wuct121 discrete mathematics numbers tutorial exercises

WUCT121 Numbers Tutorial Exercises Solutions 1

WUCT121

Discrete Mathematics

Numbers

Tutorial Exercises Solutions

1. Natural Numbers

2. Integers and Real Numbers

3. The Principle of Mathematical Induction

4. Elementary Number Theory

5. Congruence Arithmetic


Section 1: Natural Numbers

Question1 A closed operation is a rule for combining any two elements of a set which

always produces another element in the same set.

Question2 Addition, multiplication.

Question3 Multiplication, identity is 1.

Question4 Multiplicative inverse for 1 is 1. No other elements have inverses.

Question5

(a) ∈+=+ baabba ,, . Commutativity of addition

(b) ∈+=+ yxyxyx ,,44)(4 . Distributivity

(c) 603203)45()34(5 =×=××=×× . Associativity of multiplication

Question6

addition2323itydistibutiv23)158(

itycommutativ23158addition15238

vitydistributi15)158(8ityassociativ15)158(8

itycommutativ)1515()88(vitydistributi)1515()88(

15)()(8

yxyxyxxxyx

xyxxyyxxyyx

xyyxxyyx

+=++=++=++=

+++=+++=+++=+++=

+++

.

Question7 The Law of Trichotomy: If ∈ba, , then one and only one of the following

relationships hold: ba < , ba = , or ba > .

Question8 We know 251312 =+ and 13=p is a natural number. So 2512 =+ p , which

means 12 < 25.

Question9 Transitivity:

If ∈cba ,, . Then: cacbba <⇒<∧< )( , cacbba =⇒=∧= )( ,

cacbba >⇒>∧> )( .


Question10 Show if ∈<< cbabcacba ,,wherethen, .

ba < means there is a natural number p, )1(Kbpa =+

Multiplying (1) by c, we have cbcpca =+ , using Distributive Laws

We know cpr = is a natural number, by closure of multiplication.

So we have cbrca =+ .

Therefore, cbca <

Question11 The well-ordering property for Õ:

If A is any non-empty subset of Õ, then A has a least element.


Section 2: Integers and Real Numbers

Question1 Addition, subtraction, multiplication.

Question2 Addition, identity is 0. Multiplication, identity is 1.

Question3 Additive inverses, the inverse of a is –a.

Multiplicative inverse of 1 is 1, and of –1 is –1.

Question4 Not well-ordered. Consider },6,4,4,2,0,2,4,6,{ KK −−−=E , that is, the even

integers. It is a non-empty subset of Ÿ which does not have a least element. Thus the set Ÿ is

not well-ordered

Question5 1)34(267 +−×=− and ∈− 34 , therefore –67 is odd

Question6 Prime number: An integer 1>n is defined to be prime if and only if for all

positive integers r and s, if srn ×= , then 1=r or 1=s

Question7 Composite number: An integer 1>n is defined to be composite if and only if

there exist positive integers r and s, so that if srn ×= , then 1≠r and 1≠s .

Question8 Rational number: The Rational numbers are real numbers which can be written in

the form 0,,, ≠∈ bbaba

Question9 Addition, subtraction, multiplication.

Question10 Addition, identity is 0. Multiplication, identity is 1.

Question11 Additive inverses, the inverse of a is –a.

Multiplicative inverse of a, 0≠a is a1 .

Question12 No. Consider )1,0( , that is, the open interval between 0 and 1. It is a non-empty

subset of which does not have a least element. Thus the set is not well-ordered.


Section 3: The Principle of Mathematical Induction

Use Mathematical Induction to prove the following:

Question1 For all ∈n , nn 2)!1( ≥+

Let Claim(n) be: nn 2)!1( ≥+

Step 1: Claim(1) is: 12)!11( ×≥+

trueis)1(ClaimHence,RHSLHS212RHS;2!2)!11(LHS

=∴=×===+=

Step 2: Assume Claim(k), that is )1(for2)!1( ∈≥+ kkk

Prove Claim(k + 1) is true, that is )1(2)!11( +≥++ kk

RHS2

1as22)1(by2)2(

)!1()2()!2(

)!11(LHS

1 ==

≥×≥×+≥

+×+=+=

++=

+k

kkkkkk

kk

Thus Claim(k + 1) is true.

Thus by the Principle of Mathematical Induction Claim(n) is true For all ∈n

Question2 For all ∈n , x

xxxxxn

nn

sin2

2sin2cos4cos2coscos 1 =×××× −L

Let Claim(n) be: x

xxxxxn

nn

sin2

2sin2cos4cos2coscos 1 =×××× −L

Step 1: Claim(1) is: xxx

sin22sincos =

trueis)1(ClaimHence,RHSLHS

cossin2

cossin2sin2

2sinRHS;cosLHS

=∴

==== xx

xxxxx

Step 2: Assume Claim(k), that is

)1(forsin2

2sin2cos4cos2coscos 1 ∈=×××× − kx

xxxxxk

kkL

Prove Claim(k + 1) is true, that is xxxxxx k

kk

sin22sin2cos4cos2coscos 1

111

+

+−+ =×××× L


RHSsin2

2sinsin2

)22sin(sin22

2cos2sin2

)1(by2cossin2

2sin

2cos4cos2coscosLHS

1

1

1

==

×=

×=

×=

××××=

+

+

+

x

xx

xx

xx

xx

x

xxxx

k

k

k

k

k

kk

kk

k

kL



Question3 ( )( )6

12121, 222 ++=+++∈∀

nnnnn L

Let Claim(n) be: ( )( )

612121 222 ++

=+++nnnnL

Step 1: Claim(1) is: ( )( )

