(b) Here are the results of the 5 calculations for 2x2 Box on Width 12 Grid (Fig 2.2):
(c) Here are the results of the 5 calculations for 2x2 Box on Width 13 Grid (Fig 2.3):
(d) Here are the results of the 5 calculations for 2x2 Box on Width 14 Grid (Fig 2.4):
(e) Here are the results of the 5 calculations for 2x2 Box on Width 15 Grid (Fig 2.5):
From these results, it is possible to take the calculated difference of the two products, and plot this against the width of the grid (z):
Fig 2.6
4) Data Analysis
From the tables (a)-(e), it is possible to see that with a 2x2 box, the difference of the two products always equals z: the width of the grid. When plotted on a graph (Fig 2.6), the relationship is clearly visible as a perfect positive correlation. A graph is a good choice to show this relationship however, a straight line drawn between these would not be correct because with this particular problem, the values to be inputted into equations must be natural numbers i.e. Integers > 0.
5) Generalisation
Using this apparent relationship, it can be assumed that, when a 2x2 box is placed anywhere on the grid, the difference of the two products will be z for all possible widths. Therefore, the following equation should be satisfied with any real value of a and any real value of z where:
a is the top-left number in the box;
(a + 1) is the top-right number in the box, because it is always “1 more” than a;
(a + z) is the bottom-left number in the box, because it is always “the grid width (z)” more than a;
(a + z + 1) is the bottom-right number in the box, because it is always “the grid width (z) plus 1” more than a.
(a + 1)(a + z) - a(a + z + 1) = z
This means that I predict that with a 2x2 box on a width z grid, the difference of the two products will always be z, the width of the grid.
6) Testing
My formula works as shown with the following, previously unused values:
1) Where a = 9, and z = 16
2) Where a = 72 and z = 17
7) Justification
The formula can be proven to work with the following algebra:
The formula works because the “z” term is only produced in the expansion of the two brackets on the left of the minus sign, and not from the more simply factorised “a(a + z + 1)” term. The a2 term is present on both sides of the minus sign, as are the az and a terms. Therefore, they cancel each other out to leave z.
8) Conclusion
After this justification, it can now be said that for every 2x2 box on a Width z Grid, the difference of the two products will always be z.
9) Extension
Having done this, I saw that my formula would only work for 2x2 boxes on a Width z grid. To improve the usefulness of my formula, I wondered what would happen to the difference of the two products if I varied the length of the box i.e. made it a 3x3 or 4x4.
Section 3: “p x p” Box on Width 10 Grid
1) Introduction
Throughout this section, the variable p will be used to represent the length of the square box. The variable a will continue to be used for the top-left number in the box i.e. the location of the box upon the grid.
2) Method
Varying values of p will be tested to give different lengths of sides for the boxes. The lengths will range from 3 to 7. With these boxes, in 5 different random locations on the width 10 grid, the differences of the two products will be calculated. Again, I believe 5 calculations are enough to display any patterns.
3) Data Collection
Fig 3.1
Fig 3.1 is the width 10 grid, with 3x3 to 7x7 example boxes.
a) Here are the results of the 5 calculations for 3x3 Box on Width 10 Grid:
b) Here are the results of the 5 calculations for 4x4 Box on Width 10 Grid:
c) Here are the results of the 5 calculations for 5x5 Box on Width 10 Grid:
d) Here are the results of the 5 calculations for 6x6 Box on Width 10 Grid:
e) Here are the results of the 5 calculations for 7x7 Box on Width 10 Grid:
4) Data Analysis
From tables (a)-(e), it is possible to see that all the differences of the products tested are multiples of 10. It also seems that there is a pattern, where the difference equals “the length of the box minus 1” squared multiplied by 10.
5) Generalisation
Using this apparent rule, it can be assumed that for all possible locations of a box of side p on a width 10 grid, the difference of the two products is always “the length of the box minus 1” squared, multiplied by 10. Therefore the following equation should be satisfied with any real value of a, and any real value of p where:
a is the top-left number in the box;
(a + [p – 1]) is the top-right number in the box, because it is always “[p – 1] more” than a;
(a + 10[p – 1]) is the bottom-left number in the box, because it is always “10 multiplied by [p – 1]” more than a;
(a + 10[p – 1] + [p – 1]) is the bottom-right number in the box, because it is always “[p – 1] plus 10 multiplied by [p – 1]” more than a.
For simplicity, d = [p – 1]:
(a + d)(a + 10d) - a(a + 10d + d) = 10d2
a + 10[p - 1] a + 10[p - 1] + [p - 1]
This means that I predict that with a p x p box on a width 10 grid, the difference of the two products will always be 10[p – 1]2, “the length of the box minus 1” squared, multiplied by 10.
