1.01d Proof by contradiction

1. Prove by contradiction that there are no positive integers \(p\) and \(q\) such that

$$4 p ^ { 2 } - q ^ { 2 } = 25$$

1. Given that

$$y = \frac { x - 4 } { 2 + \sqrt { x } } \quad x > 0$$ show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \mathrm {~A} \sqrt { \mathrm { x } } } \quad x > 0$$ where \(A\) is a constant to be found.

1. A student attempts to answer the following question:

Given that \(x\) is an obtuse angle, use algebra to prove by contradiction that $$\sin x - \cos x \geqslant 1$$ The student starts the proof with: Assume that \(\sin x - \cos x < 1\) when \(x\) is an obtuse angle $$\begin{aligned} & \Rightarrow ( \sin x - \cos x ) ^ { 2 } < 1 \\ & \Rightarrow \ldots \end{aligned}$$ The start of the student's proof is reprinted below.
Complete the proof. Assume that \(\sin x - \cos x < 1\) when \(x\) is an obtuse angle $$\Rightarrow ( \sin x - \cos x ) ^ { 2 } < 1$$

1. (i) Kayden claims that

$$3 ^ { x } \geqslant 2 ^ { x }$$ Determine whether Kayden's claim is always true, sometimes true or never true, justifying your answer.
(ii) Prove that \(\sqrt { 3 }\) is an irrational number.

12 Prove by contradiction that 3 is the only prime number which is 1 less than a square number.

15 It is given in lines \(31 - 32\) that the square has the smallest perimeter of all rectangles with the same area. Using this fact, prove by contradiction that among rectangles of a given perimeter, \(4 L\), the square with side \(L\) has the largest area. \section*{END OF QUESTION PAPER}

8 The set \(X\) consists of all \(2 \times 2\) matrices of the form \(\left( \begin{array} { r r } x & - y \\ y & x \end{array} \right)\), where \(x\) and \(y\) are real numbers which are not both zero.

1. (a) The matrices \(\left( \begin{array} { c c } a & - b \\ b & a \end{array} \right)\) and \(\left( \begin{array} { c c } c & - d \\ d & c \end{array} \right)\) are both elements of \(X\). Show that \(\left( \begin{array} { c c } a & - b \\ b & a \end{array} \right) \left( \begin{array} { c c } c & - d \\ d & c \end{array} \right) = \left( \begin{array} { c c } p & - q \\ q & p \end{array} \right)\) for some real numbers \(p\) and \(q\) to be found in terms of \(a , b , c\) and \(d\).
   (b) Prove by contradiction that \(p\) and \(q\) are not both zero.
2. Prove that \(X\), under matrix multiplication, forms a group \(G\). [You may use the result that matrix multiplication is associative.]
3. Determine a subgroup of \(G\) of order 17.

5 In this question you may assume that if \(p\) and \(q\) are distinct prime numbers and \(\mathbf { p } ^ { \alpha } = \mathbf { q } ^ { \beta }\) where \(\alpha , \beta \in \mathbb { Z }\), then \(\alpha = 0\) and \(\beta = 0\).

1. Prove that it is not possible to find \(a\) and \(b\) for which \(\mathrm { a } , \mathrm { b } \in \mathbb { Z }\) and \(3 = 2 ^ { \frac { \mathrm { a } } { \mathrm { b } } }\).
2. Deduce that \(\log _ { 2 } 3 \notin \mathbb { Q }\).

6 Prove by contradiction that \(\sqrt { 7 }\) is irrational.

6 Prove by contradiction that there is no greatest even positive integer.

4 Millie is attempting to use proof by contradiction to show that the result of multiplying an irrational number by a non-zero rational number is always an irrational number. Select the assumption she should make to start her proof.
Tick ( \(\checkmark\) ) one box. Every irrational multiplied by a non-zero rational is irrational. □ Every irrational multiplied by a non-zero rational is rational. □ There exists a non-zero rational and an irrational whose product is irrational. □ There exists a non-zero rational and an irrational whose product is rational. □

10

1. 
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2. 
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3. Given that \(a\) and \(b\) are distinct positive numbers, use proof by contradiction to prove that $$\frac { a } { b } + \frac { b } { a } > 2$$ \section*{END OF SECTION A
   TURN OVER FOR SECTION B}

5 A student's attempt to prove by contradiction that there is no largest prime number is shown below.
If there is a largest prime, list all the primes.
Multiply all the primes and add 1.
The new number is not divisible by any of the primes in the list and so it must be a new prime. The proof is incorrect and incomplete.
Write a correct version of the proof.

1. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

The curve \(C _ { 1 }\) has equation $$y = x ^ { 4 } + 10 x ^ { 2 } + 8 \quad x \in \mathbb { R }$$ The curve \(C _ { 2 }\) has equation $$y = 2 x ^ { 2 } - 7 \quad x \in \mathbb { R }$$ Use algebra to prove by contradiction that \(C _ { 1 }\) and \(C _ { 2 }\) do not intersect.

