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We define $x \leq y$ operation as '$x'. But this is false when both $x and $x=y$ as both can not be true at the same time. But I read text books that use this expression in their proofs.

Should not exclusive or be used for this ? Is not there a logic error by defining $\leq$ using only disjunction but not xor in math books.

if you look a truth table definition of P or Q when both are 1 result is 1. But the result is 0 here. So we can not classify this as or. but using or in set union is not like this.

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    @mehdi: can you please use your registered account instead of creating new accounts for each new question?2012-11-07

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It's fine to take a disjunction (or) of two conditions which cannot occur at the same time. The fact that these two properties cannot hold at the same time is irrelevant.

If we want that $x or $x=y$, just one of them needs to be true, and the fact that we say "or" does not mean that it is even possible that both are true at the same time.

Furthermore, definitions are never wrong. Definitions could be useless, or they could be too broad or too restricted, but they are never wrong.

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    @Marc: No, when we define an element we define a set and we claim that this set is a singleton. If you "define" a minimum where it does not exist then you simply end up with an empty set. This is not a wrong definition (mathematically speaking, of course this is wrong in the sense that this is not what you want to define), just a useless one.2012-11-07
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Take the solution set {3,5} for $f(x)=(x-3)(x-5)=0$. We can say that there is a root $r$ for $f(x)$ such that $3\leq r \leq 5$.

In Probability theory it is very common to use an expression such as $P(X \leq 3)$ to specify a set of values for X (the set may be empty, has $1$ instance OR more).

So the notation makes sense indeed if used correctly.