Opposite Corners  Free Essay Example

I have been given the task to investigate the differences of the products of the diagonal opposite corners of a square on a 10×10 Grid with the numbers 1 to 100 to start with.

I will start with a 2 x 2 square on a 10 x 10 grid and discover the rule for it, then I will progress onto a 3 x 3 square on the same grid.

I will then keep on going until I eventually find the rule for any sized square on a 10 x 10 grid.

2×2 Square

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

(2 x 11) – (1 x 12) = 10

(14 x 25) – (15 x 24) = 10

(8 x 17) – (7 x 18) = 10

(20 x 29) – (19 x 30) = 10

I have discovered that the answer is always 10 I will now use algebra to see if the answer is once again 10.

n

n+1

n+10

n+11

(n+1)(n+10) – n(n+11)

(n2+11n+10) – (n2+11n)

10

As the algebraic equation also gives the answer of 10 I know it must be right. As I believe I can keep on learning throughout the investigation I will now move onto a 3×3 square on the same grid. I predict that once again all answers will be the same.

3 X 3 Square

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

(3 x 21) – (1 x 23) = 40

(6 x 24) – (4 x 26) = 40

(10 x 28) – (8 x 30) = 40

I believe the answer will always be 40 for a 3 x 3 square on this grid. So I will now use algebra to see if I am correct.

n

n+2

n+20

n+22

(n+2)(n+20) – n(n+22)

(n2+20n+2n+40) – (n2+22n)

40

This proves that the answer is always 40 when a 3 x 3 square is placed on a 10 x 10 grid.

I am now going to use a 4 x 4 square on a 10 x 10 grid. The only difference will be that I will only use algebra as numbers are very time consuming.

4 x 4 square

n

n+3

n+30

n+33

(n+3)(n+30) – n(n+33)

(n2+30n+3n+90) – (n2+33n)

90

As I am doing so well I will continue to go on to a 5 x 5 square on the 10 x 10 grid.

5 x 5 Square

n

n+4

n+40

n+44

(n+4)(n+40) – n(n+44)

(n2+40n+4n+160) – (n2+44n)

160

I am now going to create a table displaying my results so far

Square

Opposite Corners

2 x 2

10

3 x 3

40

4 x 4

90

5 x 5

160

6 x 6

???

I have noticed a pattern: the opposite corners are all square numbers of the square before multiplied by 10. Hence forth I predict that the product form the 6 x 6 square will have a difference of 250, I will now try it and see if my prediction is correct.

n

n+5

n+50

n+55

(n+5)(n+50) – n(n+55)

(n2+50n+5n+250) – (n2+55n)

250

I was correct. I will now attempt to convert my verbal rule into algebraic terms for any square on a 10 x 10 grid.

n

n+(m-1)

N+10(m-1)

N+11(m-1)

(n+(m-1))(n+10(m-1)) – n(n+11(m-1))

(n2+10n(m-1)+n(m-1)+10(m-1) 2)-(n2+11n(m-1))

10(m-1) 2

I have now discovered the rule for any square on a 10 x 10 grid I am now going to try and find a general rule which will allow me to work out on any square on any grid. As the rule for any square on a 10 x 10 grid is 10(m-1) 2 I predict that if were to use a 7 x 7 grid the rule would be 7(m-1) 2 . I will now see if I am correct. However I will once again only use algebra as numbers are too time consuming. I will use a 2 x 2 square on the grid to start with.

