A number of simple ideas for Grade 10 mathematicians around the notion of mathematical proof.

3. Let’s consider the base N (where N > 2)
We need to consider the “position-notation” system of
representing 121:
N2 N1 N0
1 2 1
So 121 really means N2 + 2N + 1
… which can always be written (N + 1)2 … which is a perfect square

5. n2 + n can be factorised:
n2 + n = n(n + 1)
So n(n+1) will always be:
If n is odd, (n+1) will be even Even number x Odd number
… which gives an even number
If (n+1) is odd, n will be even
So n2 + n will always be even

7. n3 – n can be factorised
n3 – n = n(n2 – 1) = n(n – 1)(n + 1)
It’s especially helpful to re-write the RHS thus: (n – 1)n(n + 1)
This is the product of 3 consecutive numbers
One of these numbers must be a multiple of 3
So the product, and therefore n3 – n must be divisible by 3

9. Algebraically:
The sum is: 1 + 3 + 5 + 7 + … + (2n – 1)
And this sum can be found, using the formula for
the sum of an AP (with a=1 and d=2):
Using Sn = ½n[2a + (n-1)d]
Sn = ½n[2(1) + 2(n-1)] = ½n[2 + 2n – 2] = ½n(2n)
Sn = n2 That is, Sn is always a square number

10. Using a visual proof:
The square array is
X X X X X sectioned off by the red lines
as shown
X X X X X
X X X X X These sections “capture”
1, 3, 5, 7 etc crosses
X X X X X
X X X X X
These consecutive odd
numbers of crosses make up
the square

11. P
α
That is, no matter where
you place point P, the
A B
O
angle α is always 900
Note: AB is the diameter of the
circle whose centre is at O

12. P
x y Mark in the radius OP
2 isosceles triangles are thus formed
x y
A B
O So we can mark in angles x and y
Now we add the angles in Triangle APB
The angle sum is x + x + y + y = 2x + 2y
So 2x + 2y = 1800 x + y = 900
So angle APB is always 900

13. P
That is, no matter where you
place point P, the angle α = 2
O
always
B AB is a chord of the circle with
A centre O

14. P
Mark in the radius OP
z y 3 isosceles triangles are thus formed
So we can mark in angles x, y and z
O
y Angle APB = y + z
z
x Angle AOB = 1800 - 2x
x B
A
But, looking at triangle APB,
2x + 2y + 2z = 1800
1800 – 2x = 2(y + z)
Angle AOB = 2 x Angle APB

15. That is, no matter where you
P1 place point P, the angle APB
always the same – that is
P2
=
AB is a chord of the circle, which
B splits the circle into 2 segments
A

16. P2
Mark in the centre O, and form
P1 the triangle AOB
Let the angle AOB be 2
O
Thus angle AP1B will be
angle at centre = double angle at circumference)
B
A
Angle AP2B will be for the same reason
Thus angle AP1B = Angle AP2B

17. C
That is, if AB = CD, then d1 = d2
d2 D
O
d1
AB and CD are chords of equal length
B
A

18. C P is the mid-point of AB, and Q is the
Q mid-point of CD
D AP = CQ
O Mark in OA and OC (both are radii)
Triangle OPA is congruent to triangle
B OQC
P
A
OP = OQ