# Undergraduate math - Proof for there exists a rational number between 2 real numbers

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

The Archimedean Principle:

If r belongs to a real number, then there exists an n that belongs to a natural number, so that n is larger than r.

Corollary to the Archimedean Principle:

If r belongs to a real number, then there exists a p that belongs to an integer so that p is smaller than r.

Lemma:

Let a,b belong to real numbers with a + 1 being smaller than b. Then there exists an m that belongs to an integer so that m is larger than a and smaller than b.

Proof of Lemma:

Assume a,b belong to real numbers. Assume a+1 is smaller than b.

Let A be a set containing integers k such that k is larger than a.

Since a is a real number, we know by the Archimedean Principle that there is a natural number n so that n is larger than a.

So A has at least one element. By the corollary to the Archimedean Principle, we know that there is an integer p so that p is less than a.

I understand everything except those written in bold. Lower bound means the lowest element in a set, but how is p a lower bound of set A? Set A consists of integers k that is larger than a, and there is a p that is smaller than a. Then doesn't that mean k is smaller than p as well and p does not belong to set A? How is p a lower bound of set A then?

Also, it says 'at least one' lower bound - is it possible to have more than one lower bound i.e. more than one smallest number in a set?

Thank you.

If r belongs to a real number, then there exists an n that belongs to a natural number, so that n is larger than r.

Corollary to the Archimedean Principle:

If r belongs to a real number, then there exists a p that belongs to an integer so that p is smaller than r.

Lemma:

Let a,b belong to real numbers with a + 1 being smaller than b. Then there exists an m that belongs to an integer so that m is larger than a and smaller than b.

Proof of Lemma:

Assume a,b belong to real numbers. Assume a+1 is smaller than b.

Let A be a set containing integers k such that k is larger than a.

Since a is a real number, we know by the Archimedean Principle that there is a natural number n so that n is larger than a.

So A has at least one element. By the corollary to the Archimedean Principle, we know that there is an integer p so that p is less than a.

**Then the integer p is a**That implies it must have a minimum element, say m.__lower bound of A__. So A is a set of integers with at least one element and__at least one lower bound__.I understand everything except those written in bold. Lower bound means the lowest element in a set, but how is p a lower bound of set A? Set A consists of integers k that is larger than a, and there is a p that is smaller than a. Then doesn't that mean k is smaller than p as well and p does not belong to set A? How is p a lower bound of set A then?

Also, it says 'at least one' lower bound - is it possible to have more than one lower bound i.e. more than one smallest number in a set?

Thank you.

0

reply

Report

#2

https://www.mathwords.com/l/lower_bound.htm

A quick Google on the basic defns helps?

A quick Google on the basic defns helps?

0

reply

Report

#3

(Original post by

The Archimedean Principle:

If r belongs to a real number, then there exists an n that belongs to a natural number, so that n is larger than r.

Corollary to the Archimedean Principle:

If r belongs to a real number, then there exists a p that belongs to an integer so that p is smaller than r.

Lemma:

Let a,b belong to real numbers with a + 1 being smaller than b. Then there exists an m that belongs to an integer so that m is larger than a and smaller than b.

Proof of Lemma:

Assume a,b belong to real numbers. Assume a+1 is smaller than b.

Let A be a set containing integers k such that k is larger than a.

Since a is a real number, we know by the Archimedean Principle that there is a natural number n so that n is larger than a.

So A has at least one element. By the corollary to the Archimedean Principle, we know that there is an integer p so that p is less than a.

I understand everything except those written in bold. Lower bound means the lowest element in a set, but how is p a lower bound of set A? Set A consists of integers k that is larger than a, and there is a p that is smaller than a. Then doesn't that mean k is smaller than p as well and p does not belong to set A? How is p a lower bound of set A then?

Also, it says 'at least one' lower bound - is it possible to have more than one lower bound i.e. more than one smallest number in a set?

