Math Portfolio

MATH PORTFOLIO
NUMBER OF PIECES

Kanishk Malhotra
003566-035 (May 2012)

In physics and mathematics, the ‘DIMENSION’ of a space or object is informally defined as the minimum number of coordinates needed to specify each point within it. Thus a line has a dimension of one because only one coordinate is needed to specify a point on it. A surface such as a plane or the surface of a cylinder or sphere has a dimension of two because two coordinates are needed to specify a point on it (for example, to locate a point on the surface of a sphere you need both its latitude and its longitude). The inside of a cube, a cylinder or a sphere is three-dimensional because three coordinates are needed to locate a point within these spaces.
*taken from www.Wikipedia.com

AIM: - TO INVESTIGATE THE MAXIMUM NUMBER OF PEICES OBTAINED WHEN
AN ‘N’ DIMENSIONAL OBJECT IS CUT.

1st -DIMENSION
The first dimension is best and the easiest to start with. Lets us take an example of a line segment which is a finite one dimensional object.

Now if there is 1 cut on the line segment, then

The number of pieces obtained are 2

If there are two cuts on a line segment, then
2

The number of pieces obtained are 3

If there are three cuts on the line segment, then

The number of pieces obtained are 4

If there are four cuts on the line segment, then

Then the number of pieces obtained are 5

If there are five cuts in the line segment

The number of pieces obtained are 6.
This information is represented in a tabular form below: -

3

Number of cuts
1
2
3
4
5

Pieces
2
3
4
5
6

Thus while observing the pattern it has been seen that for the number of cuts
‘N’ the pieces obtained has been ‘N+1’
Hence the conjecture for the number of pieces obtained for one dimensional object is ‘N+1’ which means that the number of pieces for a particular number of cuts will be equal to the number of cuts + 1. This conjecture according to the question has to be defined as (S).

2nd - DIMENSION

Now let us observe a finite two dimensional figure which is a circle.

Now if there is 1 cut in the circle, then

4

The number of pieces obtained are 2

If there are two cuts in the circle, then

The number of pieces obtained are 4

If there are 3 cuts in the circle, then

Then the number of pieces obtained are 7
If there are 4 cuts in the circle, then
5

The number of pieces obtained are 11

If there are 5 cuts in the circle, then

The number of pieces obtained are 16

This information has been represented in a tabular form below: -

6

Number of
Pieces for
Pieces for 2-D (R) cuts 1-D (S)
1
2
+
2
2
3
+
4
3
4
+
7
4
5
+
11
5
6
16
By analysing the above table, the pattern can be found which is recursive so the Recursive formula for 2-D figures for ‘n’ number of cuts is Rn= S(n-1) + R(n-1)
We form the conjecture for the above by a particular method which is –
Consider the general sequence with terms U1, U2, U3, U4 and U5, and then the difference array is as follows:
Sequence

U1

U2

1st difference

U3

1U1

2nd difference

U4

1 U2

U5

1U3

2U1

2U2

1U4

2U3

We can now form a conjecture that an expression for Un in terms of U1 is given byUn= U1 + (n- 1 )

1U1 +

2U1+......+

rU1+.....+

n-1U1

Applying this for the sequence 2, 4, 7, 11, 16
Sequence

1st difference

2nd difference

2

4

7

2

3

1

11

16

4

1

5

1
7

U1=2,

1U1= 2,

2U1= 1

Un= 2+ 2(n-1) + 1
= 2+ 2(n-1) +

=

=

Hence this is the conjecture for the sequence. Thus the number of pieces made by the cuts in a second dimensional object (which is a circle) can be found out by the conjecture

. This conjecture according to the question has

to be defined as (R). Thus writing it as Rn=

.

The above conjecture can hence be proved by the principle of
Mathematical Induction.
P(n): the number of pieces made by n cuts can be denoted as
Rn=

.

Thus P(1) =

= 2. Hence P(1) is true.

