I have a questions about the sum of functions

[Seda]

Okay I know that (f+g)(x)= f(x) + g(x)


I need to prove (or disprove) that the sum of surjective funcitons is surjective.

However, I keep getting stuck up on one thing. If f(x) has the codomain Y and g(x) has the codomain Z, does (f+g)(x) have the codomain Y + Z (or Y U Z)? I can very easily state for this proof that for all y (and for all z) that there exists an x where f(x) = y (and that there exists an x where g(x)=z.)


I don't know how to follow this proof through to the sum function though. SOrry I'm not good at this theory/proof stuff.

 

[Dick]

You generally only define the sum of functions if they have the same domain and codomain. And before you start sweating too hard trying to prove that the sum is surjective think of some easy examples and try to figure out why it may not be true.

 

[Seda]

You generally only define the sum of functions if they have the same domain and codomain.

Thanks, that makes my life 10 times easier. I wasn't aware of that, but it seems obvious enough.

 

[Seda]

For functions that are R-> R


f(x)=x
g(x)=-x

though both f and g are surjective, f+g(x) = 0

We can see that the sum is not subjective, since 0 is the only member of the codomain R that the function maps to.





Does that work? It seems easy enough...

 

[Dick]

It works and it's easy. That's a great combination.

 

[Seda]

Thanks a lot dick, I have one other question I need to solve. This one seems so simple is weird:


Show that a function f : R->R is increasing iff x<y,f(x)<f(y)



To me, this is like asking to prove 2+2=4, its a definition! I'm not exactly sure what my professor wants here, if I try to prove it conventionally, it ends up like this:

First I'll prove that f is increasing implies x<y, f(x)<f(y)

This is true by definition of increasing function.

Now I'll prove x<y,f(x)<f(y) implies f is an increasing function.

This is true by definition, again!



Any ideas?

 

[Dick]

If you are going to prove that you'd better find the EXACT wording of the definition of increasing. If it is 'f is increasing if for all x<y, f(x)<f(y)' then, yes, the 'proof' doesn't have much content.

 

[Seda]

Here's the definition I've been given before verbatem:

A function f:R->R is increasing iff whenever x>y,f(x)>f(y).


Now here is what I have to "prove" in this assignment, verbatem:

Show that a function f:R->R is increasing iff whenever x<y, f(x)<f(y)

As you can see, they are 99% identical. So my earilier "proof" is acceptable? Is there any other way to do the proof? If there a way to "prove" the definition?

 

[Dick]

Ok, so then you just interchange the symbols x and y in the definition and then use that y>x means the same thing as x<y and f(y)>f(x) means the same thing as f(x)<f(y) and you get the problem. It's so simple it's actually hard to write out in words. But, yes, it finally boils down to just showing that the 'theorem' means the same thing as the definition. Silly, really.