Answer:
1. We will prove by induction over n, that for an integer n, [tex]n^2-n[/tex] is divisible by 2.
For n= 1, [tex]1^2-1=0[/tex] and [tex]\frac{0}{2}=0[/tex]. Then for n=1 the property is satisfied.
Suppose as induction hypothesis that for an integer n, [tex]n^2-n[/tex] is divisible by 2, that is, [tex]\frac{n^2-n}{2}=b, \text{ for some integer b}[/tex].
Let's see with n+1.
[tex](n+1)^2-(n+1)=n^2+2n+1-n-1=n^2-n+2n+2[/tex]
But,
[tex]\frac{n^2-n+2n+2}{2}=\frac{n^2-n}{2}+\frac{2(n+1)}{2}[/tex],
by induction hypothesis
[tex]\frac{n^2-n}{2}+\frac{2(n+1)}{2}=b+(n+1)\in\mathbb{Z}[/tex]
This shows that for n+1 the property is satisfied, then we can conclude that that for an integer n, [tex]n^2-n[/tex] is divisible by 2.
2. Let [tex]A, B[/tex] sets. By definition,
[tex]A\cup B=\{x: x\in A \text{ or } x\in B\}[/tex]
Then, each element of A is in [tex]A\cup B[/tex], that is, [tex]A\subset A\cup B[/tex].
