please help me with these problems. Thank you For each of the following sets Xi,

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please help me with these problems. Thank you

For each of the following sets Xi, determine whether = N0 or = c. No justification is necessary X1 = (5, infinite). X2 = Z times Z X3 = {2x : x Z}. X4 = {(p, q) R times R : q = }. Hint. Study the meaning of the cardinal number (i.e, cardinality) of a set, as well as the meaning of N0 and c. Consider integers 24, 60, 30 and 37. List all (positive and negative) comman divisors of 24 and 60. Determine gcd(24,60). Express gcd(30,37) as a linear combination of 30 and 37 (with integer coefficients). Hint. (2) Use the Euclidean Algorithm (i.e., repeated Division Algorithm to find gcd(30,37) and then find r,s Z such that gcd(30,37) = 30r + 37s. Show your step-by-step work, as always. Let x = 3 - I and y = 4 + 2i. Compute x+y and x-y. Compute xy and x/y. Hint. All should be straightforward. Study the following claim as well as the alleged "proof": Claim. For all sets A and B, if A B and A B and A B then A B. "proof". Let A={1,2} and B {1,2,3}. Then A B; and this case, it is clear that A B. In general, if A B and A B, then B has more elements than A does, hence A B. Therefore, for all sets A and B, if B then A B. Complete the following questions concerning the above claim and "proof": Determine whether the "proof" is correct/rigorous. Identify the issues, if any. Determine whether the claim is true or false. Justify the answer in part(3). If the claim is true and the "proof" is not correct/rigorous, then provide a correct and rigorous proof. If the claim is false, give a concrete counterexample. Hint. Complete the parts as instructed. If the claim is false, then (obviously) there is no way the "proof" could be correct.

Explanation / Answer

1.

(a) c

(b) aleph_0

(c) aleph_0

(d) c

2. 24 = 2^3 * 3

60 = 2^2 * 3 * 5

The gcd is obtained from the product of the largest prime powers that appear in the factorization of *both* 24 and 60. This gcd is 2^2 * 3 = 12

The common divisors are the divisors of the gcd, so they are: 1, 2, 3, 4, 6, 12 as well as -1, -2, -3, -4, -6, -12

(b)

37 = 1*30 + 7

30 = 4*7 + 2

7 = 3*2 + 1

2 = 2 * 1 + 0

So the gcd(37,30) = 1, and we have

7 = 37 - 30

2 = 30 - 4*7 = 30 - 4*(37 - 30) = 5*30 - 4*37

1 = 7 - 3*2 = 37 -30 - 3*(5*30 - 4*37) = 13*37 -16*30

So 1 = 13*37 - 16*30

3.

x = 3 - i, y = 4 + 2i

x+y = 7 - i

x-y - -1 - 3i

xy = 12 - 4i + 6i -2i^2 = 12 - 2i + 2 = 14 + 2i

x/y = (3-i) / (4+2i) * [(4-2i)/(4-2i)] = (3-i)(4-2i)/[16 + 4] = (1/20) * (12 -6i -4i - 2) = (1/20)(10 - 10i) = .5 - .5i

4. The proof is neither correct nor rigorous. First, it attempts to use a single example to prove a theorem. The proof flatly states that "B has more elements than A" without proving this. When comparing cardinality of sets, one must usually talk about functions between sets. The author fails to take into account infinite sets, and this is where the big trouble lies.

(b) The claim is false.

(c) For a counterexample, let B = N, and let A = 2N (the natural numbers, and the even natural numbers). These sets are not equal, and A is a subset of B, but both A and B have cardinality aleph_0.

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