# $\leq$ operation and logical error

We define $x \leq y$ operation as '$x<y \ \ or \ \ x=y$'. But this is false when both $x<y$ 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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You are confused: "false when both $x<y$ 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? – Marc van Leeuwen Nov 7 '12 at 12:44
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. – coffeemath Nov 7 '12 at 12:44
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. – AD. Nov 7 '12 at 12:45
Are you the same mehdi as 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. – Asaf Karagila Nov 7 '12 at 12:52
@mehdi: can you please use your registered account instead of creating new accounts for each new question? – Willie Wong Nov 7 '12 at 13:17

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 $x<y$ 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. – mehdi Nov 7 '12 at 12:57
@mehdi: So what? If we disallow false assumptions than the entire idea behind vacuously true arguments is going to break. Remember that $\text{False}\implies\text{True}$. – Asaf Karagila Nov 7 '12 at 12:58
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. – mehdi Nov 7 '12 at 13:11
I am the same mehdi – mehdi Nov 7 '12 at 13:11
@mehdi: So your angle is that we shouldn't say $\{0\}\cup\{1\}$ is the union of these two singletons because they are disjoint? – Asaf Karagila Nov 7 '12 at 13:15

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.

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