Let us take numbers 5, 6, 7, 8: Now, 5 + 6 + 7 + 8 = 26 5 + 6 + 7 – 8 = 10 5 + 6 – 7 + 8 = 12 5 + 6 – 7 – 8 = -4 5 – 6 + 7 + 8 = 14 5 – 6 + 7 – 8 = -2 5 – 6 – 7 + 8 = 0 5 – 6 – 7 – 8 = -16 Conclusion: Regardless of the starting consecutive numbers, the same pattern of even results emerges when using all possible combinations of plus and minus signs. This is because the sum and difference of consecutive numbers will always result in an even number. Intext Questions Using our understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers. (Page 115) 1. 2a + 2b 2. 3g + 5h 3. 4m + 2n 4. 2u – 4v 5. 13k – 5k 6. 6m – 3n 7. x 2 + 2 8. b 2 + 1 9. 4k × 3j Solution: 1. Analyse the expression 2a + 2b 2a is even because it’s a multiple of 2. 2b is even because it’s a multiple of 2. The sum of two even numbers is even, so 2a + 2b is always even. 2. Analyse the expression 3g + 5h If g is odd and h is odd, then 3g is odd and 5h is odd. The sum of two odd numbers is even, e.g., 3 × 1 + 5 × 1 = 8. If g is even and h is even, then 3g and 5h are even. The sum of two even numbers is even, e.g., 3 × 2 + 5 × 2 = 16. If g is odd and h is even, then 3g is odd and 5h is even. The sum of an odd and an even number is odd, e.g., 3 × 1 + 5 × 2 = 13 Thus, 3g + 5h is not always even. 3. Analyse the expression 4m + 2n 4m is even because it’s a multiple of 2. 2n is even because it’s a multiple of 2. The sum of two even numbers is even, so 4m + 2n is always even. 4. Analyse the expression 2u – 4v 2u is even because it’s a multiple of 2. 4v is even because it’s a multiple of 2. The difference of two even numbers is even, so 2u – 4v is always even. 5. Analyse the expression 13k – 5k This simplifies to 8k. 8k is even because it’s a multiple of 2. Thus, 13k – 5k is always even. 6. Analyse the expression 6m – 3n 6m is always even. 3n can be odd (if n is odd) or even (if n is even). If n is odd, 3n is odd, and the difference of an even and an odd number is odd, e.g. 6 × 1 – 3 × 1 = 3. Thus, 6m – 3n is not always even. 7. Analyse the expression x 2 + 2 x 2 is odd or even, both. If x 2 is even, x 2 + 2 becomes even. If x 2 is odd, x 2 + 2 becomes odd. Thus, x 2 + 2 is not always even. 8. If b is even, b2 is even, and b 2 + 1 odd, e.g., 2 2 + 1 = 5. If b is odd, b 2 is odd, and b 2 + 1 is even, e.g, 3 2 + 1 = 10. Thus, b 2 + 1 is not always even. 9. Analyse the expression 4k × 3j This simplifies to 12kj. 12kj is even because it’s a multiple of 2. Thus, 4k + 3j is always even. Figure It Out (Pages 122-123)