A-Level Proof by deduction and exhaustion Practice Questions

Free A-Level Proof by deduction and exhaustion practice questions with full step-by-step worked solutions. Covers parity, even numbers, direct proof, odd numbers. Practise exam-style problems and check your method.

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.
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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
3 markseasy
Prove that the sum of any three consecutive even numbers is a multiple of 66.
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Worked solution

  1. Represent three consecutive even numbers

    2n, 2n+2, 2n+4(n∈Z)2n,\ 2n+2,\ 2n+4\quad (n\in\mathbb{Z})

    Even numbers go up in steps of 2, so consecutive even numbers are 2n, 2n+2 and 2n+4.

  2. Add them

    2n+(2n+2)+(2n+4)2n+(2n+2)+(2n+4)

    We form the total of the three even numbers.

  3. Collect like terms

    =6n+6=6n+6

    The three 2n terms give 6n, and the numbers 2 and 4 add to 6.

  4. Factorise out 6

    =6(n+1)=6(n+1)

    Taking out the common factor 6 shows the structure. n+1 is an integer.

  5. Conclude

    6(n+1) is a multiple of 66(n+1)\ \text{is a multiple of }6

    Since the sum equals 6 times an integer, it is always a multiple of 6.

Answer
6(n+1), a multiple of 66(n+1),\ \text{a multiple of }6
Question 3
4 marksintermediate
Prove that the sum of the squares of any two consecutive even numbers is a multiple of 44 but never a multiple of 88.
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Worked solution

  1. Represent two consecutive even numbers

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

    Consecutive even numbers are 2n and 2n+2.

  2. Write the sum of squares

    (2n)2+(2n+2)2(2n)^2+(2n+2)^2

    We form the quantity to analyse.

  3. Expand

    =4n2+(4n2+8n+4)=4n^2+(4n^2+8n+4)

    Expanding (2n+2)^2 gives 4n^2+8n+4; we add the first 4n^2.

  4. Simplify

    =8n2+8n+4=8n^2+8n+4

    Collecting terms gives 8n^2+8n+4.

  5. Factorise out 4

    =4(2n2+2n+1)=4(2n^2+2n+1)

    Every term has a factor of 4; taking it out gives 4 times 2n^2+2n+1.

  6. Check the parity of the bracket

    2n2+2n+1=2(n2+n)+1 is odd2n^2+2n+1=2(n^2+n)+1\ \text{is odd}

    The bracket is one more than an even number, so it is odd — it contains no extra factor of 2.

  7. Conclude

    4×(odd) is a multiple of 4 but not 84\times(\text{odd})\ \text{is a multiple of }4\ \text{but not }8

    So the sum is divisible by 4, but because the other factor is odd, it can never be divisible by 8.

Answer
4(2n2+2n+1), with an odd bracket4(2n^2+2n+1),\ \text{with an odd bracket}
Question 4
6 markshard
Prove by exhaustion that for every integer nn, n4n^4 leaves a remainder of 00 or 11 when divided by 1616.
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Worked solution

  1. Split into two cases

    n even or n oddn\ \text{even or }n\ \text{odd}

    Every integer is even or odd, so these two cases exhaust all possibilities.

  2. Case 1: even

    n=2k⇒n4=16k4n=2k\Rightarrow n^4=16k^4

    If n is even, raising 2k to the fourth power gives 16k^4.

  3. Remainder in case 1

    16k4=16(k4)+016k^4=16(k^4)+0

    This is a multiple of 16, so the remainder is 0.

  4. Case 2: odd, square first

    n=2k+1⇒n2=4k2+4k+1=4k(k+1)+1n=2k+1\Rightarrow n^2=4k^2+4k+1=4k(k+1)+1

    We first square the odd number and factor the even part.

  5. Use consecutive integers

    k(k+1)=2m⇒n2=8m+1k(k+1)=2m\Rightarrow n^2=8m+1

    k and k+1 are consecutive, so k(k+1) is even; hence n^2=8m+1 for some integer m.

  6. Square again

    n4=(8m+1)2=64m2+16m+1n^4=(8m+1)^2=64m^2+16m+1

    We square n^2 to reach n^4.

  7. Factor the multiple of 16

    =16(4m2+m)+1=16(4m^2+m)+1

    Both 64m^2 and 16m are multiples of 16, leaving remainder 1.

  8. Remainder in case 2

    n4≡1 (mod 16)n^4\equiv 1\ (\mathrm{mod}\ 16)

    So an odd number to the fourth power leaves remainder 1 on division by 16.

  9. Verify with examples

    34=81=16×5+1,44=256=16×163^4=81=16\times5+1,\quad 4^4=256=16\times16

    Checking n=3 (odd): 81 leaves remainder 1; and n=4 (even): 256 leaves remainder 0 — both match the two cases.

