Contrapositives and Monotonic FunctionsDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/