Ordering on the Set of Real Numbers .... Sohrab, Ex. 2.1(1)

[Math Amateur]

Homework Statement

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:

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

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:

Sohrab'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 ... ) ...

*** 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 ...

3. The Attempt at a Solution

I am unable to make a meaningful start on this problem ... can someone help me to get started ...

Peter

 

[Math Amateur]

I think I have made some progress with showing that ##\frac{a}{2} \gt 0## ...We have ##2 \gt 0## (can we say this? why is it valid?)

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## (definition of a as greater than 0 )

##\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

 

[fresh_42]

I think I have made some progress with showing that ##\frac{a}{2} \gt 0## ...
We have ##2 \gt 0## (can we say this? why is it valid?)

This is exercise 2.1.10 (c).

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

You can also use exercise 2.1.10 (c) again and show, that ##\frac{1}{2}<0## is impossible by ##(O_2)## and ##(O_3)\,##.

So we now have
##a, 2^{-1} \in P## (definition of a as greater than 0 )
##\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?

Yes.

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

You can use exercise 2.1.11 (1) and what you just have proven here.