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Chapter 2: Problem 153

A function \(f\) from the set of natural numbers to integers defined by \(f(n)=\left\\{\begin{array}{ll}\frac{n-1}{2}, & \text { when is odd } \\ & \text { is } \\ -\frac{n}{2}, & \text { when } n \text { is even }\end{array}\right.\) (A) one-one but not onto (B) onto but not one-one (C) one-one and onto both (D) neither one-one nor onto

Short Answer

Expert verified
The answer is (A) one-one but not onto.

Step by step solution

01

Understanding the Function Definition

The function \( f(n) \) is defined piecewise with two cases: - For odd \( n \), \( f(n) = \frac{n-1}{2} \).- For even \( n \), \( f(n) = -\frac{n}{2} \). Natural numbers include all positive integers, starting from 1. We need to determine if the function is injective (one-one) or surjective (onto).
02

Checking for Injectivity (One-One)

To check if \( f \) is injective, we assume \( f(a) = f(b) \) for any \( a, b \) in the domain of \( f \) and see if it implies \( a = b \).1. If both \( a \) and \( b \) are odd: \[ \frac{a-1}{2} = \frac{b-1}{2} \Rightarrow a-1 = b-1 \Rightarrow a = b \]2. If both \( a \) and \( b \) are even: \[ -\frac{a}{2} = -\frac{b}{2} \Rightarrow \frac{a}{2} = \frac{b}{2} \Rightarrow a = b \]3. If \( a \) is odd and \( b \) is even (or vice versa), equations yield a contradiction.Thus, \( f \) is injective.
03

Checking for Surjectivity (Onto)

To check if \( f \) is surjective, for every integer \( y \), we need to find a natural number \( n \) such that \( f(n) = y \).1. For odd \( n \), \( y = \frac{n-1}{2} \). Solving for \( n \): \[ n = 2y + 1 \] must be a natural number, \( 2y + 1 \) being odd fits.2. For even \( n \), \( y = -\frac{n}{2} \). Solving for \( n \): \[ n = -2y \] for \( y \leq 0 \). Non-natural solutions for positive integers.Thus, \( f \) is not surjective, missing positive integers as outputs.
04

Conclusion

The function \( f \) is injective but not surjective. It assigns a unique integer value to each natural number (one-one) but does not cover all integers (not onto). Therefore, the correct answer is (A) one-one but not onto.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Injective Function
An injective function, also known as a one-to-one function, has a unique input-output mapping. This means every element of the domain is paired with a unique element in the codomain. If we consider the function definition provided in the exercise, the function is injective because:
  • For odd numbers, the expression \( f(n) = \frac{n-1}{2} \) yields a unique result for each odd \( n \).
  • For even numbers, the expression \( f(n) = -\frac{n}{2} \) similarly gives a unique result for each even \( n \).
By ensuring that for any two different inputs the function provides different outputs, we demonstrate that the function is one-to-one. No two different elements ever map to the same result.
Surjective Function
A surjective function, or onto function, requires that every possible element in the function's codomain is an image of some element from its domain. In other words, the function reaches every value in its target set. For the provided exercise, the function \( f(n) \) is not surjective because:
  • The function fails to cover all positive integers within the set of integers when mapped from natural numbers. Even when \( n = 2y + 1 \), this results in positives numbers that might suffice some conditions, it doesn't allow all integers as outputs.
  • The condition \( n = -2y \) fails for positive integers.
This indicates that the function cannot achieve every integer value. Thus, it's not surjective.
Piecewise Function
A piecewise function is defined by different expressions over different parts of its domain. It allows for complex functions to be constructed from simpler building blocks. The function \( f(n) \) from the problem is piecewise, as it employs two different expressions based on whether \( n \) is odd or even:
  • For odd \( n \), use \( f(n) = \frac{n-1}{2} \).
  • For even \( n \), use \( f(n) = -\frac{n}{2} \).
This kind of definition is practical because the behavior of the function can differ while still encompassing multiple situations. Each piece of the function must handle specific input scenarios and fits within its designated interval of the domain.
Natural Numbers
Natural numbers are the set of positive integers starting from 1. They are denoted by the symbols \( \mathbb{N} \) and basically consist of numbers like 1, 2, 3, and so on. In the context of the function \( f(n) \), the domain of the function is all natural numbers, meaning it will consider inputs which are 1 and above. This inspires how the outputs of the function are calculated, especially when checking if it's injective or surjective as the sequence doesn't include zero or negative numbers.
Integer Range
The integer range refers to the set of all possible output values of the function, also known as the image. In the problem provided, the function \( f(n) \) produces values from the set of integers, denoted by \( \mathbb{Z} \). According to the piecewise definition, the range consists of particular sets of integers:
  • For odd \( n \), the results are half of certain odd sequences, which might range into negative numbers or zero.
  • For even \( n \), the results are negative even integers or zero.
These outcomes demonstrate that while the domain is limited to positive numbers, the production contributes both negative and potentially zero outcomes in the codomain.

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Most popular questions from this chapter

Domain of definition of the function \(f(x)=\frac{3}{4-x^{2}}+\log _{10}\left(x^{3}-x\right)\), is (A) \((1,2)\) (B) \((-1,0) \cup(1,2)\) (C) \((1,2) \cup(2, \infty)\) (D) \((-1,0) \cup(1,2) \cup(2, \infty)\)

The function $$ f(x)=\sin ^{-1}\left(x-x^{2}\right)+\sqrt{1-\frac{1}{|x|}}+\frac{1}{\left[x^{2}-1\right]} $$ is defined in the interval (where [-] is the greatest integer) (A) \(x \in\left(\sqrt{2}, \frac{1+\sqrt{5}}{2}\right)\) (B) \(x \in\left(1, \frac{1+\sqrt{5}}{2}\right)\) (C) \(x \in\left[\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right]\) (D) \(x \in\left(-\sqrt{2}, \frac{1+\sqrt{5}}{2}\right)\)

The distinct linear function which maps \([-1,1]\) onto \([0,2]\) is (A) \(x-1\) (B) \(x+1\) (C) \(-x+1\) (D) \(-x-1\)

If \(\alpha \in\left(0, \frac{\pi}{2}\right)\) then \(\sqrt{x^{2}+x}+\frac{\tan ^{2} \alpha}{\sqrt{x^{2}+x}}\) is always greater than or equal to (A) \(2 \tan \alpha\) (B) 1 (C) 2 (D) \(\sec ^{2} \alpha\)

Let \(f(x)=\sin x+\cos x, g(x)=x^{2}-1\). Then \(g(f(x))\) is invertible for \(x \in\) (A) \(\left[-\frac{\pi}{2}, 0\right]\) (B) \(\left[-\frac{\pi}{2}, \pi\right]\) (C) \(\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]\) (D) \(\left[0, \frac{\pi}{2}\right]\)
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