Helllo! I have the following exercise:
Let R[t] be the set of all polynomials with real coefficients, and define p ≤ q if and only if p(x) ≤ q(x) for all real values of x. Prove this defines a partial ordering of R[t], but that this partial ordering is not a linear ordering.
To be able to prove that this defines a partial ordering we need to prove reflexivity, antisymmetry and transitivity.
- Reflexivity is easy: $p(x) \le p(x)$ for all real values of $x$.
- Anti-symmetry is not hard: If $p(x) \le q(x)$ for all values of $x$, then $q(x) \le p(x)$ for all values of $x$ only if they are the same polynomial.
- Transitivity: I think it's directly implied from the $" \le "$ relation, which is transitive.
However, I have trouble proving that this is NOT a linear relation in the set $R[t]$. I'm not sure why, since it would mean that that neither $p(x) \le q(x)$ and $q(x) \le p(x)$ are true.
The only example I could think of was if $p(x) = x$ and $q(x) = x^2$, so $p(2) \le q(2)$ but $q(0.5) \le p(0.5)$ which would mean that $p(x) \le q(x)$ does not hold for all real $x$.
However, I'm confused. Judging by the definition of the relation, $p(x)$ and $q(x)$ wouldn't even be in the relation at all, so this wouldn't disprove linearity.
What I have come up with: Linearity would imply that for each two polynomials, the first one must be related to the other one or vice-versa. Since I have provided a counterexample, it does not hold.
My question is: Is this a valid counterexample? If yes, do I need to add something to my explanations of reflexivity, anti-symmetry and transitivity?
Yes, you have the right idea. A relation $\leq$ on $R[t]$ is linear (or total) if for every pair of elements $p(t), q(t) \in R[t]$, either $p(t) \leq q(t)$ or $q(t) \leq p(t)$. You've exhibited a counterexample, so you've shown that your relation is not linear.
Quick comment on notation: $R[t]$ is the set of all polynomials with coefficients in $R$ in the variable $t$, so it makes more sense to write the elements of $R[t]$ as polynomials $p(t)$, rather than $p(x)$.
Coming back to your example, it doesn't really make sense to say $p(t)$ and $q(t)$ aren't in the relation. You can think of the relation as a subset of $R[t] \times R[t]$, namely the set of all ordered pairs $(p(t), q(t))$ such that $p(t) \leq q(t)$. So it would be accurate to say that $(t, t^2)$ and $(t^2, t)$ are not in the relation, but there are ordered pairs in the relation that do contain these polynomials, like $(t, t+1)$ or $(t^2, 2t^2)$.