This type of exercise will appear frequently in the book. In each case, there is

Question

This type of exercise will appear frequently in the book. In each case, there is a proposed proof of a proposition. However, the proposition may be true or may be false. If a proposition is false, the proposed proof is, of course, incorrect. In this situation, you are to find the error in the proof and then provide a counterexample showing that the proposition is false. If a proposition is true, the proposed proof may still be incorrect. In this case, you are to determine why the proof is incorrect and then write a correct proof using the writing guidelines that have been presented in this book. If a proposition is true and the proof is correct, you are to decide if the proof is well written or not. If it is well written, then you simply must indicate that this is an excellent proof and needs no revision. On the other hand, if the proof is not well written, then you must then revise

Explanation / Answer

The proposition that x(1 – x) ¼ for all real number x is True.

However, the proof is incorrect as the contradiction method is not correctly written. The objective for the contradiction method is to prove that the assumption x(1 – x) > ¼ is not true for all real x. But the proof written here just used a chosen number x = 3 to arrive at a contradiction, which is incorrect. The correct proof is given below.

Proof: We will use contradiction method to prove the proposition. Let us assume that the proposition is false. Then there exists a real number x such that x(1 – x) > ¼. We multiply both sides by 4 to get,

4x(1 – x) > 1

4x – 4x2 > 1

0 > 4x2 – 4x + 1                 

This can be written as, 4x2 – 4x + 1 < 0

Now, 4x2 – 4x + 1 = (2x – 1)2

(2x – 1)2 0 for all x R as square of any term is always non-negative

Hence, 4x2 – 4x + 1 < 0 is not possible for any real x

The contradiction arises due the wrong assumption. Hence, the proposition is true. (Proved)

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