MHB Ordering on the Set of Real Numbers .... Sohrab, Exercise 2.1.10 (b) ....

[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 2.1.10 Part (b) ... ...

Exercise 2.1.10 Part (b) reads as follows:View attachment 7203 I am unable to make a meaningful start on Exercise 2.1.10 (b) ... can someone please help ...

PeterNOTE 1: I am not sure what assumptions Sohrab wants us to make about $$\mathbb{N}, \mathbb{N_0}, \mathbb{Z}$$ and $$\mathbb{Q}$$ for these exercises ... he has not developed/constructed the natural numbers, the integers or the rationals ... but simply named them as sets and done little more than indicate notation for them ... as follows:

View attachment 7213
View attachment 7214

So I am trying to prove the exercise using the real numbers as defined by an ordered field (which, I think, can be shown to contain a copy of each of the sets $$\mathbb{N}, \mathbb{N_0}, \mathbb{Z}$$ and $$\mathbb{Q}$$ ... )

Mind you ... after reading the start of Sohrab's Appendix A I am again a little uncertain as to what to assume when I read the following ... (mind you, this is stated well after the section where Exercise 2.1.10 (b) appears ... )https://www.physicsforums.com/attachments/7215I would, however like, as I have previously indicated, to prove the exercise using the real numbers as defined by an ordered field ...

NOTE 2: 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 7204Sohrab's definition of the usual ordering on $$\mathbb{R}$$ plus some of the properties following are as follows ...https://www.physicsforums.com/attachments/7205
View attachment 7206

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 ... View attachment 7207
https://www.physicsforums.com/attachments/7208

 

[Euge]

Note $1^2 = 1$ and $1 \neq 0$.

 

[Math Amateur]

Note $1^2 = 1$ and $1 \neq 0$.

Thanks for the help, Euge ...

I now have a way now to prove that $$1 \gt 0$$ but I am concerned because it involves what Sohrab lists as a subsequent exercise ... namely Exercise 2.1.11 (3) ... does Sohrab expect us to prove $$1 \gt 0$$ without relying on subsequent/later exercise ... can such a proof be found ... are questions that bother me ...

However ... my proof based on your hint is as follows ...
To show $$1 \gt 0$$

By trichotomy, one of $$1 \gt 0, 1 = 0$$ or $$1 \lt 0$$ holds true ...

Now ... since the field of real numbers, $$\mathbb{R}$$, is clearly not the trivial field. we have that $$1 \ne 0$$ ... ...

... so basically the proof reduces to showing that $$1 \lt 0$$ does not hold true ...Assume that $$1 \lt 0$$ holds Now ... $$1 \lt 0$$

$$\Longrightarrow 1 \cdot 1 \lt 1 \cdot 0$$ (by Exercise 2.1.11 (3) ... can we use this subsequent exercise ...)

$$\Longrightarrow 1^2 \lt 0$$ ...

But ... we know $$1^2 \gt 0$$ by Exercise 2.1.10 (a) ... Contradiction!

Therefore $$1 \lt 0$$ does not hold ...

Now we have that $$1 = 0$$ and $$1 \lt 0$$ do not hold true ...

Therefore $$1 \gt 0$$ ... ...
Is the above correct ...?

Is there a way to prove $$1 \gt 0$$ without relying on Exercise 2.1.11 (3) ...?

Peter

 

[Euge]

The point was to use part (a) of your exercise, not the next exercise.

 

[Math Amateur]

The point was to use part (a) of your exercise, not the next exercise.

Hmm ... you are right, of course ... and, indeed, the solution using Part (a) seems obvious ...

... ... don't know how I missed it ... :(...
To show $$1 \gt 0$$ ... relying only on Exercise 2.1.10 (a) and the Field Axioms for $$\mathbb{R}$$ ... Now, we have $$1 \cdot 1 = 1^2 \gt 0$$ ... ... by the definition of exponentiation, and by Exercise 2.1.10 (a) ... ...

But $$1^2 = 1 \cdot 1 = 1$$ ... ... by Field Axiom M3 ...

Therefore $$1 \gt 0$$ Is that correct?

Peter

 

[solakis1]

Hmm ... you are right, of course ... and, indeed, the solution using Part (a) seems obvious ...

... ... don't know how I missed it ... :(...
To show $$1 \gt 0$$ ... relying only on Exercise 2.1.10 (a) and the Field Axioms for $$\mathbb{R}$$ ... Now, we have $$1 \cdot 1 = 1^2 \gt 0$$ ... ... by the definition of exponentiation, and by Exercise 2.1.10 (a) ... ...

But $$1^2 = 1 \cdot 1 = 1$$ ... ... by Field Axiom M3 ...

Therefore $$1 \gt 0$$ Is that correct?
But But $$1^2 = 1 \cdot 1 = 1$$ ... ... by Field Axiom M3 .
Peter

Yes it is

But , $$1^2 = 1 \cdot 1 = 1$$ ... ... by Field Axiom M3, is not only by M3

You have to use the rule of exponents as well

Anyway the above are simple exercises of high school !

Why you have problems??