mathematics hl ib type 1

8/9/2019 Mathematics HL IB Type 1

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2/23

Math Portfolio Type I

In this portfolio it will be studied the patterns of cutting dierent obect with

dierent dimensions in the ma!imum amount of pieces with a "n# number of

cuts$ %efore starting the study two important things must be e!plained$ The

&rst one is that n will always be a positi'e integer so n()* the reason is that in

this study we will not analy+e the possibility of ha'ing negati'e cuts$ Thesecond clari&cation there is to do e!press what a cut is* a cut is an obect in the

dimension of the obect we are "cutting# ,1* which will di'ide the obect into

dierent parts$ For e!ample if we cut a cube-./0 we use planes-/0* not

lines-1/0 or points-)/0* but when we cut a circle-/0 we will use a line -1/0$ 2e

will not use lines or planes which are bent* using them would change all the

results gotten$

1 dimension

To start the in'estigation we should analy+e what is the ma!imum number of

pieces by cutting a 1 dimensional obect$ The reason we don3t start with ) is

because an obect with this amount of dimension doesn3t e!ist in practical

world as this is a point$

4 1 dimension obect is a straight line* therefore we will s5etch a line being cut

by 1* and &nally . point gi'ing the ma!imum amount of parts* but we will see

the possibility of how the line could not achie'e this$

In this picture we see the line being cut 1 time gi'ing parts$

In this picture the line is being cut time gi'ing . parts$

How many pieces?Francesco Chiocchio Page


3/23


I

n this picture the line is being cut . times gi'ing 6 parts$

4s notice when doing a number of cuts* the number of pieces is e7ual the

number of cuts 8 1* when tryin% to ha&e the ma$imum number of

pieces'

These tabulated results will help us to gi'e a general rule which relates number

of cuts -n0 and ma!imum number of segments* which we will call 9: 'alues 9 'alues$1 .. 6

;ust by loo5ing at it is clear that the general rule formula is gi'en by<

( ) n*1

If the line would ha'e cut without reaching the ma!imum number of segments

the result would ha'e been dierent$ For e!ample let us assume the second

How many pieces?Francesco Chiocchio Page .


4/23


point would been put o'er the &rst one would be done li5e this<

In this picture we used lines rather than dots because it would be complicated

to show two dots cutting in the same place$

Here* although two cuts ha'e been drawn* there are ust segments$ For this

reason it is important that in this study the ma!imum amount of segments or

parts is used to create rules$

dimensions

:ow to continue the study we ha'e to see what happens when a dimension

obect* li5e a circle* is cut dierent number of times$ I will s5etch the circle

being cut from = chords* gi'ing the ma$imum number of parts$ To create the

ma!imum parts in a circle e'ery chord must cut the biggest number of e!isting

parts$ Here we will use a circle* but we could ha'e used any other dimensional obects such as a s7uare$

In this picture we see a circle cut in by 1 chord$

How many pieces?Francesco Chiocchio Page 6


5/23


In this picture we see a circle cut in 6 by chords$

In this picture we see a circle cut in > by . chords$

How many pieces?Francesco Chiocchio Page =


6/23


In this picture we see a circle cut in 11 by 6 chords$

Finally* in this picture we see a circle cut in 1 parts by = chords$

4fter ha'ing made sure that this drawing show the ma!imum amount of parts

we can get by cutting them with dierent number of chords we can tabulate

the results to &nd using technology a general formula which relates number of

cuts -n0 and ma!imum number of parts -which we will call @0

:umber of cuts -n0 Ma!imum number of parts -@01 6. >6 11



7/23


= 1 This can be done using technology* meaning that we will use a graphical

calculator to &nd out the formula* this is done by typing the number of cuts and

ma!imum number of parts in the table labeled A1 and A -this table is found by

clic5ing 9T4T9 and after pressing B/IT on TI. or TI60

4fter ha'ing imported these numbers in the table we ha'e to decide whichdegree function it is to get a general rule$ In one dimension the general rule

