Explaining Why 1.1 and 0.95 Are Not Least Upper Bounds for Set A

[NATURE.M]

Homework Statement

Given a set A that contains all numbers of the form, n/(n+1), where n is a positive integer,
a.explain why the number 1.1 is not the lub of A.
b.explain why the number 0.95 is no the lub of A.

Homework Equations

The Attempt at a Solution

a. Suppose 1.1 is the lub of A. Then it follows, from the least upper bound axiom, that
i. 1.1 is an upper bound of A and ii. 1.1≤y, if y is an upper bound for A. Since there exists y<1.1, such as y=1.01, we can see that l.l is not the lub of A. Thus, a contradiction.

(I'm not sure if I formatted this proof accurately (as I'm fairly new to proofs)).

b.It follows in a similar manner to a., although here I will show the first property of the lub isn't satisfied. That is, the number is not a upper bound.

 

[AATroop]

Are you sure you meant ".95 is not lub the of A"? Because then you could show 1/(2) < .95 and be done with it.

 

[johnqwertyful]

Homework Statement

Given a set A that contains all numbers of the form, n/(n+1), where n is a positive integer,
a.explain why the number 1.1 is not the lub of A.
b.explain why the number 0.95 is no the lub of A.

Homework Equations

The Attempt at a Solution

a. Suppose 1.1 is the lub of A. Then it follows, from the least upper bound axiom, that
i. 1.1 is an upper bound of A and ii. 1.1≤y, if y is an upper bound for A. Since there exists y<1.1, such as y=1.01, we can see that l.l is not the lub of A. Thus, a contradiction.

(I'm not sure if I formatted this proof accurately (as I'm fairly new to proofs)).

b.It follows in a similar manner to a., although here I will show the first property of the lub isn't satisfied. That is, the number is not a upper bound.

Your proof for part a is incredibly wordy and confusing. The "least upper bound axiom" (as I know it) is that every bounded set of real numbers has a least upper bound. What you said was the definition of least upper bound. Not to mention you didn't prove anything! How do you know 1.01 is an upper bound?

Part b, you actually have to show that there exists a number of the form n/(n+1)>.95. It should be easy to find such a number, but you should give a specific example.I think you are falling into a trap of people new to proofs; trying to be as wordy and "formal" as possible. It's awkward and often you end up missing the entire point of the problem!

With math, usually the more concise the better.

 

[johnqwertyful]

Are you sure you meant ".95 is not lub the of A"? Because then you could show 1/(2) < .95 and be done with it.

That shows absolutely nothing.

 

[AATroop]

That shows absolutely nothing.

You're right, I wasn't reading it correctly. But what I was getting at was a proof by example. It's not hard to find a number greater than .95 that would disprove .95 is the LUB.

Edit: I just saw your response, and it now it seems like I just tried to copy what you said. I was trying to provide an example of a value that disprove .95 was SOMETHING (I don't know what I THOUGHT it was at the time), but that's why I was trying to provide a specific case. But you're right about the first part, his wording is too convoluted. The best way I found to start proofs was "Let x be ..." and then make conclusions from what you're given to reach a final conclusion that either proves or disproves what they want.

 

[NATURE.M]

If I can show that some upper bound of A is less than 1.1 initially, then can I mention how this does not satisfy the conditions required for lub as given by the axiom. So 1.1 cannot be the lub.

 

[AATroop]

If I can show that some upper bound of A is less than 1.1 initially, then can I mention how this does not satisfy the conditions required for lub as given by the axiom. So 1.1 cannot be the lub.

That's what you have to do. You need to find some arbitrary value of x/y/z/whatever variable you want, that shows the LUB MUST BE < 1.1.

That arbitrary value should relate to n/(n + 1) somehow. So, let x be the lub of n/(n + 1). Because we know n is blah and blah... etc.

 

[johnqwertyful]

If I can show that some upper bound of A is less than 1.1 initially, then can I mention how this does not satisfy the conditions required for lub as given by the axiom. So 1.1 cannot be the lub.

Yes but prove it.

 

[NATURE.M]

That's what you have to do. You need to find some arbitrary value of x/y/z/whatever variable you want, that shows the LUB MUST BE < 1.1.

That arbitrary value should relate to n/(n + 1) somehow. So, let x be the lub of n/(n + 1). Because we know n is blah and blah... etc.

But it wouldn't make sense to let x be the lub of n/(n+1) since the lub for the set doesn't exist.

 

[AATroop]

But it wouldn't make sense to let x be the lub of n/(n+1) since the lub for the set doesn't exist.

I'm sorry, I'm really tired, but shouldn't 1 be the LUB?

lim x-> ∞ of x/(x+1) = 1, correct?

 

[johnqwertyful]

But it wouldn't make sense to let x be the lub of n/(n+1) since the lub for the set doesn't exist.

You don't need to let x be the lub of that set, although the lub DOES exist for that set (and is equal to 1).

