Math 2027 Fundamentals
Assignment
Induitive step Suppose our desired statement holds for by Hypothesis
Goal Prove our desired statement also holds for n btl Prove 13kt 7kt is divisible by 6
After some experimenting not shown here we do som algebra
1 Prove by induction that for all natural numbers
divisible by 6 1 Base case When n t our expression is
13 7 13 7 137
the expression
whish is divisible by 6
n k 13 7 is divisible 6 the inductive
13 7k 13.7k 7.7k
13137 1377
13 1,71 tno isiinby6
bytheInductiveHypothesis has a factorof6 terms in the righthandside Rts are divisible by 6
Since both
it follows that the whole kits is divisible by 6 and hence
the lefthandside 13kt 7kt is divisible by6 This proves theGoal
By the Induction Principle it follows that for all n I N the expression 13 7 is divisible by 6 as required
Let A andAz be subsets of IR Suppose b is an upperbound for Al and
be is an upper bound for Az
Let c be an element of C Then then exist numbers
Datin the set C C A A
that b b tb
c d ta d EA and aze Az is an upper bound for C
asb forall
asb foralldzEAr
are given that
a EA and are A such that
This argument shows that c e b for every
c d taz Then c dtd fbitbe b CEC
Therefore b is an upper boundfor C as required b Let t be the maximum of b andby
Provethat t max bibe fortheunionD AVAz t is an upper bound
First Suppose de A Then deb sins b is an upperbound
Let d be an element of Butbif maxbibe t soalso dft
Theneither d e A or de Az or both forAl
Second suppose de Az Then debs so similarly d e t We have shown that d It for all de D ANAc
That is t max bi bi is an upper boundfor D as required
For each of the following sets AER write down
Sup A inf A max A and min A if they exist Noproof
handed If any of them doesnot exist whynot Sheth
briefly explain AQ Not
see E e si e 2
r min A doesnotexist First although D inf
we can’t have G min A because 52 is not an element
ofAas52isnotarationalnumber52Q andthe minimum of a set is always an element of the set Second
no nEAcanbeminAbecauseallsuchasatisfya752
so we can find a rationalnumber in A that is smallerthan
b A an where an 3 t for ne N supA 3
So no on can be max A since ant is inA andbiggerthan an
R Max A doesnotexist because thereis no largestelementof A
Max A doesnot exist
an is a strictly increasing sequenie d c 92693 946
3cA5BwhereB aER
HereA5B stxneB stx in A
for A because given any h EIR there are real numbers in A that are less than l And sup A wouldbe
a lame bound for A if it existed
Max A doesnotexist 64A so 6is not maxA And if y EA then ye 6 so there are real numbers
in A that are larger than 6 suchas 4
inf A doesnot exist There is no lower bound l
y is any number in A such as 4 1 that
Imax A does not exist
then there are are in A
andsmallerthan y so yt minA
Q4 Let A be a set of positive real numbers A is nonempty and a 0 for every a EA
Define the reiprocal set of A by Mcip A te IR there is some
such that t
Suppose that sup A 2
Use Definition 1.3.2 of the supremum of a set and its counterpart for the infimum of a set to prove that
inf frucipA
As usual he’ll have to use both parts 1 and a
of the definition of sup A
What we know
sup A 2 so by Det 1.3.7
if b is any upper bound for A then 2Eb
What we want to show that inffresip A t that is 1 t is a lower bound for recip A
is an upper bound for d f 2 for every at
11 2 is the smallest upper bound for A
t k for every t E recipA and is the largest lowerbound for nip A
if d is any lower bound for recipe A then I 3h
QU cont Let’s start by proving
1 Let t be a number in recipeA
Then there is a number at
of Def 1.3.2 Therefore
But our t Ya so t
for all tf receipt So t 3112 for all te manA This proves 1
This argument works
Let d be a lower bound for relip A Claim Then He is an upper bound for A
So t 3 I forevery te recipA
Why Suppose because
A such that this d E 2
whish is SoYaDdTakingreliproats aEd
This argument worksfor all at A Sothe Claimis true
Now by a of Det 1.3.7 since he is an upper bound
for A we see that 2 Therefore 112 3
This argument works for every lower bound l of reap A
This proves a both 1 and 11 hold for 112 We have shown that
Therefore inf rerip A 42 as required I
recip A in A
Let S 10 5,015,10 be the set of