was a polynomial with a degree 1* therefore we could hypothesi+e to use for

the general rule in dimensions a polynomial of degree * so we press on the

calculator 9tats* after we go on C4AC and we go on Duadreg were we set to use

the data in table in A1 and A * the calculator will gi'e us this answer<

This means that the general rule is gi'en by< @E1

2n ²+

1

2n+1 * where @ is the

ma!imum number of parts we can get by cutting a circle with n cuts$

To &nd out the recursi'e rule we don3t need to use a calculator or technology*

there is a longer and manual way which will be good to use because as it will

show us other features of the pattern between dierent dimensions$ This method wor5s by writing the 'alues of @ we get with n cuts after we

calculate the change between two consecuti'e 'alues and we repeat this

procedure till the changes are all e7ual

This table will e!plain it better* and it will gi'e us the 'alues of the ma!imum

possible parts we can get with n cuts* @$

n 'alues

-number of

cuts0

1 . 6 =

@ 'alues 6 > 11 1

How many pieces?Francesco Chiocchio Page >


8/23


1 -change

10

-6,0 . ->,60 6 -11,>0 = -1,110

-change

0

1 -.,0 1 -6,.0 1 -=,60

4fter ha'ing &nished the table to &nd the recursi'e rule we need to do a second

table* but in this one @ is replaced by a polynomial of degree * the choice of

the degree is gi'en by the number of changes done in the &rst tableG these two

numbers must be e7ual$

n 'alues 1 . 6 =an8bn8c a8b8c 6a8b8c a8.b8c 1a86b8c =a8=b8c

1 -change

10

.a8b =a8b >a8b a8b

-change

0

a a a

:ow we arri'e to the &nal step* we need to wor5 out a* b and c* to do so we

associate e'ery place from the &rst table to their corresponding places in the

second table* so we get e'ery number in terms of a* b and c -for e!ample

aE10* after we start sol'ing e'ery e7uation from the last change to ha'e ust

one un5nown<

aE1

aE 1

2

4t this point we 5now the 'alue of a* so we can pass to the ne!t step* wor5ing

out the 'alue with two un5nown* one of them being a<

.a8bE

.!- 1

208bE

bE

4−3

2

bE 1

2

Final step* wor5ing out the e7uation with . un5nowns* two of them being a and

b* so now 5nows 'alues

a8b8cE

1

28

1

28cE

cE1

4s a &nal answer we get1

2 n8 1

2 n81 ust li5e the pre'ious case$



9/23


4nother thing* which I belie'e is more important than this results* which we

already got with the calculator* is that in this way we clearly see that to &nd the

ne!t 'alue of @ we ha'e to do the addition of the 'alue we recei'e with 1 cut

less for @ and 9* to ma5e it more clear<

F-@*n0 is a the ma!imum number of parts we can get in dimension producedwith n cuts then* F-@*n0E F-@*n,108F-9*n,10 * to ma5e an e!ample from the

table* let us ta5e nE6 and dimension so the 6th 'alue of @ will be gi'en by

the .rd 'alue of @ and the .rd 'alue of 9

F-@*60E F-@*.08F-9*.0

E>86

E11

11 is e!actly the answer we were loo5ing for$ %ut does this mean this

conecture is true for e'ery 'alue of n?

F -@* n0E1

2 n8

1

2n81

F-9* n,10E-n,1081

F-@* n,10E 1

2-n,108

1

2-n,1081

F-@* n,108 F-9* n,10E

1

2-n,108

1

2-n,10818-n,1081E

12

n,n 8 12

8 12

n, 12

818n,181E

1

2n8

1

2n81

This conecture has been pro'en right* meaning it will wor5 for e'ery 'alue of n$

In the conecture we use F-@*n0E F-@*n,108F-9*n,10* but what would the

formula be if we don3t want to use n,1?

