Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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


share|cite|improve this question
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
up vote 12 down vote accepted

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

share|cite|improve this answer
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 $(a \lt b) \implies \neg (b \lt a)$

share|cite|improve this answer
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.

share|cite|improve this answer
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

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.