Sets

[nycmath]

See attachment. Show your steps.

[Math Tutor]

1) [tex]C=\{2,4,6,8\}[/tex], [tex]D=\{6,8,10,12\}[/tex].

The intersection keeps the elements that are in both lists. Going through [tex]C[/tex] one by one: [tex]2\notin D[/tex], [tex]4\notin D[/tex], [tex]6\in D[/tex], [tex]8\in D[/tex], so
[tex]C\cap D=\{6,8\}[/tex]

The union collects everything that is in at least one of them, each element listed once:
[tex]C\cup D=\{2,4,6,8,10,12\}[/tex]
Note [tex]6[/tex] and [tex]8[/tex] appear in both sets but are written only once — a set has no repeated elements.

2) [tex]G=\{a,e,i,o,u\}[/tex], [tex]H=\{a,e,y\}[/tex].

The symmetric difference is what is in one set but not the other:
[tex]G\,\Delta\,H=(G\setminus H)\cup(H\setminus G)[/tex]
[tex]G\setminus H=\{i,o,u\}[/tex] (remove [tex]a[/tex] and [tex]e[/tex], which are in [tex]H[/tex])
[tex]H\setminus G=\{y\}[/tex] (remove [tex]a[/tex] and [tex]e[/tex], which are in [tex]G[/tex])
[tex]G\,\Delta\,H=\{i,o,u,y\}[/tex]

Check it with the other formula, [tex]G\,\Delta\,H=(G\cup H)\setminus(G\cap H)[/tex]:
[tex](G\cup H)\setminus(G\cap H)=\{a,e,i,o,u,y\}\setminus\{a,e\}=\{i,o,u,y\}[/tex]
Same answer — the shared elements [tex]a[/tex] and [tex]e[/tex] are exactly the ones that drop out.

[nycmath]

1) [tex]C=\{2,4,6,8\}[/tex], [tex]D=\{6,8,10,12\}[/tex].

The intersection keeps the elements that are in both lists. Going through [tex]C[/tex] one by one: [tex]2\notin D[/tex], [tex]4\notin D[/tex], [tex]6\in D[/tex], [tex]8\in D[/tex], so
[tex]C\cap D=\{6,8\}[/tex]

The union collects everything that is in at least one of them, each element listed once:
[tex]C\cup D=\{2,4,6,8,10,12\}[/tex]
Note [tex]6[/tex] and [tex]8[/tex] appear in both sets but are written only once — a set has no repeated elements.

2) [tex]G=\{a,e,i,o,u\}[/tex], [tex]H=\{a,e,y\}[/tex].

The symmetric difference is what is in one set but not the other:
[tex]G\,\Delta\,H=(G\setminus H)\cup(H\setminus G)[/tex]
[tex]G\setminus H=\{i,o,u\}[/tex] (remove [tex]a[/tex] and [tex]e[/tex], which are in [tex]H[/tex])
[tex]H\setminus G=\{y\}[/tex] (remove [tex]a[/tex] and [tex]e[/tex], which are in [tex]G[/tex])
[tex]G\,\Delta\,H=\{i,o,u,y\}[/tex]

Check it with the other formula, [tex]G\,\Delta\,H=(G\cup H)\setminus(G\cap H)[/tex]:
[tex](G\cup H)\setminus(G\cap H)=\{a,e,i,o,u,y\}\setminus\{a,e\}=\{i,o,u,y\}[/tex]
Same answer — the shared elements [tex]a[/tex] and [tex]e[/tex] are exactly the ones that drop out.

Great reply. Good notes for my math files.