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.
$\leq$ operation and logical error
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1You 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 -
1The "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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0Nothing 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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1Are 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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0@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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0then if this is valid all xors are automatically or operations – 2012-11-07
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1@mehdi: can you please use your registered account instead of creating new accounts for each new question? – 2012-11-07
2 Answers
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
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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0yes 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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0@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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0if 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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0I am the same mehdi – 2012-11-07
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0@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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0"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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0I got confused from truth tables but acc. to this def. it is ok to use or. – 2012-11-07
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0@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 -
0I 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 judge – 2012-11-07
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0@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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0@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
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.