Relations and Functions - Is my answer correct?

Question

Could someone please advise if my answer is correct or incorrect? Any help will be greatly appreciated.

Given the sets $A = \{1, 2, 3\}$, $B = \{−1, 0, 1, 2\}$ and $C = \{3, 4, 5, 6\}$, indicate the members (pairs) in the following relations. Let $\Bbb N$ denote the natural numbers. (Note that in these definitions a comma ‘,’ is used as a less intrusive way of indicating predicate conjunction ‘$\land$’.)

1) $\{(x, n) \mid x \in A, n \in\Bbb N, n = 2 \cdot x\}$

Ans: $\{(1, 2), (2, 4), (3, 6)\}$

2) $\{(x, z) \mid x \in A, z \in C, x = z − 1\}$

Ans: $\{(2,3),(3,4)\}$

3) $\{(x, y, z) \mid x \in A, y \in B, z \in C, z = x + y\}$

Ans: $\{(1,2,3),(2,1,3),(2,2,4),(3,0,3),(3,1,4),(3,2,5)\}$

4) $\{(x, y, z) \mid x \in A, y \in B, z \in C, z = x \cdot y\}$

Ans: $\{(2,2,4),(3,1,3),(3,2,6)\}$

5) $\{(x, y, z) \mid x \in A, y \in B, z \in C, z = 100xy\}$

**Ans: NULL

EDIT: I have edited this post and put in the answers I believe are correct. Thank you so much.

Answer 1

Your answer to (1) is correct. Your answer to (2) is not: $4$ and $5$ are not elements of $A$, so $(4,5)$ and $(5,6)$ don’t meet the requirement that the first component of the ordered pair belong to $A$. Your answer to (3) is wrong in two ways. First, $(1,-1,0)$ and $(2,0,2)$ should not be there, since $0$ and $2$ are not in $C$: the definition requires that the third component be in $C$. Secondly, $(3,2,5)$ should be there: $3\in A$, $2\in B$, $5\in C$, and $3+2=5$, so it satisfies all of the conditions. $(2,2,4)$ should also be there. (I did not make an exhaustive search for missing triples; there might be more.)

In (4) you also have both kinds of error: $(1,-1,-1)$ should not be present, since $-1\notin C$, and $(3,2,6)$ should be. There is at least one more error of each kind, but I’ll leave them for you to find.

(5) is also wrong: no member of $C$ is a multiple of $100$, so there are no triples $(x,y,z)$ such that $x\in A$, $y\in B$, $z\in C$, and $z=100xy$. This is $\varnothing$, the empty set.

Answer 2

It seems to me that you're having some difficulties with set builder notation. Here's how it works: you have $A = \{x \in X | p(x)\}$. This means that the set $A$ is composed of all $x$ in $X$ for which $p(x)$ is TRUE, where $p(x)$ is just some logical proposition. This means you can think of the vertical bar as meaning "such that".

For instance, $\{x \in \mathbb{Z} | x$ is even$\}$ is the set of all even integers.

So if we take, for example, your second exercise, then we can put it into plain english: $\{(x, z) \mid x \in A, z \in C, x = z − 1\}$ is the set of all pairs $(x,z)$ where $x$ comes from $A$ and $z$ comes from $C$, such that $x=z-1$. The third bit is the really important condition. If you will, you can think about it as first selecting all possible pairs $(x,z)$, and then eliminating those for which that condition is not true.

Beneath is feedback on your answers.

1 - Correct.
2 - Incorrect. For $(4,5)$ and $(5,6)$, where the form is $(x,z)$ and $x\in A$, you are implying that $4,5 \in A$. This is untrue.
3 - Incorrect. For $(-1,1,0)$ and $(2,0,2)$, you are implying that $0,2 \in C$. This is untrue. Here are other combinations that you should have included: $(3,0,3),(1,2,3),(2,1,3),(3,2,5)$.
4 - Incorrect again. See above to figure out why.
5 - This should just be the empty set.