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a) "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
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a) "If m and n are irrational numbers, where m ≠ n, then mn is also irrational." Disprove this statement by means of a counter example. b) (i) Sketch the graph ... show full transcript
Worked Solution & Example Answer:a) "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
Step 1
Disprove this statement by means of a counter example.
Answer
To disprove the claim, we need to find two irrational numbers m and n such that their product mn is rational.
Let m = √2 and n = √2. Both of these numbers are irrational. However, their product:
is a rational number. This serves as a counterexample, thus disproving the statement.
Step 2
Sketch the graph of y = |x| + 3
Answer
The function y = |x| + 3 is a V-shaped graph. It is symmetric with respect to the y-axis. The vertex is at the point (0, 3) and the graph rises linearly on both sides of this point.
When x < 0, the graph slopes upward to the left, and when x > 0, it slopes upward to the right.
To sketch it:
- Start from the point (0, 3).
- Draw lines with a slope of 1 to the left and right.
Step 3
Explain why |x| + 3 ≥ |x + 3| for all real values of x.
Answer
Consider two cases based on the value of x.
- Case 1: If x ≥ 0
In this case, |x| = x, and |x + 3| = x + 3. Therefore:
- Case 2: If x < 0
Here, |x| = -x and |x + 3| = 3 - x when x is greater than -3, or |x + 3| = -x - 3 when x is less than -3. In both situations, we find that |x| + 3 ≥ |x + 3| holds true.
In summary, the inequality |x| + 3 ≥ |x + 3| is verified for all possible values of x.