# Proof by Contridiction Real Analysis

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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.

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.

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#2

(Original post by

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.

**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.

By laws of precedence 2n + 1/3 - 3n = which I'm pretty sure isn't what you mean.

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#3

**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.

Suppose that . What you aim to show is that there is actually a LESSER value that is still an upper bound for , or that contains values greater than -11/12.

So OK, if this is the supremum, then we must have that

**for all**.

But you should go ahead and solve this inequality, to find that (on the contrary) must be in a finite range of values. What happens when is outside this range? What is the explicit contradiction?

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The supremum is the LEAST upper bound.

Suppose that . What you aim to show is that there is actually a LESSER value that is still an upper bound for , or that contains values greater than -11/12.

So OK, if this is the supremum, then we must have that

But you should go ahead and solve this inequality, to find that (on the contrary) must be in a finite range of values. What happens when is outside this range? What is the explicit contradiction?

**RDKGames**)The supremum is the LEAST upper bound.

Suppose that . What you aim to show is that there is actually a LESSER value that is still an upper bound for , or that contains values greater than -11/12.

So OK, if this is the supremum, then we must have that

**for all**.But you should go ahead and solve this inequality, to find that (on the contrary) must be in a finite range of values. What happens when is outside this range? What is the explicit contradiction?

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

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

By laws of precedence 2n + 1/3 - 3n = which I'm pretty sure isn't what you mean.

**DFranklin**)Please use brackets to make it clear what your expressions are.

By laws of precedence 2n + 1/3 - 3n = which I'm pretty sure isn't what you mean.

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#5

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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.

**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.

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#6

(Original post by

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) ?

**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) ?

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

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#7

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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

**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

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#8

(Original post by

Question states n>= 2. Me, I'd probably take n = 41 here.

**DFranklin**)Question states n>= 2. Me, I'd probably take n = 41 here.

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(Original post by

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.

**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.

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(Original post by

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.

**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.

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#11

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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

**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

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.

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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.

**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.

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#13

(Original post by

oh so my proof by contridiction was correct?

**omnipotxnce**)oh so my proof by contridiction was correct?

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.

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(Original post by

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.

**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.

But still how do you show that n must be within a finite range of value?

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#15

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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?

**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?

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#17

(Original post by

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?

**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?

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.

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(Original post by

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 if this is truly a supremum. However, it holds only for . 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.

**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 if this is truly a supremum. However, it holds only for . 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.

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