Corners - Maths Investigation
Corners
Draw a grid 5 columns wide, with any number of rows above 2.
Select a square of numbers, 2x2, e.g. 7,8,12,13
Multiply together the numbers in opposite corners of the square (e.g. 7*13=91, 8*12=96)
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
I shall now begin searching for patterns, and for rules that I will then prove and use to explain patterns.
My first step shall be to make examples to compare
Ex.1
7
8
2
3
7*13=91
8*12=96
3
4
8
9
3*19=247
4*18=252
The pattern in this case would seem to be a difference of 5.
It may be a coincidence that 5 is the number of columns in the grid, and in order to test whether it is in fact a coincidence or not I shall introduce a new letter 'c' which will stand for the number of columns within the entire grid.
Algebra
n
n+1
n+c
n+c+1
n(n+c+1)=n²+nc+n
(n+1)(n+c)=n²+nc+n+c
n²+nc+n+c-(n²+nc+n) =
n²+nc+n +c -(n²-nc-n) =
c
The difference is c, the number of columns within the grid.
RULE
d = c
I shall extend this by varying the size of the square extracted from the grid.
The first stage of this will be to give an algebraic value to each .
n
n+1
n+2
n+c
n+c+1
n+c+2
n+2c
n+2c+1
n+2c+2
I shall begin my working by extrapolating the difference between products, as before.
n(n+2c+2) = n²+2cn+2n
(n+2)(n+2c) = n²+2cn+2n+4c
n²+2cn+2n+4c - (n²+2cn+2n)
n²+2cn+2n +4c -(n²-2cn-2n)
4c
I currently have insufficient data to begin searching for a rule, so I shall repeat the above process with a 4x4 square.
n
n+1
n+2
n+3
n+c
n+c+1
n+c+2
n+c+3
n+2c
n+2c+1
n+2c+2
n+2c+3
n+3c
n+3c+1
n+3c+2
n+3c+3
n(n+3c+3) = n²+3cn+3n
(n+3)(n+3c) = n²+3cn+3n+9c
n²+3cn+3n+9c-(n²+3cn+3n)
n²+3cn+3n +9c -(n²-3cn-3n)
9c
The pattern is c, 4c, 9c, which can also be written as 1c, 4c, 9c, which clearly reveals the pattern of consecutive square numbers. Not only are they consecutive square numbers, but they are each the square of 1less that the square size that produced them. This rule is written ...
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n
n+1
n+2
n+3
n+c
n+c+1
n+c+2
n+c+3
n+2c
n+2c+1
n+2c+2
n+2c+3
n+3c
n+3c+1
n+3c+2
n+3c+3
n(n+3c+3) = n²+3cn+3n
(n+3)(n+3c) = n²+3cn+3n+9c
n²+3cn+3n+9c-(n²+3cn+3n)
n²+3cn+3n +9c -(n²-3cn-3n)
9c
The pattern is c, 4c, 9c, which can also be written as 1c, 4c, 9c, which clearly reveals the pattern of consecutive square numbers. Not only are they consecutive square numbers, but they are each the square of 1less that the square size that produced them. This rule is written below in algebraic form.
RULE
d = c(s-1)²
I shall now continue my investigation by extending this theory to include numbers that increase by a different amount, for example 2,4,6,8,10 on the top row rather than 1,2,3,4,5.
In order to determine whether there is a rule or not, I shall first need to search for patterns within the square. I chose a 3x3 square because a 2x2 has a 'multiply by 1' in the rule and I may forget about this when finding a rule, possibly invalidating the rule.
2
4
6
8
0
2
4
6
8
20
22
24
26
28
30
32
2
4
6
0
2
4
8
20
22
I shall now express this in algebraic form.
n
n+2i
n+2c
n+2c+2i
n(n+2c+2i) = n²+2cn+2in
(n+2i)(n+2c) = n²+2cn+2in+4ci²
n²+2cn+2in+4ci²-(n²+2cn+2in)
n²+2cn+2in +4ci² -(n²+2cn+2in)
4ci²
In order to test this, I shall take a 4x4 square from the same grid.
