Definition talk:Operation/N-Ary Operation

I moved the big forall thing out of its even more confusing original spot, but I'm not really understanding what it's for. It seems to confuse more than it illuminates. --Dfeuer (talk) 21:36, 8 March 2013 (UTC)


 * When you define a mapping, it is worth emphasising that it is a mapping by providing an indication that it operates on all elements of its domain. Hence the forall. If it confuses you then I suggest you do some further studying before you insist on amending pages on subjects in which you are clearly out of your depth. --prime mover (talk) 21:41, 8 March 2013 (UTC)


 * That is not what it says. What it says, essentially, is that the image of the mapping is a subset of $\mathbb U$. It appears to be trying to make the point that the domain is the Cartesian product of the sets. That point is adequately covered by the link to mapping, but if you want to express it here it needs to be with a different symbolic statement. And if you want to express the fact that it's actually a mapping, that requires a different one too.


 * Just for fun: $f$ is a mapping from $S$ to $T$ iff:


 * $\forall s \in S: \exists t \in T: \forall t' \in T: ((s, t') \in f \iff t' = t)$ --Dfeuer (talk) 21:58, 8 March 2013 (UTC)


 * Not true. There might be $t'$ outside of $T$. &mdash; Lord_Farin (talk) 22:59, 8 March 2013 (UTC)


 * Actually no, it's accurate. If there is a $t'$ outside of $T$ such that $(s, t') \in f$ then $f$ may not be a mapping on a superset of $T$. But it is a mapping from $S$ to $T$.


 * The square root mapping on $\R$ for example. While $(1, 1) \in x^{1/2}$ and $(1, -1) \in x^{1/2}$, and so it is not a mapping, but when you restrict the codomain the positive reals it is. --prime mover (talk) 23:08, 8 March 2013 (UTC)


 * I was speaking in context of a relation on $S \times T$. To enforce the codomain as L_F desired, just drop the second $\in T$. If not assuming a set of ordered pairs, that will need to be handled as well, using some construction or other for that. But my actual point is that the current statement on the main page speaks to codomain image, not domain. --Dfeuer (talk) 23:43, 8 March 2013 (UTC)

Can anyone explain what he's on about? --prime mover (talk) 00:02, 9 March 2013 (UTC)


 * I'm afraid I don't get it. The presentation is clear and correct to me. It is phrasing the same thing twice for convenience, as seen by "That is" occurring. Suggest to leave the premises. &mdash; Lord_Farin (talk) 08:51, 9 March 2013 (UTC)


 * I put it back the way it was originally. The form is: "$\circ$ from $A$ to $B$ is a mapping, defined thus: for all elements in the domain of $\circ$, the image of that element is in the domain $B$ (in this case $\mathbb U$).


 * All it does is, as Df says, is state that the image of that mapping is a subset of $\mathbb U$. And, as I said earlier in different words, where's the problem? --prime mover (talk) 09:20, 9 March 2013 (UTC)