Proof by Contradiction (A Level only) (OCR A Level Maths: Pure)

Exam Questions

2 hours25 questions
1
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4 marks

Find the prime factorisation of the following numbers

(i)
100
(ii)
120

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2
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4 marks

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 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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3
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3 marks

Prove by contradiction that the sum of two consecutive integers is odd.

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4
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3 marks

Prove by contradiction that the product of two odd numbers is odd.

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5
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3 marks

Prove by contradiction that if x is even, then x squared must be even.

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6
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3 marks

Prove by contradiction that there is an infinite number of multiples of 10.

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1
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3 marks

Prove by contradiction that if x squared is odd, then x must be odd.

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2a
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2 marks

When a number is rational, it can be written in the form begin mathsize 16px style a over b end style.

(i)
Write down the condition that a and b must satisfy.

(ii)
Write down a further condition on b.
2b
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3 marks

Two numbers can be written in the form  p over q and r over s such that  p comma space q comma space r space and space s meet the necessary conditions so that the two numbers are rational.

Prove that the product of these numbers is also rational.

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3
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4 marks

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

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4a
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2 marks

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 proceeds 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 may be assumed to have no common factors.

Line 2: Squaring both sides:  begin mathsize 16px style 4 equals a squared over b squared end style
Line 3: therefore space a squared equals 2 b squared
Line 4:  
Line 5: Therefore a equals 2 m, for some integer m
Line 6: Then, a squared equals left parenthesis 2 m right parenthesis squared equals 4 m squared
Line 7: therefore space 2 b squared space equals space 4 m squared
Line 8: b squared equals 2 m squared
Line 9: So b squared is even and therefore b is also 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.

State what the error is and write the correct line down.

4b
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2 marks

Line 4 of the proof is missing.

Write down the missing line of the proof.

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5a
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3 marks
(i)
How many distinct factors does a prime number have?

(ii)
What can you say about the number of distinct factors a square number has?

(iii)
If N is the square of a prime number, then excluding N itself, write down, in terms of N, the largest factor of N.
5b
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2 marks

A composite number can be written uniquely as the product of its prime factors.i.e., any composite number N can be written uniquely as N space equals space p subscript 1 cross times p subscript 2 cross times p subscript 3 cross times..., where p subscript 1 comma space space p subscript 2 comma space p subscript 3 comma space... are the prime factors of N.

Show that a composite number N may be written in the form N equals p q, where q is an integer and p is a prime factor of N.

By expressing q in terms of the prime factors of N, be sure to explain why q must be an integer.

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6
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5 marks

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

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1
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4 marks

Prove by contradiction that if x cubed is odd, then x must be odd.

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2
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4 marks

Prove that the product of two rational numbers is rational.

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3
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4 marks

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

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4
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6 marks

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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5
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4 marks

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.

The proof contains two omissions in its argument.

Identify both omissions and correct them.

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6
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5 marks

If a positive integer greater than 1 is not a prime number, then it is called a composite number.  Prove by contradiction that any composite integer N has a prime factor less than or equal to square root of N.

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1
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4 marks

Prove by contradiction that if x to the power of n is odd, where n greater or equal than 2 is a positive integer, then x must be odd.

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2
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4 marks

Prove that the difference between two rational numbers is rational.

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3
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5 marks

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

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4
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6 marks

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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5
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4 marks

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, and b not equal to 0.  As log subscript 2 4 equals 2 space and space log subscript 2 8 equals 3, we may assume as well that a greater than b greater than 0.

Line 2:  therefore 2 to the power of a over b end exponent equals 7
Line 3:  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:  2 to the power of b equals 7 to the power of a
Line 5:  No power of 2 (all even) is equal to a power of 7 (all odd).
Line 6:  therefore 2 to the power of a not equal to 7 to the power of b unless a equals b equals 0 but this is a contradiction of the original assumption.
Line 7:  therefore log subscript 2 7 is irrational

The proof contains one mathematical error and one logical error.

Identify both errors and correct them.

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6
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5 marks

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 a over b  where a and b are both odd integers.

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7
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6 marks

Prove by contradiction that, if p comma space q comma space r space and space s are rational numbers and c is a positive non-square integer, then

p plus q square root of c equals r plus s square root of c

implies that p equals r and q equals s.  You may use without proof the fact that for any positive non-square integer n, square root of n is irrational.

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