 QUESTION

# Let f be a function in the set N of natural numbers defined by f(n) = 3n. Is f is a function from N to N. If so find the Range of function

Hint: Use the property that if f(x) is a function then for all a, b in Domain of “f”, such that a=b, we have f(a)=f(b).Range of a function is the set of all the values attainable by the function i.e. x is in Range(f) if and only if there exists y in N such that f(y) = x. Use the definition to find the Range of “f”.

We have f is defined on the set of all Natural numbers.
Now let a and b be two natural numbers such that a = b
f(a) = 3a and f(b) = 3b.
Now, we have a = b.
Multiplying both sides by 3, we get
3a = 3b, i.e. f(a) = f(b). Hence the relation f is a function on the set of natural numbers.
Now we know that Range of a function is the set of all the values attainable by the function, i.e. x is in Range(f) if and only if there exists y in N such that f(y) = x.
Now let $x\in {{R}_{f}}$
Hence from the definition, we have
$\exists n$ such that $f\left( n \right)=x$
But f(n) = 3n.
Hence, we have
x = 3n
Hence x is divisible by 3.
Also, let m be any natural number divisible by 3.
Hence there exists a natural number n such that m = 3n
Hence m = f(n).
Hence m is in ${{R}_{f}}$
Hence Range(f) is the set of all Natural numbers divisible by 3.
Hence, we have ${{R}_{f}}=\left\{ x:x\in \mathbb{N},\text{ }x\text{ is divisible by 3} \right\}$

Note: A relation is said to be a function if it maps all the points of the Domain to a unique point in the co-domain.
Consider a relation $f:A\to B$. We say f is a function if it maps all the points of A to a unique point in B. Hence a single value in A cannot be mapped to more than one points in B. Hence we have if a and b are in A and a = b, then f(a) = f(b).
The set A is called the Domain of f, and the set B is called the co-domain of f. Clearly, we have
$\text{Range}\subseteq \text{Codomain}$