In this case we would write @-n0EJ89-n0* where J is an algebraic e!pression in

terms of nG to &nd n this procedure must be used<

1

2 n8

1

2n81 EJ8n81

JE 1

2 n8

1

2n81,n,1E

1

2 n,

1

2n8$

9o @-n0E 1

2 n,

1

2n89-n0

! dimension



10/23


The continue the in'estigation we will pass to a . dimensional obect* we will

use a cubeG in some of the s5etches where I will show how the cube will be cut

to get the ma!imum number of parts the cuts aren3t clearG I will try e!plain how

I drawn it and how I thought about it to facilitate the understanding of thes5etches$ 4fter ha'ing seen the s5etches I will tabulate the result to &nd a

general formulae which will gi'e us the ma!imum number of parts* which we

will nominate P* gi'en a number of cuts$

In this picture we see the cube cut by 1 plane into parts$

In this picture we see the cube cut by planes into 6 parts

$

In this picture we see the cube cut by . planes into parts$

How many pieces?Francesco Chiocchio Page 1)


11/23


In these pictures we see the cube cut by 6 planes into 1= parts$

9een from top seen from

bottom

In this picture we see the cube cut by = planes into parts$



12/23


9een from top 9een from bottom

4s said before the last two cuts are 'ery confusing but to ma5e it easier to

understand what I did* I will e!plain the logic behind these cuts$

For the 6th cut we need to thin5 about the circle we cut before using . chords

and therefore recei'ing > pieces* though we "transform# the circle in a s7uare*

as you remember I wrote before that we didn3t need to use a circle we could

ha'e used any dimensional plane and the rules and parts would be the same$

To get the ma!imum number of parts using 6 planes in the cube I imitate the

.rd chord I drawn on the / s5etch$

In this picture we see the cube seen from the top and from the side* the red*

the blue and green line represent the 1st . planes the 6th is drawn in both

drawings e!actly li5e the .rd chord in the dimensional obects* in this way we

can get 1= pieces from the cube$

9een from top 9een from side

The =th plane does as well follow this idea of imitating the 6th chord in the /

drawing$



13/23


:ow we will see a table showing the number of cuts* planes* and the

corresponding P* number of ma!imum parts$ 4fter this we will create a general

rule using technology and the method table we used before$

:umber of cuts -n0 Ma!imum number of parts -P01 6. 6 1==

Following the procedure e!plained in the pre'ious chapter we will use the

calculator to &nd a general formula$ These numbers must be imported on the

calculator in the columns* I will use A1 and A but you can chose any$ 2e will

chose that the calculator must &nd the result with a polynomial of degree .* wechose this number because of the precedents studies in 1d and d the general

formula was polynomial in degree respecti'ely of 1 and so we hypothesis the

degree of the formula is e7ual to the dimension we are wor5ing on$

Here we see two screenshots of the calculator -TI60 showing answer$

2e might want to change the decimal into fractions for simplicity<

aE)$1>E1K bE) cE)$......E=K

4ccording to the calculator the answer is PE 1

6n.8

5

6n 81$

This formula gi'es us the ma!imum amount of parts created for a gi'en n* but

it doesn3t show us any patterns with the cutting of obects in 1/ and /$

Therefore I rather use the method I showed you before* creating a table of

'alues and &nding the change between two consecuti'e numbers$

n 'alues 1 . 6 =P 'alues 6 1=

1 -change

10

-6,0 6 -,60 > 1=,0 11 -,1=0

-change

0

-6,0 . ->,.0 6 -11,>0

. -change

.0

1 -.,0 1 -6,.0

How many pieces?Francesco Chiocchio Page 1.


