Induction proof for $x \le y $

Question

So these are $x \in \{1 ... i\}$ and $y \in \{ i + 1 ... n\}$, for $i, n \in \mathbb N$.

I want to prove it for every $x \le y $.

I know it`s easy but the solution is escaping me. I have tried with sums but it did not work out.

Answer 1

You can finish it in one line: fix $x_0\in \{1,\dots,i\}$,$y_0\in\{i+1,\dots n\}$ so $$x_0\le i < i+1 \le y_0\Rightarrow \boxed{\forall x_0,y_0\quad x_0<y_0}$$

Answer 2

Proof by induction? That is not very convenient here, but if you like.

We will prove by induction on $n$ that $x<y$ if $x\in\{1,\dots,i\}$ and $y\in\{i+1,\dots,n\}$. It follows directly that also $x\leq y$ in that situation.

If $n=0$ then $\{i+1,\dots,n\}=\emptyset$ so that $x< y$ is vacuously true (no $y$ can be found in $\emptyset$ for wich it is not true). This proves the base case.

Suppose now that the statement is true for $n$. To be shown is that it also true for $n+1$. Assuming that $y\in\{i,\dots,n,n+1\}$ we discern two cases: $y\in\{i,\dots,n\}$ or $y=n+1$.

If $y\in\{i,\dots,n\}$ then the induction hypothese immediately tells us that $x<y$.

If $y=n+1$ then $y>n\in\{i,\dots,n\}$ and $x<n$ (again the induction hypothese) and consequently $x<y$.