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N2 N1 N0
1 2 1
Let’s consider the base N (where N > 2)
We need to consider the “position-notation” system of representing 121:
So 121 really means N2 + 2N + 1
… which can always be written (N + 1)2 … which is a perfect square
n2 + n can be factorised:
n2 + n = n(n + 1)
If n is odd, (n+1) will be even
If (n+1) is odd, n will be even
So n(n+1) will always be: Even number x Odd number
So n2 + n will always be even
… which gives an even number
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
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
Using a visual proof:
X X X X X
X X X X X
X X X X X
X X X X X
X X X X X
The square array is
sectioned off by the red lines
as shown
These sections “capture”
1, 3, 5, 7 etc crosses
These consecutive odd
numbers of crosses make up
the square
A B
P
O
α
That is, no matter where you place point P, the angle α is always 900
Note: AB is the diameter of the circle whose centre is at O
A B
P
O
Mark in the radius OP
2 isosceles triangles are thus formedx
x y
y
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
A
B
P
O
That is, no matter where you place point P, the angle α = 2always
AB is a chord of the circle with centre O
Mark in the radius OP
3 isosceles triangles are thus formed
So we can mark in angles x, y and z
x
y
AB
P
O
y
xz
z
Angle APB = y + z
Angle AOB = 1800 - 2x
But, looking at triangle APB, 2x + 2y + 2z = 1800
1800 – 2x = 2(y + z)
Angle AOB = 2 x Angle APB
A
B
P1
That is, no matter where you place point P, the angle APB always the same – that is
=
AB is a chord of the circle, which splits the circle into 2 segments
P2
Mark in the centre O, and form the triangle AOB
Let the angle AOB be 2
AB
P1
O
P2
Thus angle AP1B will be angle at centre = double angle at circumference)
Angle AP2B will be for the same reason
Thus angle AP1B = Angle AP2B
A
B
C That is, if AB = CD, then d1 = d2
AB and CD are chords of equal length
D
O
d2
d1
P is the mid-point of AB, and Q is the mid-point of CD
Mark in OA and OC (both are radii)
A
B
C
D
O
Q
P
Triangle OPA is congruent to triangle OQC
AP = CQ
OP = OQ