611211112 +×+

=

( )( )


16

112111RHS;11LHS 2

=∴

=+×+

===

Step 2: Assume Claim(k), that is

( )( ) )1(for6

12121 222 ∈++

=+++ kkkkkL

Prove Claim(k + 1) is true, that is ( )( )

61)1(211)1(

)1(21 222 +++++=++++

kkkkL


( )( )

( )( )

( )( )

( ) ( )

( )

( )( )

( )( )RHS

61)1(211)1(

6322)1(

6)672(1

6)]1(612[1

6)1(6121

)1(by)1(6

121)1(21

61)1(211)1()1(21LHS

2

2

2

2222

222

=+++++

=

+++=

+++=

++++=

++++=

++++

=

+++++=

+++++=++++=

kkk

kkk

kkk

kkkk

kkkk

kkkkkk

kkkk

L

L



Question4 zzn n 314:, =−∈∃∈∀

Let Claim(n) be: zz n 314: =−∈∃

Step 1: Claim(1) is: zz 314: 1 =−∈∃

trueis)1(ClaimHence

1and13314LHS 1 ∈×==−=

Step 2: Assume Claim(k), that is )1(for314: ∈=−∈∃ kpp k

Prove Claim(k + 1) is true, that is qq k 314: 1 =−∈∃ +

RHS14where3

)14(3)1(by334

3)14(4

3444

14LHS1

1

=∈+==

+×=+×=

+−×=

+−×=

−=+

+

pqqp

p

k

k

k




Question5 45, 3 −≥∈∀ nnn

Let Claim(n) be: 453 −≥ nn

Step 1: Claim(1) is: 41513 −×≥


1415RHS;11LHS 3

≥∴=−×===

Step 2: Assume Claim(k), that is )1(for453 ∈−≥ kkk

Prove Claim(k + 1) is true, that is 4)1(5)1( 3 −+≥+ kk

RHS4)1(557as4551as745

)1(by13345

133

4)1(5)1(LHS

2

23

3

=−+=≥−+≥≥+−≥

+++−≥

+++=

−+≥+=

kk

kkkkk

kkk

kk



Question6 Prove that for every real number x and each nxxn n +≥+∈ 1)1(, .

Let Claim(n) be: ∈+≥+ xnxx n where1)1(

Step 1: Claim(1) is: xx +≥+ 1)1( 1


1RHS;1)1(LHS 1

≥∴+=+=+= xxx

Step 2: Assume Claim(k), that is )1(for1)1( ∈+≥+ kkxx k

Prove Claim(k + 1) is true, that is xkx k )1(1)1( 1 ++≥+ +

RHS0and1as)1(1

)1(1

)1(by)1)(1()1)(1(

)1(1)1(LHS

2

2

1

=≥≥++≥

+++=

++≥++=

++≥+= +

xkxk

kxxk

kxxxx

xkxk

k




Question7 Prove that 22)24(1062 nn =−++++ K for all ∈n .

The proof for Step 2:

( ) ( )( )( )

( )( )

RHS12

122

thesis by hypo2442

214241062LHS

2

2

2

=+=

++=

−++=

−++−++++=

k

kk

kk

kkK

Question8 Prove that nn 27 − is divisible by 5 for all ∈n .


( ) ( )( ) ( ) ( )( ) ( )( ) ( )( )

11

11

27|5

RHS5

275

thesis by hypo5275

27275

227275

2277

27LHS

++

++

−∴

==

+=

+=

−+=

−+=

−=

−=

kk

k

k

kkk

kkk

kk

kk

ql

l

Question9 The problem arises in Step 2. Both Claim(k) and Claim( 1+k ) are wrong. They

should be:

Claim(k): ( )( )

612121 222 ++

=+++kkkkK

Claim( 1+k ): ( ) ( )( ) ( )( )6

11221121 222 ++++=++++

kkkkK

Question10 Claim(1) hasn’t even been considered!! As a matter of fact, Claim(1) is false.


Question11 Let Claim(n) be the statement “ 112 ++ nn is a prime number”.

(a) Claim(1) is: 1311112 =++ , 13 is prime thus Claim(1) is true

Claim(2) is: 1711222 =++ , 17 is prime thus Claim(2) is true







Claim(9) is: 10111992 =++ , 101 is prime thus Claim(9) is true.

(b) No. Check Claim(10) is 11111211110102 ×==++ , 121 is not prime. No amount

of ‘examples’ will ever prove an infinite number of Claims such as these.

Question12

(a) Step 2 of the proof can go nowhere.

1

1

212

21

21

41

211LHS

+

+

+<

+++++=

k

kkK

Now what? You can’t drop the second term as the inequality goes the wrong way.

(b) The proof in Step 2 includes:

RHS2

12

2

122

hypothesisby2

1

2

12

2

1

2

141

211LHS

1

1

1

1

=

−=

−−=

+−=

+++++=

+

+

+

+

k

k

kk

kkK


Question13 Prove that 111211

111 +=⎟

⎠⎞

⎜⎝⎛ +××⎟

⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ + n

nK for all ∈n .


