(top right x bottom left) – (top left x bottom right)
= 36 x 72 – 32 x 76
= 2592 – 2432
= 160
So I successfully managed to predict the difference between the products of opposite corners in a 5 x 5 square, and therefore I was correct that the difference between the products of opposite corners goes up in square numbers multiplied by 10 for ever increase in size of the grid
Formula
If I go back to my table, I can work out a formula for the difference between the products of opposite corners in different sized squares.
I will use this to make a formula for the difference between the products of opposite corners, D, using the length or width of the square, n. I can see that, if 1 is subtracted from the length or width of the square and that answer is squared, and then multiplied by 10, the final answer is the difference between the products of opposite corners. If I put this into an algebraic formula, I would therefore get:
D = 10(n – 1)2
I can test this by substituting 2 in for n, and if the answer is 10, which was the answer when I proved that the difference between the products of opposite corners in all 2 x 2 squares equals 10, my formula is correct.
D = 10(n – 1)2 (n = 2)
D = 10(2 – 1)2
= 10 x 12
= 10 x 1 = 10
As my formula has passed this test, I will use it to work out the difference between the products of opposite corners in a 9 x 9 square:
D = 10(n – 1)2 (n = 9)
D = 10(9 – 1)2
= 10 x 82
= 10 x 64 = 640
To see if this is right, I will use an example of a 9 x 9 square and see if I get 640 again, showing that my formula can successfully predict the difference between the products of opposite corners of any sized square.
(top right x bottom left) – (top left x bottom right)
= 9 x 81 – 1 x 89
= 729 – 89
= 640
So my formula has worked; I managed to predict the difference between the products of opposite corners of a 9 x 9 square by using it.
9 x 9 Grids
In the next stage I will look at another variable: the width of the grid that contains the square. Previously it had a width of 10, but I will now try a width of 9.
Taking the highlighted 2 x 2 square, I will work out the difference between the products of opposite corners of the grid in the same way as I have been doing.
(top right x bottom left)–(top left x bottom right)
= 32 x 40 – 31 x 41
= 1280 – 1271
= 9
I will do another example of this to see if the difference between the products of opposite corners still equals 9.
(top right x bottom left) – (top left x bottom right)
= 61 x 69 – 60 x 70
= 4209 – 4200
= 9
As it is 9 once more in this example, I will now use algebra to try and prove that 9 is the difference between the products of opposite corners in all 2 x 2 squares within 9 x 9 grids, by substituting letters in for the numbers.
Let the top left number in a square equal a, and therefore:
So I can work out the difference between the products of opposite corners by:
(top right x bottom left) – (top left x bottom right)
= (a + 1) (a + 9) – a (a + 10)
= a2 + a + 9a + 9 – a2 – 10a
= a2 + 10a + 9 – a2 – 10a
= (a2 – a2) + (10a – 10a) + 9
= 9
This algebra has thus proved the difference between the products of opposite corners in a 2 x 2 square in 9 x 9 grids will always equal 9.
I will go on to investigate the difference between the products of opposite corners of 3 x 3 squares inside 9 x 9 grids:
(top right x bottom left) – (top left x bottom right)
= 17 x 33 – 15 x 35
= 561 – 525
= 36
I will do another example of this to see if the difference between the products of opposite corners is 36 again:
(top right x bottom left) – (top left x bottom right)
= 49 x 65 – 47 x 67
= 3185 – 3149
= 36
I will prove this using algebra, to show that all 3 x 3 squares in 9 x 9 grids have a difference between the products of opposite corners of 36.
Let the top left number in a square equal a, and therefore:
So I can now prove the difference between the products of opposite corners in 3 x 3 grids by:
(top right x bottom left) – (top left x bottom right)
= (a + 2) (a + 18) – a (a + 20)
= a2 + 20a + 36 – a2 – 20a
= 36
Table of all Results
If I put all my results for 10 x 10 grids and 9 x 9 grids in a table, it may be possible to see some patterns and work out some formulae.
