portfolio mathematics hl type i investigating divisibility · portfolio mathematics hl – type i...

Portfolio Mathematics HL – Type I Investigating Divisibility

Bernardo Finelli de Moraes

São Paulo, Brazil, 2008

INVESTIGATING DIVISIBILITY

In this mathematical investigation, we will examine the divisibility of the

expression nnnP x )( and its connection with the elements of Pascal’s Triangle.

The first step required by the assignment in order to do so was to factorize the

expression nn x for }5,4,3,2{x .

For 2x , we have:

)1(2 nnnn

Now, for 3x :

)1( 23 nnnn

)1)(1( nnn {since 1 are the zeroes of 12 n ,

they are also its linear factors}

And for 4x :

)1( 34 nnnn

13 n can be further factorized by finding its zeroes. In this case, one of

them is 1, since:

11101 3333 nnnn

)1)(1(1 23 bnnnn

1223 bnnnbnn

1)1()1( 23 nbnbn

01b , and 01 b {equating coefficients}

1b

Thus:

)1)(1( 24 nnnnnn

Notice that )1( 2 nn cannot be further factorized without producing a

complex polynomial, since its zeroes are complex.

Finally, for 5x :

)1( 45 nnnn

Again, it is possible to further factorize 14 n

Two possible zeroes of 14 n are clearly 1 , since:

11101 4444 nnnn

)1)(1)(1(1 24 bnnnnn

)1)(1( 22 bnnn

12234 bnnnbnn

134 bnbnn

0b {equating coefficients}

Thus:

)1)(1)(1( 25 nnnnnn

Again, it should be noted that )1( 2 n has two complex zeroes.

Now, with the expression factorized, the divisibility of nnnP x )( by x shall be

examined. However, in order to have more accurate results, the procedure will be

made with }5,4,3,2,1,3

1,0,1{x : besides the values required by this assignment, a

negative number, a non-integer value, as well as the numbers 0 and 1 will be

considered. In the same idea, the values of n are }5,4,3,2,1,3

1,0,1{n

The results can be simply found by replacing x and n with the desired values in

the expression x

nn x .

For instance, for 2x and 3n :

32

6

2

39

2

332

Since 3 (it is an integer), the expression is divisible in this case.

When 3

1x , the expression was almost

never divisible. This happens because we

raise n by 3

1 , producing a cubic root which

might not always be rational. The same

principle can be applied to any non-integer.

We can thus conclude that x .

Whenever 0x , the result is not defined

due to a division by zero. Therefore, *x .

Repeating this same procedure for }5,4,3,2,3

1,0,1{x and }5,4,3,2,3

1,0,1{n

wields the following results. Most of these calculations were facilitated through the

use of technology (green denotes integers, while red refers to non-integers):

Table 1:

Divisibility

1x 0x

31x 1x 2x

3x

4x 5x

1n 0 313 3 0 1 0 2

1 0

0n 0

0 0 0 0 0 0

31n

38 1

3

33

0

91

81

8

162

13

243

16

1n 0 0 0 0 0 0 0

2n 2

3 623 3 0 1 2 2

7 6

3n 3

8 933 3 0 3 8 2

39 48

4n 4

15 1243 3 0 6 20 63 204

5n 5

24 1553 3 0 10 40 155 624

These results will now be discussed, summarizing the characteristics that need to

be present in order to the expression to be always divisible.

When 3

1n , the expression

was almost never divisible,

because a non-integer raised

to an integer power (since

x ) will not always result in

an integer. Thus, n

If }5,3,2,1{x , it appears that

the expression is always

divisible.

Proof will be provided that the expression nnnP x )( is always divisible

when }5,3,2,1{x and n . The assignment requested the usage of mathematical

induction by showing whether )()1( kPkP is always divisible by x . The reason

that this specific proof of induction was requested by this assignment will be later

explored in this mathematical investigation. Note that )(kP is obtained by replacing

n by k in the expression nn x , with the same principle applying with )1( kP .

However, in order to prove a statement through mathematical induction, we

have to determine whether the first term, 0P , is true. But we stated that Zn , or, in

other words, n exists in the interval [,] , meaning that it has no initial term.

Thus, we will be forced to assume that Nn , so that we can define the initial

value (in this case, 0).

nP is “ )(])1[]1([ 11 kkkk is divisible by 1”

Proof: {by the principle of mathematical induction}

(1)If 0n , 100)00()00()00(])0[]0([ 11

0P is true.

(2)Let us examine Pk :

Akkkk )()1]1([ 11 , where A is an integer

Akkkk 11

A0

(3)Now, 1kP :

Akkkk ])1[]1([])2[]2([ 11

Akkkk 1122

A0

Therefore 1 kk PP .

