A simple Intro to Real Analysis question

In summary, the conversation is discussing a homework problem where the task is to prove that if F is an ordered field, then certain properties hold for any elements a, b, and c of F. The conversation includes a definition of an ordered field and the attempt at a solution involves using basic properties and questioning whether +(-a) = -a can be assumed. The response states that the closest thing in the field properties is "for all a, there exists -a such that a+(-a)=0" and if only basic properties can be assumed, then +(-a) = -a cannot be assumed.
  • #1
Jamin2112
986
12

Homework Statement



I'm asked to prove that

If F is an ordered field, then the following properties hold for any elements a, b, and c of F:

(a) a<b if and only if 0<b-a
(b) ...
...


Right now I'm working on (a)

Homework Equations



We're supposed to draw from the basic properties (closure, associativity, commutativity, etc.) and also from the following definition.

Definition 1.4. A field F is ordered if it has an ordering < so that:

For all a, b in F, exactly of of these holds: a<b, a=b, a>b.
For all a, b in F, if a<b, then a+c<b+c.
For all a, b in F, if a>0 and b>0, then a+b>0 and ab>0.

The Attempt at a Solution



So far I wrote:

(a) Assume a < b.
Using Definition 1.6, we can add c to both sides of the inequality. Let c = -a, the additive inverse of a, so that the left side of the inequality equals 0.
0 < b + (-a)


But I'm wondering, is it a "given" that + (-a) = -a? I'm wondering whether I can just change this to b - a to complete the proof (actually, half the proof because we're dealing with an "if and only if" proof) or if I need to draw from some property.
 
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  • #2
What you're asking is if the additive inverse of a is the same as the product of the additive inverse of 1 and a.

0=0*a=(1-1)a=a*1+(-1)*a=a+(-1)*a

Hence, -a = (-1)*a

This is true for any ring.
 
Last edited:
  • #3
ZioX said:
What you're asking is if the additive inverse of a is the same as the product of the additive inverse of 1 and a.

0=0*a=(1-1)a=a*1+(-1)*a=a+(-1)*a

Hence, -a = (-1)*a

This is true for any ring.

Yeah, but I just wondering whether it's a "given" that +(-a) = (-1)*a = -a, or whatever. I want to make my proof as concise as possible.
 
  • #4
The closest thing in the field properties is "for all a, there exists -a such that a+(-a)=0".

If you aren't allowed to assume anything beyond the basic properties of an ordered field, I'd say that you can't assume +(-a) = -a.
 

FAQ: A simple Intro to Real Analysis question

1. What is Real Analysis?

Real Analysis is a branch of mathematics that deals with the study of real numbers and their properties. It involves the use of rigorous mathematical proofs and techniques to understand and analyze the behavior of real-valued functions, sequences, and series.

2. What is the difference between Real Analysis and Calculus?

Although both Real Analysis and Calculus deal with the study of functions and limits, there are some key differences between the two. Real Analysis is more concerned with theoretical and abstract concepts, while Calculus focuses on practical applications. Additionally, Real Analysis is a more rigorous and formal approach to studying functions, while Calculus is often more intuitive.

3. What are some common topics in Real Analysis?

Some common topics in Real Analysis include limits, continuity, derivatives, integrals, sequences, and series. Other important topics include metric spaces, topology, and complex analysis.

4. Why is Real Analysis important?

Real Analysis is an essential tool in many branches of mathematics, physics, and engineering. It provides a foundation for studying more advanced topics in these fields and helps to develop critical thinking and problem-solving skills. Real Analysis also has practical applications in fields such as economics, statistics, and computer science.

5. What are some recommended resources for learning Real Analysis?

Some recommended resources for learning Real Analysis include textbooks such as "Principles of Mathematical Analysis" by Walter Rudin and "Real Analysis" by Royden and Fitzpatrick. Online resources such as MIT OpenCourseWare and Khan Academy also offer free lectures and practice problems for Real Analysis. Additionally, working through practice problems and seeking help from a teacher or tutor can also be beneficial in learning Real Analysis.

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