Use proof by contradiction to show that \(\sqrt { 6 }\) is irrational.

$$f(x) = (8 - 3x)^{\frac{4}{3}} \quad 0 < x < \frac{8}{3}$$

1. Show that the binomial expansion of \(f(x)\) in ascending powers of \(x\) up to and including the term in \(x^3\) is $$A - 8x + \frac{x^2}{2} + Bx^3 + ...$$ where \(A\) and \(B\) are constants to be found. [4]
2. Use proof by contradiction to prove that the curve with equation $$y = 8 + 8x - \frac{15}{2}x^2$$ does not intersect the curve with equation $$y = A - 8x + \frac{x^2}{2} + Bx^3 \quad 0 < x < \frac{8}{3}$$ where \(A\) and \(B\) are the constants found in part (a). (Solutions relying on calculator technology are not acceptable.) [4]

A student was asked to prove by contradiction that "there are no positive integers \(x\) and \(y\) such that \(3x^2 + 2xy - y^2 = 25\)" The start of the student's proof is shown in the box below. Show the calculations and statements that are needed to complete the proof. [4]

A student was attempting to prove that \(x = \frac{1}{2}\) is the only real root of $$x^3 + \frac{1}{4}x - \frac{1}{2} = 0.$$ The attempted solution was as follows. $$x^3 + \frac{1}{4}x = \frac{1}{2}$$ $$\therefore \quad x(x^2 + \frac{1}{4}) = \frac{1}{2}$$ $$\therefore \quad x = \frac{1}{2}$$ or $$x^2 + \frac{1}{4} = \frac{1}{2}$$ i.e. $$x^2 = -\frac{1}{4} \quad \text{no solution}$$ $$\therefore \quad \text{only real root is } x = \frac{1}{2}$$

1. Explain clearly the error in the above attempt. [2]
2. Give a correct proof that \(x = \frac{1}{2}\) is the only real root of \(x^3 + \frac{1}{4}x - \frac{1}{2} = 0\). [3]

The equation $$x^3 + \beta x - \alpha = 0 \quad \text{(I)}$$ where \(\alpha\), \(\beta\) are real, \(\alpha \neq 0\), has a real root at \(x = \alpha\).

1. Find and simplify an expression for \(\beta\) in terms of \(\alpha\) and prove that \(\alpha\) is the only real root provided \(|\alpha| < 2\). [6]

An examiner chooses a positive number \(\alpha\) so that \(\alpha\) is the only real root of equation (I) but the incorrect method used by the student produces 3 distinct real "roots".

1. Find the range of possible values for \(\alpha\). [7]

A circle has centre \(C\) which lies on the \(x\)-axis, as shown in the diagram. The line \(y = x\) meets the circle at \(A\) and \(B\). The midpoint of \(AB\) is \(M\). \includegraphics{figure_6} The equation of the circle is \(x^2 - 6x + y^2 + a = 0\), where \(a\) is a constant.

1. In this question you must show detailed reasoning. Show that the area of triangle \(ABC\) is \(\frac{5}{2}\sqrt{9 - 2a}\). [7]
2. 1. Find the value of \(a\) when the area of triangle \(ABC\) is zero. [1]
   2. Give a geometrical interpretation of the case in part (b)(i). [1]
3. Give a geometrical interpretation of the case where \(a = 5\). [1]

Prove that \(n\) is a prime number greater than \(5 \Rightarrow n^4\) has final digit \(1\) [5 marks]

Prove that the sum of a rational number and an irrational number is always irrational. [5 marks]

A student argues that when a rational number is multiplied by an irrational number the result will always be an irrational number.

1. Identify the rational number for which the student's argument is not true. [1 mark]
2. Prove that the student is right for all rational numbers other than the one you have identified in part (a). [4 marks]

\(a\) and \(b\) are two positive irrational numbers. The sum of \(a\) and \(b\) is rational. The product of \(a\) and \(b\) is rational. Caroline is trying to prove \(\frac{1}{a} + \frac{1}{b}\) is rational. Here is her proof: Step 1 \quad \(\frac{1}{a} + \frac{1}{b} = \frac{2}{a + b}\) Step 2 \quad \(2\) is rational and \(a + b\) is non-zero and rational. Step 3 \quad Therefore \(\frac{2}{a + b}\) is rational. Step 4 \quad Hence \(\frac{1}{a} + \frac{1}{b}\) is rational.

1. 1. Identify Caroline's mistake. [1 mark]
   2. Write down a correct version of the proof. [2 marks]
2. Prove by contradiction that the difference of any rational number and any irrational number is irrational. [4 marks]