(b)Hence prove that the difference between the squares of any two integers that differ by 2 is always a multiple of 4.(2)
(Total for Question 3 is 4 marks)
4
Let n be an integer.
(Total for Question 4 is 3 marks)
5
Let two odd numbers be written as 2n + 1 and 2m + 1, where n and m are integers.
(Total for Question 5 is 3 marks)
6
Prove that x2 + 4x + 7 > 0 for all real values of x.
(Total for Question 6 is 3 marks)
7
Let n be an integer.
(Total for Question 7 is 3 marks)
8
Two consecutive odd numbers can be written as 2n - 1 and 2n + 1, where n is an integer.
(Total for Question 8 is 3 marks)
9
Let x = 0.454545... (recurring, where the block '45' repeats forever).
(Total for Question 9 is 3 marks)
10
Show that (x2 - 9)/(x2 - x - 6) simplifies to (x + 3)/(x + 2), stating any values of x that must be excluded.
(Total for Question 10 is 3 marks)
11
Aisha attempts to prove that n2 + n is always even for any integer n. Her working is shown below: "n2 + n = n(n+1). Since n+1 is even, n(n+1) must be even."
(Total for Question 11 is 2 marks)
12
Let n be a positive integer.
(Total for Question 12 is 4 marks)
13
A student claims: "n2 + n + 41 is a prime number for every positive integer n."
(Total for Question 13 is 3 marks)
14
Two consecutive odd numbers can be written as 2n + 1 and 2n + 3, where n is an integer.
(Total for Question 14 is 4 marks)
15
The equation ax2 + bx + c = 0 has a > 0, and b2 < 4ac.
(Total for Question 15 is 4 marks)
16
Let m and n be positive integers with m > n > 0.
(Total for Question 16 is 4 marks)
17
Let n be an integer.
(Total for Question 17 is 4 marks)
18
Prove by contradiction that √2 is irrational.
(Total for Question 18 is 5 marks)
19
Every positive integer n can be written as n = 10k + r, where k is an integer and r is the units digit of n (so r is one of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9).
(Total for Question 19 is 6 marks)
20
Prove by contradiction that there is no smallest positive rational number.
(Total for Question 20 is 4 marks)
Mark scheme · A7 Algebraic Proof
Question 1
M1 writes the sum of two consecutive integers as n + (n+1), simplifying to 2n + 1 oe
A1 correct conclusion that 2n is always even, so 2n + 1 is always odd, hence the sum of two consecutive integers is always odd
Answer: n + (n+1) = 2n + 1, which is odd for every integer n
Question 2
M1 writes the sum of three consecutive integers as n + (n+1) + (n+2)
M1 simplifies correctly to 3n + 3
A1 factorises to 3(n+1) and states this is a multiple of 3 for every integer n
Answer: n + (n+1) + (n+2) = 3(n+1), a multiple of 3
Question 3
(a) M1 expands (n+2)2 correctly to n2 + 4n + 4
(a) A1 subtracts n2 to give 4n + 4 (oe 4(n+1)), answer printed
(a) Answer: 4n + 4
(b) M1 factorises their result from (a) as 4(n+1)
(b) A1 correct conclusion that since n+1 is an integer, 4(n+1) is always a multiple of 4
(b) Answer: 4n + 4 = 4(n+1), a multiple of 4
Question 4
M1 factorises n2 - n as n(n-1)
M1 identifies n and n-1 as two consecutive integers, one of which must be even
A1 correct conclusion: since one of n, n-1 is even, their product n(n-1) is always even
Answer: n2 - n = n(n-1), always even
Question 5
M1 writes the product (2n+1)(2m+1) and expands correctly to 4nm + 2n + 2m + 1
M1 factorises 4nm + 2n + 2m as 2(2nm + n + m)
A1 correct conclusion that 2(2nm+n+m) + 1 is of the form 2k + 1, so it is always odd
A1 valid conclusion: since (x+2)2 ≥ 0 for all real x, (x+2)2 + 3 ≥ 3 > 0, so x2+4x+7 > 0 always
Answer: x2 + 4x + 7 = (x+2)2 + 3 > 0 for all real x
Question 7
M1 writes three consecutive multiples of 4 as 4n, 4n+4, 4n+8 and sums to 12n + 12
A1 factorises correctly to 12(n+1)
A1 correct conclusion that 12(n+1) is always a multiple of 12, since n+1 is an integer
Answer: 4n + (4n+4) + (4n+8) = 12(n+1), a multiple of 12
Question 8
M1 uses difference of two squares: (2n+1)2 - (2n-1)2 = [(2n+1)-(2n-1)][(2n+1)+(2n-1)]
A1 simplifies correctly to (2)(4n) = 8n
A1 correct conclusion that 8n is always a multiple of 8 for integer n
Answer: (2n+1)2 - (2n-1)2 = 8n, a multiple of 8
Question 9
M1 multiplies both sides by 100 (as two digits repeat), giving 100x = 45.454545...