RHS1)1(

thesis by hypo1

11)1(

111

211

111LHS

=++=

⎟⎠⎞

⎜⎝⎛

++×+=

⎟⎠⎞

⎜⎝⎛

++××⎟

⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ +=

kk

k

kK

Question14 Prove )12(221

−=∑=

nn

i

i


RHS)12(2

222

2)12(2

)1(sisby hypothe2)12(2

22

2LHS

1

11

1

1

1

1

11

1

1

=−=

+−=

+−=

∑+−=

∑+∑=

∑=

+

++

+

+

+=

+

+==

+

=

k

kk

kk

k

ki

ik

k

ki

ik

i

i

k

i

i


Question15 Prove 14)14)(34(

11 +

=∑+−= n

nii

n

i


RHS54

1)54)(14(

)1)(14()54)(14(

1)54()54)(14(

114

)1(hypothesisby]1)1(4][3)1(4[

114

)14)(34(1

)14)(34(1

)14)(34(1LHS

1

11

1

1

=++

=

++++

=

++++

=

+++

+=

++−++

+=

∑+−

+∑+−

=

∑+−

=

+

+==

+

=

kk

kkkkkk

kkkkk

kkkk

kiiii

iik

ki

k

i

k

i

Question16 Prove 643 +− nn is divisible by 3 for all ∈n


RHS13

)1(3

(1)sisby hypothe3333

33364

644133

6)1(4)1(LHS

3

3

3

33

33

3

=∈−++==

−++=

−++=

−+++−=

+−−+++=

++−+=

kkqpp

kkq

kkq

kkkk

kkkk

kk


Question17 Use the identity 2

)1(

1

+=∑

=

nnin

i to prove that

2

11

3⎟⎟⎠

⎞⎜⎜⎝

⎛∑=∑==

n

i

n

iii

The Proof for Step 2:

RHS

2)2)(1(

4)44()1(

4)1(4

2)1(

identity)by ()1(2

)1(

(1))sisby hypothe()1(

LHS

21

1

2

22

32

32

32

1

1

1

3

1

3

1

1

3

=⎟⎟⎠

⎞⎜⎜⎝

⎛∑=

⎟⎠⎞

⎜⎝⎛ ++

=

+++=

++⎟

⎠⎞

⎜⎝⎛ +

=

++⎟⎠⎞

⎜⎝⎛ +

=

++⎟⎟⎠

⎞⎜⎜⎝

⎛∑=

∑+∑=

∑=

+

=

=

+

+==

+

=

k

i

k

i

k

ki

k

i

k

i

i

kk

kkk

kkk

kkk

ki

ii

i


Question18 Prove ∑+−

=−=

n

i

nnni1

23

)12)(12()12(


RHS3

]1)1(2][1)1(2)[1(3

)32)(1)(12(3

)352)(12(3

)]12(3)12()[12(3

]1)1(2[3)12)(12(

(1)sisby hypothe]1)1(2[3

)12)(12(

)12()12(

)12(LHS

2

2

2

1

1

2

1

2

1

1

2

=++−++

=

+++=

+++=

++−+=

−+++−=

−+++−

=

∑ −+∑ −=

∑ −=

+

+==

+

=

kkk

kkk

kkk

kkkk

kkkk

kkkk

ii

i

k

ki

k

i

k

i

Question19 Prove ∑++

=+=

n

i

nnnii1 6

)72)(1()2(


RHS6

]7)1(2][1)1)[(1(6

)92)(2)(1(6

)18132)(1(6

)]3(6)72()[1(6

)21)(1(6)72)(1(

(1)sisby hypothe)21)(1(6

)72)(1(

)2()2(

)2(LHS

2

1

11

1

1

=

+++++=

+++=

+++=

++++=

++++++=

++++++

=

∑ ++∑ +=

∑ +=

+

+==

+

=

kkk

kkk

kkk

kkkk

kkkkk

kkkkk

iiii

ii

k

ki

k

i

k

i


Question20 Prove ( ) 21231 nn =−+++ K for all ∈n .


( )

RHS)1(

(1)sisby hypothe12

1)1(231LHS

2

2

=+=

++=

−++++=

k

kk

kK

Question21 If 3,2,3 11 =≥= − bnbb nn . Prove nnb 3= for all ∈n .

We are given: )2(3)1(,2,3 11 KK =≥= − bnbb nn

Let Claim(n) be : nnb 3=

Step 1: Claim (1) is 11 3=b

trueis)1(ClaimRHSLHS

33RHS

)2(by3LHS11

∴=∴

==

== b

Step 2: Assume Claim(k), Claim( 1−k ), …,Claim(1) are all true, for some 1≥k .

That is, )3(3,3,3 11

11 KK === −− bbb k

kk

k

Prove Claim( 1+k ) is true; that is prove that 11 3 ++ = k

kb .

RHS3

(3)by 33

(1)by 3LHS

1k

1

==

×=

==

+

+

kk

kb

b

So Claim( 1+k ) is true. Thus by the Strong Principle of Mathematical Induction, Claim(n) is

true for all .∈n

Question22 If 0,2,1 11 =≥−+= − unnuu nn Prove .,2

)1(∈∀

−= nnnun

We are given: )2(0),1(2,1 11 KK =≥−+= − unnuu nn

Let Claim(n) be : .2

)1( −=

nnun

Step 1: Claim (1) is .2

)11(11

−=u



02

)11(1RHS

)2(by0LHS 1

∴=∴

=−

=

== u


That is,

)3(2

)11(1,,

2)11)(1(

,2

)1(11 KK

−=

−−−=

−= − u

kku

kku kk

Prove Claim( 1+k ) is true; that is prove that 2

)1(.2

)11)(1(1

+=

−++=+

kkkkuk .