Hence, 1kP is true whenever kP and 0P are true.

nP is true. {principle of mathematical induction}

{using *}

{which is an integer, since A is an integer}



(1)If 0n , 200)00()00()00(])0[]0([ 22

0P is true.

(2)If kP is true, then Akkkk 2)()1]1([ 22 , where A is an integer {*}

Akkkkk 2112 22

Ak 22

A

A

k

kkkkkk

kkkk

2

2

2

)1]12([)2]44([

])1[]1([])2[]2([

22

22

Thus ])1[]1([])2[]2([ 22 kkkk is divisible by 2 whenever

)(])1[]1([ 22 kkkk is divisible by 2.



Now,

{using *}

{Which is an integer, since A and k are integers}



(1)If 0n , 300)00()00()00(])0[]0([ 33

0P is true.

(2)If kP is true, then Akkkk 3)()1]1([ 33 , where A is an integer {*}

Akk 333 2

)1(3

663

)3(69)33(

693

)1]133([)2]8126([

])1[]1([])2[]2([

2

2

2223

33

kA

kA

kkkk

kk

kkkkkkkk

kkkk





Now,

{using *}

{Which is an integer, since A and k are integers}



(1)If 0n , 500)00()00()00(])0[]0([ 55

0P is true.

(2)If kP is true, then Akkkk 5)(])1[]1([ 55 , where A is an integer (*)

Akkkk 5510105 234

)614124(5

307060205

30706020)510105(

307570305

])1[]1([])2[]2([

23

23

23234

234

55

kkkA

kkkA

kkkkkkk

kkkk

kkkk




nP is true.

A main aspect in common between the numbers 2, 3 and 5 is that the three

of them are prime numbers. Therefore, one could assume that the expression will

always be divisible when x is a prime number. Additionally, the expression is also

divisible when 1x because all numbers are divisible by 1. While 1 is only divisible

by 1 and itself, it is not generally considered a prime number. Through this

investigation, we will examine the validity of our statement, proving whether it is

correct or not.

Thus, we can now make the following conjecture: “the expression nn x is

divisible by x when x is a prime number, or 1, and Nn .

Now,

+

+ +

Following this investigation on divisibility, the assignment asked us to

examine the relations between the expression )()1( kPkP and the elements of

Pascal’s triangle. This triangle can be obtained in the following way:

1st row) The number 1 is initially added.

2nd row) The number 1 is added twice below the first 1.

3rd row) The number 1 is added as the first element and the last element. The two

numbers 1 from the row above are added together, and a number 2 is thus added

below them.

4th row) Again, a number 1 is added as the first element and the last element. Each

other element of this row (in this case, two 3s) will be the sum of the two elements

above it. This process can be repeated infinitely.

1

1 1

1 2 1

1 3 3 1

Naturally, it would be complicated to perform this method to obtain Pascal’s

Triangle up to various rows, for instance, the 15th row. We can, however, use

technology to perform this task.

Using a graphic display calculator, or similar tool, one can plot a graph

of nCrXA )1( , by replacing A with the desired row’s number. The table of the

graph will then provide all the elements of Pascal’s Triangle for that row.

1)

2)

3)

4)

Example of using a GDC (in this case, a Texas Instrument – 64 Silver Edition) to obtain the elements of the 7

th row of Pascal’s

Triangle. Here, 0x refers to the

first term, 1x to the second, and

so on.

{the 4th row correspond to a

expansion of a cubic binomial)

Using this method, one can easily construct the triangle up to the 15th row:

A common term to refer to any element in any row of Pascal’s Triangle is the

binomial notation, in the form of )(n

r, where n is the row’s number (starting at 0 for

the first row) and r is the position of the element on the row, from left to right, also

starting at 0, with Nrn , . This notation will be necessary as we investigate the

relations between the elements in Pascal’s Triangle and the expression

)()1( kPkP .

Also important is the fact that all the elements of row n correspond to a

binomial expanson of degree )1( n . For instance:

32233

303

3

213

2

123

1

033

0

3

1331)(

)( )()()()(

yxyyxxyx

yxyxyxyxyx

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

1 9 36 84 126 126 84 36 9 1

1 10 45 120 210 252 210 120 45 10 1

1 11 55 165 330 462 462 330 165 55 11 1

1 12 66 220 495 792 924 792 495 220 66 12 1

1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1

1 14 91 361 1001 2002 3003 3432 3003 2002 1001 361 91 14 1

{expanding the binomial xk )1( }

In order to obtain the expression )()1( kPkP , one inserts 1k and k in

n in the expression nn x . Thus, the expression becomes:

1]...[

1)1(

1)1(

])[]([])1[]1([

)()()()()(1

2

2

1

10

xx

x

x

x

xx

xx

xx

xx

xx

xx

kkkkk

kk

kkkk

kkkk

But the first term of any row of Pascal’s triangle is 1, as well as the last term

of the row, thus 1)(0

x

and 1)( x

x. So:

x

x

x

xx

xx

xxx

x

xx

xx

x

kkk

kkkkk

)()()(

)()()(

1

2

2

1

1

1

2

2

1

1

...

