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

[Math Amateur]

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 \(\displaystyle \mathbb{R}\) 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 \(\displaystyle \mathbb{R}\) 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 \(\displaystyle \mathbb{R}\) 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

 

[Math Amateur]

I think I have made some progress with showing that \(\displaystyle \frac{a}{2} \gt 0\) ...We have \(\displaystyle 2 \gt 0\) (can we say this? How is this justified?)

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

So we now have

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

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

\(\displaystyle \Longrightarrow \frac{a}{2} \gt 0\) (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 \(\displaystyle \frac{a}{2} \lt a\) ... but how ...?Peter

 

[solakis1]

I think I have made some progress with showing that \(\displaystyle \frac{a}{2} \gt 0\) ...We have \(\displaystyle 2 \gt 0\) (can we say this? How is this justified?)

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

So we now have

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

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

\(\displaystyle \Longrightarrow \frac{a}{2} \gt 0\) (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 \(\displaystyle \frac{a}{2} \lt a\) ... 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 ??