2 x 2 square

n

N+1

n+7

N+8

(n+1)(n+7) – n(n+8)

(n2+7n+n+7) – (n2+8n)

7

I think I have already spotted a trend as this occurred when I completed the work for the 10 x 10 grid. I think the rule for any square on a 7 x 7 grid will be 7(m-1) 2

I will now try to see if I am correct.

n

n+(m-1)

n+7(m-1)

n+8(m-1)

(n+(m-1)(n+7(m-1)) – n(n+8(m-1))

n2+7n(m-1)+n(m-1)+7(m-1) 2

7(m-1) 2

The answer is exactly what I predicted I am now going to find a general rule for any square (m x m) on any grid (g x g). This is what my grid would look like…

1

2

3

4

1+g

2+g

……

……

1+2g

……

……

……

Here is the process I will use to solve the general rule:

n

n+(m-1)

n+g(m-1)

n+(m-1)(g+1)

n+(m-1)(n+g(m-1)) – n(n(m-1)(g+1))

(n2+ng(m-1)+n(m-1)+g(m-1) 2

(n2+n(g+1)(m-1)+g(m-1) 2) – (n2+n(m-1)(g+1))

g(m-1) 2

Now I have completed my aims and found the general rule I will not have to use numbers again to find the opposite corners of a square. I have done this through progressive investigation and through algebraic methods. I have taken this piece of coursework as far as I can with squares and I am therefore going to progress onto using rectangles as they are similar shapes. As I have done lots of number and algebra work finding the formula for individual squares I am simply going to find the general rule straight away for any rectangle on any grid.

The letters I will use are:

L: the length of the rectangle must be bigger than 1 and equal to or smaller than the grid

size.

H: the height of the rectangle. This is a non-restrictive measurement if it is required to be

as extra rows can be added easily to the bottom of the grid. Must be a whole positive

integer

G: The grid size which must be at least 2 x 2 upwards so that it has opposing corners. It

must also be a whole, positive integer.

n

n+1

……

n+(L-1)

n+G

……

……

……

n+2G

……

……

……

……

……

……

……

n+G(H-1)

……

……

n+G(H-1)+(L-1)

(n+(L-1))(n+G(H-1)) – n(n+G(H-1)+(L-1))

n2+nG(H-1)+n(L-1)+(G(L-1)(H-1)) – (n2+nG(H-1)+n(L-1))

G(L-1)(H-1)

I have found my rule for any rectangle on any size of grid the algebraic rule is

G(L-1)(H-1) or subtract one from the length and one from the height and multiply by the grid size. I have completed the extension of the original task and did no think I could progress any further. However I discovered one aspect that I have not even mentioned before the difference between the numbers in the grid. For instance I could instead of having; 1, 2, 3, 4, 5… I could have; 2, 4, 6, 8, 10… as I have in the last few investigations I will only use algebra as numbers as well are to time consuming.

These are the algebraic terms I will use in my formulae:

L: the length of the rectangle must be bigger than 1 and equal to or smaller than the grid

size.

H: the height of the rectangle. This is a non-restrictive measurement if it is required to be

as extra rows can be added easily to the bottom of the grid. Must be a whole positive

integer

G: The grid size which must be at least 2 x 2 upwards so that it has opposing corners. It

must also be a whole, positive integer.

S: The gap between the two numbers can be any number; positive, negative, decimals,

Fractions etc.

This is what my grid would look like

n

n+S

n+2S

……

n(g-1)S

n+GS

n+2GS

……

……

……

……

……

This is how I will work out the rule for any rectangle on any sized grid with any gap size between the numbers.

n

n+S(L-1)

n+GS(H-1)

n+S(L-1)+GS(H-1)

(n+S(L-1))(n+GS(H-1) – n(n+s(L-1)+GS(H-1))

(n2+nGS(H-1)+ns(L-1)+S2G(L-1)(H-1)) – (n2+ns(L-1)+nGS(H-1)

S2G(L-1)(H-1)

For Example

So if:

S= -0.1

G= 5

L=2

H=3

S2G(L-1)(H-1)

0.01 x 5 x 1 x 2 = 0.1

For Example

22

21.9

21.8

21.7

21.6

21.5

21.4

21.3

21.2

21.1

21.0

20.9

(21.9 x 21) – (22 x 20.9)

459.9 – 459.8

0.01

In conclusion I feel I have gone as far as possible with this investigation as I have covered all aspects of it as I can possibly think off. I am very happy with my results and have enjoyed doing this investigation.