1]1...[

Which means that the expression )()1( kPkP is the same as xk )1( but

without the first term and last term of the binomial expansion of xk )1( . The two

elements of the expansion whose coefficients are )(0

x

and )(x

xhave been

removed.

In other other words, if we relate this to Pascal’s Triangle, we will see that

the coefficients related to the first and last element, the two 1s of a row that cannot

be found by adding together the two elements directly above them, have been

removed. We will further examine in this investigation why these 1s are discarded.

As it was already stated, the notation )(n

rrefers to the coefficient for any

row of Pascal’s Triangle. So we can express the )1( xth row of the triangle in the

form of the expansion of the binom x)11( , since it is irrelevant to multiply by 1.

)()()()()()(12310

...)11(x

x

x

x

x

x

xxxx

It is noteworthy how this relates to our initial conjecture about prime

numbers, because the rth element of Pascal’s Triangle in the x row will only

appear to be a multiple of x if x is a prime number, or 1. On the other hand, this is

not valid with the first and last elements of the row, the two elements of any row

that are not found by adding the elements above (which relates with the 1s that

were discarded in )()1( kPkP ). In other words, letting p be a prime number and

A any integer:

}0)( rppAp

r

This can be easily be shown by referring to the Pascal’s Triangle that was

constructed previously. For instance:

3rd row (p=2) : 1 2 1

{1 is not multiple of 2, but 2 is}

4th row (p=3) : 1 3 3 1

{1 is not multiple of 3, but 3 is}

6th row (p=5) : 1 5 10 10 5 1

{1 is not multiple of 5, but 5 and 10 (2*5) are}

However:

11st row (p=10) : 1 10 45 120 210 252 210 120 45 10 1

{1, 45 and 252 are not multiples of 10}

Proof will be provided that )(n

ris divisible by n when n is a prime number.

The following formula gives the value of )(n

r:

)!(!

!)(

rnr

nn

r

Let p be any prime number and A an integer:

Arpr

p

)!(!

! {because the entries of Pascal’s Triangle are always integer}

Arpr

p

])[...321)(...321(

...321

We must now consider if any of the elements in the denominator can simplify the

p in the numerator.

Because p is prime, p can only be simplified by p . But p does not exist in

r ...321 , because rp . The same applies with )!( rp , as long as 0r .

Thus, p does not exist in the denominator and cannot be simplified, allowing us to

rewrite the expression as:

Arpr

pp

])[...321)(...321(

)1(...321

Arpr

pp

)!(!

)!1(

Which proves that the the coefficients of a prime row from Pascal’s Triangle

must necessarily be a multiple of p if, and only if, 0 rp

If x was not a prime, then it would be possible that a factor in the

denominator would be capable of simplifying the x in the numerator, preventing us

from putting x as an factor. For instance, if 4x and 2r .

623)2)(1(

)2)(2)(3(

)2)(1(

)4)(3(

)2)(1)(2)(1(

)4)(3)(2)(1(

)2)(1)(2)(1(

)4)(3)(2)(1(

)!2(!2

!46)(

4

2

Thus 6 is a multiple of 3 and 2, but not 4, because the 4 was simplified in the

process. As shown above, this cannot happen with a prime number.

{using *}

{which is an integer, since A and k are integers}

Now that we have investigated the divisibility of the coefficients of Pascal’s

Triangle, we are prepared to prove our initial conjecture:

We have alrealdy proven that nn x is divisible by x when 1x . We must

now prove whether it is also valid for any prime number.

nP is “the expression nn x is divisible by x when x is a prime number and

Nn ."

However, in order to prove a conjecture through the principle of

mathematical induction, one has to show that it is valid for the first available value

for n . Since 2 is the first prime number:

nP is “the expression nn 2 is always divisible by 2, with Nn ”

Proof: (by mathematical induction)

(1) If 0n , 2002

002

0P is true.

(1) If kP is true, then

Akk 22 , where A is an integer (*)

Now:

)(2

22

2)(

112

)1()1(

2

2

2

kA

kA

kkk

kkk

kk

Thus, because 0P is true and kP is true, 1kP is also true.


