Expert: Paul Klarreich - 2/11/2007

Question
Hello, Mr. Klarreich

I was wondering if you can help me with this proof?
If 0< a< b, then 0< b^-1 < a^-1.

I was thinking of ideas to start. One idea is to multiply by both side by their reciporcals and getting 0<b<a. But I dont know what to do next.

I will be really grateful if you can help me.

Answer
Questioner:   Eva L.
Category:  Calculus
 
Subject:  Order axiom proof
Question:  Hello, Mr. Klarreich

I was wondering if you can help me with this proof?
If 0< a< b, then 0< b^-1 < a^-1.

I was thinking of ideas to start. One idea is to multiply by both side by their reciprocals and getting 0<b<a. But I dont know what to do next.

I will be really grateful if you can help me.
................................................
Hi, Eva,

Your approach seems valid.  Keep in mind that inequalities are tricky things.  When you multiply an inequality by a number, you must be sure that the multiplier is positive, otherwise the 'sense' of the inequality will change.

In this case, a and b are both positive, so that's good.  Now take your 'given' inequality:

0 < a < b

and multiply both sides by  1/(ab).  Since a,b are positive, so is ab, and so is 1/ab.  [Sorry, getting sloppy with the parentheses.]

0 < a/ab < b/ab

0 < 1/b < 1/a

which is the thing you want to prove.