P2 §1.1 Proof by contradiction

Pure mathematics Year 2

Table of contents

1. Disproof by counter-example
2. Proof by deduction
3. Proof by exhaustion
4. Proof by contradiction

1. Disproof by counter-example

Examples

Exercise -

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2. Proof by deduction

Examples

Exercise -

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3. Proof by exhaution

Examples

Exercise -

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4. Proof by contradiction

A contradiction is a disagreement between two statements, which means that both cannot be true. There are statements that can be proved by contradiction and 'Proof by contradiction' will become a powerful tool for us.

 

1. To prove a statement by contradiction, we start by assuming that it is not true. (A statement that asserts the falsehood of another statement is called the negation of that statement.)
2. We then use logical steps to show that this assumption leads to something impossible, either a contradiction of the assumption, or a contradiction of a fact we know to be true.
3. Since we have followed logical steps after the assumption, the only place where something could go wrong is the assumption. Therefore, we can conclude that our assumption was incorrect and the original statement was true.

 

Examples

Example 1. Prove by contradiction that there is no greatest odd integer.

 

Solution.

 

Example 2. Prove by contradiction that if $n^2$ is even, then $n$ must be even.

 

Solution.

 

Example 3. Prove by contradiction that $ \sqrt{2} $ is an irrational number.

 

• A rational number can be written as $\frac{a}{b}$, where $a$ and $b$ are integers. (Usually, $\mathbb Q$ denotes the set of all rational numbers.)
• An irrational number cannot be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers. (Usually, $ \bar{\mathbb Q} $ or $ \mathbb Q^c $ denotes the set of all rational numbers.)

 

Solution.

 

Example 4. Prove by contradiction that there are infinitely many prime numbers.

 

Solution.

 

Remark. A variation on this proof can be found in Euclid's masterpiece The Elements, a textbook written in approximately 300 BCE but still in use in many schools in the first half of the twentieth century!

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Exercise 1 - Edexcel P2 Exercise 1A

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Exercise 2 - AQA/OCR Book 2 Exercise 1B

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In the previous section, P2 §1.3 Partial fractions, we studies partial fractions when there are two or three linear factors in the denominator.

1.1. Two linear factors

For example:

$$ \begin{align} \frac{ A }{ x + 2 } + \frac{ B }{ x - 1 } &= \frac{ A (x - 1) + B (x + 2) }{ (x + 2) (x - 1) } \\ &= \frac{ (A+B)x + (-A+2B) }{ (x + 2) (x - 1) } \\ &= \frac{ \textrm{Linear} }{ \textrm{Quadratic} } \end{align} $$

1.2. Three linear factors

For example:

$$ \begin{align} \frac{ A }{ x + 2 } + \frac{ B }{ x - 1 } + \frac{ C }{ x - 3 } &= \frac{ A (x - 1)(x - 3) + B (x + 2)(x - 3) + C (x + 2)(x - 1) }{ (x + 2) (x - 1) (x - 3) } \\ &= \frac{ A \left( x^2 - 4x + 3 \right) + B \left( x^2 - x - 6 \right) + C \left( x^2 + x - 2 \right) }{ (x + 2) (x - 1) (x - 3) } \\ &= \frac{ (A+B+C) x^2 + (-4A -B + C) x + (3A -6B -2C) }{ (x + 2) (x - 1) (x - 3) } \\ &= \frac{ \textrm{Quadratic} }{ \textrm{Cubic} } \end{align} $$

1.3. One quadratic factor and one linear factor

We may generalise this to the case where we have one quadratic factor and one linear factor.

 

A first attempt would be to consider: $$ \begin{align} \frac{ A }{ x^2 + x + 1 } + \frac{ B }{ x - 1 } &= \frac{ A (x - 1) + B \left( x^2 + x + 1 \right) }{ \left( x^2 + x + 1 \right) (x - 1) } \\ &= \frac{ B x^2 + (A + B) x + (B - A) }{ \left( x^2 + x + 1 \right) (x - 1) } \end{align} $$ However, there are three coefficients in the numerator expressed in terms of two variables $A$ and $B$. This could be problematic as we may not be able to find $A$ and $B$ that satisfy three conditions. (In the language of simultaneous equations, we need the same number of variables as the number of equations/conditions.)

