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Algebra Investigation - Grid Square and Cube Relationships

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Introduction

        Algebraic Investigation 1: Square Boxes on a 10x10 Grid        

In this first investigation, the difference in products of the alternate corners of a square, equal-sided box on a 10x10 gridsquare will be investigated. It is believed that the products and their differences should demonstrate a constant pattern no matter what dimensions are used; as long as they remain equal. In order to prove this, both a numeric and algebraic method will be used in order to calculate this difference. The numeric method will help establish a baseline set of numbers for testing, and to help in the establishment of a set of algebraic formulae for use on an n x n gridsquare.

In the example gridsquare below, the following method is used in order to calculate the difference between the products of opposite corners.

(a)

(b)

(c)

(d)

Stage A:        Top left number x Bottom right number =        (a) multiplied by (d)

Stage B:        Bottom left number x Top right number =         (c) multiplied by (b)

Stage B – Stage A:        (c)(b) - (a)(d)          =  The difference

The overall, 10 x 10 grid that is used for the first investigation will be a standard, cardinal gridsquare, which progresses in increments of 1. The formulae calculated will mainly be applicable to this grid, as other formats of gridsquares will require others formulae to provide valid results.

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This first investigation will focus only on gridsquares with equal widths and heights, which will at this stage be represented by the universal, constant term ‘w’.

The top left number in the grid (letter (a) in the above example) will be represented by the term ‘n’, which will be referred to in this manner in all proceeding investigations also.

This is only the first section of the investigation.

...read more.

Middle

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image34.png

n

n+1

n+2

n+10

n+11

n+12

n+20

n+21

n+22

Stage A:        Top left number x Bottom right number =        n(n+22) =        n2+22n

Stage B:        Bottom left number x Top right number =         (n+20)(n+2)=        n2+2n+20n+40

                                                                              =        n2+22n+40

Stage B – Stage A:        (n2+22n+40)-(n2+22n) = 40

When finding the general formula for any number (n), both answers begin with the equation n2+22n, which signifies that they can be manipulated easily. Because the second answer has +40 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 40 will always be present.

3x4 Rectangles

Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.

image35.png

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image37.png

n

n+1

n+2

n+3

n+10

n+11

n+12

n+13

n+20

n+21

n+22

n+23

Stage A:        Top left number x Bottom right number =        n(n+23) =        n2+23n

Stage B:        Bottom left number x Top right number =         (n+20)(n+3)=        n2+3n+20n+60

                                                                              =        n2+23n+60

Stage B – Stage A:        (n2+23n+60)-(n2+23n) = 60

When finding the general formula for any number (n), both answers begin with the equation n2+23n, which signifies that they can be manipulated easily. Because the second answer has +60 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 60 will always be present.

Any rectangular box with width ‘w’, and a height of 3

It is possible to ascertain from the above examples that each box follows a certain trend (each progressive increase in width shows an overall increase in the difference by +20). It is possible to use the 3x2, 3 and 4 rectangles above to find the overall, constant formula for any 3xw rectangle on a 10x10 grid, by implementing the formulae below to find a general calculation and grid rectangle.

It is additionally possible to see that the numbers that are added to n (mainly in the corners of the grids) follow certain, and constant sets of rules, which demonstrates confirmation of a pattern.

The bottom right number in the rectangle is directly linked to the top right and bottom left numbers. It is the sum of these that equal the bottom right, or:

Formula 1:                Bottom Right (BR)  =  Top Right (TR)  +  Bottom Left (BL)

As also shown by the summary boxes and examples above, the formula for the top right number remains constant, and is linked with the width, w, of the box in the following way:

Formula 2:                Top Right (TR)  =  n+ Width (w)  -  1

It is also evident from the examples calculated that the bottom left number is also linked with the height, w, of the box using a formula that remains constant:

Formula 3:                Bottom Left (BL)  =  n+ (Height (h)  -  1)  x  10

image32.png

image26.pngimage24.png

n

~

n+w-1

~

~

~

n+20

~

n+20+w-1

The above grid simplifies to form:

image32.png

image26.pngimage24.png

n

~

n+w-1

~

~

~

n+20

~

n+19+w

Stage A:        Top left number x Bottom right number         =        n(n+19+w)

=        n2+19n+nw

Stage B:        Bottom left number x Top right number         =         (n+20)(n+w-1)        

          =        n2+nw-n+20n+20w-20

                                                                =         n2+nw+19n+20w-20

Stage B – Stage A:        (n2+nw+19n+20w-20)-(n2+19n+nw)         =         20w-20

When finding the general formula for any number (n), both answers begin with the equation n2+nw+19n, which signifies that they can be manipulated easily. Because the second answer has +20w-20 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 20w-20 will always be present.