M1 subtracts x = 0.454545... from 100x to eliminate the recurring part, giving 99x = 45
A1 solves and simplifies fully to x = 45/99 = 5/11
Answer: x = 5/11
Question 10
M1 factorises the numerator, x2 - 9 = (x-3)(x+3)
A1 factorises the denominator, x2 - x - 6 = (x-3)(x+2)
A1 cancels the common factor (x-3) correctly to give (x+3)/(x+2), stating x is not equal to 3 and x is not equal to -2
Answer: (x+3)/(x+2), for x not equal to 3, -2
Question 11
B1 identifies the error: n+1 is not always even (e.g. if n is even, such as n=4, then n+1=5 is odd), so the claim 'n+1 is even' is a false general statement
B1 gives a correct justification: n and n+1 are consecutive integers, so exactly one of them must always be even, meaning their product n(n+1) is always even regardless of which one is even
Answer: The claim that n+1 is always even is false; the correct reasoning is that one of n, n+1 (consecutive integers) is always even
Question 12
M1 factorises n3 - n as n(n-1)(n+1)
M1 identifies n-1, n, n+1 as three consecutive integers
A1 valid reasoning that among any three consecutive integers, at least one is divisible by 3 and at least one is even (divisible by 2)
A1 correct conclusion that the product is divisible by both 2 and 3, hence divisible by 6
Answer: n3 - n = (n-1)n(n+1), always divisible by 6
Question 13
M1 selects a suitable value of n to test, e.g. n = 41
M1 evaluates n2 + n + 41 correctly for the chosen n, writing it in a factorisable form, e.g. n2+n+41 = n(n+1)+41, so at n=41: 41(42)+41 = 41(43)
A1 correct conclusion that 41 x 43 = 1763 has factors other than 1 and itself, so it is not prime, disproving the claim by counter-example
Answer: False: at n=41, n2+n+41 = 41 x 43 = 1763, which is not prime
Question 14
M1 expands both squares correctly, (2n+1)2 = 4n2+4n+1 and (2n+3)2 = 4n2+12n+9
A1 adds and simplifies to 8n2 + 16n + 10 (oe 2(4n2+8n+5))
A1 valid reasoning that 4n2 + 8n + 5 is always odd (since 4n2 and 8n are both even, plus the odd number 5)
A1 correct conclusion that 2 x (an odd number) is always even, but can never be a multiple of 4
Answer: (2n+1)2 + (2n+3)2 = 2(4n2+8n+5), always even but never a multiple of 4
Question 15
M1 completes the square correctly, a(x + b/(2a))2 + c - b2/(4a) = 0
A1 rearranges to isolate the squared term, a(x + b/(2a))2 = (b2 - 4ac)/(4a)
A1 valid reasoning that the left-hand side is always ≥ 0 (since a > 0), while the right-hand side is negative when b2 < 4ac (as 4a > 0)
A1 correct conclusion: a non-negative quantity cannot equal a negative quantity, so no real value of x satisfies the equation, i.e. no real roots exist
Answer: No real roots exist when b2 < 4ac and a > 0
Question 16
M1 expands (m2-n2)2 correctly to m4 - 2m2n2 + n4
M1 expands (2mn)2 correctly to 4m2n2 and adds to the previous expansion
A1 simplifies the sum correctly to m4 + 2m2n2 + n4
A1 recognises this factorises as (m2+n2)2, matching the right-hand side, so the identity holds for all m > n > 0
Answer: (m2-n2)2 + (2mn)2 = (m2+n2)2 for all m > n > 0
Question 17
M1 writes the sum of four consecutive integers as n+(n+1)+(n+2)+(n+3), simplifying to 4n + 6
A1 factorises correctly to 2(2n+3)
A1 valid reasoning that 2n + 3 is always odd (2n is even, plus the odd number 3)
A1 correct conclusion that 2 x (an odd number) is always even, but is never divisible by 4, since (2n+3) is never even
Answer: Sum = 2(2n+3), always even but never a multiple of 4
Question 18
B1 assumes, for contradiction, that √2 is rational, so √2 = p/q where p and q are integers with no common factor (in lowest terms), q not equal to 0
B1 squares both sides to get p2 = 2q2, and deduces that since p2 is even, p itself must be even
B1 substitutes p = 2k into p2 = 2q2 to obtain 4k2 = 2q2, so q2 = 2k2, and deduces q must also be even
B1 states that p and q being both even contradicts the assumption that p/q was in lowest terms (no common factor)
B1 correct final conclusion: the contradiction means the original assumption was false, so √2 is irrational
Answer: √2 is irrational (proof by contradiction)
Question 19
Level 1 (1-2): Attempts to expand n^2 = (10k+r)^2, or lists some values of r^2, but with limited progress and no valid general conclusion.
Level 2 (3-4): Expands (10k+r)^2 correctly and shows the units digit of n^2 depends only on r^2. Checks several but not all values of r from 0 to 9, or makes a minor error in the case check, but reaches a broadly correct partial conclusion.
Level 3 (5-6): Fully correct and complete argument: expands (10k+r)^2 to show the units digit of n^2 depends only on the units digit of r^2, systematically checks all ten values r = 0 to 9, correctly lists the resulting units digits (0, 1, 4, 9, 6, 5, 6, 9, 4, 1), and concludes clearly that 2, 3, 7 and 8 never appear, so a square number can never end in these digits.
The units digit of n^2 is therefore determined entirely by the units digit of r^2 (since 10(10k^2+2kr) contributes only to tens and higher place values)
Systematically evaluating r^2 for r = 0,1,2,...,9 gives: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81
Taking units digits of these values gives: 0, 1, 4, 9, 6, 5, 6, 9, 4, 1
None of these units digits is 2, 3, 7 or 8
Since every integer n has a units digit r in {0,...,9}, and all cases have been checked, no square number can ever end in 2, 3, 7 or 8
Question 20
B1 assumes, for contradiction, that a smallest positive rational number exists, and calls it q, where q > 0
B1 considers q/2, and states that q/2 is rational (since q is rational and 2 is a non-zero rational number) and q/2 > 0 (since q > 0)
B1 shows that q/2 < q, since dividing a positive number by 2 always gives a smaller positive number
B1 correct conclusion: q/2 is a positive rational number smaller than q, contradicting the assumption that q was the smallest positive rational number, so no smallest positive rational number exists
Answer: No smallest positive rational number exists (proof by contradiction)