I can see from this that the difference between the products of opposite corners in 9 x 9 grids is the square of one less than the size of the square, multiplied by 9, so I can get the formula:
D = 9(n – 1)2
I will prove this formula by working out the difference between the products of opposite corners in a 5 x 5 square within a 9 x 9 grid:
D = 9(n – 1)2 (n = 5)
D = 9(5 – 1)2
= 9 x 42
= 9 x 16 = 144
If I try an actual example of a 5 x 5 square in a 9 x 9 grid and the difference between the products of opposite corners equals 144, I have proved this formula works.
(top right x bottom left) – (top left x bottom right)
= 6 x 38 – 2 x 42
= 228 – 84
= 144
So my formula works. If I go back to my table:
I can see a pattern between my formulae for the difference between the products of opposite corners in any squares in 10 x 10 and 9 x 9 grids.
The two formulae are:
10 x 10: D = 10(n – 1)2
9 x 9: D = 9(n – 1)2
So clearly the width of the grid, w, is the same as the number that is multiplied by (n – 1)2. Therefore I can come up with this revised formula, for any sized square in any sized grid:
D = w (n – 1)2
I can test this formula by working out the difference between the products of opposite corners in a 3 x 3 square in a 7 x 7 grid. If I put these figures into my formula, I would get:
D = w (n – 1)2 (n = 3) (w = 7)
D = 7(3 – 1)2
= 7 x 22
= 7 x 4 = 28
If I draw out the start of a 7 x 7 grid, I can check this:
A 3 x 3 square out of this would be:
So the difference between the products of opposite corners in this square would be:
(top right x bottom left) – (top left x bottom right)
= 10 x 22 – 8 x 24
= 220 – 192
= 28
So my formula works in this example, as it correctly predicted the difference between the products of opposite corners.
Proving the Formula
I will now use algebra to prove my universal formula will work in a square of any size, in a grid of any width.
In a grid with w width and the top left corner of an n-sized square being a, you would get:
Top Left = a Top Right = a + (n – 1)
= a + n – 1
Bottom Left = a + w (n – 1) Bottom Right = a + w (n – 1) + (n – 1)
= a + nw – w = a + nw – w + n – 1
So the difference between the products of opposite corners of this could be calculated as follows:
(top right x bottom left) – (top left x bottom right)
= (a + n – 1) (a + nw – w) – (a (a + nw – w + n – 1))
= a2 + anw – aw + an + n2w – 2nw – a + w – (a2 + anw – aw + an – a)
= a2 + anw – aw + an + n2w – 2nw – a + w – a2 – anw + aw – an + a
= n2w – 2nw + w
= w (n2 – 2n + 1)
= w ((n – 1) (n – 1))
= w (n – 1)2
As the formula I worked out was D = w (n – 1)2 and the expression for the difference between the products of opposite corners in any square within a grid of any width cancelled down to w (n – 1)2, I have now proved that my formula will work to find the difference between the products of opposite corners in any sized square within any sized grid.
Rectangles
Now I have done as much as I can with squares, I will move onto rectangles, and have an extra variable; I now have the width of the grid, the length of the rectangle and the width of the rectangle.
I will start by looking at 2 x 3 rectangles in 10 x 10 grids. An example of one of these would be:
So I will now look at calculating the difference between the products of opposite corners in this rectangle:
(top right x bottom left) – (top left x bottom right)
= 8 x 16 – 6 x 18
= 128 – 108
= 20
I will use algebra to prove that the difference between the products of opposite corners in each 2 x 3 rectangle in a 10 x 10 grid equals 20.
Let the top left square in the rectangle equal a, and therefore:
So I can use algebra to make an expression for the difference between the products of opposite corners by:
(top right x bottom left) – (top left x bottom right)
= (a + 2) (a + 10) – a (a + 12)
= a2 + 2a + 10a + 20 – a2 – 12a
= a2 + 12a + 20 – a2 – 12a
= (a2 – a2) + (12a – 12a) + 20
= 20
So this must always be the case for 2 x 3 rectangles in 10 x 10 grids.
Next I will try 2 x 4 rectangles and see how my results differ from those of 2 x 3 rectangles, for example:
The difference between the products of opposite corners of this rectangle would equal:
(top right x bottom left) – (top left x bottom right)
= 55 x 62 – 52 x 65
= 3410 – 3380
= 30
My next stage will be to prove that all 2 x 4 rectangles in 10 x 10 grids have a difference between the products of opposite corners of 30.
