Prove/Disprove... Monotone functions and mapping sets.

[dell500]

Howdy,

This is my first post. So, howdy!

I'm currently in advance Calculus I and am currently having some homework issues.

Prove or disprove: If A is a subset of R (real #s) and f,g : A -> R are monotone, then f + g is monotone.

Now, proofs are not my cup of tea, but being a math major, I must understand and be able to read/write them without any hassle. So far, I have the definition of monotone, which is:

Suppose A belongs to R and f : A -> R,
- We say that f is monotone increasing if x1, x2 belong to A and x1 <= x2, then f(x1) <= f(x2).
- We say that f is monotone decreasing if x1, x2 belong to A and x1 >= x2, then f(x1) >= f(x2).
- We say that f is strictly monotone increasing if x1, x2 belong to A and x1 < x2, then f(x1) < f(x2).
- We say that f is strictly monotone decreasing if x1, x2 belong to A and x1 > x2, then f(x1) > f(x2).

Now, I also know that f and g both map elements of A to the R (real #s). I'm just not positive on what to do from here and how I should structure this. Any help would be great. Thanks!

Peace

 

[pka]

Think counterexample. Let \(\displaystyle A=[0,1)\)

 

[dell500]

I found that there are 3 distinct situations for this example.

1) f and g are both monotone: x1 <= x2 --> f(x1) <= f(x2) and g(x1) <= g(x2)
2) f and g are both monotone: x1 >= x2 --> f(x1) >= f(x2) and g(x1) >= g(x2)
3) a. f is monotone increasing and g is monotone decreasing: x1 <= x2 --> f(x1) <= f(x2) and x1 >= x2 --> g(x1) >= g(x2)
b. f is monotone decreasing and g is monotone increasing: x1 >= x2 --> f(x1) >= f(x2) and x1 <= x2 --> g(x1) <= g(x2)

3 fails due to your example and with A being restricted to [0,1)... is that all I need to do? That one counter example? I do believe it is correct in saying that if f(x) = x^2 and g(x)=x^3 or adding negative signs to both equations would be make f + g monotone, correct?

 

[dell500]

Also, can I just say, with this example. Since case 3 is an increasing and decreasing monotone function, this defies the f + g function to be monotone. Thus f + g is not monotone.

?

 

[pka]

The point is: a counterexample is all that is needed to disprove the statement The sum of two monotonic functions is a monotonic function. Of course, it is trivial to prove The sum of two monotonic nondecreasing/nonincreasing functions is a monotonic function.
You are in advanced calculus. That course is a great deal more rigorous that basic calculus. This should be your first lesson in realizing that fact. Be very careful in reading the exact wording of each problem. Go back and read the statement you posted. A simple counterexample is sufficient to answer exactly what you posted. There is no need to do more than what is call for by the wording of that question.
BTW. I taught advanced calculus over thirty years.

 

[dell500]

So, just showing that example and not saying anything afterwards is a sufficient answer for disproving this statement?

 

[pka]

So, just showing that example and not saying anything afterwards is a sufficient answer for disproving this statement?

Well, it is not that simple. One should explain why the example is a counterexample to the given statement.

 

[dell500]

Well I asserted that f and g are monotone, states those 3 cases, and wrote down an example for the 3rd. Then stated that example in case 3 was not monotone when the two functions were added... I thought that would be sufficient... I'm very rusty on these proofs, sorry i'm being so hard headed.