RHS2

)1(2

2)1(

(3)by 2

)1((1)by 11

LHS 1

=+

=

+−=

+−

=

−++== +

kk

kkk

kkkku

u

k

k


true for all .∈n

Question23 If .1,2,12 11 =≥+= − unuu nn Prove .12 ∈∀−= nu nn

Proof for Claim ( 1+k ):

RHS12

sisby hypothe1)12(2

givenby 12LHS

1

1

=−=

+−×=

+==

+

+

k

kk

ku

u

Question24 If 1,2,)( 11 −=≥−= − snsns nn . Prove .,)1( ! ∈∀−= ns nnn

Proof for Claim ( 1+k ):

RHS)!1()1(

sisby hypothe!)1()1(

givenby )1(LHS

1

1

=+−=

−×+−=

×+−==

+

+

k

kk

sks

k

kk

k


Question25 If 4,2,3,65 2121 ==≥−= −− uunuuu nnn . Prove .,2 ∈∀= nu nn

We are given: )2(4,2)1(3,65 2121 KK ==≥−= −− uunuuu nnn

Let Claim(n) be : .2nnu =

Step 1: Claim (1) is .211 =u


22RHS

)2(by2LHS11

∴=∴

==

== u

Claim (2) is .222 =u


42RHS

)2(by4LHS22

∴=∴

==

== u


That is,

)3(2,,2,2 11

11 KK === −− uuu k

kk

k

Prove Claim( 1+k ) is true; that is prove that 11 2 ++ = k

ku .

RHS2

22

2325

(3)by 2625

(1)by 65LHS

1

11

1

==

×=

×−×=

×−×=

−==

+

−−

+

k

k

kk

kkkk

kuu

u


true for all .∈n

Question26 If 5,1,3,65 2121 ==≥−= −− uunuuu nnn . Prove .,23 ∈∀−= nu nnn



That is,

)3(23,,23,23 111

111 KK −=−=−= −−− uuu kk

kkk

k


Prove Claim( 1+k ) is true; that is prove that 111 23 +++ −= kk

ku .

RHS23

2233

2332)23(5

esisby hypoth)23(6)23(5

givenby 65LHS

11

111

1

=−=

×−×=

×−×−−×=

−×−−×=

−==

++

−−−

+

kk

kk

kkkk

kkkkkk

kuu

u


true for all .∈n

Question27 Let ...,...,, 21 nuuu be real numbers such that 2,2,3,32 2121 ==≥+−= −− uunuuu nnn . Prove that 2=nu for all ∈n .

The proof for Step 2: Step 2: Assume Claim(k), Claim( 1−k ), …,Claim(1) are all true, for some 1≥k . That is,

)3(2,,2,2 11 KK === − uuu kk Prove Claim( 1+k ) is true; that is prove that 21 =+ku .

RHS264

esisby hypoth2322givenby 32

LHS

1

1

==+−=

×+×−=+−=

=

−

+

kk

kuu

u

So Claim( 1+k ) is true. Thus by the Strong Principle of Mathematical Induction, Claim(n) is true for all .∈n

Question28 If .4,3,3,2 2121 ==≥−= −− aanaaa nnn Prove .,2 ∈∀+= nnan .



That is,

)3(1,,1,2 11 KK +=+=+= − kakaka kk

Prove Claim( 1+k ) is true; that is prove that 3211 +=++=+ kkak .

RHS3 esisby hypoth)1()2(2

givenby 2LHS

1

1

=+=+−+×=

−==

−

+

kkk

aaa

kk

k


true for all .∈n


Section 4: Elementary Number Theory

Question1

(a) Yes. )43(286 nmnm +=+ and ∈+ nm 43 .

(b) Yes. ( ) 11232346 22 +++=++ srsr , and ∈++ 123 2sr .

(c) No. Let 3=u and 2=v . 54922 =−=− vu , which is prime.

Note: ( )( )vuvuvu +−=− 22 , which will be prime whenever 1=− vu and vu +

is prime

Question2 Divisibility: If n and d are integers and 0≠d , then n is divisible by d if

and only if kdn ×= for some ∈k .

Question3

(a) Yes. 41352 ×=

(b) Yes. ( )( )( ) ( )( )( )[ ]123133332313 +++=+++ kkkkkk and

( )( )( ) .12313 ∈+++ kkk

(c) No. 2847 =× and 3557 =× . There is no integer k so that k734 = .

(d) Yes. ( )abba 154106 =× and .15 ∈ab

Question4 The Quotient Remainder Theorem: If n and 0>d are both integers, then

there exist unique integers q and r such that rdqn += and dr <≤0 .

Question5 The Fundamental Theorem of Arithmetic: If 1and >∈ aa then a can

be factorised in a unique way in the form

kkppppa αααα ...321 321=

where kppp ,,2,1 K are each prime numbers and ∈iα for each ki ,,2,1 K= .


Question6 ⎣ ⎦ 17300 = . Eliminate multiples of 2, 3, 5, 7, 11, 13, 17.

201 202 203 204 205 206 207 208 209 210

211 212 213 214 215 216 217 218 219 220

221 222 223 224 225 226 227 228 229 230

231 232 233 234 235 236 237 238 239 240

241 242 243 244 245 246 247 248 249 250

251 252 253 254 255 256 257 258 259 260

261 262 263 264 265 266 267 268 269 270

271 272 273 274 275 276 277 278 279 280

281 282 283 284 285 286 287 288 289 290

291 292 293 294 295 296 297 298 299 300

Primes are: 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283,

293.

Question7 227, 229; 239, 241; 269, 271; 281, 283.


