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Chapter 4: Problem 16

(a) Prove that if \(n \in \mathbb{N},\) then there exists an odd natural number \(m\) and a nonnegative integer \(k\) such that \(n=2^{k} m\). (b) For each \(n \in \mathbb{N}\), prove that there is only one way to write \(n\) in the form described in Part (a). To do this, assume that \(n=2^{k} m\) and \(n=2^{q} p\) where \(m\) and \(p\) are odd natural numbers and \(k\) and \(q\) are nonnegative integers. Then prove that \(k=q\) and \(m=p\).

Short Answer

Expert verified
In this exercise, we proved that for any natural number \(n\), there exists a unique representation \(n = 2^k m\) where \(m\) is an odd natural number and \(k\) is a nonnegative integer. We demonstrated the existence of this representation by dividing \(n\) by 2 repeatedly until obtaining an odd number, and we showed the uniqueness of the representation by assuming that \(n = 2^k m = 2^q p\) and proving that \(k = q\) and \(m = p\).

Step by step solution

01

Part (a) - Existence

To prove the existence of the representation, let \(n\) be a natural number. 1. If \(n\) is odd, we can write \(n = 2^0n\). In this case, we have \(k = 0\), and \(m = n\), which is also odd. 2. If \(n\) is even, divide \(n\) by 2 until we obtain an odd number. Let the number of divisions be \(k\). Then we can write \(n = 2^k m\), where \(m\) is an odd number that we ultimately get after dividing \(n\) by 2 repeatedly for \(k\) times. Hence, we have proved the existence of the representation \(n = 2^k m\) for any given natural number \(n\).
02

Part (b) - Uniqueness

To prove the uniqueness of the representation, let \(n\) be a natural number and assume that \(n = 2^k m = 2^q p\), where \(m\) and \(p\) are odd natural numbers and \(k\) and \(q\) are nonnegative integers. Now, divide both sides of the equation by \(2^{\min(k, q)}\). Without loss of generality, let's assume \(k \leq q\). Then, we get: \[2^{q-k}m = p.\] Notice that \(p\) is an odd natural number, so it cannot be divisible by 2. But since \(q \geq k\), \(q - k \geq 0\), and \(2^{q-k}\) is an even number (except when \(k=q\), in which case \(2^{q-k} = 1\)). Thus, the left side of the equation is divisible by 2 only if \(k=q\). In this case, \(2^{q-k} = 1\), so \(m=p\). Therefore, we proved that \(k = q\) and \(m = p\), which implies that the representation \(n = 2^k m\) is unique for each natural number \(n\).

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Key Concepts

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

Proof Writing
Understanding how to write a proof is a critical skill in mathematics that allows us to verify the truth of a statement using logical reasoning. A proof is a step-by-step argument that starts from known truths, such as definitions, axioms, or previously established theorems, leading to the statement we want to prove.

In proof writing, clarity and logical coherence are key. Each step must be justified, and we must ensure that no leap in reasoning is made without proper support. A good proof often includes a direct approach, as seen in Part (a) of the exercise, where existence is proven by considering different cases: one where the natural number is odd and another where the number is even. Uniqueness, as demonstrated in Part (b), requires showing that no other valid representation can exist, often by indirect arguments or contradiction.

Moreover, proof writing can take various forms such as direct proof, proof by contrapositive, proof by contradiction, and even proof by induction. The proof provided in the step-by-step solution is an excellent example of a direct proof, where statements are deduced one after the other until the final conclusion is reached.
Odd Natural Number
An odd natural number is an integer greater than zero that cannot be exactly divided by 2. In more formal terms, an odd number can be expressed in the form of \(2k + 1\), where \(k\) is a nonnegative integer. Odd numbers always leave a remainder of 1 when divided by 2.

In the context of our exercise, the odd natural number plays a crucial role in representing any natural number \(n\) in the form \(n = 2^k m\). The odd multiplier \(m\) ensures that all natural numbers, be they even or odd, can be decomposed into a product involving a power of 2 and an odd natural number. This property is instrumental in various areas of mathematics, including number theory and computer science.
Nonnegative Integers
Nonnegative integers are exactly what the name implies: integers that are not negative. The set of nonnegative integers includes zero as well as all positive integers \(1, 2, 3, \ldots\). They are represented in mathematics by the symbol \(\mathbb{N}_0\) or sometimes simply \(\mathbb{N}\), depending on the convention, where \(0\) is sometimes included in the natural numbers.

In our proof, the power \(k\) of 2 is a nonnegative integer, highlighting that we either use a power of 2 (when \(k\) is greater than zero) or simply multiply by 1 (which is \(2^0\), since any number to the power of zero is 1). Incorporating nonnegative integers ensures that we account for all scenarios, including where no division by 2 is necessary (when \(n\) itself is an odd natural number).
Unique Representation
Unique representation in mathematics refers to expressing a mathematical object in a way that is only possible in one form under certain rules. In our number theory example, we are concerned with the unique representation of natural numbers as a product of a power of 2 and an odd natural number.

