Question

Construct functions with the following properties: (a) \(F: \mathbb{N} \rightarrow\) N such that range \((F)=\mathbb{N}\) and, for each \(n \in \mathbb{N},\) there exist exactly two solutions for the equation \(F(x)=n\). (b) \(F: \mathrm{N} \rightarrow \mathbb{N}\) such that, for each \(n \in \mathrm{N}\), there are exactly \(n\) solutions for the equation \(F(x)=n\).

Short answer

(a) Define \( F(2n) = n \) and \( F(2n+1) = n \). (b) Define \( F(m) = n \) for \( \frac{n(n-1)}{2} \leq m < \frac{n(n+1)}{2} \).

Step-by-step solution

Understanding the Problem

We need to construct two functions, each with specific requirements regarding their range and the number of solutions for each value in the range. For part (a), the function should map natural numbers \( \mathbb{N} \) to natural numbers such that each value in its range has exactly two preimages. For part (b), each value \( n \) in the range should have exactly \( n \) preimages.

Construct Function for Part (a)

To satisfy the conditions of part (a), we can define a function \( F: \mathbb{N} \rightarrow \mathbb{N} \) as follows: For every natural number \( n \), let \( F(2n) = n \) and \( F(2n+1) = n \). This ensures that for each \( n \), there are exactly two preimages, specifically \( 2n \) and \( 2n+1 \). Thus, the range of the function covers all natural numbers \( \mathbb{N} \).

Verify Function for Part (a)

Check if every \( n \in \mathbb{N} \) has exactly two solutions for \( F(x) = n \). For each \( n \), \( F(2n) = n \) and \( F(2n+1) = n \), confirming that \( n \) is reached twice, thus satisfying the condition.

Construct Function for Part (b)

To satisfy part (b), we define a different function \( F: \mathbb{N} \rightarrow \mathbb{N} \) where for each \( m \) in the form of \( \frac{n(n-1)}{2} \leq m < \frac{n(n+1)}{2} \), let \( F(m) = n \). This means, for example, for \( n = 1 \), \( 0 \leq m < 1 \), resulting in one solution; for \( n = 2 \), \( 1 \leq m < 3 \), resulting in two solutions, and so on. This construction satisfies the requirement that for each \( n \), there are exactly \( n \) solutions for \( F(x) = n \).

Verify Function for Part (b)

To confirm, observe that for any natural number \( n \), \( F(m) = n \) occurs exactly \( n \) times because \( m \) ranges from \( \frac{n(n-1)}{2} \) to \( \frac{n(n+1)}{2} - 1 \). Each increment in \( n \) introduces exactly \( n \) new values of \( m \), therefore satisfying the conditions set out in part (b).

Key concepts

Natural Numbers

Natural numbers are the set of positive integers starting from 1, 2, 3, and so on. They are an essential concept in mathematics, often denoted by the symbol \( \mathbb{N} \). Since they start from 1 and go onwards indefinitely, they do not include negative numbers or zero. Natural numbers are used for counting and ordering and form the basis of arithmetic.

In mathematical functions, especially in terms of mapping or transformations, natural numbers often serve as the domain or the range. When a function is defined as \( F: \mathbb{N} \rightarrow \mathbb{N} \), it means that both the input and output of the function are natural numbers. This is crucial in understanding how to construct functions with specific properties, such as those in the exercise provided.

When you see exercises asking to map \( \mathbb{N} \) to \( \mathbb{N} \) with certain preimage counts, understanding the behavior and characteristics of natural numbers helps to conceptualize and create such mappings.

Function Range

The range of a function is the set of all possible outputs the function can produce. In the context of discrete functions mapping natural numbers to natural numbers, the range is crucial in determining which values of the domain are mapped to which values in the codomain, i.e., \( \mathbb{N} \).

For example, a function \( F: \mathbb{N} \rightarrow \mathbb{N} \) might be constructed so that every number in the range has a specific number of preimages. Preimages are elements from the domain that map to a particular target in the range.

In Part (a) of the exercise, the task is to ensure every number in the range of the function has exactly two preimages, creating a scenario where for each \( n \), there must be exactly two solutions for \( F(x) = n \). The careful crafting of the function ensures the range consists of all natural numbers, effectively using both even and odd transformations to meet this requirement.

Preimage Solutions

Preimage solutions refer to finding inputs (from a function's domain) that correspond to a particular output (in the function's range). In other words, if you have a function \( F \) and a value \( y \) such that \( F(x) = y \), the preimage solutions are the values of \( x \) that lead to \( y \).

Understanding preimages is vital, especially when assignments in problems involve specific counts of solutions, as in the exercise. For instance, determining that a function has exactly two preimages for any given output involves defining the function such that only two values from the domain map to every individual number in the range.

In Part (b) of the exercise, the challenge is to create a function where the number of preimage solutions equals the number itself. So for each \( n \), the function must present exactly \( n \) input values (or solutions) such that \( F(x) = n \). This nuanced understanding of preimages is key to crafting and verifying that a function meets specific mapping criteria.