Question8 Find the prime factorization, gcd and lcm for the following pairs of

numbers:

(a) 368, 8619

10601374117133)369,8619(lcm

3)369,8619gcd(413369

171338619

22

2

2

=×××=−

=−×=

××=

(b) 936, 7623

7927921311732)7623,936(lcm

93)7623,936gcd(

11737623

1332936

223

2

22

23

=××××=

==

××=

××=

(c) 1375, 605

15125115)605,1375(lcm

55115)605,1375gcd(115605

1151375

23

2

3

=×=

=×=×=

×=

(d) 4968, 9000

62100023532)9000,4968(lcm

7232)9000,4968gcd(

5329000

23324968

333

23

323

33

=×××=

=×=

××=

××=

Question9 Use the Euclidean Algorithm to find the ),gcd( ba for the following pairs

of numbers. Find values ., ∈nm such that bnamba +=),gcd(

(a) a = 72, b = 63.

nmnm

63729)63,72gcd(1,1

)1(6372163729)1(

9)0,9gcd()9,63gcd()63,72gcd()1(

07963916372

+==−==∴

−×+=×−=⇒

===

+×=+×= K


(b) a = 2104, b = 21.

nmnm

2121041)21,2104gcd(501,5

)4()3(

50121)5(21045)100212104(211

100212104454211

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

4)1,4gcd(

)4,21gcd()21,2104gcd()2()1(

154214100212104

+===−=∴

×+−×=××−−=

×−=×−=

⇒⇒

===

+×=+×=

K

K

K

K

(c) a = 15, b = 10

1,1).1(101155

.5)10,15gcd(

−==∴−×+×=

=

nm

(d) a = 5, b = 9

2,125)1(91

.1)5,9gcd(

=−=∴×+−×=

=

nm

(e) a = 63, b = 24

8,3824)3(633

.12)24,63gcd(

=−=∴×+−×=

=

nm

(f) a = 336, b = 60.

11,2)11(60233612

.12)60,336gcd(

−==∴−×+×=

=

nm

(g) a = 7684, b = 4148.

50,27)50(414827768468

.68)4148,7684gcd(

−==∴−×+×=

=

nm


(h) a = 90, b = –54

2,1)2()54()1(9018

.18)54,90gcd(

−=−=∴−×−+−×=

=−

nm

Question10 Give three pairs of numbers that are relatively prime

9 and 5; 2104 and 21; 17 and 9.

Question11 Fermat Primes are given by the formula ∈+= nnFn

for,12)( 2 .

(a) Write down the first three Fermat Primes. Are they all prime?

prime25712)3(

prime1712)2(

prime512)1(

3

2

1

2

2

2

=+=

=+=

=+=

F

F

F

(b) Find F(4) and F(5). Are they prime?

not prime67004176414294967297

12)5(

prime6553712)4(

5

4

2

2

×==

+=

=+=

F

F

Question12 Mersenne Primes are given by the formula ∈−= nng n for,12)( .

T7Q11,12


(a) Find g(n) for 2, 3, 4, 5, 6, 7, 8, 9.

primenot51112)9(

prime12712)7(

prime3112)5(

prime712)3(

primenot25512)8(

primenot6312)6(

primenot1512)4(

prime312)2(

9

7

5

3

8

6

4

2

=−=

=−=

=−=

=−=

=−=

=−=

=−=

=−=

g

g

g

g

g

g

g

g

(b) Which of the values for g(n) found in part (a) are prime?

( ) ( ) ( ) ( )7,5,3,2 gggg are all prime.

(c) For what values of n do you suggest that g(n) will give a prime number?

The suggestion is that when n is prime, then ( )ng will also be prime.

(d) Find g(11). How does this compare with your opinion for part (c)?

( ) 892320471211 11 ×==−=g .

New conjecture: If ( )ng is prime, then n is prime, but if n is prime, ( )ng is not

necessarily prime.

(e) Prove that whenever n is composite, then g(n) is also composite. [Hint: Let

∈= qppqn , somefor , , and factorise g(n).]

Let 2,,4 ≥≥= qppqn

( )

1)2(

12

12

−=

−=

−=

qp

pq

nng

Let pX 2=

( ))1)(1(

121 ++++−=

−=−− XXXX

Xngqq

q

K

Now 2>X thus 11>−X and 11321 >+++++ −−− XXXX qqq K as 2≥q .

Therefore, ( ) stng = where 1≠s and 1≠t .

Hence, ( )ng is composite


Question13

(a) What can you say about ∈− nn for,13 ? Will this produce prime numbers?

Why or why not?

( )( )( )1332

133313131

21

+++=

++++−=−−

−−

K

K

n

nnn

Therefore, 13 −n is NOT prime, and never will be.

Question14 There are 367 people (pigeons, n). Including February 29 there are 366

possible birthdays (pigeonholes, k). Since n > k, by the Pigeonhole Principle, at least

two people must share the same birthday.

It does not work with 366 people, since 366 u 366.

Question15 There are seven integers (pigeons n), there are 6 even numbers from 1 to

12 inclusive (pigeonholes k). Since n > k, by the Pigeonhole Principle, at least two

numbers must be the same. However since the numbers are all different we cannot

have two the same, this means at least one must be odd.

Question16 There are 4 possible pairings which sum to 10: 64;73;82;91 ++++

The number 5 is not in any pair but every other number is in exactly one pair. This

gives 5 categories (pigeonholes k). Choosing 5 numbers (pigeons n) there could be

one from each category, e.g. 1, 2, 3, 4, 5. No pair of these adds to 10. Since n u k, we

cannot guarantee there will be two numbers whose sum is 10.To guarantee that two of

our chosen numbers sum to 10 we must choose at least 6. In this case 6 numbers

(pigeons n) from 5 categories (pigeonholes k), and n > k.

Question17 500 lines of computer code (pigeons n) in 17 days (pigeonholes k).

Now 2917500 ×> . Therefore by the generalised Pigeonhole Principle, on at least

one of the days he writes at least 30129 =+ lines of code.

