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Chapter 8: Problem 20

Use the recursive definition of product, together with mathematical induction, to prove that for all positive integers \(n\), if \(a_{1}, a_{2}, \ldots, a_{n}\) and \(b_{1}, b_{2}, \ldots, b_{n}\) are real numbers, then $$ \prod_{i=1}^{n}\left(a_{i} b_{i}\right)=\left(\prod_{i=1}^{n} a_{i}\right)\left(\prod_{i=1}^{n} b_{i}\right) . $$

Short Answer

Expert verified
In summary, we used mathematical induction to prove that for all positive integers \(n\), if \(a_{1}, a_{2}, \ldots, a_{n}\) and \(b_{1}, b_{2}, \ldots, b_{n}\) are real numbers, then the product of their individual terms is equal to the product of the two sequences: $$ \prod_{i=1}^{n}\left(a_{i} b_{i}\right)=\left(\prod_{i=1}^{n} a_{i}\right)\left(\prod_{i=1}^{n} b_{i}\right). $$ This was achieved by proving the base case (n=1), assuming the statement holds for n=k, and then using the induction hypothesis to prove the statement for n=k+1.

Step by step solution

01

Proving the base case

For n=1, we need to prove that $$ \prod_{i=1}^{1}\left(a_{i} b_{i}\right)=\left(\prod_{i=1}^{1} a_{i}\right)\left(\prod_{i=1}^{1} b_{i}\right). $$ Since there is only one factor for each product, we obtain $$ a_1 b_1 = a_1 \cdot b_1, $$ which is trivially true.
02

Induction hypothesis

Assume that the statement holds for a value n=k, such that $$ \prod_{i=1}^{k}\left(a_{i} b_{i}\right)=\left(\prod_{i=1}^{k} a_{i}\right)\left(\prod_{i=1}^{k} b_{i}\right). $$
03

Proving the induction step

Now, we need to prove that the statement holds for the next value n=k+1. Using the recursive definition of product, we need to show that $$ \prod_{i=1}^{k+1}\left(a_{i} b_{i}\right)=\left(\prod_{i=1}^{k+1} a_{i}\right)\left(\prod_{i=1}^{k+1} b_{i}\right). $$ Considering the induction hypothesis, we can rewrite this as $$ \left(\prod_{i=1}^{k}\left(a_{i} b_{i}\right)\right) \cdot \left(a_{k+1} b_{k+1}\right) = \left(\left(\prod_{i=1}^{k} a_{i}\right)\cdot a_{k+1}\right)\left(\left(\prod_{i=1}^{k} b_{i}\right) \cdot b_{k+1}\right). $$ We can then substitute our induction hypothesis into this equation: $$ \left(\left(\prod_{i=1}^{k} a_{i}\right)\left(\prod_{i=1}^{k} b_{i}\right)\right) \cdot \left(a_{k+1} b_{k+1}\right) =\left(\left(\prod_{i=1}^{k} a_{i}\right)\cdot a_{k+1}\right)\left(\left(\prod_{i=1}^{k} b_{i}\right) \cdot b_{k+1}\right). $$ We can now see that, using the associative property of products: $$ \left(\prod_{i=1}^{k} a_{i}\right) \cdot a_{k+1} \cdot\left(\prod_{i=1}^{k} b_{i}\right) \cdot b_{k+1} =\left(\left(\prod_{i=1}^{k} a_{i}\right)\cdot a_{k+1}\right)\left(\left(\prod_{i=1}^{k} b_{i}\right) \cdot b_{k+1}\right), $$ which simplifies to: $$ \left[\left(\prod_{i=1}^{k} a_{i}\right) \cdot a_{k+1}\right] \cdot \left[\left(\prod_{i=1}^{k} b_{i}\right) \cdot b_{k+1}\right] =\left(\left(\prod_{i=1}^{k} a_{i}\right)\cdot a_{k+1}\right)\left(\left(\prod_{i=1}^{k} b_{i}\right) \cdot b_{k+1}\right). $$ Thus, we have proved the induction step.
04

Conclusion

Applying the Principle of Mathematical Induction, we have demonstrated that for all positive integers \(n\), if \(a_{1}, a_{2}, \ldots, a_{n}\) and \(b_{1}, b_{2}, \ldots, b_{n}\) are real numbers, then $$ \prod_{i=1}^{n}\left(a_{i} b_{i}\right)=\left(\prod_{i=1}^{n} a_{i}\right)\left(\prod_{i=1}^{n} b_{i}\right). $$