Let the top left square in the rectangle equal a, and therefore:
So the algebraic expression for the difference between the products of opposite corners would be:
(top right x bottom left) – (top left x bottom right)
= (a + 3) (a + 10) – a (a + 13)
= a2 + 3a + 10a + 30 – a2 – 13a
= a2 + 13a + 30 – a2 – 13a
= (a2 – a2) + (13a – 13a) + 30
= 30
As I have proved the difference between the products of opposite corners in all 2 x 4 rectangles in 10 x 10 grids, I will now move on to 2 x 5 rectangles.
The difference between the products of opposite corners of this rectangle would equal:
(top right x bottom left) – (top left x bottom right)
= 5 x 11 – 1 x 15
= 55 – 15
= 40
To prove that this is the same in all 2 x 5 grids, I will have to use algebra. Let the top left square of my rectangle equal a, and therefore:
To find the difference between the products of opposite corners…
(top right x bottom left) – (top left x bottom right)
= (a + 4) (a + 10) – a (a + 14)
= a2 + 4a + 10a + 40 – a2 – 14a
= a2 + 14a + 40 – a2 – 14a
= (a2 – a2) + (14a – 14a) + 40
= 40
Therefore, all 2 x 5 rectangles in 10 x 10 grids have a difference between the products of opposite corners of 40, as I have proved.
Table of Results for Rectangles
My next job is to put all my results for rectangles in a table, to look for potential patterns and formulae.
Obviously, the difference is increasing in steps of 10 for each increase of 1 in the width, but now I shall put this into a formula relating the length and the width to the difference.
I can see that if I subtract 1 from both the length and the width and multiply these numbers together, then multiplying the product by 10, I get the difference. I can put this into an algebraic formula, like so:
D = 10 (m – 1) (n – 1)
This also fits with my formula for squares as, in squares the length and the width are equal, (m – 1) (n – 1) would equate to (n – 1)2 in a square.
I will check this formula by trying a 3 x 4 rectangle, which, according to my formula would have a difference between the products of opposite corners of:
D = 10 (m – 1) (n – 1) (m = 3) (n = 4)
D = 10 (3 – 1) (4 – 1)
= 10 x 2 x 3
= 20 x 3 = 60
If I do an example of a 3 x 4 rectangle, such as this one, I can check my formula:
The difference between the products of opposite corners of this rectangle would equal:
(top right x bottom left) – (top left x bottom right)
= 10 x 27 – 7 x 30
= 270 – 210
= 60
My formula has worked in this example.
Grids of Different Widths and Rectangles
I can see the similarities between this formula and my original formula for squares in 10 x 10 grids.
My original formula was D = 10 (n – 1)2 and this later developed to D = w (n – 1)2, with w being the width of the grid. I can predict that, in my formula for any size rectangle, D = 10 (m – 1) (n – 1), the 10 will equal the width, and therefore my theory for the formula for any size rectangle in any size grid is:
D = w (m – 1) (n – 1)
I can check this by trying a 4 x 5 rectangle in an 8 x 8 grid:
D = w (m – 1) (n – 1) (w = 8) (m = 4) (n = 5)
D = 8 (4 – 1) (5 – 1)
= 8 x 3 x 4
= 24 x 4 = 96
I can check this by drawing out an example of a 4 x 5 rectangle in an 8 x 8 grid and finding the difference between the products of opposite corners:
(top right x bottom left) – (top left x bottom right)
= 7 x 27 – 3 x 31
= 189 – 93
= 96
So my prediction that I could develop my formula to become D = w (m – 1) (n – 1) was correct in this example.
Proving the Formula
The formula now needs proving using algebra, which I will do in a similar way to how I proved my formula for squares:
Let the top left square in an n x m rectangle within a w sized grid equal a, and therefore:
Top Left = a Top Right = a + (n – 1)
= a + n – 1
Bottom Left = a + w (m – 1) Bottom Right = a + w (m – 1) + (n – 1)
= a + mw – w = a + mw – w + n – 1
If I use these expressions in my original calculation for the difference between the products of the opposite corners, I can prove my formula:
(top right x bottom left) – (top left x bottom right)
= (a + n – 1) (a + mw – w) – (a (a + mw – w + n – 1))
= a2 + amw – aw + an + mnw – nw – mw – a + w – (a2 + amw – aw + an – a)
= a2 + amw – aw + an + mnw – nw – mw – a + w – a2 – amw + aw – an + a
= mnw – nw – mw + w
= w (mn – n – m + 1)
= w (m – 1) (n – 1)
As my formula for any sized rectangle in any sized grid was D = w (m – 1) (n – 1) and this is the same expression as my algebraic expression for the difference between the products of opposite corners, I have successfully proved my formula.
Differences between the Numbers in the Grid
The next part of my investigation is to investigate the difference between the numbers in the grid; for example, in all the grids I have done so far the difference between the numbers in the grid has been 1 (it has gone up 1, 2, 3, 4…), I may now try differences of 2 (2, 4, 6, 8…) or 3 (3, 6, 9, 12…).
Here is a grid of width 10 with differences between the numbers of 2:
There is a 2 x 2 square highlighted. This is it:
So the difference between the products of opposite corners in this square would be:
(top right x bottom left) – (top left x bottom right)
= 16 x 34 – 14 x 36
= 544 – 504
= 40
I will use algebra to prove that the difference between the products of opposite corners in all 2 x 2 squares in 10 x 10 grids with a difference between the numbers in the grid of 2 is always 40.
Let the top left square equal a, and therefore:
So the algebraic expression for the difference between the products of opposite corners would be:
(top right x bottom left) – (top left x bottom right)
= (a + 2) (a + 20) – a (a + 22)
= a2 + 2a + 20a + 40 – a2 – 22a
= a2 + 22a + 40 – a2 – 22a
= 40
So all these squares have a difference between the products of opposite corners of 40.
I will now try, in a 10 x 10 grid with differences between the numbers of 2, a 3 x 3 square, to look for patterns.
The difference between the products of opposite corners in this square is:
(top right x bottom left) – (top left x bottom right)
= 76 x 112 – 72 x 116
= 8512 – 8352
= 160
I will produce an algebraic expression for the difference between the products of opposite corners in such a 3 x 3 square, to prove that I will always get the same difference. Let the top left square equal a, and therefore:
So the algebraic expression for the difference between the products of opposite corners would be:
(top right x bottom left) – (top left x bottom right)
= (a + 4) (a + 40) – a (a + 44)
= a2 + 4a + 40a + 160 – a2 – 44a
= a2 + 44a + 160 – a2 – 44a
= 160
As I have proved this, I will now try a 4 x 4 square.
The difference between the products of opposite corners, in this example, equals:
(top right x bottom left) – (top left x bottom right)
= 46 x 100 – 40 x 106
= 4600 – 4240
= 360
I will use the same method of proving this again.
Let the top left number equal a, and therefore:
So the algebraic expression for the difference between the products of opposite corners would be:
(top right x bottom left) – (top left x bottom right)
= (a + 6) (a + 60) – a (a + 66)
= a2 + 6a + 60a + 360 – a2 – 66a
= a2 + 66a + 360 – a2 – 66a
= 360
Coming up with a Formula
Now that algebra has proved that the difference between the products of opposite corners in these squares is always 360, I will put my results in a table, to find patterns and formulae.
I can see that all the Differences are multiples of 10 and as 10 is the width of the grid in these examples, from my other formulae I can predict that these two factors are related, as they were in my other formulae. I can also see that they are all square numbers multiplied by 10 and that, when the difference between the numbers in the grid is 2, they go up in every other square number multiplied by 10.
My previous formula was D = w (m – 1) (n – 1) in a grid with differences between the numbers of 1. I have noticed that the number you subtract from m or n is the same as the difference between the numbers on the grid, so I can predict that these are the same, If I use p for the difference between the numbers in the grid, I can come up with this formula:
D = w (m – p) (n – p)
I will try putting one of my examples into this to see if it is correct:
D = w (m – p) (n – p) (w = 10) (m = 2) (n = 2) (p = 2)
= 10 (2 – 2) (2 – 2)
= 10 x 0 x 0 = 0
Obviously this formula is incorrect, but I think I am close. In the above example the true difference between the products of opposite corners was 40, which, when divided by 10 (the width), is 4, 2 x 2. So I need both my brackets to equal 2, as I am finding the difference between the products of opposite corners in a square.
