Question

Using the Principle of Mathematical Induction, prove each of the following different forms of the principle; (a) Induction with a possibly negative starting point: Suppose that \(S \subseteq \mathbb{Z}\), that some integer \(n_{0} \in S,\) and that for every \(n \in \mathbb{Z},\) if \(n \in S\) and \(n \geq n_{0},\) then \(n+1 \in S .\) Then, for every integer \(n \geq n_{0}\), we have \(n \in S\). (b) Induction downward: Suppose that \(S \subseteq \mathbb{Z}\), that some integer \(n_{0} \in S\), and that for every \(n \in \mathbb{Z},\) if \(n \in S\) and \(n \leq n_{0},\) then \(n-1 \in S .\) Then, for every integer \(n \leq n_{0}\) we have \(n \in S\) (c) Finite induction upward: Let \(n_{0}, n_{1} \in \mathbb{Z}, n_{0} \leq n_{1} .\) Suppose that \(S \subseteq \mathbb{Z}, n_{0} \in S\). and for every \(n \in \mathbb{Z},\) if \(n \in S, n \geq n_{0},\) and \(n

Short answer

Each variant of induction uses different logic to cover their range in \(S\).

Step-by-step solution

Understand the Problem

The task involves proving different forms of the principle of mathematical induction. Each principle has distinct conditions and requirements that need to be explored systematically.

Prove Part (a) - Induction with a Possibly Negative Starting Point

We are given a subset \(S\) of integers such that \(n_0 \in S\). If for every integer \(n \geq n_0\) in \(S\), \(n+1 \in S\), then all integers \(n \geq n_0\) are in \(S\). The proof requires showing the base case \(n_0 \in S\) and using induction to show that if a number \(n\) is in \(S\), then \(n+1\) must also be in \(S\), thus ensuring all integers greater than or equal to \(n_0\) are in \(S\).

Prove Part (b) - Induction Downward

For induction downward, given \(S \subseteq \mathbb{Z}\) with \(n_0 \in S\) and if \(n \leq n_0\) implies that \(n-1 \in S\), then prove every integer \(n \leq n_0\) is in \(S\). Similar to upward induction but works in reverse: starting from \(n_0\), move backward, showing \(n-1\) is in \(S\). This backward method confirms \(S\) contains all integers less than \(n_0\).

Prove Part (c) - Finite Induction Upward

For the finite induction, with \(n_0 \in S\), if \(n < n_1\), then \(n+1 \in S\). Establish \(n_0\) as the base case, then prove secured transitions from \(n\) to \(n+1\) as long as \(n < n_1\). Thus, \(S\) includes all integers from \(n_0\) to \(n_1\).

Prove Part (d) - Infinite Set in Natural Numbers

Given \(S \subseteq \mathbb{N}\) is infinite and \(n+1 \in S\) implies \(n \in S\), prove \(S = \mathbb{N}\). \(S\) being infinite must imply some smallest natural number is missing, contradicting the given condition as it forces its predecessor to be included in \(S\). Thus, exposing such a node contradiction means \(S\) must cover all natural numbers.

Key concepts

Principle of Mathematical Induction

Mathematical induction is a proof technique used to establish that a certain property holds for all integers greater than or equal to a certain number. It's like a domino effect; if you can prove that the first domino falls (the base case) and that if any one domino falls, it knocks the next one over (the inductive step), then you have proven that all the dominoes will fall.

• The base case involves showing that the property is true for an initial integer, usually called \(n_0\).
• The inductive step assumes the property is true for an arbitrary integer \(n\) and then proves it for \(n+1\).

This method helps in proving statements about infinite sets by only checking two conditions, making it extremely powerful and efficient.

Induction with Negative Starting Point

When we talk about induction with negative starting points, we're expanding the traditional induction method to accommodate a wider range of integers, including negative numbers. This variation is especially useful when the sequence or set starts below zero.
You still follow the same general process of induction:

• First, confirm that a specific initial condition or base case holds. This might be for a negative or non-positive integer, \(n_0\).
• Next, assume the property is true for some \(n\) that is greater than or equal to \(n_0\).
• Finally, prove that if the property holds for \(n\), it must also hold for \(n+1\).

This approach ensures that every integer starting from \(n_0\) and moving upward possesses the desired property. It effectively extends the regular induction to include possibly negative values, thereby providing the means to prove assertions that might not have positive starting points.

Induction Downward

Induction downward, also known as reverse induction, is an interesting twist on regular induction. Instead of moving upwards, it works its way down from a known case.
Here’s how it typically unfolds:

• You confirm a base case is true, anchored at a higher point, typically called \(n_0\).
• Assume the property is true for some integer \(n\) that is less than or equal to \(n_0\).
• Then, demonstrate that if it's true for \(n\), it must also be true for \(n-1\).

Using downward induction can be particularly useful in scenarios where you need to prove properties down to negative infinity or when solving problems that naturally decrease over time.

Finite Induction Upward

Finite induction upward focuses on a set range of integers, from a starting point \(n_0\) to an upper boundary \(n_1\). It locks down a property within these limits.
The process involves:

• Verifying the property at the initial point \(n_0\).
• Assuming the property holds for a given integer \(n\) in the range \(n_0 \leq n < n_1\).
• Showing that if it holds for \(n\), then it must hold for \(n+1\).

By confirming continuity from \(n_0\) up to \(n_1\), finite induction upward ensures no gaps in properties within the span of specified integers. This method is ideal when dealing with problems finite in nature, ensuring desired conditions hold all the way through a limited segment of numbers.