[ or 3017500

=⎥⎥⎤

⎢⎢⎡ lines of code.]


Question18 Solution 1:

A person at the party can have 0 up to 9 friends at the party.

If someone has 0 friends at the party, however, then no one at the party has 9 friends.

Likewise, if someone has 9 friends at the party, then no one at the party has 0 friends.

Hence the number of possibilities for the number of friends the 10 people at the party

can have is at most 9 ( 0, 1, 2, 3, 4, 5, 6, 7, 8 or 1, 2, 3, 4, 5, 6, 7, 8, 9).

We have 10 people (pigeons n) to go into 9 friendship possibilities (pigeonholes k).

Since n > k, by the Pigeonhole Principle, at least two people at the party must have

the same number of friends.

Solution 2:

Let P be a person with the most friends, say m friends. If one of P’s friends has m

friends at the party, then we are done.

If none of P’s friends has m friends at the party, each of P’s m friends has fewer than

m friends at the party (at most, 1−m friends).

We have m friends (pigeons n) to go into 1−m friendship possibilities (pigeonholes

k).

Since n > k, by the Pigeonhole Principle, some two of P’s friends have the same

number of friends at the party.

To help you with this solution, let 5=m . Then each of P’s 5 friends has fewer than 5

friends at the party (at most, 4 friends).


Section 5: Congruence Arithmetic

Question1

(a)

( )( )( )11mod112

11mod112

11mod112

9

2

≡⇒

≡⇒

≡

.

(b) ( )

( )( )( )( )13mod93

13mod33

13mod13

13mod1273

13mod33

11

10

9

3

≡⇒

≡⇒

≡⇒

≡=

≡

.

(c) ( ) 1174 so ,1174 =−−=+ which is divisible by 11.

Therefore, ( )11mod74 −≡

Question2 Use the definition of congruence modulo to prove the following properties

for congruence modulo 3.

(a) Prove ( )3mod, aaa ≡∈∀ .

That is, we need to prove that ( )aa −|3 .

Now, 0=− aa and ( )aa −⇒ |30|3 .

Therefore, by definition, ( )3modaa ≡ .

(b) Prove ( ) ( )3mod3mod,, abbaba ≡⇒≡∈∀ .

KNOW: ( ) ( )baba −⇒≡ |33mod by definition.

PROVE: ( ) ( )abab −≡ |3 that prove is,That .3mod

( ) ( )( )

( )

)3(mod)(|33

33,|3

abab

kkabkba

bakba

≡∴−∴

∈−−=−⇒−=−−⇒

=−∈∃⇒−


(c) Prove ( ) ( ) ( )3mod3mod3mod,,, cacbbacba ≡⇒≡∧≡∈∀

KNOW: ( ) ( )baba −⇒≡ |33mod

( ) ( )cbcb −⇒≡ |33mod , by definition.

PROVE: ( )3modca ≡ . That is, prove that ( )ca −|3 .

( ) ( )( ) ( )23,|3

13,|3K

K

lcblcbkbakba

=−∈∃⇒−=−∈∃⇒−

Adding equations (1) and (2), we have

( )( )( )3mod

|33

3333

caca

lklkcalkca

lkcbba

≡∴−∴

∈++=−⇒+=−⇒

+=−+−

Question3 Let 7=n .

(a) [ ] ( )7in m is a set of numbers. ( 7 is a set of sets.)

(b) [ ] { }KK ,23,16,9,2,5,12,19,26,2 −−−−= ,

[ ] { }KK ,26,19,12,5,2,9,16,23,5 −−−−= ,

[ ] [ ] { }KK ,28,21,14,7,0,7,14,21,28,07 −−−−==

[ ] [ ] [ ] [ ] [ ] [ ] [ ]{ }6,5,4,3,2,1,07 =

(c)

(i) True. ( )7mod29 ≡ .

(ii) False. [ ]5 is a SET, so it can’t be equal to ( )7mod5 .

(iii)True. ( )7mod519 ≡ , so [ ] [ ]519 = , and [ ]55∈ .

(iv) False. 7 is a set of SETS. 3 is not a SET

(v) True. [ ] [ ] 7310 ∈= .

(vi) False. [ ]a and [ ]b are both sets of numbers. [ ]a can’t be in [ ]b .


Question4 Write out the addition and multiplication tables for 5 , 6 and 1 . [ ] [ ] [ ] [ ] [ ]{ }4,3,2,1,05 = .