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Key Concepts

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

Recursive Definition
A recursive definition forms the backbone of many mathematical proofs, including the one we are examining today. It is a way of defining a sequence or number systematically by expressing each term in relation to previous terms.
This is often used to define complex concepts in simpler terms, where each step builds upon the last. For example, consider the recursive definition of the product of a sequence. If you have a sequence of numbers, the recursive product can be represented as multiplying the first number by the product of the rest of the sequence. Let's look at it step-wise:
  • Base Case: The product of a single element sequence is just the element itself.
  • Recursive Step: If you have defined the product for a sequence of length \(n\), then for a sequence of length \(n+1\), multiply the last element of the sequence by the previously defined product.
So, in our exercise, the recursive definition of products helps us connect the dots and establish the formula for all positive integer values of \(n\). It provides a clear, logical structure to tackle problems piece by piece.
Product of Sequences
Understanding the product of sequences is crucial to this mathematical proof. When talking about the product of sequences, we're referring to multiplying terms from two sequences to get a new product sequence.
In the specific case of the exercise:
  • You have two sequences, each containing \(n\) elements, say \(a_1, a_2, \ldots, a_n\) and \(b_1, b_2, \ldots, b_n\).
  • The goal is to show that the product across these sequences \(\prod_{i=1}^{n}(a_i b_i)\) equals the separate products of each sequence multiplied together: \( (\prod_{i=1}^{n} a_i)(\prod_{i=1}^{n} b_i) \).
This concept is straightforward if you think about distributing the multiplication across each pair of terms from the sequences. For each \(i\), multiply \(a_i\) by \(b_i\), and the entire proof relies on maintaining this consistency across the sequence through the recursive definition and properties like associativity.
Associative Property
The associative property plays a vital role in proofs like this. It states that how you group numbers in a multiplication or addition problem does not change the result.
In essence, regardless of how you parenthesize products, the outcome will remain the same. In our problem:For the induction step:
  • You start with \((\prod_{i=1}^{k} a_i)(a_{k+1})\) and \((\prod_{i=1}^{k} b_i)(b_{k+1})\).
  • Using the associative property, you can group and rearrange these terms without changing the result.
  • This means \(((\prod_{i=1}^{k} a_i) \cdot a_{k+1})((\prod_{i=1}^{k} b_i) \cdot b_{k+1})\) is equivalent to \((\prod_{i=1}^{k+1} a_i)(\prod_{i=1}^{k+1} b_i)\).
Thanks to the associative property, the proof can smoothly transition through each step, allowing us to conclusively prove statements that cover infinitely many cases, showing that the products can be grouped in any way without altering the overall result. This property is a cornerstone in algebra because it provides the flexibility to manipulate expressions for ease of calculation and proof.

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Most popular questions from this chapter

Let \(a_{0}, a_{1}, a_{2}, \ldots\) be defined by the formula \(a_{n}=3 n+1\), for all integers \(n \geq 0\). Show that this sequence satisfies the recurrence relation \(a_{k}=a_{k-1}+3\), for all integers \(k \geq 1\).

A sequence is defined recursively. (a) Use iteration to guess an explicit formula for the sequence. (b) Use strong mathematical induction to verify that the formula of part (a) is correct. \(t_{k}=k-t_{k-1}\), for all integers \(k \geq 1\), \(t_{0}=0\).

Find the first four terms of each of the recursively defined sequences in 1-8. \(d_{k}=k\left(d_{k-1}\right)^{2}\), for all integers \(k \geq 1\) \(d_{0}=3\)

Refer to the sequence of Stirling numbers of the second kind. Use mathematical induction and the recurrence relation found in Example \(8.1 .11\) to prove that for all integers \(n \geq 2, \sum_{k=1}^{n}\left(3^{n-k} S_{k .2}\right)=S_{n+1.3}\)

Counting Sirings: Consider the set of all strings of \(a^{\prime} \mathrm{s}, b^{\prime} \mathrm{s}\), and \(c^{\prime} s\). a. Make a list of all of these strings of lengths zero, one, two, and three that do not contain the pattern aa. b. For each integer \(n \geq 0\), let \(s_{n}=\) the number of strings of \(a^{\prime} s, b^{\prime} s\), and \(c^{\prime} s\) of length \(n\) that do not contain the pattern \(a a\). Find \(s_{0}, s_{1}, s_{2}\), and \(s_{3}\). \(H\) c. Find a recurrence relation for \(s_{0}, s_{1}, s_{2}, \ldots .\) d. Use the results of parts (b) and (c) to find the number of strings of \(a^{\prime} s, b^{\prime} s\), and \(c^{\prime} s\) of length four that do not contain the pattern aa.
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