Let n be an integer. Which of the following always represents an even number?
A) 2n + 1
B) 2n
C) n + 1
D) 2n - 1
(Total for Question 1 is 1 mark)
2
Let n be an integer. Prove that the sum of five consecutive integers is always a multiple of 5.
(Total for Question 2 is 1 mark)
3
Let n be an integer.
(a)Show that n2 + (n+1)2 = 2n2 + 2n + 1.(1)
(b)Hence prove that the sum of the squares of two consecutive integers is always odd.(1)
(Total for Question 3 is 2 marks)
4
Let n be an integer. Prove that n2 + 3n is always even.
(Total for Question 4 is 1 mark)
5
Let a = 2n and b = 2m, where n and m are integers, so a and b are even numbers. Prove that ab is always a multiple of 4.
(Total for Question 5 is 1 mark)
6
Let two odd numbers be written as 2n + 1 and 2m + 1, where n and m are integers. Prove that their sum is always even.
(Total for Question 6 is 1 mark)
7
Prove that x2 + 2x + 5 > 0 for all real values of x.
(Total for Question 7 is 2 marks)
8
Let n be an integer. Prove that three consecutive multiples of 5 always sum to a multiple of 15.
(Total for Question 8 is 1 mark)
9
Two numbers differ by 4 and can be written as 2n + 3 and 2n - 1, where n is an integer. Prove that the difference between their squares is always a multiple of 8.
(Total for Question 9 is 2 marks)
10
Let x = 0.363636... (recurring, where the block '36' repeats forever). Prove that x = 4/11.
(Total for Question 10 is 2 marks)
11
Show that (x2 - 16)/(x2 - x - 12) simplifies to (x + 4)/(x + 3), stating any values of x that must be excluded.
(Total for Question 11 is 2 marks)
12
Ravi attempts to prove that n2 - n + 1 is always odd for any integer n. His working is shown below: "n2 - n = n(n-1). Since n-1 is even, n(n-1) must be even, so n(n-1)+1 is odd."
(Total for Question 12 is 2 marks)
13
Let n be a positive integer. Prove that n3 + n is always even.
(Total for Question 13 is 2 marks)
14
Prove that x2 - 8x + 20 > 0 for all real values of x.
(Total for Question 14 is 3 marks)
15
Prove that the equation x2 - 4x + 8 = 0 has no real roots.
(Total for Question 15 is 3 marks)
16
Show that (2n)2 + (n2 - 1)2 = (n2 + 1)2 for all positive integers n.
(Total for Question 16 is 3 marks)
17
A student claims: "n2 - n + 11 is a prime number for every positive integer n."
(Total for Question 17 is 3 marks)
18
Prove by contradiction that there is no largest even integer.
(Total for Question 18 is 4 marks)
19
Prove by contradiction that √3 is irrational.
(Total for Question 19 is 4 marks)
Mark scheme · A7D Algebraic Proof: Fluency and Exam Drill
Question 1
B1 B cao
Answer: B
Question 2
B1 cso: writes the sum as n+(n+1)+(n+2)+(n+3)+(n+4) = 5n+10 = 5(n+2), a multiple of 5
Answer: n+(n+1)+(n+2)+(n+3)+(n+4) = 5(n+2), a multiple of 5
Question 3
(a) B1 cso: expands (n+1)2 correctly to n2+2n+1 and adds n2 to give 2n2+2n+1, answer printed
(a) Answer: 2n2 + 2n + 1
(b) B1 cso: correct conclusion that 2n2+2n = 2(n2+n) is always even, so 2n2+2n+1 is always odd
B1 cso: factorises n2+3n = n(n+3) and correctly reasons that if n is even the product is even, and if n is odd then n+3 is even so the product is even, so n(n+3) is always even
Answer: n2 + 3n = n(n+3), always even
Question 5
B1 cso: ab = (2n)(2m) = 4nm, a multiple of 4 since n, m are integers
Answer: ab = 4nm, a multiple of 4
Question 6
B1 cso: (2n+1)+(2m+1) = 2n+2m+2 = 2(n+m+1), always even
A1 cso: simplifies to (x+1)2 + 4, and since (x+1)2 ≥ 0 for all real x, (x+1)2+4 ≥ 4 > 0, so x2+2x+5 > 0 always
