# Least Upper Bound and the Density of the Irrationals Theorem

• jr16
In summary: Have you shown that sqrt(7) is irrational?In summary, the given problem asks for two different upper bounds and the least upper bound (LUB) of the set {x | 1 < x < √(7) and x is irrational}. The LUB is found to be √(7) but the question asks for a proof using the density of the irrationals. The missing information of the set that x is a member of is crucial, as without it, famous set theory paradoxes can occur. However, if it is not specified that √(7) is irrational, then the density of the irrationals may be necessary to prove the LUB.
jr16

## Homework Statement

For the following set if it has an upper bound, find two different upper bounds as well as the least upper bound (LUB), justifying your answer. If the set has no upper bound, state this and justify your answer.

{x | 1 < x < √(7) and x is irrational}
(a proof requires the density of the irrationals)

## Homework Equations

density of the irrationals theorem

## The Attempt at a Solution

I stated two upper bounds as √(10) and √(37)
But where I get confused is with the LUB. Isn't it simply √(7)? Why does the question state that a proof using the density of the irrationals is necessary?

Had x been declared as rational then this would be more clear to me, since there would be no LUB and I would need to prove this using the density of the rationals.

Where am I going wrong?

hi jr

To prove that some given number is LUB, you need to prove two things. If k is the LUB First,you need to show that k is an upper bound and second, you need to prove that for any
$\epsilon > 0$, there exists some $x_1$ in the set such that $k-\epsilon <x_1$.

Can you prove these two things for $\sqrt{7}$ ? Its here you need to use the
density of irrationals...

You aren't going wrong. The answer is sqrt(7) for the LUB. The answer to LUB {x | 1 < x < √(7) and x is a element of D} for any dense set D is sqrt(7). Maybe they are just asking you to prove this without proving sqrt(7) is irrational?

jr16 said:

## Homework Statement

For the following set if it has an upper bound, find two different upper bounds as well as the least upper bound (LUB), justifying your answer. If the set has no upper bound, state this and justify your answer.

{x | 1 < x < √(7) and x is irrational}
(a proof requires the density of the irrationals)

## Homework Equations

density of the irrationals theorem

## The Attempt at a Solution

I stated two upper bounds as √(10) and √(37)
But where I get confused is with the LUB. Isn't it simply √(7)? Why does the question state that a proof using the density of the irrationals is necessary?

Had x been declared as rational then this would be more clear to me, since there would be no LUB and I would need to prove this using the density of the rationals.

Where am I going wrong?

I have the same puzzlement as you, so your confusion is reasonable. However, there's an important piece of information missing. When you wrote

{x | 1 < x < √(7) and x is irrational}

you did not specify the set that x is a member of. That's an important part of this problem. As you noted, if x is constrained to be rational, then we'll need to use the density of the rationals in the reals to solve this problem.

There's another reason to always include the set that x is part of when you're using set specification notation. That's because if you don't constrain x, you can get famous set theory paradoxes. So it's not only a notational formality, it's a logical necessity. That's kind of pedantic point; but in general it's helpful to get in the habit of always writing

{x in X | etc. }

And in this case, it's an important clue to the problem.

SteveL27 said:
I have the same puzzlement as you, so your confusion is reasonable. However, there's an important piece of information missing. When you wrote

{x | 1 < x < √(7) and x is irrational}

you did not specify the set that x is a member of. That's an important part of this problem. As you noted, if x is constrained to be rational, then we'll need to use the density of the rationals in the reals to solve this problem.

There's another reason to always include the set that x is part of when you're using set specification notation. That's because if you don't constrain x, you can get famous set theory paradoxes. So it's not only a notational formality, it's a logical necessity. That's kind of pedantic point; but in general it's helpful to get in the habit of always writing

{x in X | etc. }

And in this case, it's an important clue to the problem.

I would say {x | 1 < x < √(7) and x is irrational} is the same as saying {x is irrational | 1 < x < √(7)}. I think that is overly pedantic. I'll repeat that if you haven't shown that sqrt(7) is irrational, then the density of the irrationals may be helpful.

## 1. What is the definition of the Least Upper Bound property?

The Least Upper Bound property states that in a non-empty set of real numbers, if there is an upper bound, then there exists a least upper bound (also known as supremum) that is the smallest number that is greater than or equal to all elements in the set.

## 2. Can you give an example of a set with a least upper bound?

Yes, for example, the set of all positive real numbers has a least upper bound of 1. Any number greater than or equal to 1 will be an upper bound, but the smallest number that satisfies this property is 1.

## 3. What is the significance of the Density of the Irrationals Theorem?

The Density of the Irrationals Theorem states that between any two distinct real numbers, there exists an irrational number. This theorem is significant because it shows that the set of irrational numbers is dense in the set of real numbers, meaning that there are no "gaps" between the rational and irrational numbers. It also helps to prove the existence of certain real numbers, such as the square root of 2, which is an irrational number.

## 4. How is the Least Upper Bound property related to the Density of the Irrationals Theorem?

The Least Upper Bound property is a fundamental property of the set of real numbers, while the Density of the Irrationals Theorem is a result of this property. The existence of a least upper bound for a set of real numbers ensures that there are no gaps between the rational and irrational numbers, leading to the density of the irrationals theorem.

## 5. Can the Least Upper Bound property be extended to other number systems?

Yes, the Least Upper Bound property can be extended to other number systems, such as the complex numbers. However, it may not hold for all number systems. For example, the set of positive integers does not have a least upper bound, as there is always a larger integer that can be an upper bound.

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