Determine if the following is a partial order, and if so, is it a total order? I'm having trouble figuring out how I can solve this... I've never been good with formal proofs.
$$(\mathbb{R},\preceq), a\preceq b\iff a^{2}\leq b^{2}$$
I can easily see that it's Reflexive: $\forall a\in\mathbb{R}, a^{2}\leq a^{2}$
I'm not sure how to properly prove that it's transitive and anti-symmetric though. I get stuck here...
\begin{align}
a\preceq b,\ b\preceq c&\Rightarrow a^{2}\leq b^{2},\ b^{2}\leq c^{2}\\
&\Rightarrow a^2+b^2\leq b^2+c^2
\end{align}
And then anti-symmetric:
\begin{align}
a^2\leq b^2\wedge b^2\leq a^2&\Rightarrow a^2=b^2 ??
\end{align}
Can anybody give me any pointers on how to approach proving these things? Thanks.
 A: Hint: If you got stuck it might be the time to look for a countable example. What can you say about the case where $a=-b$?
A: For the antisymmetric property: your statement of the problem is not quite correct: you need to check whether, if $a^2 \leq b^2\; \land \; b^2 \leq a^2$, does this imply $\bf a = b$.
(Recall, any relation $\sim$ is antisymmetric on a set $A$ if and only if, for all $a, b \in A$, IF $a\sim b$ AND $b\sim a$, THEN $\bf{a = b}$. In this case, $a \sim b$ means $a^2 \leq b^2$, $b\sim a$ means $b^2 \leq a^2$, and $\bf{a = b}$ means exactly, $\bf{a  = b}$.)
Asaf is suggesting you consider a counterexample that shows antisymmetry, for example, fails:
Specifically, he asked you to consider  $a = -b$, and for a good reason:
Suppose we have that $a = -b$. If the relation were antisymmetric, then  $a^2 \leq (-b)^2 = b^2 $ and $(-b)^2 = b^2 \leq a^2$ would then imply $a = b $. But this contradicts the our supposition that $a = -b$.  Hence the relation cannot be antisymmetric.
A: Since you are using $\leq$ for $\mathbb{R}$ (i.e. the real numbers), you get $a^{2} \leq b^{2} \wedge b^{2} \leq c^{2} \Rightarrow a^{2}\leq c^{2}$ and $a^2 \leq b^2 \wedge b^2 \leq a^2 \Rightarrow a^2=b^2$ for free. You can see this more clearly if include the correct quantifiers in your proofs. For example, for proving transtivity
$$\forall a, b \in \mathbb{R}, a \preceq b \Rightarrow a^{2} \leq b^{2} \wedge b^{2} \leq c^{2} \Rightarrow a^{2}\leq c^{2}$$