Answer: x2 + 2x + 5 = (x+1)2 + 4 > 0 for all real x
Question 8
B1 cso: 5n + (5n+5) + (5n+10) = 15n + 15 = 15(n+1), a multiple of 15
Answer: 5n + (5n+5) + (5n+10) = 15(n+1), a multiple of 15
Question 9
M1 uses difference of two squares: (2n+3)2 - (2n-1)2 = [(2n+3)-(2n-1)][(2n+3)+(2n-1)]
A1 cso: simplifies correctly to (4)(4n+2) = 16n+8 = 8(2n+1), a multiple of 8
Answer: (2n+3)2 - (2n-1)2 = 8(2n+1), a multiple of 8
Question 10
M1 multiplies both sides by 100 (as two digits repeat), giving 100x = 36.3636..., and subtracts x to eliminate the recurring part, giving 99x = 36
A1 cso: solves and simplifies fully to x = 36/99 = 4/11
Answer: x = 4/11
Question 11
M1 factorises the numerator, x2 - 16 = (x-4)(x+4), and the denominator, x2 - x - 12 = (x-4)(x+3)
A1 cso: cancels the common factor (x-4) correctly to give (x+4)/(x+3), stating x is not equal to 4 and x is not equal to -3
Answer: (x+4)/(x+3), for x not equal to 4, -3
Question 12
B1 identifies the error: n-1 is not always even (e.g. if n is even, such as n=4, then n-1=3 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 13
M1 factorises n3+n as n(n2+1)
A1 cso: valid reasoning that if n is even the product is even, and if n is odd then n2 is odd so n2+1 is even, making the product even; so n3+n is always even
A1 cso: since (x-4)2 ≥ 0 for all real x, (x-4)2 + 4 ≥ 4 > 0, so x2-8x+20 > 0 for all real x
Answer: x2 - 8x + 20 = (x-4)2 + 4 > 0 for all real x
Question 15
M1 discriminant = (-4)2 - 4(1)(8) (= 16-32)
A1 = -16
A1 cso: since -16 < 0 the equation has no real roots
Answer: Discriminant = -16 < 0, so no real roots
Question 16
M1 expands (n2-1)2 correctly to n4 - 2n2 + 1
M1 adds (2n)2 = 4n2 to the previous expansion
A1 cso: simplifies the sum to n4 + 2n2 + 1, which matches the expansion of (n2+1)2, confirming the identity holds for all positive integers n
Answer: (2n)2 + (n2-1)2 = (n2+1)2 for all positive integers n
Question 17
M1 selects a suitable value of n to test, e.g. n = 11
M1 evaluates n2-n+11 correctly for the chosen n, writing it in a factorisable form, e.g. n2-n+11 = n(n-1)+11, so at n=11: 11(10)+11 = 11(11)
A1 cso: correct conclusion that 11 x 11 = 121 has factors other than 1 and itself, so it is not prime, disproving the claim by counter-example
Answer: False: at n=11, n2-n+11 = 11 x 11 = 121, which is not prime
Question 18
B1 assumes, for contradiction, that a largest even integer exists, and calls it E
B1 considers E + 2, and states that E + 2 is even, since E is even and 2 is even, so their sum is even
B1 shows that E + 2 > E, since adding a positive number always gives a larger number
B1 cso: correct conclusion: E + 2 is an even integer larger than E, contradicting the assumption that E was the largest even integer, so no largest even integer exists
Answer: No largest even integer exists (proof by contradiction)
Question 19
B1 assumes, for contradiction, that √3 is rational, so √3 = p/q where p and q are integers with no common factor (in lowest terms), q not equal to 0; squares both sides to get p2 = 3q2, and deduces that since p2 is divisible by 3, p itself must be divisible by 3
B1 substitutes p = 3k into p2 = 3q2 to obtain 9k2 = 3q2, so q2 = 3k2, and deduces q must also be divisible by 3
B1 states that p and q being both divisible by 3 contradicts the assumption that p/q was in lowest terms (no common factor)
B1 cso: correct final conclusion: the contradiction means the original assumption was false, so √3 is irrational