The right hand side is 4n^2 - n = 4*1^2 - 1 = 4*1 - 1 = 4 - 1 = 3
Assume for n = k.
3 + 11 + ... + 8k - 5 = 4k^2 - k
For n = k+1
3 + 11 + ... + 8k-5 + 8(k+1) - 5 =
(3 + 11 + ... + 8k-5) + 8(k+1) - 5 = from the assumption,
4k^2 - k + 8k + 8 - 5 =
4k^2 + 7k + 3 =
4k^2 + 7k + 3 + k + 1 - k - 1 =
4k^2 + 8k + 4 - (k+1) =
4(k+1)^2 - (k+1) =
4n^2 - n for n = k+1
Thus, we have proved for n=1 and n=k+1 given k, so we have proven for any n.
e) 1^3 + 2 ^3 + ... n^3 = (n(n+1)/2)^2
For n = 1
1^3 = 1
(n(n+1)/2)^2 = (1(1+1)/2)^2 = (2/2)^2 = 1^2 = 1
Assume for n=k 1^3 + 2 ^3 + ... k^3 = ((k(k+1)/2)^2
Then, for n=k+1,
1^3 + 2 ^3 + ... k^3 +(k+1)^3 =
(1^3 + 2 ^3 + ... k^3) + (k+1)^3 =
((k(k+1)/2)^2 + (k+1)^3 =
(k+1)^2 * (k/2)^2 + (k+1)^2*(k+1) =
(k+1)^2 * (k^2/4 + k + 1) =
(k+1)^2 * 1/4 * (k^2 + 4k + 4) =
(k+1)^2 * 1/4 * (k + 2)^2 =
((k+1)(k+2)/2)^2 =
(n(n+1)/2)^2 for n = k+1
Thus, we have proved for n=1 and n=k+1 given k, so we have proven for any n.
7. Show n^3 + 5n + 6 is divisible by 3 for n >= 1
For n = 1, 1^3 +5(1) + 6 = 1 + 5 + 6 = 12, and 12 is divisible by 3.
Assume for n=k
k^3 + 5k + 6 is divisible by 3, or k^3 + 5k + 6 = 3z for some integer z.
Then, for k+1
(k+1)^3 + 5(k+1) + 6 =
k^3 + 3k^2 + 3k + 1 + 5k + 5 + 6 =
k^3 + 5k + 6 + 3k^2 + 3k + 1 + 5 =
(k^3 + 5k + 6) + (3k^2 + 3k + 6) =
3z + 3(k^2 + k + 2) =
3(z + k^2 + k + 2)
Thus, this is divisible by 3.
Thus, we have proved for n=1 and n=k+1 given k, so we have proven for any n.
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