2
4
6
8
0
2
4
6
8
20
22
24
26
28
30
32
Expressed in algebraic form, with irrelevant sections removed, it would look like this:
n
n+3i
n+3c
n+3c+3i
n(n+3c+3i) = n²+3cn+3in
(n+3i)(n+3c) = n²+3cn+3in+9ci²
n²+3cn+3in+9ci²-(n²+3cn+3in)
n²+3cn+3in +9ci² -(n²+3cn+3in)
9ci²
3x3 square: 4ci²
4x4 square: 9ci²
4 and 9 are both square numbers (similar to the pattern on page 2)
4 and 9 can be replaced with (s-1)², making the rule:
RULE
d = (s-1)²ci²
The logical way forward seems to be to change the square to a rectangle, for which I already have a theory. I believe that the (s-1)² in the rule, being (s-1) (s-1), is both of the sides of the square - each s represents the length of a side. Based upon this theory, using x and y to represent the horizontal and vertical dimensions of the square, the rule would be d = (x-1)(y-1)ci². I shall now attempt to prove or disprove this, and if it turns out to be incorrect, I shall endeavour to discover the correct rule.
I shall begin testing the theoretical rule by using it to find a solution, then finding the solution without a rule. If I am successful, both of the solutions should match.
2
4
6
8
0
2
4
6
8
20
22
24
26
28
30
32
I shall use a 3x4 rectangle (3 grid squares wide, 4 grid squares tall).
The rule, with numbers substituted in now reads as (3-1)(4-1)*4*2², which is more simply written as 2*3*4*4, the product of which is 96.
Working it out without the use of a rule gives the following result:
2*30=60
6*26=156
56-60=96
This does not prove the rule, but it does give the impression that it is correct and so proceeding to prove it algebraically would not be a complete waste of time.
Algebraically, the rectangle would look something like this:
n
n+(x-1)i
n+(y-1)ci
n+(y-1)ci+(x-1)i
I will substitute extra letters in to take the places of (x-1) and (y-1), because this simplifies the process of multiplying together the corners. a will represent (x-1) and b will represent (y-1).
So the rectangle now reads:
n
n+ai
n+bci
n+bci+ai
n(n+bci+ai) = n²+nbci+nai
(n+ai)(n+bci) = n²+nbci+nai+abci²
n²+nbci+nai+abci²-(n²+nbci+nai)
n²+nbci+nai +abci² -(n²+nbci+nai)
abci²
I can now convert the a and b back into (x-1) and (y-1), and this should give me a general rule.
abci² = (x-1)(y-1)ci²
RULE
d = (x-1)(y-1)ci²
Another way to extend the investigation is to multiply by a number (e.g. 2) as an interval, rather than adding a specified amount (known as i in the previous rules). In order to do this, I shall draw a grid of numbers with what I will now call, for lack of a better term, a multiplication interval.
2
4
8
6
32
64
28
256
512
024
2048
4096
8192
6384
32768
Now I will take a square from this grid and work out formulae for each of the corners.
2
4
64
28
n
n*m^(x-1)
n*m^(y-1)c
n*m^((y-1)c+(x-1))
I shall now once again substitute a and b for (x-1) and (y-1).
n
n*m^a
n*m^bc
n*m^(a+bc)
Unfortunately, I am finding myself unable to prove this algebraically due to too many powers. I am not sure how to multiply powers to things and leave a formula. Using a CFX-9850G, however, I was able to prove that the products of each pair of opposite corners are the same, no matter what the multiplier (m) is, but this does not give me any algebra to write down.
In addition, I performed some preliminary experimentation with sequences of square and triangular numbers, but to no avail. Interesting patterns frequently appeared, however I was unable to explain any of them. I believe that, given sufficient time, a person could and probably has found rules for such sequences. This person, however, is not me.
I have since reverted to working with algebra to find other formulae, and noticed that I have a rule for he contents of each corner of the square. Since only the corners are used, (this investigation could extended to use more than just the corners, but in this case has not been) I have therefore discovered each of the formulae for an addition interval (another term I made up, meaning that the sequence of numbers increases by the same number each time) that I believe would be most useful if I was ever to repeat or further extend this investigation. The formulae for the corners are:
n
n+(x-1)i
n+(y-1)ci
n+(y-1)ci+(x-1)i
A very brief conclusion
The overall rule for addition intervals (or whatever they're really called) is d = (x-1)(y-1)ci².
The rule for multiplication intervals (or whatever these are really called) is d = 0
More detailed conclusions are given to each section, and so this will end approximately... now.
Appendix