6) Testing
My formula works as shown with the following, previously unused values:
1) Where a = 12, and p = 8: I predict 10 x [p – 1]2 = 10 x 49 = 490
2) Where a = 2, and p = 9: I predict 10 x [p – 1]2 = 10 x 64 = 640
7) Justification
The formula can be proven to work with the following algebra (where d = [p – 1]):
The formula works because the “10d2” term is only produced in the expansion of the two brackets on the left of the minus sign, and not from the more simply factorised “a(a + 10d + d)” term. The a2 term is present on both sides of the minus sign, as is the 11ad term. Therefore, they cancel each other out to leave “10d2”.
8) Conclusion
After this justification, it can now be said that for every possible p x p box on a Width 10 Grid, the difference of the two products will always be 10(p – 1)2.
9) Extension
Having done this, I saw that my formula would only work for square boxes on a width 10 grid. To improve the usefulness of my formula, I wondered what would happen to the difference of the two products if I varied the length of the box and the width of the box i.e. made it a 2x3, 2x4, 3x5, 3x2, 4x7, 4x9 etc.
Section 4: “p x q” Box on Width 10 Grid
1) Introduction
Throughout this section, the variable q will be used to represent the length of the box extending down the grid and the variable p will be used to represent the length of the box extending across the grid. The variable a will continue to be used for the top-left number in the box i.e. the location of the box upon the grid.
2) Method
Random varying values and combinations of p and q will be tested to give different lengths and widths of sides for the boxes. The lengths will range from 2 to 5. With these boxes, in 5 different random locations on the width 10 grid, the differences of the two products will be calculated. Again, I believe 5 calculations are enough to display any patterns.
3) Data Collection
Referring back to Fig 3.1, it shows examples of the varying box sizes. The only difference is, the boxes may now be made into rectangles.
a) Here are the results of the 5 calculations for 2x4 Box on Width 10 Grid:
b) Here are the results of the 5 calculations for 3x6 Box on Width 10 Grid:
c) Here are the results of the 5 calculations for 5x3 Box on Width 10 Grid:
4) Data Analysis
From ‘Section 3’, I proved the formula, “Difference = 10(p – 1)2”. If (p – 1) is the variable representing the side of the square, then the area calculated from that is (p – 1)2, or (p – 1)(p – 1). Because each of these brackets represents the side of the square, when it becomes a rectangle, it is safe to assume that the area will be:
(p – 1)(q – 1)
Bearing this in mind whilst analysing this data helped me realise that the difference of the two products, say in a 5x3 box, was 4x2x10 = 80, and in a 3x6 box, was 2x5x10 = 100. The relationship was such that the difference was the “length of the box minus 1”, times the “width of the box minus 1” times 10. I noticed the “10” from, again, all the differences being multiples of 10.
5) Generalisation
Using this apparent relationship, it can be assumed that for all possible locations of a box of one side p, and the other q, on a width 10 grid, the difference of the two products is always “the length of the box minus 1” times the “width of the box minus 1” multiplied by 10. Therefore the following equation should be satisfied with any real value of a, any real value of p, and any real value of q where:
a is the top-left number in the box;
(a + [p – 1]) is the top-right number in the box, because it is always “[p – 1] more” than a;
(a + 10[q – 1]) is the bottom-left number in the box, because it is always “10 multiplied by [q – 1]” more than a;
(a + 10[q – 1] + [p – 1]) is the bottom-right number in the box, because it is always “10 times [q – 1]” plus “[p – 1]” more than a.
For simplicity, d = [p – 1], and e = [q – 1]:
(a + d)(a + 10e) - a(a + 10e + d) = 10de
a + 10[q – 1] a + 10[q - 1] + [p - 1]
This means that I predict that with a p x q box on a width 10 grid, the difference of the two products will always be 10(p – 1)(q – 1), “the length of the box minus 1” multiplied by “the width of the box minus 1”, multiplied by 10.
6) Testing
My formula works as shown with the following, previously unused values:
1) Where a = 12, p = 4, and q = 5: I predict 10 x [p – 1][q – 1] = 10 x 3 x 4 = 120
2) Where a = 16, p = 6 and q = 7: I predict 10 x [p – 1][q – 1] = 10 x 5 x 6 = 300
7) Justification
The formula can be proven to work with the following algebra (where d = [p – 1], and e = [q – 1]):
The formula works because the “10de” term is only produced in the expansion of the two brackets on the left of the minus sign, and not from the more simply factorised “a(a + 10e + d)” term. The “a2” term is present on both sides of the minus sign, as is the “10ae” term and the “ad” term. Therefore, they cancel each other out to leave “10de”.
8) Conclusion
After this justification, it can now be said that for every possible p x q box on a Width 10 Grid, the difference of the two products will always be 10(p – 1)(q – 1).
9) Extension
Having done this, I saw that my formula would only work for boxes on a width 10 grid. To improve the usefulness of my formula, I wondered what would happen to the difference of the two products if I varied the length of the box, the width of the box and the grid width.
