Question 23 of 26intermediate🔧 ApplyLong Answer3 marks

Use proof by contradiction to prove that if for an integer aa, a2a^2 is even, then so is aa.

Correct Answer

We want to prove: If a2a^2 is even, then aa is even.

Proof by contradiction: Let us assume that aa is not even, that is, aa is odd. If aa is odd, then aa can be written in the form a=2n+1a = 2n + 1 for some integer nn.

Now,

a2=(2n+1)2=4n2+4n+1=2(2n2+2n)+1a^2 = (2n + 1)^2 = 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1

This shows that a2a^2 is of the form 2k+12k + 1 (where k=2n2+2nk = 2n^2 + 2n), which means a2a^2 is odd.

But we are given that a2a^2 is even. Thus, we have arrived at a contradiction because a2a^2 cannot be both even and odd simultaneously. Therefore, our assumption that aa is odd is false. Hence, aa must be even.

∴ aa is even.

Exercise: EXERCISE A1.6 | Q: 3 | (Page 21)
For More Understanding

Explanation

This question is from Exercise A1.6, Question 3. The proof follows the standard contradiction method: assume the negation of what we want to prove (that aa is odd), derive a consequence that contradicts the given condition (a2a^2 is even), and conclude the original statement must be true. The hint in the textbook guides students to express an odd number as 2n+12n + 1.

Solution Steps

  1. Step 1: Assume the negation — let aa be odd, so a=2n+1a = 2n + 1 for some integer nn.

  2. Step 2: Calculate a2=(2n+1)2=4n2+4n+1=2(2n2+2n)+1a^2 = (2n+1)^2 = 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1, which is odd.

  3. Step 3: This contradicts the given that a2a^2 is even.

  4. Step 4: Conclude that aa must be even.

PreviousNext