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Homework Sheet – Algebraic Proof
1.
A
x2 – 1 = (x-1)(x+1)
B
3x2 + 2x -1
C
s = ut +
D
7x + 1 = 5x - 4
1
2
at2
In the table above, which of A, B, C or D is:
(a) an equation
(b) a formula
(c) an identity
(d) an expression?
2.
Amy says that for every whole number, 6n – 1 is always a prime number.
Give a counter example to show that she is wrong.
3.
Prove that the answer to every line of the pattern below is 6
3x4-1x6
4x5-2x7
5x6-3x8
6x7-4x9
..
..
..
..
4.
Prove that the sum of two odd numbers is even.
5.
Prove that the product of two odd numbers is odd.
6.
For any three consecutive numbers prove that the product of the first and third numbers is
always one less than the square of the middle number.
7.
Below is Joe’s attempt at forming a proof that 1 = 2
Let a = b

ab = b2
(multiply by b)
 ab – a2 = b2 – a2
(subtract a2)
 a(b - a) = (b – a)(b+ a)
(factorise)

a=b+a
(divide by b-a)

a=a+a
(a = b)

a = 2a
(a + a = 2a)

1=2
(divide by a)
Which step in Joe’s argument is not allowed?
Algebraic Proof
Q
Answers and Mark Scheme
Answer
Mark
1(a)
D
B1
1(b)
C
B1
1(c)
A
B1
1(d)
B
B1
When n = 6, 6n – 1 = 35
M1
35 is not a prime number
A1
General line is (n + 2)(n + 3) - n(n + 5)
M1
= (n2 + 5n + 6) – (n2 + 5n)
A1
2.
3.
= 6 as required
4.
Let x and y be integers
 2x and 2y are even integers
M1
2x + 1 and 2y + 1 are odd numbers
M1
(2x + 1) + (2y + 1) = 2x + 2y + 2
5.
= 2(x + y + 1) which is a multiple of 2 – hence
A1
even
E1
Let x and y be integers
 2x and 2y are even integers
M1
2x + 1 and 2y + 1 are odd numbers
M1
(2x + 1) x (2y + 1) = 4xy +2x + 2y + 1
M1
= 2(2xy + x + y) + 1
A1
2(2xy + x + y) is a multiple of 2 –
hence even
2(2xy + x + y) + 1 is odd
E1
Comments
or n = 11 etc
Q
6.
Answer
Mark
Comments
Let the first of the three numbers be x
Then next consecutive number is x + 1
M1
M1 for setting up expressions
(x + 1)2 = x2 + 2x + 1
M1
The product of the first and third number is
x(x + 2) = x2 + 2x
M1 for multiplying products as
required
A1
A1 for x2 + 2x + 1 and x2 + 2x
And final consecutive number is x + 2
The square of the middle number is
7.
Hence the product of the first and third
numbers is one less than the square of the
middle number.
E1
Dividing by (b – a) is division by zero, since
a = b, and hence not allowed.
E1