Section 5: “p x q” Box on Width z Grid
1) Introduction
Throughout this section, the variable q will be continue to be used to represent the length of the box extending down the grid and the variable p will continue to be used to represent the length of the box extending across the grid. The variable z will continue to be used for the width of the grid and the variable a will continue to be used for the top-left number in the box i.e. the location of the box upon the grid.
2) Method
Varying values and combinations of p and q will be tested to give different lengths and widths of sides for the boxes. Simultaneously, varying values of z will be tested to give different widths of grids. With these boxes, in 5 different random locations on the width z grid, the differences of the two products will be calculated. Again, I believe 5 calculations are enough to display any patterns.
3) Data Collection
Referring back to Figs 2.1 to 2.5, these will continue to be used as examples of grid widths 11 to 15. Fig 3.1 shows examples of the varying box sizes, except that they can be rectangular.
a) Here are the results of the 5 calculations for 2x4 Box on Width 11 Grid:
b) Here are the results of the 5 calculations for 3x6 Box on Width 13 Grid:
c) Here are the results of the 5 calculations for 5x3 Box on Width 14 Grid:
4) Data Analysis
As found in ‘Section 4’, (p – 1)(q – 1) is the formula for an area relating to the box. When the grid width was 10, as in the last section, the formula for the difference of the two products was 10(p – 1)(q – 1). Being 10, I guessed that the ‘10’ referred to the grid width.
As it happened, I noticed that all the differences tested in this section were multiples of their respective grid widths i.e. 112 is a multiple of 14 and 33 is a multiple of 11. I also noticed that the (p – 1)(q – 1) was still correct. The “length of the box minus 1”, times the “width of the box minus 1”, times the “grid width” equals the difference.
5) Generalisation
Using this apparent relationship, it can be assumed that for all possible locations of a box of one side p, and the other q, on a width z grid, the difference of the two products is always “the length of the box minus 1” times the “width of the box minus 1” times the “grid width”. Therefore the following equation should be satisfied with any real value of a, any real value of p , any real value of q, and any real value of z where:
a is the top-left number in the box;
(a + [p – 1]) is the top-right number in the box, because it is always “[p – 1] more” than a;
(a + z[q – 1]) is the bottom-left number in the box, because it is always “z multiplied by [q – 1]” more than a;
(a + z[q – 1] + [p – 1]) is the bottom-right number in the box, because it is “z multiplied by [q – 1]” plus “[p – 1]” more than a.
For simplicity, d = [p – 1], and e = [q – 1]:
(a + d)(a + ze) - a(a + ze + d) = zde
a + z[q – 1] a + z[q - 1] + [p - 1]
This means that I predict that with a p x q box on a width z grid, the difference of the two products will always be z(p – 1)(q – 1), “the length of the box minus 1” multiplied by “the width of the box minus 1”, multiplied by the “grid width”.
6) Testing
My formula works as shown with the following, previously unused values:
1) Where a = 12, p = 4, q = 5, and z = 17: I predict z x [p – 1][q – 1] = 17 x 3 x 4 = 204
2) Where a = 16, p = 6, q = 7, and z = 23: I predict z x [p – 1][q – 1] = 23 x 5 x 6 = 690
7) Justification
The formula can be proven to work with the following algebra (where d = [p – 1], and e = [q – 1]):
The formula works because the “10de” term is only produced in the expansion of the two brackets on the left of the minus sign, and not from the more simply factorised “a(a + 10e + d)” term. The “a2” term is present on both sides of the minus sign, as is the “10ae” term and the “ad” term. Therefore, they cancel each other out to leave “10de”.
8) Conclusion
After this justification, it can now be said that for every possible p x q box on a Width z Grid, the difference of the two products will always be z(p – 1)(q – 1).
Investigation Conclusion and Evaluation
From the simple study of a 2x2 square box on a width 10 grid, I have been able to progress all the way to the formula to find the difference of the two products on any rectangular box on any width grid. The 2x2 box given in the project brief was useful because, now, I can see that it gives the number “1” in (p – 1)(q – 1). This is, therefore, very useful for spotting patterns i.e. the differences with different grid widths. However, when advancing into more complex areas of the project, like the last section where three different variables were altered, spotting patterns became more difficult and knowledge from the prior sections was required to find the formula.
Overall, the best way of presenting the results was in tables because there was an obvious pattern of values in the “Difference” column, because they were all in one line. A graph or diagram would not have been as suitable because the patterns would not have been as apparent. However, a graph was useful in proving that the grid width has a bearing on the difference with a 2x2 box. It showed a perfect positive correlation which meant that the formula was visibly true for all real grid width values. (N.B. “Real”, in this investigation, meant a number that could be used within the problem itself – as it happens, these were natural numbers.)
As a result of this investigation, I can now present a formula which has been proven to work for all real values inputted:
z(p – 1)(q – 1)
