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If $m$ and $n$ are irrational numbers, where $m \neq n$, then $mn$ is also irrational - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 2

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If $m$ and $n$ are irrational numbers, where $m \neq n$, then $mn$ is also irrational. Disprove this statement by means of a counter example. (b) (i) Sketch the gr... show full transcript

Worked Solution & Example Answer:If $m$ and $n$ are irrational numbers, where $m \neq n$, then $mn$ is also irrational - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 2

Step 1

Disprove this statement by means of a counter example.

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Answer

To disprove the statement, we need to find a pair of irrational numbers $m$ and $n$ such that $m \neq n$ , but their product $mn$ is rational. Consider:

Let $m = \sqrt{2}$ and $n = \frac{\sqrt{2}}{2}$ . Both $m$ and $n$ are irrational numbers (since roots of non-perfect squares and non-rational multiples represent irrational numbers). Now calculate:

$mn = \sqrt{2} \times \frac{\sqrt{2}}{2} = \frac{2}{2} = 1,$ which is rational.

Thus, this counterexample disproves the statement.

Step 2

Sketch the graph of $y = |x| + 3$

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Answer

The graph of $y = |x| + 3$ is a V-shaped graph that opens upwards. The vertex of the graph is located at the point (0, 3).

When $x < 0$ , $y = -x + 3$ .
When $x \geq 0$ , $y = x + 3$ .

To sketch:

Plot the vertex at (0, 3).
Draw the line $y = -x + 3$ for $x < 0$ (sloping downwards).
Draw the line $y = x + 3$ for $x \geq 0$ (sloping upwards).

This will create a symmetrical graph about the y-axis.

Step 3

Explain why $|x| + 3 \geq |x + 3|$ for all real values of $x$

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Answer

To explain why this inequality holds, we can analyze two cases for $x$ :

Case 1: When $x \geq 0$ :
- Here, $|x| = x$ . Therefore, the left side becomes $x + 3$ .
- For $|x + 3|$ , since $x + 3 \geq 3 > 0$ , we get $|x + 3| = x + 3$ .
- Thus, $x + 3 \geq x + 3$ is always true.
Case 2: When $x < 0$ :
- Here, $|x| = -x$ . Therefore, the left side becomes $-x + 3$ .
- For $|x + 3|$ $∣ x + 3∣$ , we need to evaluate this term based on whether $x + 3 \geq 0$ $x + 3 \geq 0$ or $x + 3 < 0$ $x + 3 < 0$ :
  - If $x + 3 \geq 0$ , then $|x + 3| = x + 3$ , leading to: $-x + 3 \geq x + 3\Rightarrow 0 \geq 2x\Rightarrow x \leq 0,$ which holds.
  - If $x + 3 < 0$ , then $|x + 3| = -x - 3$ , leading to: $-x + 3 \geq -x - 3\Rightarrow 3 \geq -3,$ which is clearly true.

Overall, in both cases, $|x| + 3 \geq |x + 3|$ holds for all values of $x$ .

If $m$ and $n$ are irrational numbers, where $m \neq n$, then $mn$ is also irrational - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 2

Worked Solution & Example Answer:If $m$ and $n$ are irrational numbers, where $m \neq n$, then $mn$ is also irrational - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 2

Disprove this statement by means of a counter example.

Sketch the graph of $y = |x| + 3$

Explain why $|x| + 3 \geq |x + 3|$ for all real values of $x$

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