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I'm reading about lattices and order relations. I came up with a property that says.

$a \land b \lt a$ and $a \land b \lt b$ iff $a$ and $b$ are incomparable.

This confuses me up a litle because I think $a \land b \lt a$ and $a \land b \lt b$ is always true since $\land$ denotes the maximum lower bound. Can anyone show me a proof and explain to me why the above is true?

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    What is the infimum of $\{a,b\}$ when $a$ and $b$ are comparable? – 2017-02-11
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    is $a \land b $, isn't it? – 2017-02-11
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    Correct, your answer is not what I expected. Let me rephrase: what is $a\land b$ when $a$ and $b$ are comparable? – 2017-02-11
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    Im sory, in that case what I know is that It's an element $x$ that satisfies $x \le a$ and $ x \le b$ – 2017-02-11
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    OK. Let me try to guide you to a more enlightening point of view. What does it mean to say that $a$ and $b$ are comparable? – 2017-02-11
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    Thanks for your help. From what I know, if $a$ and $b$ are comparable then $a$ and $b$ belong to the binary relation, in this case they belong to the relation $\le$. – 2017-02-11
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53477/discussion-between-git-gud-and-pablo-estrada). – 2017-02-11

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Easiest examples are positive integers, comparison is by divisor and strict divisor.

2 and 3 are not comparable. The gcd is 1, which (strictly) divides both.

2 and 6 are comparable. the gcd is 2. This strictly divides 6, but does not strictly divide 2 because it is equal to 2.