Proof by Contridiction Real Analysis

Watch
omnipotxnce
Badges: 6
Rep:
?
#1
Report Thread starter 3 weeks ago
#1
S = {2n + 1/3 − 3n : n ∈ N, n ≥ 2} . The questions asks me to show that -11/12 is not a supremum.

So far i have:

Let sup(s) = -11/12 so 2n + 1 / 3 - 3n <= -11/12,
9n = 45 so n = 5

and now im stuck in my proof in what i do with this n value and how to complete the contridiction.

Thanks in advance.
0
reply
DFranklin
Badges: 18
Rep:
?
#2
Report 3 weeks ago
#2
(Original post by omnipotxnce)
S = {2n + 1/3 − 3n : n ∈ N, n ≥ 2} . The questions asks me to show that -11/12 is not a supremum.

So far i have:

Let sup(s) = -11/12 so 2n + 1 / 3 - 3n <= -11/12,
9n = 45 so n = 5

and now im stuck in my proof in what i do with this n value and how to complete the contridiction.

Thanks in advance.
Please use brackets to make it clear what your expressions are.

By laws of precedence 2n + 1/3 - 3n = 2n + \frac{1}{3} - 3n which I'm pretty sure isn't what you mean.
0
reply
RDKGames
Badges: 20
Rep:
?
#3
Report 3 weeks ago
#3
(Original post by omnipotxnce)
S = {2n + 1/3 − 3n : n ∈ N, n ≥ 2} . The questions asks me to show that -11/12 is not a supremum.

So far i have:

Let sup(s) = -11/12 so 2n + 1 / 3 - 3n <= -11/12,
9n = 45 so n = 5

and now im stuck in my proof in what i do with this n value and how to complete the contridiction.

Thanks in advance.
The supremum is the LEAST upper bound.

Suppose that \sup (S) = -\dfrac{11}{12}. What you aim to show is that there is actually a LESSER value that is still an upper bound for S, or that S contains values greater than -11/12.

So OK, if this is the supremum, then we must have that \dfrac{2n+1}{3-3n} \leq - \dfrac{11}{12} for all n\in\mathbb{N}.

But you should go ahead and solve this inequality, to find that (on the contrary) n must be in a finite range of values. What happens when n is outside this range? What is the explicit contradiction?
0
reply
omnipotxnce
Badges: 6
Rep:
?
#4
Report Thread starter 3 weeks ago
#4
(Original post by RDKGames)
The supremum is the LEAST upper bound.

Suppose that \sup (S) = -\dfrac{11}{12}. What you aim to show is that there is actually a LESSER value that is still an upper bound for S, or that S contains values greater than -11/12.

So OK, if this is the supremum, then we must have that \dfrac{2n+1}{3-3n} \leq - \dfrac{11}{12} for all n\in\mathbb{N}.

But you should go ahead and solve this inequality, to find that (on the contrary) n must be in a finite range of values. What happens when n is outside this range? What is the explicit contradiction?
So then, let n = 6, this returns the value of -13/15 which is > -11/12.
As sup (S) is the least upper bound by definition, and since -13/15 > -11/12, there is a value in S greater than sup (S) ?
(Original post by DFranklin)
Please use brackets to make it clear what your expressions are.

By laws of precedence 2n + 1/3 - 3n = 2n + \frac{1}{3} - 3n which I'm pretty sure isn't what you mean.
Apologies I dont know how to use the syntax on this forum to properly write my equations but ill use brackets where necessary to avoid future confusion.
0
reply
DFranklin
Badges: 18
Rep:
?
#5
Report 3 weeks ago
#5
(Original post by omnipotxnce)
So then, let n = 6, this returns the value of -13/15 which is > -11/12.
As sup (S) is the least upper bound by definition, and since -13/15 > -11/12, there is a value in S greater than sup (S) ?

Apologies I dont know how to use the syntax on this forum to properly write my equations but ill use brackets where necessary to avoid future confusion.
I dont expect people to know how to do the formatting but brackets are always available and often necessary when providing equations using plain text.
0
reply
RDKGames
Badges: 20
Rep:
?
#6
Report 3 weeks ago
#6
(Original post by omnipotxnce)
So then, let n = 6, this returns the value of -13/15 which is > -11/12.
As sup (S) is the least upper bound by definition, and since -13/15 > -11/12, there is a value in S greater than sup (S) ?
Sure.

It is not immediately clear to me that -13/15 > -11/12 just from looking at it, so perhaps you could've picked a better value, or show explicitly that this relation holds.
Last edited by RDKGames; 3 weeks ago
0
reply
DFranklin
Badges: 18
Rep:
?
#7
Report 3 weeks ago
#7
(Original post by RDKGames)
Sure, but much simpler to use n=0 here and note that we get 3/2 which is > -11/12. So clearly this cannot be an upper bound.

