[discia] GCSE Maths › Algebra › Proof

Proof

discia.org 10 questions · 35 marks
Name Date
  1. Either paper 3 marks

    \(k = n^2 + 7n + 23\)

    Idris says, "\(k\) will be a prime number for all integer values of \(n\) from 1 to 8."

    Show that Idris is wrong.

    You must show that your value of \(k\) is not prime.

    Show answer
    for example \(n = 7\) gives \(k = 121 = 11 \times 11\), so \(k\) is not prime (any \(n\) from 1 to 8 with a non-prime \(k\) shown by a factor pair is also correct)
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  2. Either paper 3 marks

    Here is the term-to-term rule for a sequence.

    Multiply by 2, then add 10

    The first three terms of the sequence are

    \(a + 2\) \(2a + 14\) \(4a + 38\)

    Show that the sum of the first four terms is a multiple of 5.

    Show answer
    The fourth term is \(8a + 86\). The sum of the four terms is \(15a + 140\), which is \(5(3a + 28)\), so the sum is a multiple of 5.
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  3. Either paper Higher4 marks
    (a)

    The smallest of eleven consecutive integers is \(k\).

    Write an expression, in terms of \(k\), for the sum of the eleven integers. Give your answer in its simplest form.

    (b)

    Prove that the sum of any eleven consecutive integers is always a multiple of 11.

    Show answer
    (a) \(11k + 55\)
    (b) \(11k + 55 = 11(k + 5)\). \(k + 5\) is a whole number, so the sum is 11 times a whole number, which is what being a multiple of 11 means.
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  4. Either paper Higher4 marks
    (a)

    Two whole numbers differ by 11. The smaller of them is \(k\).

    Multiply out the brackets and simplify \((k + 11)^2 - k^2\).

    (b)

    Prove that the difference between the squares of any two whole numbers that differ by 11 is always a multiple of 11.

    Show answer
    (a) \(22k + 121\)
    (b) \(22k + 121 = 11(2k + 11)\). \(2k + 11\) is a whole number, so the difference is 11 times a whole number.
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  5. Either paper Higher4 marks
    (a)

    Expand and simplify \(r(r + 16) + 64\).

    (b)

    Hence prove that \(r(r + 16) + 64\) is a square number for every whole number \(r\).

    Show answer
    (a) \(r^2 + 16r + 64\)
    (b) \(r^2 + 16r + 64 = (r + 8)^2\), and \(r + 8\) is a whole number, so the expression is the square of a whole number.
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  6. Either paper Higher4 marks
    (a)

    Factorise \(m^2 + 3m\).

    (b)

    Prove that \(m^2 + 3m\) is always even when \(m\) is a whole number.

    Show answer
    (a) \(m(m + 3)\)
    (b) \(m^2 + 3m = m(m + 3)\). The two factors differ by 3, which is odd, so one of them is even and the other is odd. A product with an even factor is even.
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  7. Either paper Higher4 marks

    \(m\) and \(n\) are integers.

    (a)

    Prove that \((2m+1)^2 - (2n-1)^2 = 4(m+n)(m-n+1)\)

    (b)

    Marcus says the result in part (a) shows that the difference of the squares of any two odd numbers must be a multiple of 16. Is Marcus correct? You must give reasons for your answer.

    Show answer
    (a) \((2m+1)^2 - (2n-1)^2 = 4m^2 + 4m + 1 - (4n^2 - 4n + 1) = 4m^2 + 4m - 4n^2 + 4n = 4(m+n)(m-n+1)\)
    (b) No, Marcus is not correct: \(3\) and \(1\) are both odd, and \(3^2 - 1^2 = 9 - 1 = 8\), which is not a multiple of 16.
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  8. Either paper 1 mark

    \(a\) is a prime number and \(b\) is an even number.

    \(n = 5a + b + 7\)

    Which statement about \(n\) must be true? Write down the correct statement.

    A: always odd

    B: always prime

    C: always even

    D: could be even or odd

    Show answer
    could be even or odd
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  9. Either paper Higher4 marks

    Here is a quadrilateral \(ABCD\). The diagonal \(AC\) is drawn and \(DA = DC\).

    Angle \(BAC = 54^\circ\), angle \(BCA = 49^\circ\) and angle \(CAD = 45^\circ\).

    ABCD

    Prove that \(ABCD\) is not a cyclic quadrilateral.

    You must give a reason for each angle you work out.

    Show answer
    Angle \(ABC = 77^\circ\) and angle \(ADC = 90^\circ\), which add to \(167^\circ\), less than \(180^\circ\). Opposite angles do not sum to \(180^\circ\), so \(ABCD\) is not cyclic.
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  10. Either paper Higher4 marks
    (a)

    \(2p - 3\) is an odd number for every whole number \(p\).

    Expand and simplify \((2p - 3)^2\).

    (b)

    Prove that the square of any odd number is odd.

    Show answer
    (a) \(4p^2 - 12p + 9\)
    (b) \(4p^2 - 12p + 9 = 2(2p^2 - 6p + 4) + 1\), which is two times a whole number plus one, so it is odd.
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discia.org Proof · sheet 1
Answers Proof · sheet 1
  1. for example \(n = 7\) gives \(k = 121 = 11 \times 11\), so \(k\) is not prime (any \(n\) from 1 to 8 with a non-prime \(k\) shown by a factor pair is also correct)
  2. The fourth term is \(8a + 86\). The sum of the four terms is \(15a + 140\), which is \(5(3a + 28)\), so the sum is a multiple of 5.
  3. (a) \(11k + 55\)
    (b) \(11k + 55 = 11(k + 5)\). \(k + 5\) is a whole number, so the sum is 11 times a whole number, which is what being a multiple of 11 means.
  4. (a) \(22k + 121\)
    (b) \(22k + 121 = 11(2k + 11)\). \(2k + 11\) is a whole number, so the difference is 11 times a whole number.
  5. (a) \(r^2 + 16r + 64\)
    (b) \(r^2 + 16r + 64 = (r + 8)^2\), and \(r + 8\) is a whole number, so the expression is the square of a whole number.
  6. (a) \(m(m + 3)\)
    (b) \(m^2 + 3m = m(m + 3)\). The two factors differ by 3, which is odd, so one of them is even and the other is odd. A product with an even factor is even.
  7. (a) \((2m+1)^2 - (2n-1)^2 = 4m^2 + 4m + 1 - (4n^2 - 4n + 1) = 4m^2 + 4m - 4n^2 + 4n = 4(m+n)(m-n+1)\)
    (b) No, Marcus is not correct: \(3\) and \(1\) are both odd, and \(3^2 - 1^2 = 9 - 1 = 8\), which is not a multiple of 16.
  8. could be even or odd
  9. Angle \(ABC = 77^\circ\) and angle \(ADC = 90^\circ\), which add to \(167^\circ\), less than \(180^\circ\). Opposite angles do not sum to \(180^\circ\), so \(ABCD\) is not cyclic.
  10. (a) \(4p^2 - 12p + 9\)
    (b) \(4p^2 - 12p + 9 = 2(2p^2 - 6p + 4) + 1\), which is two times a whole number plus one, so it is odd.
discia.org Proof · sheet 1