Question

6. Prove that for an integer n, n2 - n is divisible by 2. 7. Prove A SAUB, for all sets A,B

Answer 1

Answer:

1. We will prove by induction over n, that for an integer n, $n^2-n$ is divisible by 2.

For n= 1, $1^2-1=0$ and $\frac{0}{2}=0$. Then for n=1 the property is satisfied.

Suppose as induction hypothesis that for an integer n, $n^2-n$ is divisible by 2, that is, $\frac{n^2-n}{2}=b, \text{ for some integer b}$.

Let's see with n+1.

$(n+1)^2-(n+1)=n^2+2n+1-n-1=n^2-n+2n+2$

But,

$\frac{n^2-n+2n+2}{2}=\frac{n^2-n}{2}+\frac{2(n+1)}{2}$,

by induction hypothesis

$\frac{n^2-n}{2}+\frac{2(n+1)}{2}=b+(n+1)\in\mathbb{Z}$

This shows that for n+1 the property is satisfied, then we can conclude that that for an integer n, $n^2-n$ is divisible by 2.

2. Let $A, B$ sets. By definition,

$A\cup B=\{x: x\in A \text{ or } x\in B\}$

Then, each element of A is in $A\cup B$, that is, $A\subset A\cup B$.