 

Our second (and correct) attempt is to consider: $$ \begin{align} \frac{ Ax + B }{ x^2 + x + 1 } + \frac{ C }{ x - 1 } &= \frac{ (Ax + B) (x - 1) + C \left( x^2 + x + 1 \right) }{ \left( x^2 + x + 1 \right) (x - 1) } \\ &= \frac{ ( A + B + C) x^2 + (-A + B + C) x + (- B + C) }{ \left( x^2 + x + 1 \right) (x - 1) } \end{align} $$ where the number of equations and that of variables are equal. The final result is of the form $$ \frac{ Ax + B }{ \textrm{Quadratic} } + \frac{ C }{ \textrm{Linear} } = \frac{ \textrm{Quadratic} }{ \textrm{Quadratic} \times \textrm{Linear} } = \frac{ \textrm{Quadratic} }{ \textrm{Cubic} } $$

 

1.4. A general case

We may generalise this further to: $$ \begin{align} \frac{ \textrm{Polynomial of order $n$} }{ \textrm{Polynomial of order $n+1$} } = \frac{ \textrm{Polynomial of order $p-1$} }{ \textrm{Polynomial of order $p$} } + \frac{ \textrm{Polynomial of order $q-1$} }{ \textrm{Polynomial of order $q$} } \end{align} $$ where $p+q=n+1$.

• The polynomial of order $n$ has $n+1$ coefficients.
• The polynomial of order $p-1$ has $p$ coefficients.
• The polynomial of order $q-1$ has $q$ coefficients.
• Thus, the number of coefficients on the LHS, $n+1$, matches the number of variables on the RHS, $p+q$.

An example: $$ \begin{align} \frac{ 11x^4 - 7 x^3 + 5 x^2 + x - 3 }{ \left( x^3 + x^2 + x + 1 \right) \left( x^2 + x + 1 \right) } &= \frac{ A x^2 + B x + C }{ x^3 + x^2 + x + 1 } + \frac{ D x + E }{ x^2 + x + 1 } \end{align} $$

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2. Partial fractions with repeated factors

We will study partial fractions with repeated factors with an example.

 

Step 1. For partial fractions with repeated factors, it is essentially the case where we have a quadratic factor and a linear factor in the denominator. For example, $$ \begin{align} \frac{ 2x + 9 }{ (x - 5)(x + 3)^2 } &= \frac{ 2x + 9 }{ (x - 5) \left( x^2 + 6x + 9 \right) } \\ &= \frac{A}{x-5} + \frac{Bx + C}{x^2 + 6x + 9} \\ &= \frac{ A (x + 3)^2 + (Bx + C) (x - 5) }{ (x - 5) \left( x^2 + 6x + 9 \right) } \end{align} $$

By the method of substitution, we find: $$ \begin{align} \textrm{(i)}&& x &= -3 &&\Rightarrow& & -8(-3B + C) = 2(-3) + 9 = 3 \\ \\ \textrm{(ii)}&& x &= 5 &&\Rightarrow& & 64A = 2(5) + 9 = 19 \\ \\ \textrm{(iii)}&& x &= 0 &&\Rightarrow& & 9A - 5C = 9 \end{align} $$ We see that:

• Unlike linear factors, the coefficients appear together and we need to treat them as simultaneous equations.
• The numerator on the RHS indicates the first two convenient choices but not the third one. We may choose any convenient number for it, usually $x=0,\pm 1$ and so on.

This gives

$$ \begin{align} A &= \frac{19}{64} \\ \\ C &= \frac{9(A-1)}{5} = \frac95 \left( - \frac{45}{64} \right) = - \frac{81}{64} \\ \\ B &= - \frac13 \left( -\frac38 - C \right) = - \frac13 \left( -\frac{24}{64} + \frac{81}{64} \right) = - \frac{19}{64} \end{align} $$ and thus, for partial fractions, we obtain : $$ \begin{align} \Rightarrow \qquad \frac{ 2x + 9 }{ (x - 5)(x + 3)^2 } &= \frac{19}{64(x-5)} - \frac{19x + 81}{ 64( x + 3 )^2 } \end{align} $$

 

Step 2. We can simplify the second term with the repeated factor as follows. $$ \begin{align} \frac{19x + 81}{ 64( x + 3 )^2 } &= \frac{19(x+3-3)+ 81}{ 64( x + 3 )^2 } \\ &= \frac{19(x+3) + 24}{ 64( x + 3 )^2 } \\ &= \frac{19}{64(x+3)} + \frac{3}{8(x+3)^2} \end{align} $$ and thus the partial fractions now read: $$ \begin{align} \frac{ 2x + 9 }{ (x - 5)(x + 3)^2 } &= \frac{19}{64(x-5)} - \frac{19}{64(x+3)} - \frac{3}{8(x+3)^2} \end{align} $$

