Question

Decide which of the following are true statements. Provide a short justification for those that are valid and a counterexample for those that are not:

\begin{array} { l } { \text { (a) Two real numbers satisfy } a < b \text { if and only if } a < b + \epsilon \text { for every } \epsilon > 0 \text { . } } \\ { \text { (b) Two real numbers satisfy } a < b \text { if } a < b + \epsilon \text { for every } \epsilon > 0 \text { . } } \\ { \text { (c) Two real numbers satisfy } a \leq b \text { if and only if } a < b + \epsilon \text { for every } \epsilon > 0 \text { . } } \end{array}
(a) Two real numbers satisfy a0 . (b) Two real numbers satisfy a0 . (c) Two real numbers satisfy a≤b if and only if a0 . ​

Answer 1

Statement (a) is false with a counterexample provided, while statements (b) and (c) are true with justifications based on the properties of real numbers and inequalities.

Step-by-step explanation:

Let's address each statement one by one:

(a) False. While it is true that if a is less than b, adding any positive ε to b will still result in a being less than b+ε, the reverse is not necessarily true. As a counterexample, take a = 1 and b = 1. For any ε > 0, a < b + ε holds true, yet a < b is false since both are equal.

(b) True. If a < b + ε for every ε > 0, then it implies a must be strictly less than b. This is because if a were equal to or greater than b, there would exist some ε < a - b (assuming a > b) such that a >= b + ε which contradicts the given condition.

(c) True. The statement essentially defines the non-strict inequality a ≤ b. It's true that if a ≤ b, then a will always be less than b plus any positive ε, including when a = b.