Question 11 of 26intermediate🔧 ApplyLong Answer3 marks

Prove that the sum of two consecutive odd numbers is divisible by 4.

Correct Answer

Let the two consecutive odd numbers be 2n+12n + 1 and 2n+32n + 3, where nn is an integer. Any odd number can be expressed in the form 2n+12n + 1, and consecutive odd numbers differ by 2. The sum of these numbers is:

(2n+1)+(2n+3)=4n+4=4(n+1)(2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1)

Since 4(n+1)4(n + 1) clearly has 4 as a factor, where (n+1)(n + 1) is an integer, the sum is divisible by 4. This completes the proof.

Exercise: EXERCISE A1.3 | Q: 1 | (Page 11)
For More Understanding

Explanation

This question is from Exercise A1.3, which asks students to prove statements by listing all steps with reasons. The proof follows the deductive approach shown in Example 11 of the textbook, where each step builds logically from the previous one.

The key insight is representing consecutive odd numbers algebraically (2n+12n+1 and 2n+32n+3), then showing their sum simplifies to 4(n+1)4(n+1), which is clearly divisible by 4.

Solution Steps

  1. Step 1: Let the two consecutive odd numbers be 2n+12n + 1 and 2n+32n + 3, where nn is an integer.

  2. Step 2: Find the sum:

    (2n+1)+(2n+3)=4n+4=4(n+1)(2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1)

  3. Step 3: Since 4(n+1)4(n + 1) has 4 as a factor, the sum is divisible by 4.

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