Contrapositives and Monotonic Functions

Date: 03/17/2003 at 12:09:24
From: B.Y. Vinay Kumar
Subject: Monotonic functions

We usually define an increasing function as (say) 

   if x1<=x2   implies f(x1) <=f(x2)  then f is said to be increasing

BUT will it make any difference if we define it as 

   if f(x1) <=f(x2)  implies x1<=x2  then f is increasing

I just wanted to know whether the converse holds.

Date: 03/18/2003 at 04:08:59
From: Doctor Jacques
Subject: Re: Monotonic functions

Hi,

There may be some confusion regarding the definition of an increasing 
function, i.e. whether or not we consider a constant function as 
increasing. To avoid ambiguity, we will only use "non-decreasing" 
and "strictly increasing."

Your first definition describes a non-decreasing function, i.e. it is 
also satisfied by a constant function. But the second definition is 
not satisfied by a constant function: we always have f(x1) <= f(x2), 
whatever x1 and x2 may be. You can, however, define a non-decreasing 
function by:

  f(x1) < f(x2)  --> x1 < x2

In the case of a constant function, this is verified, since the LHS is 
always false, and p -> q is true whenver p is false.

To describe a strictly increasing function, we could write:

  x1 < x2  --> f(x1) < f(x2)

and this is equivalent to:

  f(x1) <= f(x2) --> x1 <= x2

The logic behind these equivalences is based on the concept of 
contraposition.

The contrapositive of a logic statement p --> q is (~q) --> (~p), 
where ~ stands for negation. A statement and its contrapositive are 
equivalent. You can see an example of this kind of reasoning in:

   Contrapositive
   http://mathforum.org/library/drmath/view/55706.html 

Let us start with the first definition (non-decreasing function) :

  x1 <= x2  --> f(x1) <= f(x2)

The contrapositive is:

  ~(f(x1) <= f(x2)) --> ~(x1 <= x2)

Now, in a totally ordered set, ~(a <= b) is the same as b > a. Our 
statement becomes:

  f(x1) > f(x2) --> x1 > x2

    or

  f(x2) < f(x1) --> x2 < x1

In this statement, x1 and x2 stand for any numbers, so we may 
interchange the names of these variables. We get:

 f(x1) < f(x2) --> x1 < x2

as shown before. The logic for a strictly increasing function is 
entirely similar.

Note that, as x1 = x2 trivially implies f(x1) = f(x2), we could even 
define a non-decreasing function by the (apparently) weaker statement:

 x1 < x2 --> f(x1) <= f(x2)

which is equivalent to:

 f(x1) < f(x2) --> x1 <= x2

Does this help?  Write back if you'd like to talk about this some 
more, or if you have any other questions.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/