all integers that are divisible by 5 Prom thats is countably
by finding a suitably function f IN S Explain briefly why yourfunition f is one to one and onto
J t Is it to its This function is given by the
illustrated here IN I 2 3 4 5 6 7 8
Lat f IN S bythefunction
its to formula
f is one to one
even n isodd
two Different cats always chase differentdogs
Forexample 4 and 7 are sent to 10and 15 and10 I 15 F is onto because every number in S is matched
with a natural number in theformula andsketch above Every dog is chased by a cat
For example the number 15 in S is f 7 the cat 7 chases the dog 15
if n is if
different natural number to the same number in S
because it never sends
b lat t be the set of all rational numbers in lowest terms such that the denominator is
divisible by 5
T Pq PEZqeN qisdivisibleby5
Prove that and P q have no common factors T is countable
First T is a subset of the rational numbers because every number Pig in T is a rational number
Theorem 1.5.7 page 29 says that if A E B Third by Theorem 1.5.61 page 27 Q is countable
and B is countable then A is either countable or finite
Sobytheorem1.5.7withAT andBQ
Now T is not finite because T containsinfinitely many
different numbers For example the numbers
that eithrtiscounth
artisfinite
are all in T F
Therefore T must be countable as required D
Show using the E N definition of
a Sarath work The last line of our proof should that for suitably large values of n
convergence
of a sequence that lim
a your scratch work and 45
In your answer include both
b your write up following the template on page
Ian al I 477
Work on the
We want to make E
that is that is
n E This will betrue for all n
expression
Programming Help
Qb b Writeup using the template from page 45
Choose a natural number NEN with N E 1
Let a o be arbitrary
To verify that this choice of N is appropriate
take an arbitrary ne IN with n
Then n NsE1 so n E1
This argument works for all
above n 3N
as required I
Definition We say a sequence can is eventually
in a set A if there exists a natural number N E IN
Give IIIII.EE Ia
an example of a sequence can that is eventually
in the set A 3
Manypossible answers
thatis IT T 2T 2T 33,3
an 3 for all nein
You have a
Shih 3,013,3 3
fray choice of the first finitely manyforms on
But Sinn A has only one element 3 all theterms an
from some on onwards must be equal to 3 Intheexamples above we coulduse N 1 N 3 andN 5
b Give an example of a sequence bn that is not
eventually in the set A Many possibly answers
end with 3,3 3 an 4YY
33 doesn’t any sequence an that
or Can 1,4 9 25,36
For example
thatis an 4forallatIN
that is Can t that is fan n
a Show that the sequanie Can cont eventually in the set A 1,2
ite is scratch work to understandthe question notrequired in you
First fee terms
931513 83 213
ay 151 914 2x 95155 2
as 1516 116 156 97 It517 177 157
The set A 1,23 is a closed interval Itincludes itsendpoint 1 and2
t iii I an 23 RFhm
Intuition it looks as if the numbers an are decreasing
a 3a sa 3 andasifdiadayamnotinA
butas 96,97 are all in A
Solution We must show there is a natural number
suchthat ant I I for all n Let N 5
TakinIN Son35
First an I 5h I since
En to at Ifat
Second sinn n 5 we havethef andso
So an satisfies I cant 2 Therefore this ant 1,2
Thisargumentworks for all n N in 1,23 This shows that can ith is eventually
an ItInE151s
d Use the E N definition of convergence of a sequence either Def 2.2.3 or Det 2.2.313 to prom thefollowing statement
If land is a sequent that is eventually in theset A 33
then fan converges to 3
Note We must prom this statement for every smh signenin Can
not just for one particular sequence can Solution using Dut 2.7.3
Let Ian be a sequence that is eventually in the set A 133 Def 2.2.3 page 43 says that Can converges to a
if there exists an N E IN such that
n N itfollowsthat Ian al E Leta 3
Let s 0 by an arbitrary positive number
Since Ian is eventually in A 33
by definition there exists we may choose a natural number N such that
an E 3 for all n I N
That is such that an
Take meIN such that n 3 N
3 for all n s thisN
C sinter a 3 skathisan 3
Ian al Ian 31 1331
Therefore Idn al CEworks for all n 3N
This argument
And this argument works for all E
Thus we have shown that Can 3 as required
