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mathmatt

Member

Registered: 2012-02-11

Posts: 1

Proposition: If x>0, then x^-1 >0.

Hello,

I am currently working through John Erdman's "Problem Text in Advanced Calculus" in preparation for graduate school entrance exams. 
In Appendix H, Problem H.1.10 claims:

Proposition: If x > 0, then x^-1 > 0.

My question concerns the efficiency of my proof.  Using the axioms laid out by Erdman, it seems impossible to prove H.1.10 without first proving a later proposition and two lemmas.  In the "For Students" section, Erdman encourages readers to 'consider the organization of the material.'  Is there a more straight forward proof that I have missed which does not require the lemmas and later proposition or did Erdman (un)intentionally present proposition H.1.10 before H.1.13? 

I understand this is a rather specific question, but any feedback would be highly appreciated.  I have spent much time considering an alternate solution, yet have found none.  While I enjoyed reading your discussions on the proposition, which you will not allow me to provide a link for, each approach assumes axioms Erdman has not.  The textbook can be found easily through a google search for "Erdman A Problem Text in Advanced Calculus."

Sorry if these proofs are a little cumbersome due to the naming of each theorem justifying a conclusion.  Since many of these claims are commonly held 'truths,'  I want to be sure I'm making no unreasonable assumptions.  My proof of H.1.10 is as follows:

H.1.13

Prop: Show that x<0 if and only if -x >0.

Proof: 
    The assumption x<0, by definition of <, is equivalent to

which, by H.1.5, means -x>0.
                                                                                    QED.

Lemma 1:

Proof:
     We proceed by contradiction.  Suppose x=0 whenever x>0.  Then by H.1.5, the assumption x>0 tells us x is in P.  However, this is a contradiction, since the axiom of trichotomy states x cannot be both equal to zero and in P.  So, x !=0.
                                                                                                                                                                         QED.
Lemma 2:

If x>0 and y<0, then xy<0.

Proof:
     By H.1.5, x>0 means x is in P.  By H.1.13, y<0 implies -y>0.  That is, -y is also in P.  Since P is closed under multiplication, x(-y) is in P, or by G.4.4, -(xy) is in P.  So,

by the definition of <.
                                                                                                                                                                                      QED.

H.1.10

Proposition: If x>0, then x^-1>0.

Proof:
     Firstly, by Lemma 1, x != 0.  By axiom III, every element in R different from 0 has a multiplicative inverse.  So, x^-1 exists and is nonzero.  It remains to show that x^-1 > 0.  We proceed by contradiction.

We begin by observing that by the definition of multiplicative inverse and H.1.9,

So, x*x^-1 is in P by H.1.5.  Now, suppose x^-1 < 0.  Then, by the assumption that x>0 and Lemma 2, we see x*x^-1 < 0.  That is, x*x^-1 is NOT in P, a contradiction.  So, x^-1 > 0.
                                                                                                                                                                                         QED.

Thanks!

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: Proposition: If x>0, then x^-1 >0.

Hi mathmatt

I do not see where the theorem H.1.13 comes in your proof. Could you tell me what to look at in your proof?

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