 

[AATroop]

You don't need to let x be the lub of that set, although the lub DOES exist for that set (and is equal to 1).

Right, you can even say "Assume x is the LUB" and then show why that would not hold if x was the LUB (but in this case, you should NOT do this- we're only saying this for future proofs).

 

[johnqwertyful]

Right, you can even say "Assume x is the LUB" and then show why that would not hold if x was the LUB (but in this case, you should NOT do this- we're only saying this for future proofs).

Agreed.

 

[NATURE.M]

I'm sorry, I'm really tired, but shouldn't 1 be the LUB?

lim x-> ∞ of x/(x+1) = 1, correct?

Oh yeah, I overlooked that fact.

 

[NATURE.M]

Yes but prove it.

To prove there is an upper bound of A less than 1.1, would I just simply demonstrate that there exists an upper bound, a of the set A, such that a≥m, for all mεA. Then, a≥n/(n+1). Then would I just pick a value of a arbitrarily.

 

[johnqwertyful]

To prove there is an upper bound of A less than 1.1, would I just simply demonstrate that there exists an upper bound, a of the set A, such that a≥m, for all mεA. Then, a≥n/(n+1). Then would I just pick a value of a arbitrarily.

You're overthinking this.


1.01 is an upper bound of the set, although it is not the most convenient example. 1 is an upper bound for the set and 1.1>1. It should be easy to show that 1>n/(n+1) (hint:multiply).

 

[NATURE.M]

You're overthinking this.


1.01 is an upper bound of the set, although it is not the most convenient example. 1 is an upper bound for the set and 1.1>1. It should be easy to show that 1>n/(n+1) (hint:multiply).

Your probably right. It seems I'm worrying to much about the formalism and all that sort. Anyways, that clarifies it. Thanks.

 

[johnqwertyful]

Your probably right. It seems I'm worrying to much about the formalism and all that sort. Anyways, that clarifies it. Thanks.

Please write out your proof. I'd like to see and critique it, if you are interested.

 

[NATURE.M]

I'm defiantly interested, but do you mind if we do it tomorrow? I have class early tomorrow morning, so after, I'll post my proof.

 

[NATURE.M]

*definitely

 

[johnqwertyful]

Sure. Can't say I'll get to it for a bit, but I'll look for it if I get a chance.

 

[Ray Vickson]

I'm sorry, I'm really tired, but shouldn't 1 be the LUB?

lim x-> ∞ of x/(x+1) = 1, correct?

Why did you hand the OP the solution to the problem? He is the one who is supposed to solve it, according to the Forum rules.

 

[johnqwertyful]

Why did you hand the OP the solution to the problem? He is the one who is supposed to solve it, according to the Forum rules.

I don't think that hands OP the solution. AA simply stated a relevant limit.

 

[NATURE.M]

Okay rough version of my solution for a.

We can see that 1 is an upper bound for the set A, since 1≥n/(n+1)*. Since 1<1.1, then by the least upper bound axiom, 1.1 is not the least upper bound of the set A.

*for 1≥ n/(n+1), I'm not sure if I should simplify that to 1≥0. But how is this significant to the result?
Although this doesn't seem like a proof.

 

[NATURE.M]

I have an alternative version which I think is better.

Lets suppose 1.1 is the sup of A. Then ,we know the set A is bounded above by 1.1 since 1.1>n/(n+1), for n=1,2,3... Now, if 1.1 is the sup of A, then each number m<1.1 is not an upper bound of A. That is, if m=1<1.1, then 1 is not a upper bound of A. However, 1≥n/(n+1), thus we have a contradiction.

-I think I need to justify 1≥n/(n+1), but I'm not sure how to go about it.

 

[johnqwertyful]

Simple: multiply

 

[NATURE.M]

1≥n/(n+1) ⇔ n+1≥n

(if we multiply by n+1)

 

[johnqwertyful]

Your proof is still somewhat wordy. Although a lot better.

"Then ,we know the set A is bounded above by 1,1. since 1.1>n/(n+1), for n=1,2,3... "
Why include this? It's irrelevant.

"Now, if 1.1 is the sup of A, then each number m<1.1 is not an upper bound of A. That is, if m=1<1.1, then 1 is not a upper bound of A. "
Surely there's a quicker way of saying this.

 

[NATURE.M]

Wait, don't I still need to justify 1>n/(n+1)?

 

[johnqwertyful]

It's correct, but try to be concise.

I would write it:

Lets suppose 1.1 is the sup of A.Then every number less than 1.1 is not an upper bound. But n+1>n for all n; or 1>n/(n+1). Thus showing 1 is an upper bound for A, so 1.1 can't be the sup.

 

[johnqwertyful]

Wait, don't I still need to justify 1>n/(n+1)?

No more than you need to justify n+1>n.

 

[NATURE.M]

Okay I see the logic now. I overlooked the statement n+1>1, as being of no significance.
Thanks..