+ [ ]0 [ ]1 [ ]2 [ ]3 [ ]4

[ ]0 [ ]0 [ ]1 [ ]2 [ ]3 [ ]4

[ ]1 [ ]1 [ ]2 [ ]3 [ ]4 [ ]0

[ ]2 [ ]2 [ ]3 [ ]4 [ ]0 [ ]1

[ ]3 [ ]3 [ ]4 [ ]0 [ ]1 [ ]2

[ ]4 [ ]4 [ ]0 [ ]1 [ ]2 [ ]3

¥ [ ]0 [ ]1 [ ]2 [ ]3 [ ]4

[ ]0 [ ]0 [ ]0 [ ]0 [ ]0 [ ]0

[ ]1 [ ]0 [ ]1 [ ]2 [ ]3 [ ]4

[ ]2 [ ]0 [ ]2 [ ]4 [ ]1 [ ]3

[ ]3 [ ]0 [ ]3 [ ]1 [ ]4 [ ]2

[ ]4 [ ]0 [ ]4 [ ]3 [ ]2 [ ]1

[ ] [ ] [ ] [ ] [ ] [ ]{ }5,4,3,2,1,06 =

+ [ ]0 [ ]1 [ ]2 [ ]3 [ ]4 [ ]5

[ ]0 [ ]0 [ ]1 [ ]2 [ ]3 [ ]4 [ ]5

[ ]1 [ ]1 [ ]2 [ ]3 [ ]4 [ ]5 [ ]0

[ ]2 [ ]2 [ ]3 [ ]4 [ ]5 [ ]0 [ ]1

[ ]3 [ ]3 [ ]4 [ ]5 [ ]0 [ ]1 [ ]2

[ ]4 [ ]4 [ ]5 [ ]0 [ ]1 [ ]2 [ ]3

[ ]5 [ ]5 [ ]0 [ ]1 [ ]2 [ ]3 [ ]4

¥ [ ]0 [ ]1 [ ]2 [ ]3 [ ]4 [ ]5

[ ]0 [ ]0 [ ]0 [ ]0 [ ]0 [ ]0 [ ]0

[ ]1 [ ]0 [ ]1 [ ]2 [ ]3 [ ]4 [ ]5

[ ]2 [ ]0 [ ]2 [ ]4 [ ]0 [ ]2 [ ]4

[ ]3 [ ]0 [ ]3 [ ]0 [ ]3 [ ]0 [ ]3

[ ]4 [ ]0 [ ]4 [ ]2 [ ]0 [ ]4 [ ]2

[ ]5 [ ]0 [ ]5 [ ]4 [ ]3 [ ]2 [ ]1

[ ]{ }01 = .

+ [ ]0 ¥ [ ]0

[ ]0 [ ]0 [ ]0 [ ]0

(a) The element [ ]0 is the only element in 5 that doesn’t have a multiplicative

inverse

(b) The elements [ ] [ ] [ ] [ ]4,3,2,0 don’t have multiplicative inverses in 6 .

(c) If [ ] 6∈m does have an inverse under multiplication, then 1)6,gcd( =m .


Question5 Let .∈n Prove that ( )nbaba mod][][ ≡⇔=

To prove that [ ] [ ] ( )nbababa mod,, ≡⇔=∈∀ , there are two things to prove:

1. [ ] [ ] ( )nbababa mod,, ≡⇒=∈∀

2. ( ) [ ] [ ]banbaba =⇒≡∈∀ mod,,

Proof of 1.KNOW: [ ] [ ] )1(Kba = .

PROVE: ( )nba mod≡ .

Let [ ]ax∈ .

Then ( )nax mod≡ , by definition of [ ]a .

Since [ ] [ ]ba = , [ ]bx∈ also. Then ( )nbx mod≡ .

Now, ( ) ( )nxanax modmod ≡⇒≡ by the symmetric property.

Also, ( )nbx mod≡ .

Therefore, by the transitive property ( )nba mod≡ .

Proof of 2. ( ) [ ] [ ]banbaba =⇒≡∈∀ mod,, ,there are two things to prove:

i. [ ] [ ]ba ⊆

ii. [ ] [ ]ab ⊆

Proof of i:

KNOW ( ) )1(mod Knba ≡

PROVE: [ ] [ ]ba ⊆ .

Let [ ] ( )naxax mod≡⇒∈ by definition

( )nbx mod≡⇒ by transitivity (and (1))

[ ]bx∈⇒

Therefore, [ ] [ ]ba ⊆ by a typical element argument.

Proof of ii:

KNOW: ( ) ( )nabnba modmod ≡⇒≡ by symmetry

PROVE: [ ] [ ]ab ⊆ .

Let [ ] ( )nbxbx mod≡⇒∈ by definition

( )nax mod≡⇒ by transitivity

[ ]ax∈⇒

Therefore, [ ] [ ]ab ⊆ by a typical element argument


Question6

(a)

( )( ) ( )( ) ( )( )( )11mod44

11mod14

11mod111mod454

11mod911mod204

11mod5164

11

10

5

3

2

≡⇒

≡⇒

≡≡⇒

≡≡⇒

≡=

Therefore [ ]1410 ∈ and [ ]4411∈ , where [ ] [ ] 114,1 ∈

(b)

( )( ) ( )( ) ( )( )( )11mod33

11mod13

11mod111mod453

11mod511mod273

11mod593

11

10

5

3

2

≡⇒

≡⇒

≡≡⇒

≡≡⇒

≡=

Therefore [ ]1310 ∈ and [ ]3311 ∈ , where [ ] [ ] 113,1 ∈ .

(c) Conjecture: Let 110, <<∈ xx .

( )11mod110 ≡x and ( )11mod11 xx ≡ .

As a matter of fact, this result is true for all values of x, where ( ) 111,gcd =x .


Question7

(a) [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ][ ]{ }8,7,6,5,4,3,2,1,09 =

