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Proof by Contradiction (Edexcel A Level Maths: Pure): Exam Questions

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1
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State whether the following are rational or irrational quantities.

For those that are rational, write them in the form begin mathsize 16px style a over b end style, where a space and space b are integers and begin mathsize 16px style a over b end style is a fraction in its simplest terms.

(i) square root of 2

(ii) ln space 3

(iii) begin mathsize 16px style fraction numerator 4 square root of 2 over denominator square root of 18 end fraction end style

(iv) fraction numerator 3 space ln space 2 over denominator ln space 32 end fraction

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2
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Two consecutive integers are given by n and open parentheses n plus 1 close parentheses.

Use algebra to prove by contradiction that the sum of two consecutive integers is odd.

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3
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Let open parentheses 2 n plus 1 close parentheses and open parentheses 2 m plus 1 close parentheses represent two different odd numbers, where n comma space m element of straight natural numbers and n not equal to m.

Use algebra to prove by contradiction that the product of two different odd numbers is odd.

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4
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An even number is given by x equals 2 p, where p element of straight natural numbers.

Use algebra to prove by contradiction that x squared is even.

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1
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Given that x squared is odd, use proof by contradiction to show, using algebra, that x is odd.

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2a
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A student is attempting to answer the following exam question:

“Prove by contradiction that square root of 2 is an irrational number. You may use without proof the fact that if a number n squared is even, then n must also be even.”

The student’s proof is as follows:

Line 1:

Assume square root of 2 is a rational number.  Therefore, it can be written in the form begin mathsize 16px style square root of 2 equals a over b end style, where a space and space b are integers with b not equal to 0, and where a and b have no common factors.

Line 2:

Squaring both sides gives 4 equals a squared over b squared

Line 3:

Multiply both sides by b squared gives a squared equals 2 b squared

Line 4:

 

Line 5:

This means a can be written as a equals 2 m, for some integer m

Line 6:

Squaring gives a squared equals left parenthesis 2 m right parenthesis squared equals 4 m squared

Line 7:

Substituting a squared equals 4 m squared into a squared equals 2 b squared gives space 4 m squared equals 2 b squared

Line 8:

Dividing both sides by 2 gives 2 m squared equals b squared

Line 9:

This shows that b squared is even, and therefore b must be even

Line 10:

It has been shown that both a and b are even, so they share a common factor of 2.

Line 11:

This is a contradiction of the assumption that a and b have no common factors.

Line 12:

Therefore, square root of 2 is irrational.

There is an error within the first three lines of the proof.

Find the error and write down the correct line of the proof.

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2b
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Line 4 of the proof is missing.

Complete this line of the proof.

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3
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Prove by contradiction that a triangle cannot have more than one obtuse angle.

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4
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"There are an infinite number of positive multiples of 10."

A proof by contradiction starts as follows:

Proof

Assume there are a finite number of positive multiples of 10.

This means there is a largest multiple of 10, written as 10 k, where k element of straight natural numbers.

Consider the number 10 k plus 10.

Write the statements needed to complete the proof.

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5
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Given that x cubed is odd, use proof by contradiction to show, using algebra, that x is odd.

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1
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Prove by contradiction that there are an infinite number of even numbers.

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2
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Prove by contradiction that square root of 11 is an irrational number.  You may use without proof the fact that if n squared is a multiple of 11, then n is a multiple of 11.

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3
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Below is a proof by contradiction that there is no largest multiple of 7.

Line 1:

Assume there is a number, S, say, that is the largest multiple of 7

Line 2:

S equals 7 k

Line 3:

Consider the number S plus 7

Line 4:

S plus 7 equals 7 k plus 7

Line 5:

therefore S plus 7 equals 7 left parenthesis k plus 1 right parenthesis

Line 6:

So S plus 7 is a multiple of 7

Line 7:

This is a contradiction to the assumption that S is the largest multiple of 7

Line 8:

Therefore, there is no largest multiple of 7

Both lines 2 and 6 are incomplete.

Complete these lines of the proof.

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4
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Prove by contradiction that there are an infinite number of powers of 2.

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5
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Given that x to the power of n is odd where n is a positive integer, use proof by contradiction to show, using algebra, that x is odd.

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1
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Prove by contradiction that there are an infinite number of prime numbers.

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2
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Below is a proof by contradiction that log subscript 2 7 is irrational.

Line 1: 

Assume log subscript 2 7 is a rational number.  Therefore it can be written in the form log subscript 2 7 equals a over b, where a and b are integers with no common factors, and b not equal to 0. Note that a over b greater than 1 (as a over b equals log subscript 2 7 greater than log subscript 2 2 equals 1) so we can assume that a greater than b greater than 0.

Line 2: 

Rearranging log subscript 2 7 equals a over b gives 2 to the power of a over b end exponent equals 7

Line 3: 

Raising both sides to the power b gives open parentheses 2 to the power of a over b end exponent close parentheses to the power of b equals 7 to the power of b

Line 4: 

Line 5: 

Line 6: 

The only possible powers are zeros,  a equals b equals 0, but this is a contradiction of the original assumption

Line 7: 

Therefore log subscript 2 7 is irrational

Lines 4 and 5 are missing.

Complete these lines of the proof.

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3
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Without solving the equation directly, use algebra to prove by contradiction that the solutions to the equation 

3 x squared plus 10 x minus 8 equals 0

cannot be written in the form x equals a over b  where a and b are both odd integers.

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4
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A composite number, N, is a positive integer greater than 1 that is not prime. A composite number has at least two prime factors and can be written as a product of its prime factors, p subscript 1, p subscript 2, ..., p subscript k, as follows:

N equals p subscript 1 cross times p subscript 2 cross times... cross times p subscript k

where k greater or equal than 2.

Prove by contradiction that any composite integer, N, must have at least one prime factor that is less than or equal to square root of N.

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5
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Prove by contradiction that square root of k, where k is a prime number, is an irrational number. 

You may use without proof the fact that any positive integer may be written uniquely as a product of its prime factors.

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