Question

Suppose \(x_{k}

Short answer

Both (a) and (b) are proven by induction on \(n\).

Step-by-step solution

Base Case for Sum Inequality

To prove (a), starting with the base case where \(n = 1\). Given \(x_{1} < y_{1}\), it is clear that \(\sum_{k=1}^{1} x_{k} = x_{1} < y_{1} = \sum_{k=1}^{1} y_{k}\). Thus, the base case holds.

Inductive Step for Sum Inequality

Assume for an arbitrary positive integer \(n\) that \(\sum_{k=1}^{n} x_{k} < \sum_{k=1}^{n} y_{k}\). Consider \(n+1\): we want to show that \(\sum_{k=1}^{n+1} x_{k} < \sum_{k=1}^{n+1} y_{k}\). Using the assumption, we have, \(\sum_{k=1}^{n+1} x_{k} = \sum_{k=1}^{n} x_{k} + x_{n+1} < \sum_{k=1}^{n} y_{k} + y_{n+1} = \sum_{k=1}^{n+1} y_{k}\) because \(x_{n+1} < y_{n+1}\). This completes the induction for part (a).

Base Case for Product Inequality

To prove (b), starting with the base case where \(n = 1\). Given \(x_{1} < y_{1}\) and both are positive (as stated), it follows that \(\prod_{k=1}^{1} x_{k} = x_{1} < y_{1} = \prod_{k=1}^{1} y_{k}\). Thus, the base case holds.

Inductive Step for Product Inequality

Assume for an arbitrary positive integer \(n\) that \(\prod_{k=1}^{n} x_{k} < \prod_{k=1}^{n} y_{k}\), given each \(x_{k} < y_{k}\) and positive. For \(n+1\), our goal is to prove \(\prod_{k=1}^{n+1} x_{k} < \prod_{k=1}^{n+1} y_{k}\). We know, \(\prod_{k=1}^{n+1} x_{k} = \prod_{k=1}^{n} x_{k} \cdot x_{n+1} < \prod_{k=1}^{n} y_{k} \cdot y_{n+1} = \prod_{k=1}^{n+1} y_{k}\), by using the inductive hypothesis and the fact that \(x_{n+1} < y_{n+1}\) with both \(x_{n+1}, y_{n+1} > 0\). This completes the proof for part (b).

Key concepts

Ordered Field

An ordered field is a set where you can do basic arithmetic operations like addition and multiplication, and where you can also compare elements using inequalities. Examples include the set of real numbers and rational numbers. In an ordered field, we have a special order relation, often denoted by "<", which allows us to compare the sizes of different elements. For instance, you can say something like \( x < y \) within the field. This concept is foundational because it assures that the usual rules for inequalities hold when you're adding or multiplying elements. Understanding ordered fields helps in establishing and proving mathematical properties using inequalities.

Sum Inequality

The sum inequality states that if you have multiple terms and each term in one sequence is less than the corresponding term in another sequence, their totals will follow the same inequality rule. In simpler terms, if

• \( x_1 < y_1 \)
• \( x_2 < y_2 \)
• ...,
• \( x_n < y_n \)

then the sum of all \( x_i \) from 1 to \( n \) will be less than the sum of all \( y_i \) from 1 to \( n \). We prove this using the principle of mathematical induction by starting with the simplest case and building up a general case.
This is vital because it ensures that if every step along the way is smaller, the entire path (or sum) will consistently remain smaller too.

Product Inequality

The product inequality involves multiplying sequences of terms such that each term in one sequence is less than the corresponding term in the other sequence, while all terms are positive. When this is true, the product of one sequence will be less than the product of the other. So if

• \( x_1 < y_1 \)
• \( x_2 < y_2 \)
• ...,
• \( x_n < y_n \)

and all these values are greater than zero, then \( x_1 \times x_2 \times ... \times x_n < y_1 \times y_2 \times ... \times y_n \). The positivity condition ensures that no negative value flips the inequality sign.
This principle confirms that as long as each component in a sequence of multiplicands is smaller and positive, their overall product will be smaller too.

Inductive Hypothesis

The inductive hypothesis is a key component in the proof by induction. It involves making an assumption that a given statement is true for some arbitrary case \( n = k \). For example, if we're proving a statement for all natural numbers \( n \), we assume it's true for \( n = k \) in the inductive hypothesis. This assumption forms the bridge from the base case to the case \( n = k+1 \).
In our exercise, we used the inductive hypothesis to show that:

• For the sum inequality, if it's true up to \( n \), then it must be true for \( n+1 \) by adding the next smaller corresponding terms.
• For the product inequality, if the statement is true up to \( n \), then multiplying through with the next smaller and positive corresponding terms keeps the product inequality intact.

This logic elegantly extends truths in sequences across all natural numbers, effectively validating properties for infinite sets.