Consider the following theorem and proof. Theorem : The number 2 is not rational

Question

Consider the following theorem and proof.

Theorem:   The number 2 is not rational number.

Proof: Let's suppose 2 is a rational number.

Then we can write 2 = a/b where a, b are whole numbers, b not zero.

We additionally assume that  a/b is simplified to lowest terms, i.e., a and b have no common factors.

Both of the numbers a and b cannot be even.

From the equality 2 = a/b it follows that 2 = a2/b2 .

So,   a2 = 2 · b2. Showing that the square of a is an even number since it is two times something.

From this we know that a itself is also an even number.

In symbols, a = 2k where k is a natural number.

If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:

Showing that b2 is even because it is 2 times a number.

This implies that b itself is also even.

Which of the following is a true statement about the proof?

The proof structure is generally a proof by contradiction.

This proof uses the contrapositive of the statement to prove the theorem.

This proof structure is generally a direct proof.

This proof uses the inverse and implication structure for proofs.

Explanation / Answer

The correct answer is Option A

The proof structure is generally a proof by contradiction.

What is the Contrdiction?

At the starting of the proof, we assume that sqrt(2) can be written in the lowest forms as a/b. But in the end, we have seen that the both a and b are even which means that they were not in its lowest form as a/b, hence our initial assumption is wrong so by contradiction sqrt(2) is an irrational number

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