Let f, g : N - N such that f(n + 1) = f(n) + f(a) for all n ∈ N and g be any arbitrary function. Wh

Q: Wait, why is $f(a)$ a constant? Isn't $a$ a variable?

In the equation $f(n+1) = f(n) + f(a)$, the variable is $n$. The value $a$ is a fixed element in the domain $\mathbb{N}$. Since $a$ is fixed, $f(a)$ is a fixed output value, which acts as a constant for all $n$.

Q: Why does $f(n+1) - f(n) = f(a)$ imply $f$ is strictly increasing?

Since the codomain is $\mathbb{N}$, $f(a)$ must be at least $1$. If $f(n+1) - f(n) = C$ where $C \ge 1$, then $f(n+1) = f(n) + C$. Since $C \ge 1$, $f(n+1) > f(n)$ for all $n$, which is the definition of a strictly increasing function.

Q: Why does $g$ being onto not guarantee $f \circ g$ is one-one?

The onto property of $g$ only ensures that every element in the codomain of $g$ is mapped to by at least one element in its domain. It does not prevent multiple elements from mapping to the same value. If $g$ is not one-one, then $f(g(x_1)) = f(g(x_2))$ can occur even if $x_1 \neq x_2$, breaking the one-one property of the composite function.

Question

Let $f, g : N 	o N$ such that $f(n + 1) = f(n) + f(a)$ $\forall n \in N$ and $g$ be any arbitrary function. Which of the following statements is NOT true?

Options:

(A) f is one-one

(B) If fog is one-one, then g is one-one

(C) If g is onto, then fog is one-one

(D) If f is onto, then f(n) = n$\forall n \in N$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are dealing with functions $f, g: \mathbb{N} 	o \mathbb{N}$ and the functional equation:
$$f(n+1) = f(n) + f(a)$$

The variable $n$ ranges over the natural numbers $\mathbb{N}$, and $a$ is a fixed natural number. Consequently, $f(a)$ is a constant. Let us define this constant as $C = f(a)$.

The equation simplifies to:
$$f(n+1) = f(n) + C$$

### Decoding the Functional Equation

This recurrence relation is the hallmark of an **arithmetic progression**. The difference between any two consecutive terms is always the constant $C$.

Since the codomain of $f$ is $\mathbb{N}$, the output $f(a)$ must be a natural number, implying $C \ge 1$. Because $C \ge 1$, the function $f$ is strictly increasing.

As $n$ increases, $f(n)$ must also increase. This confirms that $f$ can never repeat a value, making it a **one-one function**. Thus, Statement 1 is true.

### The Logic of Composition

Consider Statement 2: "If $f \circ g$ is one-one, then $g$ is one-one." This is a fundamental property of function composition.

Suppose $g(x_1) = g(x_2)$. Applying $f$ to both sides yields:
$$f(g(x_1)) = f(g(x_2))$$

By definition, this is $(f \circ g)(x_1) = (f \circ g)(x_2)$. Since $f \circ g$ is one-one, it must be that $x_1 = x_2$. Thus, $g$ is necessarily one-one, and Statement 2 is true.

### The Trap of the Onto Function

Now, we examine Statement 3: "If $g$ is onto, then $f \circ g$ is one-one." We can disprove this using a counter-example.

Suppose $g$ is a function that maps both $1$ and $2$ to $1$, and maps $n$ to $n-1$ for $n > 2$. This function $g$ is onto because it covers all natural numbers, but it is not one-one because $g(1) = g(2) = 1$.

Now, consider the composite function $(f \circ g)(n) = f(g(n))$. For $n=1$ and $n=2$:
$$(f \circ g)(1) = f(g(1)) = f(1)$$
$$(f \circ g)(2) = f(g(2)) = f(1)$$

Since $(f \circ g)(1) = (f \circ g)(2)$, the composite function $f \circ g$ is not one-one. Therefore, Statement 3 is **false**.

### Final Verification

Finally, we evaluate Statement 4: "If $f$ is onto, then $f(n) = n$ for all $n \in \mathbb{N}$." We established that $f(n) = f(1) + (n-1)C$.

If $f$ is onto, it must cover every natural number. If $C > 1$, the function would skip values (e.g., if $C=2$, it would only hit odd or even numbers depending on $f(1)$).

To cover every natural number, we must have $C=1$ and $f(1)=1$. This leads to:
$$f(n) = 1 + (n-1)(1) = n$$

Thus, Statement 4 is **true**.

Animated Solution for Mathematics - Functions: Let $f, g : N \to N$ such that $f (n + 1) = f (n) + f (a)$ $\forall n \in N$ and $g$ be any arbitrary function. Which of the following statements is NOT true?

Select Answer:

Visualized Solution

Analyzing the Functional Equation

Deduce the Nature of $f$

$f$ is Strictly Increasing

Checking Statement 1

Checking Statement 2

Proof for Statement 2

Checking Statement 4

$f (n) = n$

Checking Statement 3

The Counter-example

Conclusion for Statement 3

The Sigma Insight: Classification of Functions

Analyzing the Setup

Decoding the Functional Equation

The Logic of Composition

The Trap of the Onto Function

Final Verification

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