Use proof by contradiction to prove that if for an integer , is even, then so is . [Hint : Assume is not even, that is, it is of the form , for some integer , and then proceed.]

Understand the Question Proof by Contradiction Strategy: To prove a statement by contradiction, we assume the opposite of what we want to prove (the negation of the conclusion) and show that it leads to a logical impossibility or contradicts a given fact. Given: For an integer , is even. To Prove: is even. Assumption: Assume is not even (i.e. is odd, ). If squaring leads to an odd number, it contradicts that is even, proving that must be even. Step 1 · Assume the Opposite Assume to the contrary that is not even. Then must be an odd integer of the form: Step 2 · Evaluate Squaring both sides Let . Since is an integer, is also an integer. This shows that is an odd integer. Step 3 · Reach a Contradiction This contradicts the given fact that is even. Therefore, our assumption that is odd is false. Hence, must be even. Answer > Hence proved that if is even, then is even. Common Mistakes Assuming the Wrong Statement: Assuming is odd instead of assuming is odd. In proof by contradiction, you assume the negation of the conclusion ( is not even), not the given hypothesis ( is even). Forgetting the Integer Condition: Forgetting to state that is an integer, which is essential to formally establish that is an odd integer.

Appendix 1: Proofs in Mathematics | A1.6

Question 3

Use proof by contradiction to prove that if for an integer $a$ , $a^2$ is even, then so is $a$ .

[Hint : Assume $a$ is not even, that is, it is of the form $2n + 1$ , for some integer $n$ , and then proceed.]

Solution

Understand the Question

Proof by Contradiction Strategy: To prove a statement by contradiction, we assume the opposite of what we want to prove (the negation of the conclusion) and show that it leads to a logical impossibility or contradicts a given fact.
Given: For an integer $a$ , $a^2$ is even.
To Prove: $a$ is even.
Assumption: Assume $a$ is not even (i.e. $a$ is odd, $a = 2n + 1$ ). If squaring $a$ leads to an odd number, it contradicts that $a^2$ is even, proving that $a$ must be even.

Step 1 · Assume the Opposite

Assume to the contrary that $a$ is not even.

Then $a$ must be an odd integer of the form: $a = 2n + 1 \quad \text{for some integer } n$

Step 2 · Evaluate $a^2$

Squaring both sides

\begin{aligned} a^2 &= (2n + 1)^2 \\ &= (2n)^2 + 2(2n)(1) + 1^2 \\ &= 4n^2 + 4n + 1 \\ &= 2(2n^2 + 2n) + 1 \end{aligned}

Let $k = 2n^2 + 2n$ . Since $n$ is an integer, $k$ is also an integer.

$a^2 = 2k + 1$

This shows that $a^2$ is an odd integer.

Step 3 · Reach a Contradiction

This contradicts the given fact that $a^2$ is even.

Therefore, our assumption that $a$ is odd is false.

Hence, $a$ must be even.

Answer

Hence proved that if $a^2$ is even, then $a$ is even.

Common Mistakes

Assuming the Wrong Statement: Assuming $a^2$ is odd instead of assuming $a$ is odd. In proof by contradiction, you assume the negation of the conclusion ( $a$ is not even), not the given hypothesis ( $a^2$ is even).
Forgetting the Integer Condition: Forgetting to state that $k = 2n^2 + 2n$ is an integer, which is essential to formally establish that $2k + 1$ is an odd integer.

Question 3

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