MHB Ordering on the Set of Real Numbers .... Sohrab, Exercise 2.1.10 (1) ....

[Math Amateur]

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

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 (a) ... ...

Exercise 2.1.10 Part (a) reads as follows:

View attachment 7199I am unable to make a meaningful start on Exercise 2.1.10 (a) ... can someone please help ...

PeterNOTE: Sohrab defines [FONT=MathJax_AMS]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 7200Sohrab's definition of the usual ordering on [FONT=MathJax_AMS]R plus some of the properties following are as follows ...View attachment 7201
https://www.physicsforums.com/attachments/7202Hope 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/7211
https://www.physicsforums.com/attachments/7212

 

[Math Amateur]

After a little reflection ... I have come to the following solution ...To show that $$a \in \mathbb{R} \ \{ 0 \} \Longrightarrow a^2 \gt 0$$ ... ...First consider $$a \gt 0$$ ...

$$a \gt 0$$

$$\Longrightarrow a \in P$$ (see definition of $$P$$ in scanned text in above post, essentially $$P$$ is the positive reals ...)

$$\Longrightarrow a \cdot a \in P $$

$$\Longrightarrow a^2 \in P$$

$$\Longrightarrow a^2 \gt 0 $$
Next, consider $$a \lt 0$$ ... $$a \lt 0$$

Put $$a = -b$$ ... then ...

$$(-b) \cdot (-b) $$

$$= (-1) \cdot (b) \cdot (-1) \cdot (b)$$ (see Exercise 2.1.1 (2) ... scanned text below)

$$= (-1) \cdot (-1) \cdot (b) \cdot (b)$$ (commutativity of multiplication in a field ... see scanned text below ...)

$$= (1) \cdot (b) \cdot (b)$$ (Exercise 2.1.1 (4) ... see scanned text below ... )

= $$b^2$$ where $$b$$ is positive and so $$b^2 \gt 0$$ ...Is that correct?

Please critique the proof ...

Note: Not at all sure I treated $$a \lt 0$$ validly or well ...

Peter

 

[solakis1]

After a little reflection ... I have come to the following solution ...To show that $$a \in \mathbb{R} \ \{ 0 \} \Longrightarrow a^2 \gt 0$$ ... ...First consider $$a \gt 0$$ ...

$$a \gt 0$$

$$\Longrightarrow a \in P$$ (see definition of $$P$$ in scanned text in above post, essentially $$P$$ is the positive reals ...)

$$\Longrightarrow a \cdot a \in P $$

$$\Longrightarrow a^2 \in P$$

$$\Longrightarrow a^2 \gt 0 $$
Next, consider $$a \lt 0$$ ... $$a \lt 0$$

Put $$a = -b$$ ... then ...

$$(-b) \cdot (-b) $$

$$= (-1) \cdot (b) \cdot (-1) \cdot (b)$$ (see Exercise 2.1.1 (2) ... scanned text below)

$$= (-1) \cdot (-1) \cdot (b) \cdot (b)$$ (commutativity of multiplication in a field ... see scanned text below ...)

$$= (1) \cdot (b) \cdot (b)$$ (Exercise 2.1.1 (4) ... see scanned text below ... )

= $$b^2$$ where $$b$$ is positive and so $$b^2 \gt 0$$ ...Is that correct?

Please critique the proof ...

Note: Not at all sure I treated $$a \lt 0$$ validly or well ...

Peter

The 1st part of your proof is correct but not complete

For the 2nd proof

Proof:

1) 0>a.................assumption

2) (0-a)εP................By definition (a-b)εP<=> a>b

3) [0+(-a)]εP..............By definition 2.1.4 (subtraction)

4) (-a)εP...............By axiom A3

5) (-a)(-a)εP...............By axiom O2

Now we prove that (-a)(-a) = a^2 by using the equality axioms of the real Nos

6) [(-1)a][(-1)a]..............By using exercise 2.1.1(2)

7) [((-1)a)(-1)]a..............By axiom M2

8) [(a(-1))(-1)]a..............By axiom M1

9) [a((-1)(-1)]a..............By axiom M2

10) [a.1]a...............By exercise 2.1.1(4)

11) a.a.................By axiom M3

12) a^2.................By definition 2.1.4

13) But $$a^2=a^2-0$$
You can prove that ,I suppose.
So we have ,(a^2-0)εP => a^2>0