Question

Prove the Bernoulli inequalities: For any element \(\varepsilon\) of an ordered field, (i) \((1+\varepsilon)^{n} \geq 1+n \varepsilon\) if \(\varepsilon>-1 ;\) (ii) \((1-\varepsilon)^{n} \geq 1-n \varepsilon\) if \(\varepsilon<1 ; n=1,2,3, \ldots\)

Short answer

Both inequalities are proven by induction for all positive integers \(n\).

Step-by-step solution

Base Case for Part (i)

Verify the inequality for the base case \(n=1\). Substitute \(n = 1\) into the inequality \((1+\varepsilon)^{n} \geq 1+n\varepsilon\):\[(1+\varepsilon)^{1} = 1+\varepsilon \geq 1+1\cdot \varepsilon = 1+\varepsilon\]This holds true, so the base case \(n=1\) is satisfied.

Inductive Step for Part (i)

Assume the inequality \((1+\varepsilon)^k \geq 1+k\varepsilon\) is true for some integer \(k\geq 1\). Now, prove it for \(k+1\):\[(1+\varepsilon)^{k+1} = (1+\varepsilon)^k(1+\varepsilon) \geq (1+k\varepsilon)(1+\varepsilon)\]Expanding the right side using distribution:\[= 1 + k\varepsilon + \varepsilon + k\varepsilon^2 = 1 + (k+1)\varepsilon + k\varepsilon^2\]Since \(k\varepsilon^2 \geq 0\) for \(\varepsilon > -1\), we have:\[(1+\varepsilon)^{k+1} \geq 1 + (k+1)\varepsilon\]Thus, by induction, the inequality holds for all \(n\geq 1\).

Base Case for Part (ii)

Verify the inequality for the base case \(n=1\). Substitute \(n = 1\) into the inequality \((1-\varepsilon)^{n} \geq 1-n\varepsilon\):\[(1-\varepsilon)^{1} = 1-\varepsilon \geq 1-1\cdot \varepsilon = 1-\varepsilon\]This holds true, hence the base case \(n=1\) is satisfied.

Inductive Step for Part (ii)

Assume the inequality \((1-\varepsilon)^{k} \geq 1-k\varepsilon\) is true for some integer \(k\geq 1\). Now, prove it for \(k+1\):\[(1-\varepsilon)^{k+1} = (1-\varepsilon)^{k}(1-\varepsilon) \geq (1-k\varepsilon)(1-\varepsilon)\]Expanding the right-hand side using distribution:\[= 1 - k\varepsilon - \varepsilon + k\varepsilon^2 = 1 - (k+1)\varepsilon + k\varepsilon^2\]Because \(k\varepsilon^2 \geq 0\) for \(\varepsilon < 1\), we get:\[(1-\varepsilon)^{k+1} \geq 1 - (k+1)\varepsilon\]Therefore, by induction, the inequality holds for all \(n\geq 1\).

Key concepts

Mathematical Induction

Mathematical induction is a powerful technique used to prove statements that are true for all natural numbers. It works by first showing that a statement holds for a base case, usually when the variable representing a number is equal to 1. After that, we assume it is true for some arbitrary positive integer, say \(k\), and then prove it holds for \(k + 1\). If both of these steps are successful, we can conclude that the statement is true for all natural numbers.

In the context of Bernoulli inequalities, we start by verifying them for \(n = 1\). This is our base case, and we see that the inequalities naturally hold as both sides are equal. Then, we move on to assume that the inequalities are true for some integer \(k\) and use basic algebraic manipulations to show they hold for \(k + 1\).

This step-by-step proof not only establishes the validity of these inequalities but also highlights the utility of mathematical induction in verifying statements for infinitely many instances. It's a highly structured process, ensuring that we don't miss any cases when proving general mathematical statements.

Inequalities

Inequalities are mathematical expressions involving the symbols \(>\), \(<\), \(\geq\), or \(\leq\). They show the relative size or order of two values. Understanding inequalities involves comparing two expressions and determining which one is larger or if they are equal under certain conditions.

The Bernoulli inequalities are classic examples of how inequalities are used to define relationships between numbers raised to powers. Inequality (i), \((1 + \varepsilon)^{n} \geq 1 + n\varepsilon\), deals with instances where \(\varepsilon > -1\), while inequality (ii), \((1 - \varepsilon)^{n} \geq 1 - n\varepsilon\), focuses on \(\varepsilon < 1\). These inequalities give us a sense of how exponential growth compares to linear growth under specific conditions.

When tackling inequalities, it's crucial to consider all given constraints. In Bernoulli's case, the importance lies in the conditions placed on \(\varepsilon\). Also, achieving equality doesn't stop at just testing the base criteria, but we must ensure the inequality holds as we incrementally increase \(n\), which is efficiently handled by mathematical induction.

Ordered Fields

An ordered field is a field that is equipped with a total order. This order allows for the comparison of any two elements within the set. An ordered field satisfies certain properties, such as:

• For any elements \(a\), \(b\), and \(c\), if \(a > b\), then \(a + c > b + c\).
• If \(a > b\) and \(c > 0\), then \(a \cdot c > b \cdot c\).

These properties allow for the meaningful comparison and ordering of numbers, forming the basis for many mathematical operations and inequalities.

In the case of Bernoulli inequalities, being in an ordered field helps us understand how \(\varepsilon\) behaves relative to other terms and verify that the inequalities hold under given conditions. For instance, the manipulations performed in the step-by-step solution rely heavily on the ordered nature of the field, ensuring that expressions maintain their inequalities as you progress through the calculation. This ordered structure also ensures the predictable behavior of operations such as addition and multiplication within the inequalities.