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

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#1
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 lower bound of A. So A is a set of integers with at least one element and at least one lower bound. That implies it must have a minimum element, say m.

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
1 month ago
#2
https://www.mathwords.com/l/lower_bound.htm
A quick Google on the basic defns helps?
0
1 month ago
#3
(Original post by 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 lower bound of A. So A is a set of integers with at least one element and at least one lower bound. That implies it must have a minimum element, say m.

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.
p is not in A, but it is still a lower bound for it.

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
#4
(Original post by 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 lower bound of A. So A is a set of integers with at least one element and at least one lower bound. That implies it must have a minimum element, say m.

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.
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
1 month ago
#5
(Original post by 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?
Yes, but p is an integer so it's the one you're interested in.
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