Thank you.

**ChloeYeo**)The Archimedean Principle:

If r belongs to a real number, then there exists an n that belongs to a natural number, so that n is larger than r.

Corollary to the Archimedean Principle:

If r belongs to a real number, then there exists a p that belongs to an integer so that p is smaller than r.

Lemma:

Let a,b belong to real numbers with a + 1 being smaller than b. Then there exists an m that belongs to an integer so that m is larger than a and smaller than b.

Proof of Lemma:

Assume a,b belong to real numbers. Assume a+1 is smaller than b.

Let A be a set containing integers k such that k is larger than a.

Since a is a real number, we know by the Archimedean Principle that there is a natural number n so that n is larger than a.

So A has at least one element. By the corollary to the Archimedean Principle, we know that there is an integer p so that p is less than a.

**Then the integer p is a**That implies it must have a minimum element, say m.__lower bound of A__. So A is a set of integers with at least one element and__at least one lower bound__.I understand everything except those written in bold. Lower bound means the lowest element in a set, but how is p a lower bound of set A? Set A consists of integers k that is larger than a, and there is a p that is smaller than a. Then doesn't that mean k is smaller than p as well and p does not belong to set A? How is p a lower bound of set A then?

Also, it says 'at least one' lower bound - is it possible to have more than one lower bound i.e. more than one smallest number in a set?

Thank you.

Which is why they introduce m to be the lower bound *and* inside A.

Its like saying that if x is in the set [1,2] then clearly x>0 which means 0 is a lower bound, but it is not in the set itself.

There are many lower bounds, but only one greatest lower bound (called infimum).

Last edited by RDKGames; 1 month ago

0

reply

**ChloeYeo**)

The Archimedean Principle:

If r belongs to a real number, then there exists an n that belongs to a natural number, so that n is larger than r.

Corollary to the Archimedean Principle:

If r belongs to a real number, then there exists a p that belongs to an integer so that p is smaller than r.

Lemma:

Let a,b belong to real numbers with a + 1 being smaller than b. Then there exists an m that belongs to an integer so that m is larger than a and smaller than b.

Proof of Lemma:

Assume a,b belong to real numbers. Assume a+1 is smaller than b.

Let A be a set containing integers k such that k is larger than a.

Since a is a real number, we know by the Archimedean Principle that there is a natural number n so that n is larger than a.

So A has at least one element. By the corollary to the Archimedean Principle, we know that there is an integer p so that p is less than a.

**Then the integer p is a**That implies it must have a minimum element, say m.

__lower bound of A__. So A is a set of integers with at least one element and__at least one lower bound__.I understand everything except those written in bold. Lower bound means the lowest element in a set, but how is p a lower bound of set A? Set A consists of integers k that is larger than a, and there is a p that is smaller than a. Then doesn't that mean k is smaller than p as well and p does not belong to set A? How is p a lower bound of set A then?

Also, it says 'at least one' lower bound - is it possible to have more than one lower bound i.e. more than one smallest number in a set?

Thank you.

A lower bound is ANY NUMBER that is less than or equal to all the elements in a given set, e.g. 5,6,7,8 are all lower bounds of {8,9}.

So does this mean a is also a lower bound of set A? So are there two lower bounds of set A found so far - p and a?

0

reply

Report

#5

(Original post by

Ohh I think I kind of get it now.

A lower bound is ANY NUMBER that is less than or equal to all the elements in a given set, e.g. 5,6,7,8 are all lower bounds of {8,9}.

So does this mean a is also a lower bound of set A? So are there two lower bounds of set A found so far - p and a?

**ChloeYeo**)Ohh I think I kind of get it now.

A lower bound is ANY NUMBER that is less than or equal to all the elements in a given set, e.g. 5,6,7,8 are all lower bounds of {8,9}.

So does this mean a is also a lower bound of set A? So are there two lower bounds of set A found so far - p and a?

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top