“If m and n are irrational numbers, where m ≠ n, then mn is also irrational.” Disprove this statement by means of a counter example - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 2

The graph of the function y = |x| + 3 is a V-shaped graph that is symmetrical about the y-axis with its vertex at the point (0, 3). The graph goes upward to the left and right from this vertex, starting at the point (0, 3).

• For x < 0: y = -x + 3 (line with a negative slope)
• For x ≥ 0: y = x + 3 (line with a positive slope)

Hence, the graph is drawn by connecting points which create a V-shape, indicating values of y increasing as x moves away from 0.

To explain why |x| + 3 ≥ |x + 3| for all real x, we can analyze two cases:

• Case 1: When x ≥ 0 In this case, |x| = x, so: $|x| + 3 = x + 3$ Also, |x + 3| = x + 3. Therefore, $|x| + 3 = |x + 3|$ This means that |x| + 3 is equal to |x + 3|.

• Case 2: When x < 0 Here, |x| = -x, so: $|x| + 3 = -x + 3$ For |x + 3|, we need to evaluate:

1. If x + 3 ≥ 0 (which is true when x ≥ -3), then |x + 3| = x + 3, leading to: $-x + 3 ≥ x + 3$, which simplifies to: $0 ≥ 2x$, a condition always true for x < 0.
2. If x + 3 < 0 (x < -3), then |x + 3| = - (x + 3) = -x - 3, leading to: $-x + 3 ≥ -x - 3$, which simplifies to: $3 ≥ -3$, hence always true.

This analysis shows that the inequality holds for all real values of x.