 

Step 3. In general, for partial fractions with repeated factors, we find: $$ \begin{align} \frac{ ax^2 + bx + c }{ (x - 5) (x + 3)^2 } = \frac{ A }{ x - 5 } + \frac{ B }{ (x + 3) } + \frac{ C }{ (x + 3)^2 } \end{align} $$

A generalisation

If more factors are repeated more times, e.g. $$ \begin{align} \frac{ 2x^4 + 9x - 7 }{ (x - 5)^2 (x + 3)^3 } &= \frac{ Ax + b }{ (x - 5)^2 } + \frac{ Cx^2 + dx + e }{ (x + 3)^3 } \\ &= \frac{ A(x-5) + 5a + b }{ (x - 5)^2 } + \frac{ C(x+3)^2 + D(x+3) + E }{ (x + 3)^3 } \\ &= \frac{ A }{ x - 5 } + \frac{ B }{ (x - 5)^2 } + \frac{ C }{ x + 3 } + \frac{ D }{ x + 3)^2 } + \frac{ E }{ (x + 3)^3 } \end{align} $$

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3. Examples

Example 1. Show that $$ \frac{ 11x^2 + 14x + 5 }{ (x+1)^2 (2x+1) } $$ can be written in the form $$ \frac{ A }{ x + 1 } + \frac{ B }{ (x+1)^2 } + \frac{ C }{ 2x + 1 }, $$ where $A,B$ and $C$ are constants to be found.

 

Solution.

 

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4. Edexcel P2 Exercise 1E

Question 1. (E) $$ \textrm f(x) = \frac{ 3x^2 + x + 1 }{ x^2 (x+1) }, \qquad x \ne 0, \quad x \ne -1. $$ Given that $ \textrm f(x) $ can be expressed in the form $$ \frac{ A }{ x } + \frac{ B }{ x^2 } + \frac{ C }{ x + 1 }, $$ find the values of $A,B$ and $C$. [4 marks]

 

Solution.

 

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Question 2. (E) $$ \textrm g(x) = \frac{ - x^2 - 10x - 5 }{ x^2 (x+1) }, \qquad x \ne -1, \quad x \ne 1. $$ Find the values of the constants $D,E$ and $F$ such that $$ \textrm g(x) = \frac{ D }{ x+1 } + \frac{ E }{ (x+1)^2 } + \frac{ F }{ x - 1 }. $$ [4 marks]

 

Solution.

 

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Question 3. (E) Given that, for $ x < 0 $, $$ \frac{ 2x^2 + 2x - 18 }{ x(x-3)^2 } \equiv \frac{P}{x} + \frac{Q}{x-3} + \frac{R}{(x-3)^2}, $$ where $P,Q$ and $R$ are constants, find the values of $P,Q$ and $R$. [4 marks]

 

Solution.

 

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Question 4. (E) Show that $$ \frac{ 5x^2 - 2x - 1 }{ x^3 - x^2 } $$ can be written in the form $$ \frac{C}{x} + \frac{D}{x^2} + \frac{E}{x-1}, $$ where $C,D$ and $E$ are constants to be found. [4 marks]

 

Solution.

 

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Question 5. (E) $$ \textrm p(x) = \frac{ 2x }{ (x+2)^2 }, \qquad x \ne -2. $$ Find the values of the constants $A$ and $B$ such that $$ \textrm p(x) = \frac{A}{x+2} + \frac{B}{(x+2)^2} $$ [4 marks]

 

Solution.

 

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Question 6. (E) $$ \frac{ 10x^2 - 10x + 17 }{ (2x+1)(x-3)^2 } \equiv \frac{A}{2x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}, \qquad x > 3. $$ Find the values of the constants $A,B$ and $C$. [4 marks]

 

Solution.

 

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Question 7. (E) Show that $$ \frac{ 39x^2 + 2x + 59 }{ (x+5)(3x-1)^2 } $$ can be written in the form $$ \frac{A}{x+5} + \frac{B}{3x-1} + \frac{C}{ (3x-1)^2 }, $$ where $A,B$ and $C$ are constants to be found. [4 marks]

 

Solution.

 

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Question 8. (P) Express the following as partial fractions: $$ \begin{align} &\textbf{(a)} \qquad \frac{ 4x + 1 }{ x^2 + 10x + 25 } \\ \\ &\textbf{(b)} \qquad \frac{ 6x^2 - x + 2 }{ 4x^3 - 4x^2 + x } \end{align} $$

 

Solution.