Q7 d Solution using Def 2.2.313
Dat 2.2.313 page 43 says that Ian converges to a
Let cant be a sequin that is eventually in the set A 33 if given any neighbourhood Kla a E ate of a
there exists a point in the sequence after which all of theterms an am in Vaa
That is if for every s o there exists an N e IN such that an E la s at E for all n N
Let a o be an arbitrary positive number
Then VaCa V 13 3 E 3 E Since Ian is eventually in A 33
by definition there exists we may choose a natural number N such that
anE 3E3E sodnEY3
an E 3 for all n I N
3 for all n s thisN Take meIN such that n N Then an 3
That is such that an
works for all n 3N Thus we have shown that Can
3 as required contingd
Rewrite Def 2.2.3 in terms of the concept of
a sequence beingeventually in a set A StateyoursetAclearly
e Solution
A sequeninfan converges to a realnumber a for every positive number E
there exists an N e IN such that
whenever n N it follows that tan al CE
Tiiiiiatiiji
Take a e IR and let E o be an arbitrary positive number
Saying Ian at s is the same as saying ane Ca E ate Given our 9 0 let
Saying There exists an Nt IN such that
at E whenever n N it follows that fan a
is the same as saying number N such that there exists a natural natural SN
An E A a E at El for all numbers n
which is the definition.tn Q7 of Can iseventually in A Cagats
if given any 270
the sequence an is eventually in A fa 9,9 9
So we can rewrite Def 2.7.3
as follows
Rewritten Def 2.2.3 A sequent fan converges to a
e Solution
Note Not required for 07 but
we can also rewrite the
of convergente of a sequence in
a sequence being eventually in a set A as follows
Rewrite Def 2.2.313 in terms of the concept of
a sequence beingeventually in a set A StateyoursetAclearly
there exists a point in thesequence afterwhich all
Det 2.7.3B
it given any E neighbourhood Vela
otherdefinition Def 2.2.31
A sequeninfan converges to a of d
Let A Vala la
Rewritten Def 2.2.35 A sequenu Can converges to a
Vaca of a in Va la
an is the print
Take a e IR and let E o be an arbitrary positive number
if given any E neighbourhood the sequence an is eventually
thesequence in
Give an example of each or state that the request
is impossible When a request is impossible explain briefly
why that is the case You may use definitions on theorems from the textbook in your explanation
A countable set W such that sup W is not an elementof W
bAcountablesetA afinitesetB anda onetoone andonto function f A B so that A B
Impossible
There cannot be a one to one function f from an infinite set
to a finite set Whynot Let N bethenumberofelements
of the finite set Choose Ntl distant elements 111 24 3 anti of the infinity set Then the Nt 1
can’t all be distinct since thefinite set here only N elements
So f is notone to one
Bonus How do we know the countable set A is infinite
A is countable so there is a one to one andonto function
W an when an 3
Many possible examples including t
elements flu flu flint of the finite
from IN to A We know IN is infinite so by the argument above A is not finite So A is infinite
A convergent sequence and for which infinitely many
of the terms an are equal to 8
an and du and ar
Many possibleexamples 8 8,8 8
including Chatis n where
an8forall 8 8 t8,818,858,84,88th
Code Help
Q8 d A divergent sequence Can for which infinitely many
Can and Can
Many possible examples
1,8 10,8 100,8 8 8 8 8,8 8
of the terms an are equal to 8
e A convergent sequence that is eventually in the
8 0,8 0,0 8 0,00,8 90,0 0,8 open interval 11,3 and eventually in the open interval
Impossible
If Ian is eventually in 11,3 thenthere is some NE IN
an e 1,3 that is lean 3 forallin N ButthenanC5 foralln3N
So an 4 5,7 for all n 3 N
So there cannot be an N E IN such that ant
for all n I Na if there were we’dhave foreach n max NNr
that both an 25 and an 5 which is impossible f AsetBsuchthat BanditspowersetPB have
the same cardinality B NP B Impossible
ByCantor’sTheorem Theorem 1.6.27 there doesnotexist
a function f B P1B that is onto But by Definition 1.5.2
if B PIB thenthere is a function f B PCB thatis oneto one and onto whichwouldcontradict Cantor’s Theorem
So there is no such set B
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