I will try multiplying m and n by p as well as subtracting p:
So:
D = w (pm – p) (pn – p) (w = 10) (m = 2) (n = 2) (p = 2)
= 10 (4 – 2) (4 – 2)
= 10 x 4 x 4 = 40
So I have found a formula that covers every one of the variables I have looked at, the width of the grid (w), the length and the width of the rectangle (m and n) and the difference between the numbers on the grid (p):
D = w (pm – p) (pn – p)
I can simplify this formula by extracting p from both brackets, getting:
D = wp2 (m – 1) (n – 1)
I will now try this formula on a completely different example, a 4 x 5 rectangle on a grid with a width of 8 and differences between the numbers on the grid of 3.
D = wp2 (m – 1) (n – 1) (w = 8) (m = 4) (n = 5) (p = 3)
= 8 x 32 (4 – 1) (5 – 1)
= 8 x 9 x 3 x 4 = 864
I will draw an example of this rectangle in the same grid to check this answer:
So the difference between the products of opposite corners in this rectangle is:
(top right x bottom left) – (top left x bottom right)
= 42 x 102 – 30 x 114
= 4284 - 3420
= 864
My formula has worked once again, as I got the same difference between the products of opposite corners using my formula; but now I must use algebra to prove it.
Proving the Formula
In a grid with the differences between the numbers of p, with a width of w, in a rectangle with sides of m and n, let the top left corner equal a, and therefore:
Top Left = a Top Right = a + p (m – 1)
= a + pm – p
Bottom Left = a + wp (n – 1) Bottom Right = a + wp (n – 1) + p (m – 1)
= a + wpn – pw = a + wpm – pw + pm – p
If I use these expressions in my original calculation for the difference between the products of the opposite corners, I can prove my formula:
(top right x bottom left) – (top left x bottom right)
= (a + pm – p) (a + wpn – pw) – (a (a + wpn – pw + pm – p))
= a2 + anpw – apw + apm + p2mnw – p2mw – pa – p2wn + p2w – (a2 + apnw – apw +
[apm – ap)
= a2 + anpw – apw + apm + p2mnw – p2mw – pa – p2wn + p2w – a2 – apnw + apw –
[apm + ap)
= – p2mw + p2wnm – p2wn + p2w
= w (p2 – p2m – p2mn – p2n)
= wp2 (1 + mn – m – n)
= wp2 (m – 1) (n – 1)
As my formula for the difference between the products of opposite corners in a square with m and n lengths within a w sized grid with differences between the numbers of p was
D = wp2 (m – 1) (n – 1) and my algebraic expression for the difference between the products of opposite corners cancelled down to wp2 (m – 1) (n – 1), I have successfully proved that my formula will work in any rectangle in any grid.
Validity and Conclusion
Validity – This formula is only valid in grids in which the numbers will fit into the grid alignment, that is numbers that go up by consecutive amounts – because otherwise it would be impossible to have a difference between the numbers in the grid, which is p in the formula; therefore it would not be possible to complete the formula. The numbers in the grid do not have to be integers though, as long as the numbers go up consecutively, as this example shows:
D = wp2 (m – 1) (n – 1) (w = 10) (m = 2) (n = 3) (p = 0.5)
= 10 x 0.52 (2 – 1) (3 – 1)
= 10 x 0.25 x 1 x 2 = 5
If I then put this in an example of a grid and work out the difference between the products of opposite corners, I can check that the formula works in this example in which the numbers in the grid are not all integers:
(top right x bottom left) – (top left x bottom right)
= 1.5 x 5.5 – 0.5 x 6.5
= 8.25 – 3.25
= 5
So my formula is valid with any numbers, integers or not, in the grid, so long as they go up by consecutive amounts
In conclusion, I have investigated the difference between the products of opposite corners in rectangles such that, from starting with the information that the difference between the products of opposite corners in a 2 x 2 square in a 10 x 10 number grid equals 10, I have come up with a formula with 4 variables for the difference between the products of opposite corners in rectangles within number grids that is valid in number grids that go up consecutively. This formula is: D = wp2 (m – 1) (n – 1)