{Note that )1()1( kk x is 1kP }

Now that we know that our conjecture is true for the first prime, let us

proceed with its final proof:

nP is “the expression nn x is divisible by x when x is a prime number and

Nn ."

Proof: (by mathematical induction)

(1) If 2x , nn 2 is always divisble by 2, as was shown before.

2P is true.

(2) Let us examine the following expression:

kkkkk

kkkkk

x

x

xx

xx

xx

x

x

x

x

xx

xx

xx

x

)()()(

)()()()()(

1

2

2

1

1

1

2

2

1

10

...1)1(

...)1(

But because x is a prime number and 0 rx , x must divide all the

binomial coefficients and, therefore, the entire RHS.

However, if x can divide the RHS, it must also divide the LHS, thus:

1)1( xx kk is divisible by x {Note that 1)1( xx kk is equal to

)()1( kPkP }

(3) Now, assuming that kP is true, let us examine 1kP :

)1()1()1()1( kkkk xx

)()1]1([

)()(1)1(

kkkk

kkkk

xxx

xxx

But, as shown in (2), 1)1( xx kk is a multiple of x . Also, )( kk x is equal

to kP , which, if true, must also be divisible by x . Therefore, if we rewrite these

expressions as multiples of x :

BxAxkk x )1()1( , where A and B are integers

)()1()1( BAxkk x

Thus )1()1( kk x is divisible by x whenever )( kk x is divisible by x .

Therefore, 1kP will be true whenever kP is true and 2P is true.


(divisible)

(divisible, even though 4 is not a prime)

(divisible, even though Nn )

By investigating the properties of Pascal’s Triangle and examinating the

importance of the expression )()1( kPkP , which was indicated by the form of

proof by induction requested by this assignment, we have now proven to be correct

our conjecture. Indeed, nn x will always be divisible by x whenever x is a prime

number and Nn . Naturally, it is possible that Zn , but since we had to

determine a first value to prove our conjecture by induction, we cannot conclude

that it is valid for any integer.

Let us observe the divisibility of the expression with further examples of

prime numbers (or even other numbers), using technology to facilitate the

calculations:

3127

2184

7

32187

7

333,7

7

nx (divisible)

122640 13

1594320

13

31594323

13

333,13

13

nx (divisible)

268435455 29

536870910

29

2536870912

29

222,29

29

nx (divisible)

49995000 2

99990000

2

10000100000000

2

100001000010000,2

2

nx

155 4

620

4

5-625

4

555,4

4

nx

4194303- 2

8388606-

2

8388608- 2

2

2)2(2,23

23

nx

To conclude, we must now consider the converse of our conjecture. Our

conjecture is:

“For Nn , if x is a prime number or 1, the expression nn x is divisible by x ”

Since a converse is a logical implication with the propositions reversed:

“For Nn , if the expression nn x is divisible by x , then x is a prime or 1”

In other words, the converse states that only prime numbers and the number

1 are valid values for x that will make the expression nn x a multiple of x .

If we were to prove whether the converse holds we could express it in the form of

an “if and only if” statement, and then determine whether one statement implies the

other.

x is a prime number or 1 nn x is divisible by x

) This has been proven in this portfolio: if x is a prime number or 1, then it

necessarily will make the expression nn x divisible by x .

) The converse states that if nn x is divisible by x , then it is implied that x is a

prime number or 1.

We have demonstrated through this portfolio that this is indeed wrong: Recall when

4x , 5n

155 4

620

4

5-625

4

555,4

4

nx

Which is divisible, even though 4 is not a prime. Note, however, that if 4x , 2n ,

then the result (2

7 ) is not an integer.

In any case, we have provided one counter example, which is enough to disprove

the converse.

This assignment did not ask for proof of the converse, but a simple

explanation to why the converse does not hold can be easily provided. If we were

to check the validity of our converse, we could recall when we proved that the

binomial coefficient )(x

r is divisible by x if x is a prime number. But what if x was

a coprime to r , meaning that the greatest common divisor of x and r was 1?

When x is not a coprime to r , then one of the prime factors of x can be simplified

by one of the prime factors of r . But when they are coprime, this does not

happen, and a similar, albeit with some differences, result to when x is prime

occurs. Thus, since the bionomial coefficient can be a multiple of non-prime

numbers, our conjecture can also apply to non-prime numbers, hence proving that

the converse is wrong.

Therefore, the conclusion of this portfolio is that if x is a prime number or 1,

then nn x will necessarily be divisible by x ; however, just because nn x is

divisible by x does not necessarily imply that x is a prime number or 1.

portfolio mathematics hl type i investigating divisibility · portfolio mathematics hl – type i...

Documents