4 x Width Rectangles

image26.pngimage24.png

4x2 Rectangles

Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.

image02.png

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image38.png

n

n+1

n+10

n+11

n+20

n+21

n+30

n+31

Stage A:        Top left number x Bottom right number =        n(n+31) =        n2+31n

Stage B:        Bottom left number x Top right number =         (n+30)(n+1)=        n2+1n+30n+30

                                                                              =        n2+31n+30

Stage B – Stage A:        (n2+31n+30)-(n2+31n) = 30

When finding the general formula for any number (n), both answers begin with the equation n2+31n, which signifies that they can be manipulated easily. Because the second answer has +30 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 30 will always be present.

4x3 Rectangle

Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.

image02.png

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image39.png

n

n+1

n+2

n+10

n+11

n+12

n+20

n+21

n+22

n+30

n+31

n+32

Stage A:        Top left number x Bottom right number =        n(n+32) =        n2+32n

Stage B:        Bottom left number x Top right number =         (n+30)(n+2)=        n2+2n+30n+60

                                                                              =        n2+32n+60

Stage B – Stage A:        (n2+32n+60)-(n2+32n) = 60

When finding the general formula for any number (n), both answers begin with the equation n2+32n, which signifies that they can be manipulated easily. Because the second answer has +60 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 60 will always be present.

4x4 Rectangles

Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.

image40.png

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image41.png

n

n+1

n+2

n+3

n+10

n+11

n+12

n+13

n+20

n+21

n+22

n+23

n+30

n+31

n+32

n+33

Stage A:        Top left number x Bottom right number =        n(n+33) =        n2+33n

Stage B:        Bottom left number x Top right number =         (n+30)(n+3)=        n2+3n+30n+90

                                                                              =        n2+33n+90

Stage B – Stage A:        (n2+33n+90)-(n2+33n) = 90

When finding the general formula for any number (n), both answers begin with the equation n2+33n, which signifies that they can be manipulated easily. Because the second answer has +90 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 90 will always be present.

Any rectangular box with width ‘w’, and a height of 3

It is possible to ascertain from the above examples that each box follows a certain trend (each progressive increase in width shows an overall increase in the difference by +30). It is possible to use the 4x2, 3 and 4 rectangles above to find the overall, constant formula for any 4xw rectangle on a 10x10 grid, by implementing the formulae below to find a general calculation and grid rectangle.

It is additionally possible to see that the numbers that are added to n (mainly in the corners of the grids) follow certain, and constant sets of rules, which demonstrates confirmation of a pattern.

The bottom right number in the rectangle is directly linked to the top right and bottom left numbers. It is the sum of these that equal the bottom right, or:

Formula 1:                Bottom Right (BR)  =  Top Right (TR)  +  Bottom Left (BL)

As also shown by the summary boxes and examples above, the formula for the top right number remains constant, and is linked with the width, w, of the box in the following way:

Formula 2:                Top Right (TR)  =  Width (w)  -  1

It is also evident from the examples calculated that the bottom left number is also linked with the height, w, of the box using a formula that remains constant:

Formula 3:                Bottom Left (BL)  =  n+ (Height (h)  -  1)  x  10

image32.png

image24.pngimage26.png

n

~

n+w-1

~

~

~

~

~

~

n+30

n+30+w-1

The above grid simplifies to form:

image32.png

image26.pngimage24.png

n

~

n+w-1

~

~

~

~

~

~

n+30

n+29+w

Stage A:        Top left number x Bottom right number         =        n(n+29+w)

=        n2+29n+nw

Stage B:        Bottom left number x Top right number         =         (n+30)(n+w-1)        

          =        n2+nw-n+30n+30w-30

                                                                =         n2+nw+29n+30w-30

Stage B – Stage A:        (n2+nw+29n+30w-30)-(n2+29n+nw)         =         30w-30

When finding the general formula for any number (n), both answers begin with the equation n2+nw+29n, which signifies that they can be manipulated easily. Because the second answer has +30w-30 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 30w-30 will always be present.

Part B:Changing the Height, ‘h’

2 x w

n

~

n+w-1

n+10

~

n+9+w

Overall, constant difference of 2 x w grids:        10w-10

3 x w

n

~

n+w-1

~

~

~

n+20

~

n+19+w

Overall, constant difference of 3 x w grids:        20w-20

4 x w

n

~

n+w-1

~

~

~

~

~

~

n+30

n+29+w

Overall, constant difference of 4 x w grids:        30w-30

From the trend clearly displayed in both the algebraic summary boxes and the overall difference equation above, it is possible to create a sample algebra box for a grid size of any h x w:

image08.png

image09.png

image03.pngimage11.png

nimage04.png

~

n+w-1

~

~

~

n+10(h-1)

~

n+10h-10+w-1

Which simplifies easily into:

image08.png

image09.png

image03.pngimage11.png

nimage04.png

~

n+w-1

~

~

~

n+10h-10

~

n+10h+w-11

Stage A:        Top left number x Bottom right number         

=        n(n+10h+w-11)

=        n2+10hn+nw-11n

Stage B:        Bottom left number x Top right number         

=         (n+w-1)(n+10h-10)

=        n2+10hn-10n+nw+10hw-10w-n+10h+10

                        =         n2+10hn-11n+nw+10hw-10w+10h+10

Stage B – Stage A:        (n2+10hn-11n+nw+10hw-10w+10h+10)-(n2+10hn+nw-11n)         

=         10hw-10w-10h+10

When finding the general formula for any number (n) and any width (w), both answers begin with the equation n2+10hn-11n+nw, which signifies that they can be manipulated easily. Because the second answer has +10hw-10w+10h+10 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 10hw-10w+10h+10 will always be present.