Now assume that P(k) is true.
Therefore Rk=
To prove: - P(k+1) is true
Proof: For this part we would have to use recursive rule: Rn= R (n-1) + n
8

Substituting (k+1) in the place of n,
R (k+1) = Rk + (k+1)
R (k+1) =

+ k+1

=
=
=
= P (k+1)
Hence by the principle of Mathematical Induction P (n) is true for all value of Z+

Hence the conjecture can be written in the form of: =
(R)

+ n+1
(X)

..................................................... equation (1)

(S)

(X) can also be written as
Therefore equation (1) becomes

=
(R)

+ n+1
(X)

(S)

3rd - DIMENSION
Now let us observe a finite three dimensional object which is a cube.

9

Now if there is one cut in the cube, then

Number of pieces obtained are 2

If there are two cuts in the cube, then

Number of pieces obtained are 4

10

If there are three cuts in the cube, then

Number of pieces obtained are 8

If there are four cuts in the cube, then

Number of pieces obtained are 15

11

If there are five cuts in the cube, then

Number of pieces obtained are 26

This information has been represented in a tabular form below: Number of cuts 1
2
3
4
5

Pieces for
1-D (S)
2
3
4
5
6

Pieces for 2-D (R)

Pieces for 3-D (P)

2
4
7
11
16

2
4
8
15
26

+
+
+
+

By analysing the above table, the pattern can be found which is recursive, so the Recursive formula for 3-D figures for ‘n’ number of cuts is

Pn = Pn-1 + Rn-1
12

The conjecture for the sequence above is formed by a particular pattern which is: Consider the general sequence with terms U1, U2, U3, U4 and U5, and then the difference array is as follows:
Sequence

U1

U2

1st difference

U3

1U1

U4

1 U2

2nd difference

U5

1U3

2U1

2U2

1U4

2U3

We can now form a conjecture that an expression for Un in terms of U1 is given byUn= U1 + (n- 1 )

1U1 +

2U1+......+

rU1+.....+

n-1U1

Applying this to the sequence 2, 4, 8, 15, 26
Sequence

2

4

1st difference

8

2

2nd difference

4

2

1

U1 = 2,

1

2

U1 = 2,

7

3

3rd difference

U1 = 2,

15

3

26

11

4

1

U1 = 1

13

Un = 2 + 2(n-1) + 2
= 2+ 2(n-1) + 2

+

+

= 2n +
= 2n + (n-1) (n-2) +
= 2n +
=

=
Un =
Hence this is the conjecture for the sequence. Thus the number of pieces made by the cuts in a second dimensional object which is a circle can be found out by the conjecture

. This conjecture according to the question had to be

defined as (P). Hence the conjecture can be written as Pn =
The above conjecture can be proved by the principle of Mathematical
Induction.
w(n): - the number of pieces obtained by n cuts on a cube which is denoted by
Pn = w(1): -

= 2, which is true.

Hence w(1) is true.
Assume that w(k) is true.
P(k) =
14

To prove w(k+1) is true
P(k+1) =
For this part we would need to use the recursive rule to prove it further, which is: Pn = R(n-1) + P(n-1)
Where P is the number of pieces in 3-D objects,
R(n-1) is that value of R which is one less cut in a 2-D object.
P (n-1) is that value of P which is one less cut in a 3-D object.
Substituting n with (k+1) in the recursive rule: P (k+1) = Rk + Pk
P(k+1) =
=
=
P(k+1) =
= w(k+1)
Hence by the principle of mathematical induction w(n) is true for all values of
+

Z.

