Final Answer:
The negation for the given statement is (d) "There is a greatest negative real number." The proof is as follows: Suppose there is a greatest negative real number, say -b. Then, by assumption, a < -b and, for every negative real number x, a < x. However, choosing a new real number, -c, which is both negative and greater than a, the inequality a < -c < 0 contradicts the assumption that a is the greatest negative real number. Therefore, the supposition is false, and the given statement "There is no greatest negative real number" is true.
Step-by-step explanation:
The original statement asserts that there is no greatest negative real number. Its negation claims the existence of a greatest positive real number. To prove the given statement, we utilize proof by contradiction. We assume the negation is true, meaning there is a greatest positive real number, and then demonstrate that this assumption leads to a contradiction.
Let's represent the supposed greatest positive real number as 'a.' The proof starts by assuming that there is a number greater than 'a,' denoted as 'b.' The contradiction arises by showing that 'b' is both negative and greater than 'a,' leading to an inconsistency. This contradiction refutes the initial assumption and confirms the truth of the original statement – that there is no greatest negative real number.
In conclusion, the proof demonstrates that assuming the existence of a greatest positive real number leads to a logical contradiction. Therefore, the original statement, stating the absence of the greatest negative real number, is proven to be true. This type of proof method helps establish mathematical assertions by showing that the opposite assumption leads to inconsistencies.