RESULTS OF INVESTIGATING PATTERNS

 

RESULTS OF INVESTIGATING PATTERNS                                     Noah Nishihara Home Group T09

I remember clearly when I was in the primary school stage of my schooling. It was very difficult for me to find the patterns in which a series of numbers/arithmetic or geometric sequences were given. I realised why after studying algebra. It was because I wasn’t familiar with the usage of unknown numbers (variables and constants) which came up from equations to expressions. I learned that unknown numbers can be treated just like any other number. From that moment, I was very clear on the fact that mathematical study is actually the act of ‘performing’ an operation using addition, subtraction, multiplication, and division. There is little difference between arithmetic and algebra. With algebra, we need to combine letters and numbers in what is known as algebraic expressions. We operate variables and numbers to solve real world problems in a fast, convenient and mathematical way. To use algebra in real problems, we have to translate real situations into expressions and equations.

We investigated patterns in order to become more aware of the types of patterns which mathematicians are interested in describing. We provided formal mathematical proof using logic and algebra for the patterns we discovered. Patterns are important. General ability tests have patterns questions. Primary school students study number patterns. Without algebra, there is no mathematics. Patterns are so important that the number of words beginning with pattern is endless. Pattern recognition does not refer to pattern as something mathematical but as something which occurs again and again.

PATTERN #1
12+1=1×2		22+2=2×3		32+3=3×4		Conjecture:		42+4=4×5
PROOF:		PROOF:		PROOF:		(IN GENERAL TERMS)		PROOF:
LHS=12+1		LHS=22+2		LHS=32+3		n2+n=n(n+1)		LHS=42+4
=1+1		=4+2		=9+3		PROOF:		=16+4
=2		=6		=12		LHS= n2+n		=20
RHS=1×2		RHS=2×3		RHS=3×4		RHS= n(n+1)		RHS=4×5
=2		=6		=12		=n2+n		=20
⇒LHS=RHS		⇒LHS=RHS		⇒LHS=RHS		⇒LHS=RHS		⇒LHS=RHS
Q.E.D.		Q.E.D.		Q.E.D.		Q.E.D.		Q.E.D.
52+5=5×6		(-42)+(-4)=(-4)×3		102+10=10×11		To prove: n2+n=n(n+1)
PROOF:		PROOF:		PROOF:		RHS= n(n+1)
LHS=52+5		LHS=(-42)+(-4)		LHS=102+10		= n×n+n×1, distributive law
=25+5		=16-4		=100+10		= n2+n
=30		=12		=110		=LHS
RHS=5×6		RHS=(-4)×3		RHS=10×11
=30		=12		=110
⇒LHS=RHS		⇒LHS=RHS		⇒LHS=RHS
Q.E.D.		Q.E.D.		Q.E.D.
ANOTHER PATTERN
Pattern 1: 12+2=1×3	Pattern 2: 22+4=2×4		Pattern 3: 32+6=3×5		42+8=4×6		52+10=5×7
PROOF:	PROOF:		PROOF:		PROOF:		PROOF:
LHS=12+2	LHS=22+4		LHS=32+6		LHS=42+8		LHS=52+10
=1+2	=4+4		=9+6		=16+8		=25+10
=3	=8		=15		=24		=35
RHS=1×3	RHS=2×4		RHS=3×5		RHS=4×6		RHS=5×7
=3	=8		=15		=24		=35
⇒LHS=RHS	⇒LHS=RHS		⇒LHS=RHS		⇒LHS=RHS		⇒LHS=RHS
Q.E.D.	Q.E.D.		Q.E.D.		Q.E.D.		Q.E.D.
Conjecture:	(-92)+(-18)=(-9)(2+(-9))		0.52+2×0.5=(0.5)(2+(0.5))		To prove:		2+2={}
(IN GENERAL TERMS)	PROOF:		PROOF:		n2+2n=n(n+2)		PROOF:
n2+2n=n(2+n)	LHS=(-92)+(-18)		LHS=0.52+2×0.5		RHS= n(n+2)		LHS=2+2
PROOF:	=81-18		=0.25+1		= n×n+n×2, distributive law		RHS={}
LHS= n2+ 2n	=63		=1.25		= n2+2n		=22
RHS= n(2+n)	RHS=(-9)(2+(-9))		RHS=(0.5)(2+(0.5))		= LHS		⇒LHS=RHS
=2n+ n2	=-18+81 = 63		=1+0.25 =1.25		Proven		Q.E.D.
⇒LHS=RHS	⇒LHS=RHS		⇒LHS=RHS
Q.E.D.	Q.E.D.		Q.E.D.

