Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
It turns out not to matter, but to prove transitivity, you can use the transitivity of $\le$ itself. – Trevor Wilson Feb 24 '13 at 23:04
I'm not sure if that will be enough for my prof though... He's been doing some examples akin to how I attempted to approach the solution, but I can't figure out how. Maybe I'll have to ask him tomorrow. – agent154 Feb 24 '13 at 23:07
What might not be enough, proving transitivity? It's definitely not, because that's only part of being a partial order. – Trevor Wilson Feb 24 '13 at 23:08
For transitivity, now subtract $b^2$ from both sides and you have exactly what you want to prove. +1 for showing where you are stuck. Others, please note, this enabled answers that addressed OP's specific problem. – Ross Millikan Feb 26 '13 at 3:17
up vote 8 down vote accepted

Hint: If you got stuck it might be the time to look for a countable example. What can you say about the case where $a=-b$?

share|cite|improve this answer
(count er example) – Zev Chonoles Feb 24 '13 at 23:01
@Zev: If I were a female you could have said "Count her example". Another tacky pun lost amidst the genitals of math.SE users... – Asaf Karagila Feb 24 '13 at 23:01
well - $b=-a\dots$ but I don't see how that helps my situation. – agent154 Feb 24 '13 at 23:07
@agent154: Is it true that $b\preceq a$ or $a\preceq b$? What about both? – Asaf Karagila Feb 24 '13 at 23:08
agent154: for the antisymmetric portion: 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 $\bf a = b$ – amWhy Feb 24 '13 at 23:31

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.

share|cite|improve this answer
Now I can use the tacky joke, count-her-example. Although it was my example, so I suppose I still can't use that. Drats. :-) – Asaf Karagila Feb 24 '13 at 23:43
@Asaf: Hehehe... :-) Seriously, if you'd like for me to delete this ( I tried to very clearly to credit you. I just didn't want to make for a very long comment.) If I disclosed too much, given your desire to "hint", I will happily delete. – amWhy Feb 24 '13 at 23:47
No, it's fine. I gave a hint, the user essentially requested for more. Your answer is just an elaboration. – Asaf Karagila Feb 24 '13 at 23:54
Boldface = Countermeasures? – Asaf Karagila Feb 25 '13 at 1:07
@Asaf: countermeasure to the incorrect assumption of the OP and counter-measure to the incorrect second conclusion given by Code-Guru (the implication that you get antisymmetry for free, and the misunderstanding of what is required for antisymmetry reflected therein). Over the top? – amWhy Feb 25 '13 at 1:13

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

$$\forall a, b \in \mathbb{R}, a \preceq b \Rightarrow a^{2} \leq b^{2} \wedge b^{2} \leq c^{2} \Rightarrow a^{2}\leq c^{2}$$

share|cite|improve this answer
about antisymmetry, please see my answer and my second comment below my answer. – amWhy Feb 25 '13 at 1:15
Your answer is wrong. $x\mapsto x^2$ is not order-preserving. – Asaf Karagila Feb 25 '13 at 1:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.