Help with a proof for the Reals.

[daon]

How can I show that given two reals x,y and that x<y that there is a Real number z s.t. x<z<y.

By some definitions, I know x<y implies y-x is a positive real, and that x^-1 > y^-1 if x and y are the same sign. At this point, I also know that basic algebra works for the real numbers, so all axioms that work for the integers work for the Reals.

I have exhausted a pile of loose-leaf trying to figure a way to solve this. Any help is appreciated!

 

[tkhunny]

Has it struck you that (x+y)/2 is Real?

 

[daon]

Has it struck you that (x+y)/2 is Real?

No, it did not!

But if I assume BWOC that there does not exist a Real z s.t. x<z<y, is is sufficient to say that:

x is a real and y is a real, so x+y is a real
and (x+y) is a real and 2^-1 is a Real so (x+y)2^-1 is a real.

And I know x<y, so
x+y < 2y
(x+y)2^-1 < (2y)2^-1
(x+y)2^-1 < (2*2^-1)y
(x+y)2^-1 < y

also,
x<y
2x < x+y
(2x)2^-1 < (x+y)2^-1
(2*2^-1)x < (x+y)2^-1
x < (x+y)2^-1

Now is this a contradiction proof, or can I just say I found such a Real? I don't know what actually proves it properly.

Thanks for the insight BTW!

 

[tkhunny]

I didn't like this: "if x and y are the same sign"

Generally, constructing a desired value is sufficient. Are you SURE the Reals are Closed for the four elementary arithmetic operations? You're done, in my view.

 

[pka]

I like this proof.
\(\displaystyle \L
\begin{array}{l}
a < b\quad \Rightarrow \quad \frac{a}{2} < \frac{b}{2} \\
\frac{a}{2} + \frac{a}{2} < \frac{b}{2} + \frac{a}{2}\quad \Rightarrow \quad a < \frac{{b + a}}{2} \\
\frac{a}{2} + \frac{b}{2} < \frac{b}{2} + \frac{b}{2}\quad \Rightarrow \quad \frac{{b + a}}{2} < b \\
\quad \Rightarrow \quad a < \frac{{b + a}}{2} < b \\
\end{array}\)

 

[daon]

Awesome, thanks for the help.

I'm trying to show that R₊ has no least element through this lemma, so I figured this would be the right way to go about it:

Assume BWOC that R₊ has a least element.
By the Well Ordering Principle, if R₊ has a least element, let r₀ < 1 denote that element.

We know that 0 is not an element of R₊.
We know that 1 is.

I want to show that there is a real number greater than 0 but less than r₀.

By the lemma, there exists a real number 0 < (r₀+0)2^-1 < r₀ \(\displaystyle \Rightarrow\) 0< 2^-1r₀ < r₀

Also 2^-1r₀ is an element of R₊ because by the defn of greater-than, 2^-1r₀ - 0 is an element of R₊. Thus 2^-1r₀ is also.

This a contradiction since our assumtion that r₀ is a least element is incorrect, thus R₊. has no least element.

Done?

 

[pka]

“By the Well Ordering Principle, if R<SUB>+</SUB> has a least element”
BE CAREFUL! R<SUB>+</SUB> is not well ordered in the usual ordering.
In fact, you are proving that R<SUB>+</SUB> is not well ordered.
If c>0 then 0<c/2<c. So Given any \(\displaystyle c \in R_ + \quad \Rightarrow \quad \exists \left( {\frac{c}{2}} \right)\left( {0 < \frac{c}{2} < c} \right)\).
Given any positive number there is a smaller positive number.

 

[daon]

“By the Well Ordering Principle, if R<SUB>+</SUB> has a least element”
BE CAREFUL! R<SUB>+</SUB> is not well ordered in the usual ordering.
In fact, you are proving that R<SUB>+</SUB> is not well ordered.
If c>0 then 0<c/2<c. So Given any \(\displaystyle c \in R_ + \quad \Rightarrow \quad \exists \left( {\frac{c}{2}} \right)\left( {0 < \frac{c}{2} < c} \right)\).
Given any positive number there is a smaller positive number.

Alright, the WOP is still a bit confusing to me.

So we can't use the WOP for anything other than a subset of the integers?

 

[pka]

In the study of basic analysis there is the following definition:
A set is said to be well ordered if each non-empty subset has a first term.
Then there is an axiom that states: The positive integers are well ordered with respect less than. That axiom is known as the Well Ordering Principle for the positive integers.

Do you see why the negative integers are not well ordered?

 

[daon]

In the study of basic analysis there is the following definition:
A set is said to be well ordered if each non-empty subset has a first term.
Then there is an axiom that states: The positive integers are well ordered with respect less than. That axiom is known as the Well Ordering Principle for the positive integers.

Do you see why the negative integers are not well ordered?

Yes, I get it thanks. Although couldn't you denote the negative integers as such:

Z_<0 = {n \(\displaystyle \in\) N | -(n)}


You said A set is said to be well ordered if each non-empty subset has a first term.

And the first term of the sequence will be -1, so couldn't that be viewed as a 'first element'? Or does the definition of order require that the first element be a least element?

 

[pka]

And the first term of the sequence will be -1, so couldn't that be viewed as a 'first element'? Or does the definition of order require that the first element be a least element?

This is a perfect illustration of the danger of doing online mathematics.
The fact is, we may not have the same mathematical vocabulary.
Usually in set theory we understand that ‘order’ is a relation in a set. The usual order on the reals is ‘less-than or equal to’. Saying that z is the first term of a set C means that z proceeds any term in C. Thus 1 is the first term in Z<SUB>+</SUB> but –1 is the last term in Z<SUB>-</SUB>; Z<SUB>-</SUB> has no first term.

 

[daon]

Okay, I understand now, thanks again