It’s the same process as for a simpler relation; the notation just gets a bit more cumbersome.
To show that $R$ is reflexive, you’d have to show that $(i,j)\,R\,(i,j)$ for every $(i,j)$. Now translate that into more basic terms, using the definition of $R$: you’d have to show that for each $(i,j)$, either
$$i<i\text{ and }i\le j\le j$$ or
$$i<i\text{ and }i\le j\le j\;.$$
Since these are the same statement, you'd have to prove that $i<i$ and $i\le j\le j$ for each $(i,j)$. But you can do at most half of this: if $i\le j$, then it’s true that $i\le j\le j$, but it’s never true that $i<i$. Thus, $R$ is not reflexive: it’s not true that $(i,j)\,R\,(i,j)$ for each $(i,j)$. In fact, it’s not true for any $(i,j)$, so $R$ is irreflexive.
I’m going to leave symmetry for you; it’s pretty easy if you think about the form of the definition of $R$, and transitivity is a bit messier.
To show that $R$ is transitive, you’d have to show that for any $(i,j),(k,\ell)$, and $(m,n)$, if $(i,j)\,R\,(k,\ell)$ and $(k,\ell)\,R\,(m,n)$, then $(i,j)\,R\,(m,n)$. Here again your automatic first step should be to translate this:
- either $i<k$ and $k\le j\le\ell$, or $k<i$ and $i\le\ell\le j$, and
- either $k<m$ and $m\le\ell\le n$, or $m<k$ and $k\le n\le\ell$,
and the desired conclusion is
- either $i<m$ and $m\le j\le n$, or $m<i$ and $i\le n\le j$.
You could systematically investigate all of the combinations: $i<k$ and $k<m$, $i<k$ and $m<k$, $k<i$ and $k<m$, and $k<i$ and $m<k$. Or you could try to look ahead to see whether one combination seems especially problematic. Specifically, what if $$i<k\text{ and }k\le j\le\ell\tag{1}$$ and $$m<k\text{ and }k\le n\le\ell\;?\tag{2}$$ Is there any guarantee from $(1)$ and $(2)$ that either $i<m$ or $m<i$? No, because it might be the case that $i=m$. Is this possibility actually consistent with $(1)$ and $(2)$? If $i=m$ they say that
$$i<k\text{ and }k\le j\le\ell\text{ and }k\le n\le\ell\;,$$
which is certainly possible. Thus, we can’t in general infer $(i,j)\,R\,(m,n)$ from $(i,j)\,R\,(k,\ell)$ and $(k,\ell)\,R\,(m,n)$: if $i=m<k$ and $k\le j,n\le\ell$, $(i,j)\,R\,(k,\ell)$ and $(k,\ell)\,R\,(m,n)$ will be true, but $(k,\ell)\,R\,(m,n)$ won’t be. In particular, if $(m,n)=(i,j)$ and $i<k$ and $k\le j\le\ell$, we get a counterexample to transitivity of $R$.
(Actually, if you show first that $R$ is irreflexive and symmetric, you can take a shortcut to show that it can’t be transitive. Just find two different pairs $(i,j)$ and $(k,\ell)$ such that $(i,j)\,R\,(k,\ell)$. Then by symmetry $(k,\ell)\,R\,(i,j)$, so if $R$ were transitive, we could deduce that $(i,j)\,R\,(i,j)$. However, ... )
