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Chapter 5: Problem 1

There are infinitely many stations on a train route. Suppose that the train stops at the first station and suppose that if the train stops at a station, then it stops at the next station. Show that the train stops at all stations.

Short Answer

Expert verified
By mathematical induction, the train stops at all stations because it stops at the first station, and stopping at any station implies stopping at the next one.

Step by step solution

01

Understand the given information

The train stops at the first station. Additionally, if the train stops at a station, it will also stop at the next station.
02

Define a logical basis

We need to show that the train stops at every station. Let's assume for the purpose of induction that the stations are labeled consecutively as Station 1, Station 2, Station 3, and so on.
03

Use mathematical induction setup

Use the principle of mathematical induction. Start by proving the base case: The train stops at the first station.
04

Base case

The base case is given because it is stated in the problem that the train stops at the first station (Station 1).
05

Inductive step

Assume the train stops at station n. According to the problem, if the train stops at a station n, it will stop at station n+1. Therefore, if the train stops at station n, it stops at station n+1.
06

Conclusion by induction

By the principle of mathematical induction, if the train stops at Station 1 and if stopping at a station implies stopping at the next station, then the train stops at all stations.

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Key Concepts

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

Infinite Sequence
An infinite sequence is a series where the terms continue indefinitely. In this exercise, the train stations can be thought of as forming an infinite sequence: Station 1, Station 2, Station 3, and so on. This means there is no final station, and the train route extends endlessly. Understanding this helps us grasp the problem better. Since the train stops at each consecutive station and there are no end stations, the train must stop at each station in the infinite sequence. This is fundamental to setting up our proof by induction.
Base Case
The base case is the foundation of a proof by induction. Here, we start with Station 1. It is given in the problem that the train stops at the first station.
The base case establishes that our initial assumption is true for the first element (Station 1) of our sequence.
This simple step is crucial because it starts our chain of logical reasoning.
If we can show that the first element is true, we can move on to the next steps in our proof.
Inductive Step
The inductive step is where we prove that if the train stops at any given station, it will stop at the next station as well.
This part of the proof involves assuming that the train stops at station n (our inductive hypothesis).
Then, according to the problem鈥檚 statement, if the train stops at station n, it must also stop at station n+1.
This logical step allows us to extend our proof from one station to the next. Here鈥檚 the process in simple terms:
  • Assume the train stops at station n.
  • Show that it must then stop at station n+1.
By proving this step, we show that all stations in the sequence will be visited by the train.
Proof by Induction
Proof by induction is a mathematical technique used to prove statements about infinite sequences or series. It involves two key steps: the base case and the inductive step.
Here's a quick recap of the process:
  • Base Case: Prove the statement is true for the initial element of your sequence. For our problem, it鈥檚 Station 1.
  • Inductive Step: Assume the statement is true for an arbitrary element (like station n) and then prove it is true for the next element (station n+1).
If both steps hold, we conclude that the statement is true for all elements in the sequence. In our example, this means the train stops at every station in the infinite sequence.
The proof by induction leverages the basis and the inductive step to cover all scenarios in an infinite set, ensuring the accuracy and completeness of our argument.

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Most popular questions from this chapter

Devise a rule of inference for verification of partial correctness of statements of the form $$ \begin{array}{c}{\text { if condition } 1 \text { then }} \\ {S_{1}} \\\ {\text { else if condition } 2 \text { then }} \\ {\qquad \begin{array}{c}{S_{2}} \\ {\vdots} \\ {\text { else }} \\\ {S_{n}}\end{array}}\end{array} $$ where \(S_{1}, S_{2}, \ldots, S_{n}\) are blocks.

Describe a recursive algorithm for multiplying two non- negative integers \(x\) and \(y\) based on the fact that \(x y=2(x \text { . }\) \((y / 2) )\) when \(y\) is even and \(x y=2(x \cdot\lfloor y / 2\rfloor)+x\) when \(y\) is odd, together with the initial condition \(x y=0\) when \(y=0\)

a) Determine which amounts of postage can be formed using just 4 -cent and 11 -cent stamps. b) Prove your answer to (a) using the principle of math- ematical induction. Be sure to state explicitly your inductive hypothesis in the inductive step. c) Prove your answer to (a) using strong induction. How does the inductive hypothesis in this proof differ from that in the inductive hypothesis for a proof using mathematical induction?

Let \(a_{1}, a_{2}, \ldots, a_{n}\) be a list of \(n\) distinct real numbers. How many comparisons are needed to form two sublists from this list, the first containing elements less than \(a_{1}\) and the second containing elements greater than \(a_{1} ?\)

Sometimes we cannot use mathematical induction to prove a result we believe to be true, but we can use mathematical induction to prove a stronger result. Because the inductive hypothesis of the stronger result provides more to work with, this process is called inductive loading. We use inductive loading in Exercise \(74-76\) . Suppose that we want to prove that $$ \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n}} $$ for all positive integers \(n .\) a) Show that if we try to prove this inequality using mathematical induction, the basis step works, but the inductive step fails. b) Show that mathematical induction can be used to prove the stronger inequality $$ \frac{1}{2} \cdot \frac{3}{4} \dots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n+1}} $$ for all integers \(n\) greater than \(1,\) which, together with a verification for the case where \(n=1\) , establishes the weaker inequality we originally tried to prove using mathematical induction.
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