It is not immediately clear to me that -13/15 > -11/12
Question states n>= 2. Me, I'd probably take n = 41 here.
0
reply
RDKGames
Badges: 20
Rep:
?
#8
Report 3 weeks ago
#8
(Original post by DFranklin)
Question states n>= 2. Me, I'd probably take n = 41 here.
Yes shortly after posting that I realised that I should probably check whether it's even allowed to have zero.
0
reply
omnipotxnce
Badges: 6
Rep:
?
#9
Report Thread starter 3 weeks ago
#9
(Original post by DFranklin)
I dont expect people to know how to do the formatting but brackets are always available and often necessary when providing equations using plain text.
I understand that but both RDKGames and I managed to read it perfectly clear as its not a very difficult equation to read in plain text if youre not pedantic about it, youve literally been the opposite of helpful on both threads, so if youre not gonna contribute in a postive way please stop replying
0
reply
omnipotxnce
Badges: 6
Rep:
?
#10
Report Thread starter 3 weeks ago
#10
(Original post by RDKGames)
Sure.

It is not immediately clear to me that -13/15 > -11/12 just from looking at it, so perhaps you could've picked a better value, or show explicitly that this relation holds.
okay i clearly have no understanding of this topic I wont lie to you. I really dont understand how to get from the start to the end of the question
0
reply
RDKGames
Badges: 20
Rep:
?
#11
Report 3 weeks ago
#11
(Original post by omnipotxnce)
okay i clearly have no understanding of this topic I wont lie to you. I really dont understand how to get from the start to the end of the question
What's the confusion?

We assumed that -11/12 was a supremum (an upper bound) but then we have shown that there is a value in the set S which is greater than this upper bound, so this contradicts the assumption that -11/12 is an upper bound.
0
reply
omnipotxnce
Badges: 6
Rep:
?
#12
Report Thread starter 3 weeks ago
#12
(Original post by RDKGames)
What's the confusion?

We assumed that -11/12 was a supremum (an upper bound) but then we have shown that there is a value in the set S which is greater than this upper bound, so this contradicts the assumption that -11/12 is an upper bound.
oh so my proof by contridiction was correct?
0
reply
RDKGames
Badges: 20
Rep:
?
#13
Report 3 weeks ago
#13
(Original post by omnipotxnce)
oh so my proof by contridiction was correct?
You had the right idea, I think, but it was a bit hard to follow.

As I said in my last posts, you needed to solve the inequality to find that it is only satisfied for certain values, and not just n=5. Why did you solve the equality case when we wanted to solve an inequality??

But then you basically picked n=6, which is outside of our certain values and found that S has a term which is bigger than the supremum. Hence contradiction.
0
reply
omnipotxnce
Badges: 6
Rep:
?
#14
Report Thread starter 3 weeks ago
#14
(Original post by RDKGames)
You had the right idea, I think, but it was a bit hard to follow.

As I said in my last posts, you needed to solve the inequality to find that it is only satisfied for certain values, and not just n=5. Why did you solve the equality case when we wanted to solve an inequality??

But then you basically picked n=6, which is outside of our certain values and found that S has a term which is bigger than the supremum. Hence contradiction.
I think it was supposed to be 9n < 45 then n < 5 and thats when i chose n = 6 for the contridiction.

But still how do you show that n must be within a finite range of value?
0
reply
RDKGames
Badges: 20
Rep:
?
#15
Report 3 weeks ago
#15
(Original post by omnipotxnce)
I think it was supposed to be 9n < 45 then n < 5 and thats when i chose n = 6 for the contridiction.

But still how do you show that n must be within a finite range of value?
You need to solve \dfrac{2n+1}{3-3n} \leq -\dfrac{11}{12}. This is an A-Level type problem.
0
reply
omnipotxnce
Badges: 6
Rep:
?
#16
Report Thread starter 3 weeks ago
#16
(Original post by RDKGames)
You need to solve \dfrac{2n+1}{3-3n} \leq -\dfrac{11}{12}. This is an A-Level type problem.
so we have 1 < n <= 5 and this shows that it has a finite range? so when i used n = 6 and it gave me a smaller upper bound than -11/12, thats what proved the contridiction?
0
reply
RDKGames
Badges: 20
Rep:
?
#17
Report 3 weeks ago
#17
(Original post by omnipotxnce)
so we have 1 < n <= 5 and this shows that it has a finite range? so when i used n = 6 and it gave me a smaller upper bound than -11/12, thats what proved the contridiction?
To be honest, you still sound confused so I recommend you read through my post #3 in this thread. The inequality must hold for all n\geq 2 if this is truly a supremum. However, it holds only for 2 \leq n \leq 5. So we investigate what happens when n is outside this region. Sure, n=6 gives a value and we find that it is greater than the supremum. A contradiction.

I am running low on energy for this question so I am going to leave it here. I cannot keep repeating things over and over.
0
reply
omnipotxnce
Badges: 6
Rep:
?
#18
Report Thread starter 3 weeks ago
#18
(Original post by RDKGames)
To be honest, you still sound confused so I recommend you read through my post #3 in this thread. The inequality must hold for all n\geq 2 if this is truly a supremum. However, it holds only for 2 \leq n \leq 5. So we investigate what happens when n is outside this region. Sure, n=6 gives a value and we find that it is greater than the supremum. A contradiction.

I am running low on energy for this question so I am going to leave it here. I cannot keep repeating things over and over.
i understand now thank you
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Should there be a new university admissions system that ditches predicted grades?

No, I think predicted grades should still be used to make offers (617)
33.75%
Yes, I like the idea of applying to uni after I received my grades (PQA) (768)
42.01%
Yes, I like the idea of receiving offers only after I receive my grades (PQO) (362)
19.8%
I think there is a better option than the ones suggested (let us know in the thread!) (81)
4.43%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise