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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.)2011-01-03

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

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    I am a big $f$an o$f$ first answering exactly the question that was asked and then seeing where things stand. +1 for this.2011-01-03
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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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    I just didn't notice you got the question correctly, it's possible because of my English :-)2011-01-03
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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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    @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...)2011-01-03