Can I use induction to prove this transitive relation problem?

Question

Let π be a permutation of the set {1,2,...,n}. We say that an ordered pair (i,j) ∈ {1,2,...,n} × {1,2,...,n} is an inversion of π if i < j and π(i) > π(j).

Let $\pi$ be a permutation of the set $\{1,2,\dots,n\}$ and let $I(\pi)$ denote the set of all inversions, regarded as a relation on $\{1,2,\dots,n\}$ is transitive.

To solve this problem, my logics is that

The definition of transitive relation is

A relation R on a set A is called transitive if whenever (a,b)∈R and (b,c)∈R, then (a,c) ∈ R, for all a,b,c ∈ A.

also, the definition of inversion above says

if i < j then

if i < j and π(i) > π(j).

So, I can set up an induction proof.

with base case. p(2)

i =1 j=2

then

π(1) > π(2)

And then use the inductive proof step?

Am I on the right track at all?

Answer 1

Given $i \sim j$ and $j \sim k$ ,we have \begin{eqnarray*} i \sim j \Rightarrow i\pi(j) \end{eqnarray*} and \begin{eqnarray*} j \sim k \Rightarrow j\pi(k) \end{eqnarray*} so \begin{eqnarray*} i\pi(k). \end{eqnarray*} So $i \sim k$ and therefore the relation is transitive.