 

More Number Patterns	Conjectures	Proving identities
Pattern #1	Pattern #2	Pattern #3	Pattern #4
n2+n-2=(n-1)(n+2)	n2+n-6=(n-2) (n+3)	n2+n-12=(n-3)(n+4)	n2-n-2=(n+1) (n-2)
PROOF:	PROOF:	PROOF:	PROOF:
RHS =(n-1)(n+2)	RHS =(n-2) (n+3)	RHS =(n-3)(n+4)	RHS =(n+1) (n-2)
= n2+2n-n-2	= n2+3n-2n-6	= n2+4n-3n-12	= n2-2n+n-2
= n2+n-2	= n2+n-6	= n2+n-12	= n2-n-2
LHS = n2+n-2	LHS = n2+n-6	LHS = n2+n-12	LHS = n2-n-2
⇒LHS=RHS	⇒LHS=RHS	⇒LHS=RHS	⇒LHS=RHS
Q.E.D.	Q.E.D.	Q.E.D.	Q.E.D.
Pattern #5	Pattern #6
n2-n-6=(n+2) (n-3)	n2-n-12=(n+3) (n-4)
PROOF:	PROOF:
RHS =(n+2) (n-3)	RHS =(n+3) (n-4)
= n2-3n+2n-6	= n2-4n+3n-12
= n2-n-6	= n2-n-12
LHS = n2-n-6	LHS = n2-n-12
⇒LHS=RHS	⇒LHS=RHS
Q.E.D.	Q.E.D.

 

Extension (Conjectures)
Pattern 1: LHS= n2+(n+1)2+[n×(n+1)]2	Pattern 2: RHS= [n+2]2	Pattern 3: n2+x2=[x+1]2	To prove: n2+{(n2-1)/2}2={(n2-1)/2+1} 2
= n2+n2+2n+1+(n2+n)2	= [n+2] [n+2]	RHS= x2+2x+1	RHS= {(n2-1)/2+2/2} 2
= 2n2+2n+1+1+n4+2n3+n2	(expanding brackets)	So, n2+x2= x2+2x+1	= {(n2-1+2)/2} 2
= n4+2n3+3n2+2n+1	= n2+4n+4	n2=2x+1	= {(n2+1)/2} 2
RHS= [n(n+1)+1]2	= LHS	x= (n2-1)/2	= (n4+2n2+1)/4
= [n2+n+1]2	⇒LHS=RHS	Substitute into: and:	LHS=4n2/4+ (n4-2n2+1)/4
= [(n2+n)2+2(n2+n)+1]	Proven Q.E.D.	n2+{(n2-1)/2}2={(n2-1)/2+1} 2	=(n4+2n2+1)/4
= n4+2n3+n2+2n2+2n+1	PROOF:		⇒LHS=RHS
= n4+2n3+3n2+2n+1	LHS=(-92)+(-36)+4		Proven Q.E.D.
⇒LHS=RHS	=81-32
Q.E.D.	=49
	RHS=(-9+2) 2
	=(-9+2) (-9+2)
	=81-18-18+4
	=49
	⇒LHS=RHS
	Q.E.D.

 

True or false?
x4=4x2	x4=x2+ x2	2x=x2	(2x)3=2x3	(x+2)2=x2+4
44=4×42	74=72+ 72	2×8=82	(2×16)3=2×163	(5+2)2=52+4
256=64	2401=98	16=64	32768=8192	49=100
Not true	Not true	Not true	Not true	Not true
(-10)4=(-10)×(-10)2	x4-2x2=0	x2-2x=0
10000=-1000	x2(x2-2)=0	x(x-2)=0
Not true	x=0, x=±	x=0, x=2
x4-4x2=0
x2-(x2-4)=0	e.g.	(x+1)(x3-1)=	To prove:
x×x (x-2)(x+2)=0	x4=4x2	(x2-1)(x2+x+1)	RHS= (x2-1)(x2	LHS=(x+1)(x3-1)
x=0, x=2, x=-2	24=4×22  16=16	Check: (2+1)(23-1)=	+x+1)	=x4-x +x3 -1
It will only work in the		(22-1)(22+2+1)	= x4+x3+x2-x2-x-1	= x4+x3 -x-1
above cases.		3×7=3×7	= x4+x3 -x-1	⇒LHS=RHS
		21=21	= LHS	Proven

Only one identity of these six is true.

