Proof by deduction and exhaustion Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Proof by deduction and exhaustion questions. See exactly how to solve problems on parity, even numbers, direct proof, odd numbers.

parityeven numbersdirect proofodd numbersproductsconsecutive integers
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Prove that the sum of any two even numbers is even.

Worked solution

  1. Represent the two even numbers

    2m, 2n(m,n∈Z)2m,\ 2n\quad (m,n\in\mathbb{Z})

    Any even number is a multiple of 2, so we can write two DIFFERENT even numbers as 2m and 2n. Using different letters means the proof works for every pair, not just one example.

  2. Add them together

    2m+2n2m+2n

    We write down the sum of our two general even numbers. This is the quantity we need to show is even.

  3. Factorise out the 2

    2m+2n=2(m+n)2m+2n=2(m+n)

    Taking a factor of 2 out is the key move. Whatever m+n turns out to be, it is a whole number.

  4. Conclude

    2(m+n) is a multiple of 22(m+n)\ \text{is a multiple of }2

    Because the result is 2 times an integer, it must be even. This proves the statement for ALL even numbers, not just a few examples.

Answer
2(m+n), which is even2(m+n),\ \text{which is even}
Question 2
2 markseasy
Prove that the sum of any two odd numbers is even.

Worked solution

  1. Represent the two odd numbers

    2m+1, 2n+1(m,n∈Z)2m+1,\ 2n+1\quad (m,n\in\mathbb{Z})

    Every odd number is one more than an even number, so an odd number is written 2m+1. We use different letters m and n so the two odd numbers can be different.

  2. Add them together

    (2m+1)+(2n+1)(2m+1)+(2n+1)

    We form the sum of the two general odd numbers, which is what we must analyse.

  3. Collect like terms

    =2m+2n+2=2m+2n+2

    The two +1 terms combine to give +2. Everything on the right is now even-looking.

  4. Factorise out the 2

    =2(m+n+1)=2(m+n+1)

    Taking out a factor of 2 shows the structure. m+n+1 is an integer whatever m and n are.

  5. Conclude

    2(m+n+1) is a multiple of 22(m+n+1)\ \text{is a multiple of }2

    A multiple of 2 is even, so the sum of two odd numbers is always even.

Answer
2(m+n+1), which is even2(m+n+1),\ \text{which is even}
Question 3
2 markseasy
Prove that the product of any two odd numbers is odd.

Worked solution

  1. Represent the two odd numbers

    2m+1, 2n+1(m,n∈Z)2m+1,\ 2n+1\quad (m,n\in\mathbb{Z})

    We write each odd number in the general form 2k+1, using different letters so they need not be equal.

  2. Multiply them

    (2m+1)(2n+1)(2m+1)(2n+1)

    We form the product, which is the quantity we must show is odd.

  3. Expand the brackets

    =4mn+2m+2n+1=4mn+2m+2n+1

    Multiplying out carefully gives four terms. Remember to multiply every term in the first bracket by every term in the second.

  4. Factorise the even part

    =2(2mn+m+n)+1=2(2mn+m+n)+1

    The first three terms all share a factor of 2. Pulling it out leaves a clear ‘2 times an integer, plus 1’ shape.

  5. Conclude

    2(2mn+m+n)+1 is one more than a multiple of 22(2mn+m+n)+1\ \text{is one more than a multiple of }2

    Anything of the form 2k+1 is odd, so the product of two odd numbers is always odd.

Answer
2(2mn+m+n)+1, which is odd2(2mn+m+n)+1,\ \text{which is odd}
Question 4
2 markseasy
Prove that the sum of an even number and an odd number is odd.

Worked solution

  1. Represent the numbers

    2m, 2n+1(m,n∈Z)2m,\ 2n+1\quad (m,n\in\mathbb{Z})

    We write the even number as 2m and the odd number as 2n+1, using different letters so they are independent.

  2. Add them

    2m+(2n+1)2m+(2n+1)

    We form the sum, which is the quantity to analyse.

  3. Factorise the even part

    =2(m+n)+1=2(m+n)+1

    The two even terms combine into 2(m+n); the +1 is left over. This is the tell-tale shape of an odd number.

  4. Conclude

    2(m+n)+1 is one more than a multiple of 22(m+n)+1\ \text{is one more than a multiple of }2

    A number of the form 2k+1 is odd, so even plus odd is always odd.

Answer
2(m+n)+1, which is odd2(m+n)+1,\ \text{which is odd}
Question 5
2 markseasy
Prove that the sum of any three consecutive integers is a multiple of 33.

Worked solution

  1. Represent three consecutive integers

    n, n+1, n+2(n∈Z)n,\ n+1,\ n+2\quad (n\in\mathbb{Z})

    Consecutive integers go up in steps of 1, so we write them starting from a general integer n. This covers every possible run of three.

  2. Add them

    n+(n+1)+(n+2)n+(n+1)+(n+2)

    We form the total of the three numbers, which is what we must show is a multiple of 3.

  3. Simplify

    =3n+3=3n+3

    Collecting the n terms gives 3n and the numbers 1 and 2 add to 3.

  4. Factorise out 3

    =3(n+1)=3(n+1)

    Taking out the common factor 3 shows the structure clearly. n+1 is an integer.

  5. Conclude

    3(n+1) is a multiple of 33(n+1)\ \text{is a multiple of }3

    Because the sum equals 3 times an integer, it is always a multiple of 3.

Answer
3(n+1), a multiple of 33(n+1),\ \text{a multiple of }3

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