  10. Conclude

    n4≡0 or 1 (mod 16)n^4\equiv 0\ \text{or}\ 1\ (\mathrm{mod}\ 16)

    Both cases are covered, so by exhaustion every fourth power leaves remainder 0 or 1 when divided by 16.

Answer
n even⇒16∣n4;n odd⇒n4=16(⋯ )+1n\ \text{even}\Rightarrow 16\mid n^4;\quad n\ \text{odd}\Rightarrow n^4=16(\cdots)+1
Question 5
8 markschallenging
Prove that (a2+b2+c2)(x2+y2+z2)≥(ax+by+cz)2(a^2+b^2+c^2)(x^2+y^2+z^2)\ge (ax+by+cz)^2 for all real numbers, and state when equality holds.
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Worked solution

  1. State the target

    (a2+b2+c2)(x2+y2+z2)≥(ax+by+cz)2(a^2+b^2+c^2)(x^2+y^2+z^2)\ge (ax+by+cz)^2

    This is the Cauchy–Schwarz inequality in three dimensions. We prove it via an exact identity.

  2. Expand the left product

    ∑a2∑x2=a2x2+a2y2+a2z2+b2x2+b2y2+b2z2+c2x2+c2y2+c2z2\textstyle\sum a^2\sum x^2=a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2

    We multiply out the two brackets, giving nine terms.

  3. Expand the right square

    (ax+by+cz)2=a2x2+b2y2+c2z2+2(axby+bycz+axcz)(ax+by+cz)^2=a^2x^2+b^2y^2+c^2z^2+2(axby+bycz+axcz)

    We square the linear combination, obtaining three squared terms and three cross terms.

  4. Form the difference

    D=(a2+b2+c2)(x2+y2+z2)−(ax+by+cz)2D=(a^2+b^2+c^2)(x^2+y^2+z^2)-(ax+by+cz)^2

    We compute the left minus the right and hope to recognise a sum of squares.

  5. Cancel the matching squares

    a2x2, b2y2, c2z2 cancela^2x^2,\ b^2y^2,\ c^2z^2\ \text{cancel}

    These three terms appear on both sides and disappear in the difference.

  6. Write the surviving terms

    D=a2y2+b2x2+b2z2+c2y2+a2z2+c2x2−2(abxy+bcyz+acxz)D=a^2y^2+b^2x^2+b^2z^2+c^2y^2+a^2z^2+c^2x^2-2(abxy+bcyz+acxz)

    Six mixed square terms remain, minus twice the three cross terms.

  7. Group into three pairs

    (a2y2−2abxy+b2x2)+(b2z2−2bcyz+c2y2)+(a2z2−2acxz+c2x2)(a^2y^2-2abxy+b^2x^2)+(b^2z^2-2bcyz+c^2y^2)+(a^2z^2-2acxz+c^2x^2)

    Each pair of square terms is matched with a cross term to complete a square.

  8. Write as three squares

    D=(ay−bx)2+(bz−cy)2+(az−cx)2D=(ay-bx)^2+(bz-cy)^2+(az-cx)^2

    This is Lagrange's identity: the difference is exactly a sum of three squares.

  9. Use non-negativity

    (ay−bx)2,(bz−cy)2,(az−cx)2≥0(ay-bx)^2,(bz-cy)^2,(az-cx)^2\ge 0

    Each square is non-negative, so their sum is non-negative.

  10. Deduce D≥0D\ge0

    D≥0D\ge 0

    Since D is a sum of squares, it is at least 0.

  11. Rearrange

    (a2+b2+c2)(x2+y2+z2)≥(ax+by+cz)2(a^2+b^2+c^2)(x^2+y^2+z^2)\ge (ax+by+cz)^2

    D\ge0 is exactly the required inequality.

  12. State equality condition

    ay=bx, bz=cy, az=cxay=bx,\ bz=cy,\ az=cx

    Equality needs all three squares to vanish, which means (a,b,c) and (x,y,z) are proportional.

  13. Interpret geometrically

    equality  ⟺  (a,b,c)∥(x,y,z)\text{equality}\iff (a,b,c)\parallel(x,y,z)

    Cauchy–Schwarz becomes an equality precisely when the two vectors are parallel.

  14. Verify an example

    (1,2,2)⋅(2,1,2): 9⋅9=81≥82=64(1,2,2)\cdot(2,1,2):\ 9\cdot9=81\ge 8^2=64

    Checking a=(1,2,2), x=(2,1,2): left is 9\times9=81, right is (2+2+4)^2=64, and 81\ge64.

  15. Conclude

    (a2+b2+c2)(x2+y2+z2)≥(ax+by+cz)2(a^2+b^2+c^2)(x^2+y^2+z^2)\ge (ax+by+cz)^2

    Therefore the inequality holds for all real numbers, with equality iff the vectors are proportional.

Answer
D=(ay−bx)2+(bz−cy)2+(az−cx)2≥0D=(ay-bx)^2+(bz-cy)^2+(az-cx)^2\ge 0

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