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Prove the following.
Show that the function f : N → N, defined by f(x) = 5x - 3, ∀ x ∈ N, is one-one function but not…
Show that the function $f : N \to N$, defined by $f(x) = 5x - 3, \forall x \in N$, is one-one function but not onto function.
To show: $f \text{ is one-one function but not onto function}$
Answer
Answer
$f(x) = 5x - 3$ is one-one but not onto.
- One-one: let $x_1, x_2 \in N$ with $f(x_1) = f(x_2)$.
- Then $5x_1 - 3 = 5x_2 - 3$, so $5x_1 = 5x_2$ (adding 3 to both sides).
- Hence $x_1 = x_2$ (dividing by 5), so $f$ is one-one.
- Not onto: take $1 \in N$ (codomain). If $f(x) = 1$, then $5x - 3 = 1$, so $x = \frac{4}{5}$.
- But $\frac{4}{5} \notin N$, so no element of the domain has image 1 and the range is not all of $N$.
- Hence $f$ is one-one function but not onto function.