Proving a^2 is Always Positive

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Homework Help Overview

The discussion revolves around proving that \( a^2 \) is always positive, with participants exploring various approaches to establish this claim within the context of real analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to prove the statement by considering two cases based on the sign of \( a \). Some participants question the completeness of the assumptions and the clarity of the arguments presented. There is a suggestion to simplify the proof by directly multiplying both sides of the inequality, which leads to further discussion about the implications of multiplication in the context of real number axioms.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original proof attempts and suggesting alternative approaches. There is recognition of the need for clearer assumptions and definitions, particularly regarding the implications of multiplication in the context of the problem.

Contextual Notes

Participants note that the original poster is in a beginning real analysis course, which influences their understanding of algebraic structures and the implications of certain operations.

jaqueh
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Homework Statement


prove a^2 is positive

I have tried this, but i don't know if it's conclusive:

i) Suppose a≥0
then a+1≥1
then a(a+1)≥a
then a^2+a≥a
then a^2≥0
ii) Suppose a≤0
then 1+a≤1
then a(a+1)≥a since a is negative, by a lemma
then a^2+1≥a
then a^2≥0
 
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Seems pretty ok. You aren't stating all of your assumptions and lemmas here, though. And there is a typo in your second argument.
 
ok great, yeah i'll be more specific and i do see my typo: a+1 instead of 1+a
 
jaqueh said:
ok great, yeah i'll be more specific and i do see my typo: a+1 instead of 1+a

I actually meant turning a(a+1) into a^2+1, but I'm sure that's not news.
 
oh gosh i didn't even realize i had typed that, haha well thanks anyways.

btw you are always incredibly helpful, and for that i am really grateful!
 
Why don't you just multiply both sides immediately by a?

i.e.

Case 1: a≥0
a2≥0

Case 2: a≤0
a2≥0

Seems a lot more direct and it's the same idea.
 
because you don't know what a*a means yet
 
jaqueh said:
because you don't know what a*a means yet

What do you mean, exactly?
 
well I am taking a beginning real analysis course and basically we're building the basic structure of algebra from scratch from a couple of real number axioms.
 
  • #10
jaqueh said:
well I am taking a beginning real analysis course and basically we're building the basic structure of algebra from scratch from a couple of real number axioms.

OK. But saying you don't know the implication of a*a means you don't know the implication of a*(a+1) either.

You're already making an implicit assumption when you reverse the sign of the inequality upon multiplying by negative a. Multiplying a≤0 by a assumes no more than multiplying a+1≤1 by a.
 
  • #11
the multiplying done by the negative constant was proved in a lemma but you're right, it wasn't known before. and when i did a(a+1) i guess i knew that it was a*a+a, but i didn't know what a*a was still so i had to describe it using the inequalities
 

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