14/23


To &nd out the general rule formula we should do a second table and replacing

P 'alues by the polynomial an.8bn8cn8d* but because we already 5now the

'alues of a* b* c and d doing it would not help us$

From this table though there is a clear pattern which will help us with all the

in'estigation* we see that the change between two consecuti'e 'alues of

ma!imum parts which can obtained with n cuts* gi'es us the 'alue of the

dimension of the obect we are cutting minus 1* for e!ample 1 in this table

corresponds to @ 'alues and corresponds to 9 'alues$ 2e could ha'e

noticed it also on the &rst table that we did* but it may ha'e been ust a

coincidence* while now although not pro'en it seems li5e a real pattern$

4nother way to &nd a P 'alue is to apply a conecture* which is 'ery similar to

the conecture we found in /* F-P*n0EF-P*n,108F-@*n,10 where F-P*n0 is thema!imum number of parts we can get by cutting a .d obect with n cuts$

This formula is gi'en by the table on top* an e!ample lets use nE* so for the

ne!t term of n -.0 we will be e!pecting P to be according to the table and the

general rule and the drawings* let us see this conecture to see if it gi'es us the

same answer$

F-P*.0EF-P*08F-@*0E

686E

Therefore the answer is the right one* but this isn3t pro'ing the conecture is

right* it might ust be for coincidence that it gi'es us in this case the same

results as the general rule* let us try to pro'e it for any positi'e 'alues of n$

F-P*.0E 1

6n.8

5

6n 81$

F-P*n,10E1

6-n,10.8

5

6-n,10 81

F-@*n,10E 1

2-n,108

1

2-n,1081

F-P*n,108F-@*n,10E

1

6-n,10.8

5

6-n,10 818

1

2-n,108

1

2-n,1081E

1

6n. ,

3

6n8

3

6n L

1

68

5

6n ,

5

6 8 1 8

1

2n,n 8

1

28

1

2n ,

1

28

1E

1

6n.8

5

6n8 1



15/23


This calculations show us that we could use this conecture is true* we could try

tough to not use any :,1 and ust ha'e a formula with F-P*n0 and f-@*n0* where

P e7uals @ 8 $

PE 1

6n.8

5

6n 81$

@E 1

2-n08

1

2-n081

EP,@E

1

6n.8

5

6n 81,

1

2-n08

1

2-n081

1

6n.,

1

2n8

1

3n$

4s said in the pre'ious chapter @EJ89* so we could write down the e7uation

where PE8J89 which would gi'e this<

PE 1

6n.,

1

2n8

1

3n 8

1

2 n,

1

2n 8n 8 1E

1

6n.8

5

6n 81$

" dimensions

In this section we will continue the study in a more theoretical way* here we will

analy+e a 6 dimensions obect$ 4t this time we do not 5now how to show in a

picture 6 dimensions* but there has been some hypothetical hyperspaces which

are in 6 / such as Binstein spacetime$ Nsing the data we ha'e the pre,

collected* we can wor5 out when 6/ obect will be separetaed with n cuts* how

many part we can get* the solutions D$

The &rst step is &nding the 'alues of D* to do this we will use the table showing

the dierences we used in . / and we can hypothesi+e that the 'alues of P are

e7ual to the change of two consecuti'e 'alues of D* the hypothesis is gi'en

that for any other dimension this is true$

n 'alues 1 . 6 =P 'alues 6 1=

1 -change

10

-6,0 6 -,60 > 1=,0 11 -,1=0

-change

0

-6,0 . ->,.0 6 -11,>0

. -change

.0

1 -.,0 1 -6,.0

How many pieces?Francesco Chiocchio Page 1=


16/23


Our new table will start with for n e7ual to 1 the 'alue of D will be * as for an

obect in any dimension* with 1 cut* the obect will be di'ided into * this part

will be e!plained in the last chapter more accurately$

The second 'alue of D* so for nE* will be the &rst 'alue 8 the &rst 'alue of P*

therefore for nE D E 8E6$ 2e continue till we ha'e the 'alues for the &rst= cuts* which are shown in the following table<