+ [ ]0 [ ]1 [ ]2 [ ]3 [ ]4 [ ]5 [ ]6 [ ]7 [ ]8

[ ]0 [ ]0 [ ]1 [ ]2 [ ]3 [ ]4 [ ]5 [ ]6 [ ]7 [ ]8

[ ]1 [ ]1 [ ]2 [ ]3 [ ]4 [ ]5 [ ]6 [ ]7 [ ]8 [ ]0

[ ]2 [ ]2 [ ]3 [ ]4 [ ]5 [ ]6 [ ]7 [ ]8 [ ]0 [ ]1

[ ]3 [ ]3 [ ]4 [ ]5 [ ]6 [ ]7 [ ]8 [ ]0 [ ]1 [ ]2

[ ]4 [ ]4 [ ]5 [ ]6 [ ]7 [ ]8 [ ]0 [ ]1 [ ]2 [ ]3

[ ]5 [ ]5 [ ]6 [ ]7 [ ]8 [ ]0 [ ]1 [ ]2 [ ]3 [ ]4

[ ]6 [ ]6 [ ]7 [ ]8 [ ]0 [ ]1 [ ]2 [ ]3 [ ]4 [ ]5

[ ]7 [ ]7 [ ]8 [ ]0 [ ]1 [ ]2 [ ]3 [ ]4 [ ]5 [ ]6

[ ]8 [ ]8 [ ]0 [ ]1 [ ]2 [ ]3 [ ]4 [ ]5 [ ]6 [ ]7

¥ [ ]0 [ ]1 [ ]2 [ ]3 [ ]4 [ ]5 [ ]6 [ ]7 [ ]8

[ ]0 [ ]0 [ ]0 [ ]0 [ ]0 [ ]0 [ ]0 [ ]0 [ ]0 [ ]0

[ ]1 [ ]0 [ ]1 [ ]2 [ ]3 [ ]4 [ ]5 [ ]6 [ ]7 [ ]8

[ ]2 [ ]0 [ ]2 [ ]4 [ ]6 [ ]8 [ ]1 [ ]3 [ ]5 [ ]7

[ ]3 [ ]0 [ ]3 [ ]6 [ ]0 [ ]3 [ ]6 [ ]0 [ ]3 [ ]6

[ ]4 [ ]0 [ ]4 [ ]8 [ ]3 [ ]7 [ ]2 [ ]6 [ ]1 [ ]5

[ ]5 [ ]0 [ ]5 [ ]1 [ ]6 [ ]2 [ ]7 [ ]3 [ ]8 [ ]4

[ ]6 [ ]0 [ ]6 [ ]3 [ ]0 [ ]6 [ ]3 [ ]0 [ ]6 [ ]3

[ ]7 [ ]0 [ ]7 [ ]5 [ ]3 [ ]1 [ ]8 [ ]6 [ ]4 [ ]2

[ ]8 [ ]0 [ ]8 [ ]7 [ ]6 [ ]5 [ ]4 [ ]3 [ ]2 [ ]1


(b) All of the statements are False.

(i) In [ ] [ ]513,9 =− since, ( )9mod513 ≡− , [ ] [ ]53 ≠ , so [ ] [ ]133 −≠ .

(ii) ( )9mod09 ≡− , so [ ]09∈− (in 9 ); ( )9mod119 ≡ , so

[ ] [ ]119 ≡ in 9 . ]19[9∉−∴

(iii) 9 is a set of sets NOT a set of numbers, so 92 ∉

(iv) ( )5mod94 ≡ is true, but [ ]54∈ (in 9 ) is false as [ ]44∈ (in 9 ).

We therefore have ‘ FT ⇒ ’, which is always False

(v) The identity of 9 under addition is [ ]0 .

[ ]1 is the identity under multiplication

(vi) There are three elements of 9 that do not have multiplicative inverses: [ ]0 ,

[ ]3 and [ ]6 ..

Question8 Every natural number m can be written in the form

0121

1 101001010 dddddm kk

kk +×+×++×+×= −

− K , where .∈id is the digit

in the th10i position of m.

Examples : 61021005100077526 +×+×+×=

210010041000710000557402 +×+×+×+×=

(a) Let 206257 =+++=S .

(i) ( )9mod77 ≡ , ( ) ( )9mod7100079mod11000 ≡×⇒≡

(ii) ( )9mod55 ≡ , ( ) ( )9mod510059mod1100 ≡×⇒≡

(iii) ( )9mod22 ≡ , ( ) ( )9mod21029mod110 ≡×⇒≡

(iv) Now, ( )9mod66 ≡ .

Thus, ( ) ( )( )9mod62576102100510007 +++≡+×+×+×

Therefore, ( )9mod207526 ≡ .


(b) 1820475 =++++=S . ( )3mod0≡S .

210041000710000557402 +×+×+×= .

Now,

( )3mod25 ≡ , ( ) ( )3mod21000053mod110000 ≡×⇒≡

( )3mod17 ≡ , ( ) ( )3mod1100073mod11000 ≡×⇒≡

( )3mod14 ≡ , ( ) ( )3mod110043mod1100 ≡×⇒≡

( )3mod22 ≡ .

So,

( )( )( )( )( )( )3mod0

3mod63mod2112

3mod210041000710000557402

≡≡

+++≡+×+×+×≡

Therefore, ( )3mod57402 S≡ .

(c) Let ∈m . Then m can be written in the form

0121

1 101001010 dddddm kk

kk +×+×++×+×= −

− K

where ∈id is the digit in the th10i position of m.

Let S be the sum of the digits of m. That is, 0121 dddddS kk +++++= − K .

Consider ( )3mod10iid × , where ki ≤≤0 .

At best, we know ( )3modii dd ≡ by the reflexive property.

Also, we know ( )3mod110 ≡i . Therefore, ( )3mod110 ×≡× ii

i dd .

That is, ( ) kidd ii

i ≤≤≡× 0for 3mod10 .

Thus, we have

( )( )( )( )( )3mod

3mod3mod101001010

0121

0121

1

Sddddd

dddddm

kk

kk

kk

≡+++++≡

+×+×++×+×≡

−

−−K

K

The proof for ( )9modSm ≡ is similar.

(d) We have just shown that any number is divisible by 3 if the sum of its digits is

also divisible by 3.

Similarly, any number is divisible by 9 if the sum of its digits is also divisible by

9.

wuct121 discrete mathematics numbers tutorial exercises

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