Therefore this formula can be written as –
..........................Equation (2)
P
Thus

Y

X

S

can be written as

Thus equation (2) can be written as

=
15

4th dimension
Observing the pattern of the parts obtained in 1-D, 2-D and 3-D we can also get the number of pieces obtained in 4th dimension by ‘n’ cuts even if 4th dimension is unknown.
The number of pieces can be found out by the rule mentioned in 3-D which is: Qn = P (n-1) + Q (n-1)
Where Q is the number of parts in 4-D object,
P (n-1) is that value of P which is one cut less in the 3-D object
Q (n-1) is that value of Q which is one less in the 4-D object.
DIMENSIONS
CUTS
1
2
3
4
5

1 (S)

2 (R)

3 (P)

2
4
4
5
6

2
4
7
11
16

2
4
8
15
26

4 (Q)
+
+
+
+

2
4
8
16
31

4th dimension 1st cut – it will always remain 2 throughout in all the dimensions as it has remained constant in the first three dimensions.
2nd cut – As stated in the rule above that Qn = P (n-1) + Q (n-1), therefore the pieces obtained in the second cut will be P (n-1) which is 2 + Q (n-1) which is 2.
Thus there will be four pieces obtained from two cuts.
3rd cut - Will be P (n-1) which is 4 + Q (n-1) which is also 4 and thus results to be 8 pieces. 4th cut- Will be P (n-1) which is 8 + Q (n-1) which is also 8 and thus results to be 16 pieces. 5th cut- Will be P (n-1) which will be 15 + Q (n-1) which will be 16 and thus will result to be 31.

16

The information about the pieces obtained by 4-D is represented in the table below: Cuts
1
2
3
4
5

Pieces
2
4
8
16
31

The conjecture for the sequence above is formed by a particular pattern which is: Consider the general sequence with terms U1, U2, U3, U4 and U5, and then the difference array is as follows:
Sequence

1st difference

U1

U2

U3

1U1

2nd difference

U4

1 U2

2U1

U5

1U3

2U2

1U4

2U3

We can now form a conjecture that an expression for Un in terms of U1 is given byUn= U1 + (n- 1 )

1U1 +

2U1+......+

rU1+.....+

n-1U1

17

Applying this to the sequence 2, 4, 8, 16, 31

Sequence

2

4

1st Difference

8

2

4

2nd Difference

8

2

3rd Difference

1U1

= 2,

31

15

4

7

2

4th Difference

U1 = 2,

16

3

1

2U1

= 2,

3U1

= 2,

4U1

=1

Un = 2 +2 (n-1) +
= 2n + (n-1)(n-2) +
= 2n +
=
=
Hence this is the conjecture for the sequence. Thus the number of pieces made by the cuts in a second dimensional object which is a circle can be found out by

18

the conjecture

.This conjecture according to the question

had to be defined as (Q). Hence the conjecture can be written as
Qn =

.

The above conjecture can be proved by the principle of Mathematical
Induction.
M (n): - the number of pieces obtained by n cuts which is denoted by
Qn =
M (1) =

=2

Hence M (1) is true
Assume that M (k) is true
Qk =
To prove M (k+1) is true
Q (K+1) =

=

For this part we would need to use the recursive rule to prove it further, which is: Qn = P(n-1) + Q(n-1)
Substituting n with (k+1) in the recursive rule: Q (k+1) = Pk + Qk
=
Q (k+1) =

= M (k+1)

19

Hence by the principle of mathematical induction M (n) is true for all values of
+

Z.
...EQ 3
Q

Z

Y

X

S

Can be written as
Thus the equation (3) can be written as
=
Q

Z

Y

X

S

CONCLUSION
As per the combination rule when ‘m’ things are selected out of ’n’ ways, it is denoted as: n

Cm =

Consider this example for the number of cuts and dimensions where ‘n’ are the number of cuts and ‘m’ are the dimensions
Consider m = 2 n C2 =
=
=

which is equal to X (

Thus we can write equation (3) stated above as: Q = nC4 + nC3 + nC2 + nC1 +1

20

Q

+1

This formula gives us the number of pieces obtained by
‘n’ cuts in the fourth dimension.
Therefore observing the pattern, the general formula for obtaining the number of pieces by ‘n’ cuts when m dimension figure is involved is: -

21