Conditional existence of real numbers

Are there such numbers $a$ and $b$ that if $a < b$, then $a > b$ ?

Thanks.

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As TonyK points out, any $a$ and $b$ such that $a\not\lt b$ will do. In such a case, $a\lt b$ will imply whatever you want. Others interpreted it as meaning $a\lt b$ and $a\gt b$. Could you please clarify? (The title makes me think that you intended it as written, but given the potential misinterpretations I think clarification would help.) – Jonas Meyer Jan 3 '11 at 11:05

Yes. Take for instance a = 1, b = 0. (Or perhaps this isn't what you meant?)

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Despite my answering in another sense, I disagree with the downvote. This is a correct answer to the question as asked and may clarify the OP's intent. – Ross Millikan Jan 3 '11 at 11:07
um,I don't understand why $1 < 0$ ? – VelvetThunder Jan 3 '11 at 12:12
I understand after reading Yasir Arsanukaev answer. +1 to both. – VelvetThunder Jan 3 '11 at 12:15
I am a big fan of first answering exactly the question that was asked and then seeing where things stand. +1 for this. – Pete L. Clark Jan 3 '11 at 12:58

No. The relation less than (as opposed to less than or equal to) is defined as strict, so $(a \lt b) \implies \neg (b \lt a)$

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This answer would've been true, if I'd asked "Are there such numbers $a$ and $b$ that $a < b$ and $a > b$ ?" See my answer. – Yasir Arsanukaev Jan 3 '11 at 13:47
@Yasir Arsanukaev: You are right, and I apologize for not reading it correctly. I acknowledged that in my comment to TonyK's answer. – Ross Millikan Jan 3 '11 at 13:47
I just didn't notice you got the question correctly, it's possible because of my English :-) – Yasir Arsanukaev Jan 3 '11 at 14:07

I don't really think so. The condition $a < b$ implies that $a \ngeq b$ because $<$ is a total order on $\mathbb{R}$. If you're not familiar with orderings, they are just relations that are reflexive, anti-simmetric and transitive (see Wikipedia. A set is said to be totally ordered if every element of the set can be compared to another with the ordering given.

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It is not a well-ordering. You mean to say that it is a strict total order. The relation you defined with the modulus is not anti-symmetric, so it is not a partial order. – Jonas Meyer Jan 3 '11 at 10:38
@Jonas Meyer: you're right, I meant a total order, I'll correct and take off the example (dang, I thought it could be THAT easy...) – Andy Jan 3 '11 at 10:41