Algebraic Investigation 3: Boxes on a ‘g’ x ‘g’ Grid, with Increments of ‘s’

The second investigation comprises of two main parts

Part A: Changing Grid Size, ‘g’

After completing all the stages previously detailed, it is now a logical decision to consider and investigate changing the actual dimensions of the overall grid.

Part B: Changing Increment Size, ‘s’

Another part of the investigation will be to analyse the overall difference formula in the event that a ‘step-size’ larger of 1 is used. In the diagram below, s is representing the jump (increment) size.

image43.png

image44.png

image44.pngimage44.pngimage44.png

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

y

Part A:Changing Grid Size:5 x 5 Grid

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Experimental number box:

image02.png

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image45.png

Algebraic box:

n

n+w-1

n+5(h-1)

n+5h-5+w-1

Which (when brackets are multiplied out) simplifies to:

n

n+w-1

n+5h-5

n+5h+w-6

Stage A:        Top left number x Bottom right number         

=        n(n+5h+w-6)

=        n2+nw+5hn-6n

Stage B:        Bottom left number x Top right number         

=         (n+5h-5)(n+w-1)        

=        n2+nw-n+5hn+5hw-5h-5n-5w+5

                        =         n2+nw+5hn-6n+5hw-5h-5w+5

Stage B – Stage A:        (n2+nw+5hn-6n+5hw-5h-5w+5)-(n2+nw+5hn-6n)         

=         5hw-5h-5w+5

When finding the general formula for any number (n), any height (h), and any width (w) both answers begin with the equation n2+nw+5hn-6n, which signifies that they can be manipulated easily. Because the second answer has +5hw-5h-5w+5 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 5hw-5h-5w+5 will always be present.

Testing:

Using the experimental number box above, it is possible to prove that the formula works, and is correct. The box had a height and width of 2x2, and was based on the 5x5 grid in question. As the number box clearly demonstrates, a difference of 5 should be present. This means the algebraic equation should be:

Difference:        (5x2x2)-(5x2)-(5x2)+5        =        5

Therefore, the formula gained has been proven to be correct and valid for calculating the difference.

6 x 6 Grid

< ...read more.

Conclusion

To obtain the overall algebraic boxes, for any cube size, it is necessary to refer back to the previously stated formulae for calculating the individual terms that should be present in the corners. It is therefore possible to use the grids to find the overall, constant formula for any cube/cuboid dimensions, by implementing the formulae below to find a general calculation and grid rectangles.

Top Face of Cube ‘c x c x c’

n

~

n+sw-s

~

~

~

n+ghs-sg

~

n+sw-s+ghs-sg

Bottom Face of Cube ‘c x c x c’

n+c2d-25

~

n+c2d-c2+sw-s

~

~

~

n+c2d-c2+ghs-gs

~

n+ c2d- c2+sw+ghs-gs-s

Difference in Cube:        

Stage A:        (TF) Top left number x (BF) Bottom right number

=        n(n+c2d-c2+sw+ghs-gs-s)

=        n2+c2dn-c2n+nsw+ghns-gns-ns

Stage B:        (TF) Bottom left number x (BF) Top right number

=         (n+ghs-gs)(n+c2d-c2+sw-s)        

=        n2+c2dn-c2n+nsw-ns+ghns+c2dghs-c2ghs+s2ghw-ghs2-gns-c2dgs+c2gs-s2gw+gs2

=         n2+c2dn-c2n+nsw+ghns-gns-ns+c2dghs-c2ghs+s2ghw-ghs2-c2dgs+c2gs-s2gw+gs2

Stage B – Stage A:        (n2+c2dn-c2n+nsw+ghns-gns-ns+c2dghs-c2ghs+s2ghw-ghs2-c2dgs+c2gs-s2gw+gs2) - (n2+c2dn-c2n+nsw+ghns-gns-ns)

=         c2dghs- c2ghs- c2dgs+ c2gs+ghws2-ghs2-gws2+gs2

When finding the general formula for any cube dimensions, both answers begin with the equation n2+c2dn-c2n+nsw+ghns-gns-ns, which signifies that they can be manipulated easily. Because the second answer has + c2dghs-c2ghs+s2ghw-ghs2-c2dgs+c2gs-s2gw+gs2 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of c2dghs-c2ghs+s2ghw-ghs2-c2dgs+c2gs-s2gw+gs2 will always be present.

Using this formula, it is possible to calculate the difference for any cube size, with any dimentions, increment size and gridsize.

n

n+1

n+10

n+11

...read more.

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