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Let $A=\{1,2,3,4\}, B=\{1,5,9,11,15,16\} $ and $f=\{(1,5),(2,9),(3,1),(4,5),(2,11)\}$ Are the following statements true ?

$1.)$ $f$ is a function from $A$ to $B.$

$2.)$ $f$ is a relation from $A$ to $B$

For the first part as $(2,9)\ \text{and} \ (2,11)$ is repeated ,i think its not a function , not sure about the second.

I would like simple explanation.

I have studied maths up to $12th$ grade, thanks.

R K
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  • What is your definition of "relation from $A$ to $B$", and what part of the definition are you having trouble with? – Trevor Wilson Jun 01 '15 at 00:08
  • I am new to this topic, i know that relation is subset of cartesian product. – R K Jun 01 '15 at 00:15
  • Then you can check whether or not every element of $f$ is an element of the cartesian product $A \times B$. (Note that it's not necessary to write out all of the elements of $A \times B$ in order to do this.) – Trevor Wilson Jun 01 '15 at 00:18

2 Answers2

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In a function, one element cannot have two images, just like when you see yourself in a Mirror, you are creating a function (bijection) from yourself to yourself, but can any of your organ have two images. No. In your case $2$ is taking up to images, so no it is not a function, and a relation is any subset of $A \times B$ where $A$ is a domain and $B$ is codomain, so yeah it is a relation.

Bhaskar Vashishth
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A function $f$ from a set $A$ to $B$ is a relation from $A$, the domain and $B$, the co-domain that satisfies:

  • Every element $a$ in $A$ is related to some element $b$ in $B$.
  • In addition, no element $a$ in $A$ is related to more than one element $b$ in $B$.

So for (1) you are correct, since $2$ in $A$ is related to more than one element in $B$, you found that $2$ is related to $9$ and $11$, so $f$ is not a function for (1).

Let $A$ and $B$ be sets. A relation $R$ is a subset of $A\times B$.

$A\times B$ is the set that contains all the ordered pair $(a,b)$ such that $a$ is from $A$ and $b$ is from $B$.

It is unnecessary to list out the ordered pairs contained in $A\times B$ since there are too many elements in $A\times B$. What we can do instead is to check the ordered pairs in $f$ and see whether we can pair a given element in $A$ with the appropriate element in $B$.

If all the ordered pairs in $f$ do belong in $A\times B$, then it is true that it is a relation; otherwise, if you find one ordered pair in $f$ that is not in $A\times B$, then it is not true that it is a relation.

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