Question

Prove that (i) if \((\forall \varepsilon>0) a \geq b-\varepsilon,\) then \(a \geq b ;\) (ii) if \((\forall \varepsilon>0) a \leq b+\varepsilon,\) then \(a \leq b\).

Short answer

Both statements assert that limits imply the direct inequality \(a \geq b\) for (i) and \(a \leq b\) for (ii).

Step-by-step solution

Understand the Statements

The problem requires proving two implications using the concept of limits. In essence, it's arguing the idea of bounding a number using a very small value \(\varepsilon\) to infer a tighter inequality between \(a\) and \(b\).

Prove Part (i)

To prove: if \((\forall \varepsilon>0)\ a \geq b-\varepsilon\), then \(a \geq b\).Assume for the sake of contradiction that \(a < b\). Then there exists some \(\varepsilon = b-a > 0\) such that \(b - \varepsilon = a\). However, according to the assumption, for all \(\varepsilon > 0\), \(a \geq b - \varepsilon\). Substituting for \(\varepsilon = b-a\), it leads to \(a \geq a\), which contradicts \(a < b\). Thus, \(a \geq b\) must hold.

Prove Part (ii)

To prove: if \((\forall \varepsilon>0)\ a \leq b+\varepsilon\), then \(a \leq b\).Assume for the sake of contradiction that \(a > b\). Then there exists some \(\varepsilon = a - b > 0\) such that \(b + \varepsilon = a\). However, according to the assumption, for all \(\varepsilon > 0\), \(a \leq b + \varepsilon\). Substituting for \(\varepsilon = a-b\), it leads to \(a \leq a\), which contradicts \(a > b\). Thus, \(a \leq b\) must hold.

Key concepts

Inequality

The concept of inequality is central to mathematics, often describing how one quantity relates to another. Inequalities show that one value is less than or greater than another. In our problem, we are dealing with inequalities in the form of \( a \geq b \) or \( a \leq b \). These symbols mean that \( a \) is larger than or equal to \( b \), or that \( a \) is smaller than or equal to \( b \), respectively.

Inequalities can be strict or non-strict. A strict inequality like \( a < b \) means \( a \) is definitely less than \( b \), while a non-strict inequality \( a \leq b \) means \( a \) could be equal to \( b \). Inequalities are not only limited to real numbers but can involve anything from whole numbers to complex expressions.

Knowing how to prove inequalities is essential. It often involves showing that assuming the opposite leads to a contradiction. Besides this logical approach, solving inequalities typically requires algebraic manipulation and logical reasoning to compare the sizes of different values. Ensuring the conditions fit the required inequalities often involves checking multiple scenarios and understanding the implications of each.

Contradiction Method

Contradiction is a powerful mathematical proof method. To use it, you start by assuming the opposite of what you want to prove. If this assumption leads to a contradiction, then the original statement must be true. This approach is elegant because it often circumvents complex direct proofs.

In the given exercise, when proving that \( a \geq b \) given \( a \geq b - \varepsilon \) for all \( \varepsilon > 0 \), the contradiction method is applied. Suppose \( a < b \). By choosing \( \varepsilon = b - a \), we create a situation that should logically fit the inequality \( a \geq b - \varepsilon \).

• Substitute \( \varepsilon = b - a \)
• Simplify to get \( a \geq a \), which is always true.

However, this leads to both \( a < b \) and \( a \geq a \), a contradiction, implying \( a \geq b \) is indeed correct.

This is often a preferred method as it usually uncovers logical flaws in false assumptions quickly, assuming one can find a clear choice of \( \varepsilon \) as was done here. Such proofs are almost puzzle-like, useful for arguments that might otherwise seem circular or involve complex direct reasoning.

Epsilon-Delta Argument

The epsilon-delta argument is common in calculus for explaining limits and continuity in a rigorous manner. It introduces the notion of \( \varepsilon \) (epsilon) which represents a small positive number, to define how close two quantities need to be for a condition to hold. The delta part of the argument often specifies a range of inputs.

In our exercise, \( \varepsilon \) is used to tighten the bound between \( a \) and \( b \). The proposal is that no matter how small \( \varepsilon \) is, the inequality \( a \geq b - \varepsilon \) (or \( a \leq b + \varepsilon \)) still holds.

• Think of \( \varepsilon \) as a shrinking buffer.
• As \( \varepsilon \to 0^+ \), \( a \) must be very close to \( b \), leading to \( a \geq b \).

This mimics the foundational idea in calculus where you prove continuity or limit behavior by controlling how close outputs get in response to tiny changes in inputs, a crucial skill for handling advanced mathematical problems.

In essence, epsilon-delta provides a framework for discussing how fine we can make the gap between numbers to affirm statements about their relation.