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We define $x \leq y$ operation as '$x

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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    You are confused: "false when both $x and $x=y$ as both can not be true at the same time". So you say there is a problem in a case that cannot exist. So what?2012-11-07
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    The "or" here is the inclusive or usually used in mathematics, which for "P or Q" means "either P or Q or both". But you're correct in saying that $x \le y$ is also equivalent to the xor, in this case.2012-11-07
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    Nothing is wrong about that definition, we do not need to use an exclusive-or but it happens that both conditions cannot be true at the same time.2012-11-07
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    Are you the same mehdi as http://math.stackexchange.com/users/43997/mehdi? I am asking because that user had a gravatar similar to yours at some point, if I recall correctly.2012-11-07
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    @marc, yes 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.2012-11-07
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    then if this is valid all xors are automatically or operations2012-11-07
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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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    yes but when we use or there is some possibility that both may be true although we are not sure. here we are sure that both cannot be true.2012-11-07
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    @mehdi: So what? If we disallow false assumptions than the entire idea behind vacuously true arguments is going to break. Remember that $\text{False}\implies\text{True}$.2012-11-07
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    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.2012-11-07
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    I am the same mehdi2012-11-07
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    @mehdi: So your angle is that we **shouldn't** say $\{0\}\cup\{1\}$ is the union of these two singletons *because they are disjoint*?2012-11-07
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    "Logical disjunction is an operation on two logical values, typically the values of two propositions, that produces a value of false if and only if both of its operands are false."2012-11-07
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    I got confused from truth tables but acc. to this def. it is ok to use or.2012-11-07
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    @mehdi: Yes. And to say $x\leq y$ if and only if $x or $x=y$ is **not** a logical disjunction? Is it not *false* if and only if *both assumptions* are false?2012-11-07
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    I got it thanks. truth tabels made me think like when 1-1 it must be 1. but i must only look 0-0 case to judge2012-11-07
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    @AsafKaragila: It is not to the point of this question, but a definition _can_ be wrong in many ways, leading to something that is not well defined. Example: defining something as the smallest element in a set that could turn out to be empty, or otherwise fail to have a minimum.2012-11-07
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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.