n 'alues 1 . 6 =D 'alues 6 1 .1P 'alues

1 -change

10

6 1=

-change

0

-6,0 6 -,60 > 1=,0 11 -,1=0

. -change.0

-6,0 . ->,.0 6 -11,>0

6 -change

60

1 -.,0 1 -6,.0

2ith this table we can &nd a general rule for the 6/ obects* using the

calculator* or by hand$ I will enter the n 'alues and the D 'alues on the A1 A

table in the calculator and after e!pecting a result of a function of degree 6* the

reason of why we are e!pecting is that in the pre'ious dimension their

respecti'e general function3s degree was corresponding to the number of

dimensions$

2e can change the numbers into decimals for simplicity<

)$)61>E1K6 ,)$)....E,1K1 )$6=...E11K6 )$=....E>K1

The calculator gi'es this result<1

24n6,

1

12n.8

11

24n 8

7

12n 81$

The recursi'e formula to &nd the 'alues of D is the same as the one we used

to create the table<

F-D*n0EF-D*n,108F-P*n,10 where F-D*n0 is the ma!imum number of parts we

can get by cutting a 6d obect with n cuts$



17/23


In the pre'ious chapters* after the conecture I always added an e!ample

showing that for a gi'en 'alue of n the result was the same using this

conecture or the drawings$ For this dimension it is pointless doing this as the

'alue we ha'e* are calculated than5s to the formula* but we can still chec5 the

formula mathematicallyF-D*n0E

1

24n6,

1

12n.8

11

24n 8

7

12n 81$

F-P*n,10E 1

6-n,10.8

5

6-n,108 1

F-D*n,10E 1

24-n,106,

1

12-n,10.8

11

24-n,10 8

71

12-n,10 81$

F-D*n,108F-P*n,10E

1

24-n,106,

1

12-n,10.8

11

24-n,10 8

7

12-n,10 81 8

1

6-n,10.8

5

6-n,108

1E1

24n6 ,

4

24n. 8

6

24n ,

4

24n8

1

24 ,

1

12n. 8

3

12n,

3

12 n 8

1

128

11

24n ,

22

24n 8

11

24 8

7

12n ,

7

12 8 1 8

1

6n. ,

3

6n8

3

6n L

1

68

5

6n ,

5

6 8 1 E

1

24n6,

1

12n.8

11

24n 8

7

12n 81

4s we did in the pre'ious chapter we will try to create a formula which gi'es us

the 'alue of D* using the e7uation found the pre'ious chapter and a secondary

e7uation which we will wor5 on* so in this case we ha'e to &nd DE8P* we will

wor5 out by using simple calculations$

F-D*n0E 1

24n6,

1

12n.8

11

24n 8

7

12n 81

F-P*n0 E1

6-n0.8

5

6-n08 1

ED,PE

1

24n6,

1

12n.8

11

24n 8

7

12n 81 ,

1

6-n0.8

5

6-n08 1E

1

24n6,

13

12n.8

11

24n ,

1

4 n$

How many pieces?Francesco Chiocchio Page 1>


18/23


4s stated before PE8J89* so we could rewrite D as<

DE 1

24n6,

13

12n.8

11

24n ,

1

4 n 8

1

6n.,

1

2n8

1

3n 8

1

2 n ,

1

2

n8n81

2here the green is G red is G blue is J and blac5 is 9$

Further studies

2e analy+ed the pattern of cutting an obect of / dimensions* / being a

positi'e integer* n times* though our study &nished with / being e7ual to 6G my

7uestion is what would the conecture be or the general rule be when we will

use a / dimension?

