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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.

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2  
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
1  
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

3 Answers 3

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 $a=-b$?

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5  
(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
1  
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.

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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

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

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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

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