Problem 56
Question
A function \(f\) from the set of natural numbers to integers defined by, $$ f(n)= \begin{cases}\frac{n-1}{2}, & \text { when } n \text { is odd } \\\ -\frac{n}{2}, & \text { when } n \text { is even }\end{cases} $$ (A) neighter one-one nor onto (B) one-one but not onto (C) onto but not one-one (D) one-one and onto both
Step-by-Step Solution
Verified Answer
(D) one-one and onto both.
1Step 1: Analyze the Conditions
Examine the given function which is piecewise defined. It has different expressions for odd and even inputs: - If \( n \) is odd: \( f(n) = \frac{n-1}{2} \) - If \( n \) is even: \( f(n) = -\frac{n}{2} \). These two conditions change how the function behaves across the set of natural numbers.
2Step 2: Check for One-One (Injective) Property
A function is one-one if distinct elements in the domain map to distinct elements in the codomain. For \( n_1 \) and \( n_2 \), check both parts:- If both \( n_1 \) and \( n_2 \) are odd, then \( \frac{n_1 - 1}{2} = \frac{n_2 - 1}{2} \Rightarrow n_1 = n_2 \).- If both are even, \( -\frac{n_1}{2} = -\frac{n_2}{2} \Rightarrow n_1 = n_2 \).- If one is odd and the other is even, try specific examples to see if \( f(n_1) = f(n_2) \): - Example with \( n_1 = 3 \), \( f(3) = 1 \) and \( n_2 = 4 \), \( f(4) = -2 \), which are different; thus suggesting different outputs.Therefore, the function is potentially one-one, but examine further.
3Step 3: Check Onto (Surjective) Property
A function is onto if every integer (codomain) is mapped by some natural number (domain). Analyze the range produced by each case:- Odd \( n \) yields \( \frac{n-1}{2} \), generating non-negative integers.- Even \( n \) yields \( -\frac{n}{2} \), generating negative integers.For example, try making \( f(n) = 0 \). This is only possible if \( n = 1 \) for odd (\( f(1) = 0 \)). Thus, every integer is covered by either the odd or even condition.
4Step 4: Conclusion
By examining injectivity and surjectivity, determine that:
- As distinct inputs map to distinct outputs, and all integers are covered, the function is both one-one and onto.
Therefore, the function satisfies both conditions, further confirming that it is both one-one and onto.
Key Concepts
Injective FunctionSurjective FunctionSet of Natural Numbers
Injective Function
An injective function is often referred to as a one-to-one function. This means that every element in the domain of the function maps to a unique element in the codomain. Let's break it down:
- Similarly, for even numbers \(n_1\) and \(n_2\) using \(-\frac{n}{2}\), the outputs remain distinct. Thus, the function checks out as injective for both odd and even cases.
- If two different elements in the domain go to the same element in the codomain, the function is not injective.
- In simple terms, no two different inputs should give the same output.
- Similarly, for even numbers \(n_1\) and \(n_2\) using \(-\frac{n}{2}\), the outputs remain distinct. Thus, the function checks out as injective for both odd and even cases.
Surjective Function
Surjective functions are also called onto functions. An onto function means every element in the codomain has a pre-image in the domain. In simpler terms, the function covers every possible value in the codomain.
Let’s dive into this a bit deeper:
Let’s dive into this a bit deeper:
- For the function given in the exercise, we have two different expressions based on whether the input is odd or even.
- For odd natural numbers, the function \(\frac{n-1}{2}\) can generate all non-negative integers.
- For even natural numbers, the function \(-\frac{n}{2}\) produces all negative integers.
Set of Natural Numbers
The set of natural numbers is denoted by \(\mathbb{N}\). This set is crucial to understanding functions that map from natural numbers to other sets, like integers.
What makes natural numbers special?
What makes natural numbers special?
- Natural numbers include all positive integers starting from 1, i.e., \(1, 2, 3, ...\).
- They are the building blocks for understanding how functions interact with numbers.
- Although not always explicitly stated, zero can sometimes be considered a natural number too, depending on context.
Other exercises in this chapter
Problem 54
If \(f(x)=\frac{\cos ^{2} x+\sin ^{4} x}{\sin ^{2} x+\cos ^{4} x}\) for \(x \in R\), then \(f(2002)=\) (A) (B) 2 (C) 3 (D) 4
View solution Problem 55
Let \(f(x)=x+1\) and \(\phi(x)=x-2\), then the values of \(x\) satisfying \(|f(x)+\phi(x)|=|f(x)|+|\phi(x)|\) are (A) \((-\infty, 1)\) (C) \((-\infty,-2)\) (B)
View solution Problem 61
If \(2 f(x)+3 f\left(\frac{1}{x}\right)=x^{2}-1\), then \(f(x)\) is (A) a periodic function (B) an even function (C) an odd function
View solution Problem 62
If \(f\left(x_{1}\right)-f\left(x_{2}\right)=f\left(\frac{x_{1}-x_{2}}{1-x_{1} x_{2}}\right)\) for \(x_{1}, x_{2} \in[-1,1]\) then \(f(x)\) is (A) \(\log \left(
View solution