I will start by wor5ing out a conecture< I will &rst show what e'ery conecture

we used for the pre'ious dimension was* after I will wor5 out a general rule$

/imension Conecture -where n is the number of

chords 0 F-@*n0E F-@*n,108F-9*n,10 where F-@*n0 is a the

ma!imum number of parts we can get in

dimension produced with n cuts then. F-P*n0EF-P*n,108F-@*n,10 where F-P*n0 is the

ma!imum number of parts we can get by

cutting a .d obect with n cuts$

6 F-D*n0EF-D*n,108F-P*n,10 where F-D*n0 is thema!imum number of parts we can get by

cutting a 6d obect with n cuts

From this table we can immediately notice that to &nd the ma!imum number of

parts done with -n0 cuts in a gi'en dimension we ust need to ta5e the sum of

the ma!imum number of parts we can obtain in the same dimension with -n,10

cuts and the ma!imum number of parts we can obtain in the gi'en dimension

minus one with -n,10 cuts$

This ma5es us understand that the conecture for any dimension is simply< F-/*

n0EF-/*n,108F-/,1* n,10* where F-D*n0 is the ma!imum number of parts we

can get by cutting a / d obect with n cuts and / stands for the number of

dimension* which has to be a positi'e integer$

The problem of this formula is that the only ways to 5now the ma!imum

number of parts in a dimension you ha'e to 5now the number of cuts minus 1

in the same dimension and the number of cuts minus 1 in the dimension,1$



19/23


Imagine you want to calculate the ma!imum number of parts you can obtain by

cutting a 1)) dimensional obect using 1)) cuts$ This is not impossible* but

'ery long$

To ma5e it easier therefore I want to create a formula where it is possible todirectly calculate the ma!imum number of parts ust by 5nowing how many

cuts and in what /imension

It is clear that the formulas to &nd the ma!imum number of parts for

hyperspaces in dierent dimensions ha'e something in common* for e!ample

their formula all ha'e a degree which is e7ual to the dimension* and the

coeQcient of the highest degree is always1

D !G we can notice that they all

end up with a constant of 1$ I &rst researched all their similitude to &nd a

general formula* but after a while* I changed path$

@ather than loo5ing at the formula for each dimension I started to loo5 at the

ma!imum number of parts I could obtain from each dimension and there I

found a formula$

To e!plain it* I will need a lot of graphical support* I will mainly use e!cel tables$

First I will show e'ery ma!imum result for the &rst cuts in dimensions$ The

data it ta5en by adding* ust li5e the recursi'e rule said< P -/* n0EP-/*n,108P-/,1* n,10

I will start now by tal5ing about patters which can be easily noticed$ The &rst

one is that when the / ≥ n the results are ust n -green cells0$ The second

pattern which can be seen is that when /En,1 the result will gi'e n

,1 -yellowcells0$


Tape+ une citation prise dans

document ou la synthSse

un passage intUressant$ Vousou'e+ placer la +one de te!te

importe oW dans le document$

tilise+ longlet Outils de +one

e te!te pour modi&er la mise

n forme de la +one de te!te de

citation$X

Table 4


20/23


From here we can deduce that for a cut* the dierence from the green part to

the yellow part is 1$

Aet us analy+e further the changed between dimensions for a gi'en 'alue of

cuts$Aet us watch at the column for the .rd cut and its dierences* we start from the

dierence from the green +one to the yellow +one* so it always starts with 1$

/ierences< -,>E01 G ->,6E0 . G -6,1E0.$

I will now show a table of dierences* this will show the dierence between the

ma!imum number of parts as you increase of dimension<

4 'alue* for e!ample 1 cut in ) dimension* is gi'en by the dierence of the

'alues of 1 Cut 1 / and 1 Cut ) / found in table 1$

Aoo5ing at the dierences from dimension to dimension in the same column we

can &nd a pattern* this brings us to the similarity between the dierences in a

column to the Pascal triangle and the binomial theorem$Pascal triangle represents a series of number that follow a rule which says that

e'ery 'alue of a line is e7ual to the sum of the two 'alues abo'e it<

4 G %

C G / G B

2here /E48%

if we start from 1<

1

1 G 1

1 G G 1

1 G . G . G 1

How many pieces?Francesco Chiocchio Page )

Table %

:E)

:E1

:E

:E.


21/23


2e can e!press any 'alue of the Pascal triangle by using the binomial theorem*

using the formula N !

