Proving a property of real numbers

In summary, the conversation is about a homework problem from Apostol's Calculus Vol. 1, which involves proving the existence of a real number z between two given real numbers x and y. The conversation touches on concepts such as algebra, inequalities, supremum, infimum, and the Archimedean property. The solution involves proving that inf S = x and using the property (x+y)/2 to show that there are no gaps between x and y. The conversation also discusses the difficulty of proving "intuitively obvious" theorems and the usefulness of the property (x+y)/2 in this proof.
  • #1
neutrino
2,094
2

Homework Statement



Given x<y for some real numbers x and y. Prove that there is at least
one real z satisfying x<z<y


Homework Equations



This is an exercise from Apostol's Calculus Vol. 1. The usual laws of
algebra, inequalities, a brief discussion on supremum, infimum and
the Archimedean property preceeded this exercise.


The Attempt at a Solution



I'm trying to solve as many problems from the first chapter, which I had
just glossed over during a first read. Proving theorems in a rigorous
fashion is not exactly my forte. It becomes all the more difficult when
I'm asked to prove something "intuitively obvious."

I think I have the solution in fragments.

Let S be the set of all numbers greater than x. x is a lower bound for
this set, and therefore S has a greatest lower bound - inf S.

[tex]inf S \geq x [/tex]

[The following argument assumes that x = inf S]

Consider some [tex]y \in S[/tex]. I can always find at least one [tex]z \in S[/tex] such that z < y, because if there are no z's satisfying that inequality then y would be inf S.

The only problem here seems to be to prove that inf S = x. The first choice is a proof by contradiction.

Assume inf S > x...:uhh:

Am I on the right track?



 
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  • #2
How about proving some stuff about (x+y)/2?
 
  • #3
Dick said:
How about proving some stuff about (x+y)/2?

:redface: Yes, thank you...I really don't know why I was thinking about lower bounds and stuff.

Was my earlier "proof" correct? It will be complete when I prove x = inf S, right?
 
  • #4
The thing is that I think you might find this property useful for proving x=inf S.
 
  • #5
Dick said:
The thing is that I think you might find this property useful for proving x=inf S.

:confused:

I thought you were suggesting this...

x<(x+y)/2
y>(x+y)/2

Therefore x<(x+y)/2<y - this being a much shorter solution than what I had written down earlier. I'm not sure how (x+y)/2 relates to inf S.
 
  • #6
Well, x is a lower bound of S. Suppose there is another lower bound y and y>x. Knowing there is a z such that x<z<y gives you a problem since then z is in S. Trying to prove inf S=x leads you to prove things that are an awful lot like what you are trying to prove to begin with.
 
  • #7
Ah...
Assume inf S > x. This implies that inf S belongs to S.

But then x<(inf S + x)/2 < inf S and it also belongs to S, which would imply inf S is not the infimum. A contradiction!

But trying to prove that x = inf S this way seems redundant since it relies on the fact that for all real x and y, x<(x+y)/2<y, which was what I set to prove in the first place.

And that's what I think you tried to convey in your last post. :biggrin:
 
  • #8
Yes. If there were a gap in the real numbers between x and y, then inf S would be y. So you want to show there are no gaps before you prove inf S=x.
 
  • #9
Got it now. Thank you.
 
  • #10
Dick said:
Well, x is a lower bound of S. Suppose there is another lower bound y and y>x. Knowing there is a z such that x<z<y gives you a problem since then z is in S.
?? No, it isn't. If y is a lower bounds on S, x< y, then x is NOT in s and neither is the z above.

Trying to prove inf S=x leads you to prove things that are an awful lot like what you are trying to prove to begin with.

In fact, "infimum" cannot have anything to do with this property: If x and y are rational numbers then there exist a rational number between them (again (x+y)/2 is a rational number) but bounded sets of rational numbers do not necessarily have rational infimum.
 
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  • #11
HallsofIvy said:
?? No, it isn't. If y is a lower bounds on S, x< y, then x is NOT in s and neither is the z above.

I didn't say x in is S. S is defined to be the set of all y such that y>x. I find it difficult to accept that z is not in S if x<z<y.

In fact, "infimum" cannot have anything to do with this property: If x and y are rational numbers then there exist a rational number between them (again (x+y)/2 is a rational number) but bounded sets of rational numbers do not necessarily have rational infimum.

I really don't think you read the thread.
 

1. What does it mean to prove a property of real numbers?

Proving a property of real numbers means demonstrating that a specific statement or formula holds true for all real numbers, using logical reasoning and mathematical techniques.

2. Why is it important to prove properties of real numbers?

Proving properties of real numbers is important because it allows us to establish the validity of mathematical statements and formulas, and to make accurate predictions and calculations in various fields such as science, engineering, and finance.

3. What are some common techniques used to prove properties of real numbers?

Some common techniques used to prove properties of real numbers include mathematical induction, direct proof, proof by contradiction, and proof by contrapositive.

4. Can all properties of real numbers be proven?

No, not all properties of real numbers can be proven. Some properties may be true but cannot be proven using current mathematical techniques, while others may be unprovable due to their complexity or the limitations of our mathematical systems.

5. How can we apply properties of real numbers in real-life situations?

Properties of real numbers can be applied in a variety of real-life situations, such as calculating interest rates, measuring distances, and predicting the behavior of physical systems. They are also essential in fields like statistics, economics, and computer science.

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