MHB Ordering on the Set of Real Numbers .... Sohrab, Exercise 2.1.12 ....

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I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

I need help with Exercise 2.1.12 Part (1) ... ...

Exercise 2.1.12 Part (1) reads as follows:

View attachment 7195

I am unable to make a meaningful start on Exercise 2.1.12 (1) ... can someone please help ...

PeterNOTE: Sohrab defines $$\mathbb{R}$$ as a field with binary operations of addition and multiplication ... he then goes on to define subtraction, division and exponentiation as follows:View attachment 7196Sohrab's definition of the usual ordering on $$\mathbb{R}$$ plus some of the properties following are as follows ... (but note that Exercises 2.1.10 and 2.1.11 precede Exercise 2.1.12 and so, I think, must be taken as given properties for the purposes of Exercise 2.1.12 ... ) ...View attachment 7197
View attachment 7198

Hope someone can help ...

Peter
*** EDIT *** I am concerned that Exercises 2.1.1 and 2.1.2 contain properties of addition, multiplication and inverses that flow directly form the properties of $$\mathbb{R}$$ as a field, ... ... and these properties could possibly be useful in the exercise ... so I am providing Sohrab's description of the field of real numbers and the exercises that follow it, namely Exercises 2.1.1 and 2.1.2 ... https://www.physicsforums.com/attachments/7209
https://www.physicsforums.com/attachments/7210
 
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I think I have made some progress with showing that $$\frac{a}{2} \gt 0$$ ...We have $$2 \gt 0$$ (can we say this? How is this justified?)

and so $$2^{-1} \gt 0$$ by Exercise 2.1.11 (5) (see scanned text in above post)

So we now have

$$a, 2^{-1} \in P$$ ... (by definition of a as greater than zero)

$$\Longrightarrow a \cdot 2^{-1} \in P $$ (Order axiom $$O_2$$ ... see scanned text in above post)

$$\Longrightarrow \frac{a}{2} \gt 0$$ (by definition of division ... see scanned text in above post )Is that correct?

If it is a valid and good proof ... then we still need to show $$\frac{a}{2} \lt a$$ ... but how ...?Peter
 
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Peter said:
I think I have made some progress with showing that $$\frac{a}{2} \gt 0$$ ...We have $$2 \gt 0$$ (can we say this? How is this justified?)

and so $$2^{-1} \gt 0$$ by Exercise 2.1.11 (5) (see scanned text in above post)

So we now have

$$a, 2^{-1} \in P$$ ... (by definition of a as greater than zero)

$$\Longrightarrow a \cdot 2^{-1} \in P $$ (Order axiom $$O_2$$ ... see scanned text in above post)

$$\Longrightarrow \frac{a}{2} \gt 0$$ (by definition of division ... see scanned text in above post )Is that correct?

If it is a valid and good proof ... then we still need to show $$\frac{a}{2} \lt a$$ ... but how ...?Peter

To prove 2>0 ,you need the definition : 1+1=2

Then having the above definition and that 1>0 we have:

0<1 and 0<1 => 0+0<1+1.........by exercise 2.1.11 (2)

0<2........... by the above definition and axiom (A3)

The rest of your proof is correct

Now to prove a/2<a

proof:

1) 0<a...............assumption

2) 0<a => 0+a<a+a...........by exercise 2.1.11(1)

3) a<a.1+a.1.............axioms (A3) and (M3)

4) a<a(1+1)..............by axiom (D)

5) a<a.2 ..............by above definition

Can you prove the rest of the exercise ??
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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