The proof of uniqueness is vital because it establishes that this representation is deterministic 鈥� there's exactly one way to represent any natural number in this form. This idea of unique representation has far-reaching implications in topics such as the Fundamental Theorem of Arithmetic, where each integer greater than 1 is expressed uniquely as a product of prime numbers.

The step-by-step solution for Part (b) cleverly utilizes the properties of odd numbers and powers of 2 to show that, if two such representations existed, they would necessarily be equivalent, thereby proving uniqueness. This is an essential property that helps in the understanding of the structure of the set of natural numbers.

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Most popular questions from this chapter

Prove or disprove each of the following propositions: (a) For each \(n \in \mathbb{N}, \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}\). (b) For each natural number \(n\) with \(n \geq 3\), $$ \frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\cdots+\frac{1}{n(n+1)}=\frac{n-2}{3 n+3} $$ (c) For each \(n \in \mathbb{N}, 1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+n(n+1)=\frac{n(n+1)(n+2)}{3}\).

The Future Value of an Ordinary Annuity. For an ordinary annuity, \(R\) dollars is deposited in an account at the end of each compounding period. It is assumed that the interest rate, \(i,\) per compounding period for the account remains constant. Let \(S_{t}\) represent the amount in the account at the end of the \(t\) th compounding period. \(S_{t}\) is frequently called the future value of the ordinary annuity. So \(S_{1}=R\). To determine the amount after two months, we first note that the amount after one month will gain interest and grow to \((1+i) S_{1} .\) In addition, a new deposit of \(R\) dollars will be made at the end of the second month. So $$ S_{2}=R+(1+i) S_{1} $$ (a) For each \(n \in \mathbb{N},\) use a similar argument to determine a recurrence relation for \(S_{n+1}\) in terms of \(R, i,\) and \(S_{n}\). (b) By recognizing this as a recursion formula for a geometric series, use Proposition 4.16 to determine a formula for \(S_{n}\) in terms of \(R, i,\) and \(n\) that does not use a summation. Then show that this formula can be written as $$ S_{n}=R\left(\frac{(1+i)^{n}-1}{i}\right) $$ (c) What is the future value of an ordinary annuity in 20 years if \(\$ 200\) dollars is deposited in an account at the end of each month where the interest rate for the account is \(6 \%\) per year compounded monthly? What is the amount of interest that has accumulated in this account during the 20 years?

For the sequence \(a_{1}, a_{2}, \ldots, a_{n}, \ldots,\) assume that \(a_{1}=1, a_{2}=1,\) and that for each \(n \in \mathbb{N}, a_{n+2}=\frac{1}{2}\left(a_{n+1}+\frac{2}{a_{n}}\right)\). (a) Calculate \(a_{3}\) through \(a_{6}\). (b) Prove that for each \(n \in \mathbb{N}, 1 \leq a_{n} \leq 2\).

(a) Verify that \(\left(1-\frac{1}{4}\right)=\frac{3}{4}\) and that \(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)=\frac{4}{6}\). (b) Verify that \(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)=\frac{5}{8}\) and that \(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)\left(1-\frac{1}{25}\right)=\frac{6}{10}\) (c) For \(n \in \mathbb{N}\) with \(n \geq 2,\) make a conjecture about a formula for the product \(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right) \cdots\left(1-\frac{1}{n^{2}}\right) .\) (d) Based on your work in Parts (4a) and (4b), state a proposition and then use the Extended Principle of Mathematical Induction to prove your proposition.

In Section \(3.1,\) we defined congruence modulo \(n\) for a natural number \(n,\) and in Section \(3.5,\) we used the Division Algorithm to prove that each integer is congruent, modulo \(n,\) to precisely one of the integers \(0,1,2, \ldots, n-1\) (Corollary 3.32). (a) Find the value of \(r\) so that \(4 \equiv r(\bmod 3)\) and \(r \in\\{0,1,2\\}\). (b) Find the value of \(r\) so that \(4^{2} \equiv r(\bmod 3)\) and \(r \in\\{0,1,2\\}\). (c) Find the value of \(r\) so that \(4^{3} \equiv r(\bmod 3)\) and \(r \in\\{0,1,2\\}\). (d) For two other values of \(n,\) find the value of \(r\) so that \(4^{n} \equiv r(\bmod 3)\) and \(r \in\\{0,1,2\\}\) (e) If \(n \in \mathbb{N},\) make a conjecture concerning the value of \(r\) where \(4^{n} \equiv r(\bmod 3)\) and \(r \in\\{0,1,2\\} .\) This conjecture should be written as a self-contained proposition including an appropriate quantifier. (f) Use mathematical induction to prove your conjecture.
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