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Chapter 4: Problem 8

Can each natural number greater than or equal to 4 be written as the sum of at least two natural numbers, each of which is a 2 or a 3 ? Justify your conclusion. For example, \(7=2+2+3,\) and \(17=2+2+2+2+3+3+3\).

Short Answer

Expert verified
Yes, every natural number greater than or equal to 4 can be written as the sum of at least two natural numbers, each of which is a 2 or a 3. By using mathematical induction, we confirmed this for the base cases of 4, 5, and 6, and showed that if it is true for any natural number k (k 鈮� 4), it is also true for k+1.

Step by step solution

01

Induction Base Case

First, let's verify the numbers 4, 5, and 6 as base cases for our induction. For 4, we can represent it as the sum of 2s: \(4 = 2 + 2\). For 5, we can represent it as the sum of 2s and 3s: \(5 = 2 + 3\). For 6, we can represent it as the sum of 3s: \(6 = 3 + 3\). These base cases hold true, so we can proceed with the induction step.
02

Induction Hypothesis

Now, we will assume that any natural number k (k 鈮� 4) can be represented as the sum of 2s or 3s.
03

Induction Step

We will show that if the statement is true for k, it is also true for k+1. Since we know k can be represented as the sum of 2s or 3s, we can represent k + 1 in the following ways: 1) If k is represented by the sum of 2s only then, the sum looks like \(k = 2 + 2 + ... + 2\) (a number of 2s). So, for k + 1, we can replace two 2s with three 3s, giving \(k + 1 = 3 + 3 + 2 + 2 + ... + 2\) (k + 1 can be represented as a sum of two 3s and the remaining 2s). 2) If k is represented by the sum of 3s only then, the sum looks like \(k = 3 + 3 + ... + 3\) (a number of 3s). So, for k + 1, we can add one more 2 to the sum, giving \(k + 1 = 2 + 3 + 3 + ... + 3\) (k + 1 can be represented as a sum of one 2 and the remaining 3s). 3) If k is represented by the sum of both 2s and 3s then, the sum looks like \(k = 2 + 2 + ... + 3 + 3 + ...\) (some 2s and 3s in the sum). So, for k + 1, again, we can add one more 2 to the sum, giving \(k + 1 = 2 + 2 + 2 + ... + 3 + 3 + ...\) (k + 1 can be represented as a sum of additional one 2 along with the existing 2s and 3s). In each case, we can represent k + 1 as the sum of 2s or 3s. Thus, by induction, every natural number greater than or equal to 4 can be represented as the sum of 2s or 3s.

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Key Concepts

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

Natural Numbers
Natural numbers, often denoted as the set \( \mathbb{N} \), are the most basic numerical building blocks used in mathematics. They represent whole, positive numbers starting from 1 and continuing indefinitely: 1, 2, 3, ..., and so on. This set does not include zero, negative numbers, fractions, or decimals.

For simplicity's sake, we can think of natural numbers as the count of objects. If you have 3 apples, you are dealing with natural numbers. Furthermore, they are used to number things in real life鈥攍ike numbering the pages of a book or the houses on a street. In our exercise, we deal with natural numbers that are above 3, looking into how we can express these as sums of other natural numbers.
Proof by Induction
Proof by induction is a powerful method used widely in mathematics to establish that a given proposition is true for all natural numbers. It consists of two primary steps. The first is the 'Base Case,' where the proposition is verified for the initial value in the sequence of natural numbers. For instance, if we are considering numbers 4 and onward, then 4 would be our base case.

The second step is the 'Inductive Step,' where we assume that the proposition holds for a particular natural number \( k \) and then prove that it must also hold for the next number, \( k+1 \). If both of these steps are successful, we can conclude that the proposition is true for all natural numbers greater than or equal to our base case, thereby proving an infinite number of cases with a finite amount of work.
Sum Representation
In mathematics, representing a number as a sum of other numbers is a familiar concept. This often involves breaking down a complex number into simpler, more understandable components. Sum representation plays a crucial role in many areas, such as in solving equations or understanding numerical patterns.

In our exercise, we explore the idea of expressing any natural number greater than or equal to 4 as a sum of 2s and 3s. This notion of sum representation not only simplifies numbers into more manageable portions but also demonstrates the additive structure of natural numbers鈥攈ow they can be constructed from simpler elements.
Inductive Reasoning
Inductive reasoning is a logical process in which multiple premises, all believed true or found true most of the time, are combined to obtain a specific conclusion. It is different from deductive reasoning in that it allows for the possibility that the conclusion may not necessarily be 100% guaranteed, even if all of the premises are true.

In our context, inductive reasoning is used to extend a truth from specific cases (the base case and the induction hypothesis) to a general rule (for all natural numbers \( k \) greater than or equal to 4). This reasoning is crucial in proving statements in number theory and arithmetic, as it allows us to make generalized claims based on a pattern observed in specific instances.

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Most popular questions from this chapter

For which natural numbers \(n\) is \(n !>3^{n} ?\) Justify your conclusion.

(a) Can a finite, nonempty set be inductive? Explain. (b) Is the empty set inductive? Explain.

Can each natural number greater than or equal to 6 be written as the sum of at least two natural numbers, each of which is a 2 or a 5 ? Justify your conclusion. For example, \(6=2+2+2,9=2+2+5,\) and \(17=2+5+5+5\).

The quadratic formula can be used to show that \(\alpha=\frac{1+\sqrt{5}}{2}\) and \(\beta=\frac{1-\sqrt{5}}{2}\) are the two real number solutions of the quadratic equation \(x^{2}-x-1=0\). Notice that this implies that $$ \begin{array}{l} \alpha^{2}=\alpha+1, \text { and } \\ \beta^{2}=\beta+1 \end{array} $$ It may be surprising to find out that these two irrational numbers are closely related to the Fibonacci numbers. (a) Verify that \(f_{1}=\frac{\alpha^{1}-\beta^{1}}{\alpha-\beta}\) and that \(f_{2}=\frac{\alpha^{2}-\beta^{2}}{\alpha-\beta}\). (b) (This part is optional, but it may help with the induction proof in part (c).) Work with the relation \(f_{3}=f_{2}+f_{1}\) and substitute the expressions for \(f_{1}\) and \(f_{2}\) from part (a). Rewrite the expression as a single fraction and then in the numerator use \(\alpha^{2}+\alpha=\alpha(\alpha+1)\) and a similar equation involving \(\beta .\) Now prove that \(f_{3}=\frac{\alpha^{3}-\beta^{3}}{\alpha-\beta}\).(c) Use induction to prove that for each natural number \(n,\) if \(\alpha=\frac{1+\sqrt{5}}{2}\) and \(\beta=\frac{1-\sqrt{5}}{2},\) then \(f_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta} .\) Note: This formula for the \(n^{t h}\) Fibonacci number is known as Binet's formula, named after the French mathematician Jacques Binet ( \(1786-1856\) ).

Use mathematical induction to prove each of the following: (a) For each natural number \(n, 3\) divides \(\left(4^{n}-1\right)\). (b) For each natural number \(n, 6\) divides \(\left(n^{3}-n\right)\).
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