Let \(A = \left \{ x ∈N | x^2 < 37 \right \}\) and \(B = \left \{ 3k + 1 |k ∈ \left \{ 1, 2, 3, 4 \right \} \right \}\) be sets.

1. List the elements of \(A\) and \(B\), and the subsets of \(B\).

2. For any sets \(A\), \(B\), and \(C\), prove or disprove the claim: If \(A ∈ B\), \(B ∈ C\), then \(A ∈ C\).

3. Show \(\left \{ 12a + 4b |a, b ∈ Z \right \} = \left \{ 4c |c ∈Z \right \}\)

EighthMersennePrime Aug 23, 2022

#1**+1 **

I assume \(N\) is for natural numbers and \(Z\) is for integers.

1.

A = {1, 2, 3, 4, 5, 6}

B = {4, 7, 10, 13}

Subsets of B are any combination of the elements of B (excluding some).

2.

If A is contained within B, and B is contained with C, A is contained within C.

This is, I suppose, one of those annoying proofs which I take for granted and see no way of prooving it other than it is logical. It is here I should also note that I have never studied set theory, and I'm probably getting everything wrong .

3.

{4(3a+b)|a,b \(\in\)Z} = {4c|c\(\in\)Z}

{3a+b|a,b\(\in\)Z}={c|c\(\in\)Z}

3a+b\(\in\)Z

c\(\in\)Z

(I don't know what to do after that....)

**Again, I've never studied this, so apologies in advance for faulty logic everywhere.**

WhyamIdoingthis Aug 23, 2022