Ready for your P2s
n2-n=n(n-1)	n3-n=n(n2-1)	[n(n-1)]2	n2-(n-1)2=2n-1
Check: 52-5=5(5-1)	=n(n+1)(n-1)	n2(n-1)2	n2-(n+1)2=-2n-1
20=20	If n is even, then the	22(2-1)2	the above identities
(-10)2-(-10)=(-10)((-10)-1)	(n+1) is odd, and the	4×1=4	prove, no matter the
110=110	(n-1) is odd as well.	As n2-n, proved earlier,	order of the odd or
Let n be 5: 5(5-1)=20	n3-n is always a multiple	has a factor of 2,	even no. the difference
(n-1) makes it even	of six. The lowest even number is 2. The lowest odd	this [n(n-1)]2, will have a factor of 4.	between their square
Let n be 6: 6(6-1)=30	no. is 3. Hence, within the three factors stated above,	the square of every even no. is a multiple of 4	no’s outcome- is always an odd no.
If n is even, it is still even after multiplying with odd number	At least one must be even	any other factors will not not be considered,	when an even no is taken from 1, the out
No matter what n is, in	and another odd. The third	as 4×anything is a multiple of four	come is always odd.
n(n-1), the result will	no. can be even or odd.		However,
always be an even	n3-n is always a multiple of 6		(n+1)2-(n-1)2
number. n2-n will always be an even number.	n2+pn=n(n+p)	n2+(p+q)n+pq= (n+p)(n+q)	= n2+2n+1- n2+2n-1 =4n
Even numbers multiplied by odd	PROOF:	PROOF:	when the pairs are both
numbers will always have an	RHS= n(n+p)	RHS= (n+p)(n+q)	odd or both even, the
answer that is even.	=n2+pn, distribution law	= n2+(p+q)n+pq, distribution law	difference of their square no. will be even.
⇒All even numbers are divisible	=LHS	=LHS	The statement has
by 2 and therefore have a factor of	⇒LHS=RHS	⇒LHS=RHS	been disproved
2.	Proven	Proven	by this case.

I have investigated the patterns in the form of algebraic expressions and equations and widened my knowledge in this area. This investigation allowed me to realise that all sorts of different types of patterns exist in algebra and other learning areas. The algebraic expressions or identities of this booklet are only basic when compared with others. As our study of algebra expands, we will be likely to encounter more complex patterns. An investigation of powerful and highly abstract patterns from higher algebra would be an investigation that can lead on from this one. Even at the highest level, all of the work rests on a foundation of basic algebra. My main task currently is to completely master the key acts, laws, and rules. Some may consider this study of basic algebra patterns as being less important than other aspects of algebra. This study is very useful and important. With just a little bit of informed knowledge and experience of algebra, patterns will become easier to identify. Algebra is like a window, to be opened out towards the richness of the mathematical world. By following the teacher’s introduction, identifying patterns is not hard. They are all simple and tidy.

Using the layout method instructed on the booklet, I was able to become more familiar with the terms ‘conjecture’, ‘identity’, QED, etc. Word processing mathematics work is one of the difficulties that I face.By practicing several times, the speed and efficiency will probably increase. Other types of software obviously require a certain amount of experience before they will benefit the user. The formatting is also one of the aspects which I cannot be satisfied on.

‘The use of language’ and ‘the role of insight in connection with mathematics and the discovery of proofs’ are questions to do with interpretation. By using huge amounts of gathered global data and analysing them for perhaps, patterns, new ‘proven knowledge’ may be created which would answer these unanswerable questions.

Pascal’s triangle is used in many fields including probability. It is the pattern found between the coefficients of expanded binomials raised to the power of 2, 3, 4 etc. There are many more examples of patterns in our life.

The examples in this booklet are all linear or squared binomials or trinomials. The amount of terms can be increased and their power can be raised to different degrees. However, this would make it harder to handle. The patterns here are the sum of two or three terms (on the left hand side) and the product (on the right hand side). If the square of the unknown number has a coefficient other than 1, the pattern would become harder to identify/describe for beginners. It is fortunate that we are given the simple cases.

The reason we must investigate patterns is to recognise the importance of them in mathematics, and particularly algebra. It makes our understanding of algebra easier.It supports our learning of mathematics in the early stages and helps people to predict.