My message on discrete stclairdrake.netematics explains:

A relation is a subset of a Cartesian product and also a function is a special type of relation.

You are watching: A relation is a special type of function

But it would certainly make much more sense come me if a role described a relation together a subset the the Cartesian product.

My think being:Given a function, f(x) = y, we have the right to compute a set of (x,y) collaborates within the Cartesian plain. And this set of works with would be the relation the is the subset that the Cartesian product.

Am i confused? could anyone help explain just how a role IS a relation?

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request Sep 8 "16 at 5:19

tim_xyztim_xyz

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A duty is a certain kind the relation.

The best means to understand this is v the assist of an example. So, let us take a pair of to adjust - set $A$ = 1, 2, 3 and collection $B$ = $a$, $b$, $c$. Hence the set $A \times B$ will have 9 elements.

We can select $2^9 = 512$ different subsets of $A \times B$. Every of this subset is a relation between $A$ and $B$. So, $\phi$ is a relation, $(1, a), (1, c), (2, b)$ is additionally a relation. $A$ is referred to as the domain and also $B$ is dubbed the co-domain.

A duty is also a subset the $A \times B$ (hence a relation), yet it has actually constraints. For every element in collection $A$, there have to be precisely one aspect in set $B$. An ext concretely, for every aspect $x$ in set A, there is exactly one $(x, y)$ in $f$ for some $y \epsilon B$

For example, $(1, a), (2, b), (3, b)$ is a function, yet $(1, a), (2, b)$ is not because there is no entry of the form (3, *). Also, $(1, a), (1, b), (3, b), (2, b)$ is likewise not a duty because 1 has actually two values it is mapping to.

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edited Sep 8 "16 in ~ 6:37

answer Sep 8 "16 at 5:38

Pratyush RathorePratyush Rathore

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A duty is a binary relation where the an initial value that the pair is unique.

If $A$ and $B$ are sets, then a binary relationship $R$ is just a collection of bag $(a,b) \in A \times B$. A function is a special type of relation where given any kind of $a \in A$ there is just one $b \in B$ such the $(a,b)$ is $R$. That is, both $(a,b_1)$ and also $(a,b_2)$ cannot be in $R$ unless $b_1 = b_2$.

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For a function $f : A \to B$, we deserve to express $f$ as the collection of facets $(a,b) \in A \times B$ wherein $b = f(a)$. In collection builder notation,

$$f = \(a,f(a)) : a \in A,\ f(a) \in B\.$$

Since any type of subset the $A \times B$ is a relation from $A$ come $B$, a role is certainly a relation.

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edited Sep 8 "16 at 5:45

reply Sep 8 "16 at 5:36

Alexis OlsonAlexis Olson

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Think the the relationship "is much less than" in $\1,2,3\^2$. That relation is $\stclairdrake.netrmLess=\(1,2),(1,3),(2,3)\$, however you normally will not write $(x,y)\in\stclairdrake.netrmLess$, however you will write $x

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