# Functions-7

### From Homeworkwiki

**If a, b ∈ {1, 2, 3, 4}, then which of the following are functions in the given set?**

**(a) f _{1} = {(x,y) : y = x + 1} (b) f_{2} = {(x,y) : x + y > 4}**

**(c) f**

_{3}= {(x,y) : y < x} (d) f_{4}= {(x,y) : x + y = 5}**Also, in case of a function give its range.**

**Solution:** If we express f_{1}, f_{2}, f_{3} and f_{4} as set of ordered pairs, then we have

f_{1} = {(1,2), (2,3), (3,4)}

f_{2} = {(1,4), (4,1), (2,3), (3,2), (2,4), (4,2), (3,4), (4,3)}

f_{3} {(2,1), (3,1), (4,1), (3,2), (4,2), (4,3)} and

f_{4} = {(1,4), (2,3), (3,2), (4,1)}

(a) We have: f_{1} = {(1,2), (2,3), (3,4)}.

We observe that an element 4 of the given set has not appeared in the first place of any ordered pair of f_{1}. So,

f_{1} is not a function from the given set to itself.

(b) We have f_{2} = {(1,4), (4,1), (2,3), (3,2), (2,4), (4,2), (3,4), (4,3)}

We observe that 2, 3, 4 have appeared more than once as first components of the ordered pairs in f_{2}. So, f_{2} is

not a function.

(c) We have f_{3} = {(2,1), (3,1), (4,1), (3,2), (4,2), (4,3)}

We observe that 3 and 4 have appeared more than once as first components of the ordered pairs in f_{3}. So, f_{3} is

not a function.

(d) We have f_{4} = {(1,4), (2,3), (3,2), (4,1)}

We observe that each element of the given set has appeared as first components in one and only one ordered pair of f_{4}. So,

f_{4} is a function in the given set. In this case,

Range of f {1, 2, 3, 4}