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Step-by-step Explanation
Step 1: Understand the Given Functional Equation for f
We are given that for every natural number $n$,
$$f(n+1) = f(n) + f(1).$$
This implies that the difference between consecutive values of $f$ is the constant $f(1)$. Such a function follows an arithmetic progression form.
Step 2: Determine the General Form of f
Assume $f(1) = a$ for some $a \in \mathbb{N}$. Then:
$f(2) = f(1) + a = a + a = 2a,$
$f(3) = f(2) + a = 2a + a = 3a,$
Continuing similarly, we get $f(n) = na$ for any $n \in \mathbb{N}.$
Step 3: Check if f is One-one
If $a \neq 0$ (i.e., $f(1) \neq 0$), then for $n_1 \neq n_2$,
$$f(n_1) = n_1 a \quad \text{and} \quad f(n_2) = n_2 a.$$
Since $n_1 \neq n_2,$ it follows that $n_1 a \neq n_2 a$ (because $a \in \mathbb{N}$). Hence $f(n_1) \neq f(n_2),$ implying $f$ is injective (one-one).
Step 4: Analyze the Statement “If f is onto, then f(n) = n”
For $f$ to be onto $\mathbb{N}$, every natural number $m$ must appear as $f(n)$ for some $n\in \mathbb{N}.$ Since $f(n) = n a,$ having $f(n) = n a$ cover all natural numbers requires $a=1.$ Thus,
$$f(n) = n \quad \text{for all } n \in \mathbb{N}.$$
This shows the third statement is true.
Step 5: Consider the Statement “If fog \, \text{is one-one, then g is one-one}”
For the composite function $f \circ g$ to be one-one, it must hold that $g(x_1) \neq g(x_2)$ whenever $x_1 \neq x_2$; otherwise, $f(g(x_1)) = f(g(x_2)).$ Hence if $f \circ g$ is injective and $f$ is injective, then $g$ must also be injective. So the fourth statement is true.
Step 6: Check the Statement “If g is onto, then f \circ g \text{ is one-one}”
Being onto (surjective) for $g$ means every element in the codomain of $g$ has a preimage. However, surjectivity of $g$ alone does not guarantee that $f \circ g$ is injective. We needed $g$ to be one-one or to have other conditions satisfied for $f \circ g$ to be one-one. Hence the statement
“If g is onto, then $f \circ g$ is one-one”
is not necessarily true. This matches the question’s assertion that this particular statement is “NOT true.”
Conclusion
The incorrect statement is indeed: “If g is onto, then $f \circ g$ is one-one.”