T !

( N

−T

)!

where T is the position of the 'alue you want to

&nd -) to :0$

2e can now notice that the dierence between a 'alue which is a power of two

- in the green +one0 and another 'alue is the sum of the 'alues in the Pascal

triangles line :En$

Imagine now that we are loo5ing for the ma!imum number of parts we can get

from cutting a 6 dimensional obect times* from the table we can see that the

result is =>* but how could we do it without the table?

2e said that when / ≥ n the result is n* so it is part of the green +one$* so to

calculate a 'alue which is not a power of * we can simply subtract the sum of

all the dierences using Pascal triangle* to be more clear* for this e!ample we

ha'e n being si!* so we can loo5 the 'alues of the line :E of the Pascal

triangle< 1 G G 1=G ) G 1= G G 1$

:ow as we now that our /En, * we 5now that the cell we are loo5ing for itsfounded two abo'e the green +one$

To &nd the total dierence therefore we will use the &rst numbers of the

serie< 1 and * and subtract the addition to the number in the green +one*

which correspond to $

,-180E6,>E=>

This is what we would get by loo5ing at the table$

To create a real general formula we should write something where from a

power of is subtracted the sum from 1 to a 'alue of a line of the Pascal

triangle$ 4ny 'alue of the Pascal triangle can be e!pressed by the binomial

theorem$ 2e seen that for cutting the 6 / obect with chords we needed the

sum of the &rst two 'alues of the Pascal triangle of the line :E* 1 and which



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written in binomial way are6!

0!(6 !) and

6 !

1 !(5 !) $ 9o the formula would loo5

something li5e this< , ∑i=0

1

6 !

i ! (6−i )!

%ut now let us generali+e it for any dimension or cuts< n, ∑i=0

n− D−1n !

i ! (n−i ) !

#eometrical e$planation

/uring all the study we loo5ed at the ma!imum number of parts of ahyperspace in dierent dimensions we could get by cutting it* we seen somepatterns but ne'er e!plained it* so in this chapter I will focus on thegeometrical e!planation of the results and the patters$

First we must 5now that points in an hyperspace are determined as n E Y !i * !* !. *Z[ * for istance in dimension e'ery point can be determined using twonumbers - ! G y 0* in ./ a plane is identi&ed as - ! G y G + 0 or in 6 /* for instanceBinstein space time* the points are obtained - ! G y G + Gt0$

2hen we cut the hyperspace for n ≤ /* e'erything we do* is we &rst cut the\!3 'alues in two* after the \y3 and so on* depending how many dimensions weare wor5ing on$

For e!ample let us ta5e a s7uare in dimension* it is de&ned as - J * 0 whenwe do the &rst cut we create two parts which are de&ned as - J1 * 0 and - J *

0 when we cut the s7uare with a second chord we will ha'e 6 parts which canbe de&ned li5e so < - J1 * 1 0*- J1 * 0*- J * 10*- J * 0:ow we do not ha'e any other dimensions to split into two so our .rd chord will*when trying to obtain the ma!imum number of parts* di'ide by two as manyparts as it possibly can* which we 5now is .$ This will gi'e us a > parts which

can be de&ned li5e this <

- J1 * 1 0* - J1 * 1 03* - J1 * 0* - J1 * 03* - J * 10*- J * 103 * - J * 0

The ne!t chord will cut 6 of these parts by two so we would ha'e somethig li5ethis <- J1 * 1 0* - J1 * 1 03* - J1 * 1 033* - J1 * 0* - J1 * 033* - J1 * 0333 *- J * 10* - J *

103 *- J * 1033* - J * 10333 * - J * 0$ This procedure can be applied to any 'alue of n$2e can apply the same rule for a . dimensional obect such as a cube* which is

de&ned as - ! * y * +0 to cut all the dimension by we need . cuts* and weobtain . parts<



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mathematics hl ib type 1

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