What are the functions satisfying f(x+y)=f(x)+f(y)+1 and f(1)=0?

[iceblits]

hmm.. Ok I know that since its continuous a point exists everywhere which guarantees the existence of there being a rational number right?..I guess what I am trying to say after that is that since an irrational number is a sequence of rational numbers..then I can take 1 minus those rational numbers. Like pi is (3, .1, .04, .001..) = pi so I can do (3-1, .1-1, .4-1, .001-1...) = pi-1 by using the formula right?

 

[iceblits]

maybe if i add like... f(3)+f(.1)+f(.04)+f(.005)...=pi-1?

 

[Dick]

hmm.. Ok I know that since its continuous a point exists everywhere which guarantees the existence of there being a rational number right?..I guess what I am trying to say after that is that since an irrational number is a sequence of rational numbers..then I can take 1 minus those rational numbers. Like pi is (3, .1, .04, .001..) = pi so I can do (3-1, .1-1, .4-1, .001-1...) = pi-1 by using the formula right?

I understand what you are saying and your understanding is correct. BUT the logic is a little hashed. You don't need f continuous to say there is a sequence of rational numbers approaching any irrational. There just is. That's a different theorem and has nothing to do with f. If limit(q_i)=c then why can you say limit f(q_i)=f(c)? And what does limit f(q_i) equal?

 

[LCKurtz]

Why don't you finish the first question first?

Agreed. I thought he understood the last step but I guess he isn't quite finished. Sorry, I should have waited to post the "extra credit" problem.

 

[iceblits]

I understand what you are saying and your understanding is correct. BUT the logic is a little hashed. You don't need f continuous to say there is a sequence of rational number numbers approaching any irrational. There just is. That's a different theorem and has nothing to do with f. If limit(q_i)=c then why can you say limit f(q_i)=f(c)? And what does limit f(q_i) equal?

But if f was not continuous wouldn't we not be able to guarantee the existence of every irrational number? ohh i guess it wouldn't matter for the formula...

anyways: q_i is the sequence of rationals so isn't the limit going to equal an irrational? so you can say that f(q_i)=f(c) because the sequence q_i converges to c? Haha am I still far off?..I feel that I am missing something obvious here..

 

[Dick]

But if f was not continuous wouldn't we not be able to guarantee the existence of every irrational number? ohh i guess it wouldn't matter for the formula...

anyways: q_i is the sequence of rationals so isn't the limit going to equal an irrational? so you can say that f(q_i)=f(c) because the sequence q_i converges to c? Haha am I still far off?..I feel that I am missing something obvious here..

Yeah, you are still somehow missing it. A limit of rationals can also approach a rational but that's not even important. You are saying correct things but for the wrong reason. Remember f is given to be continuous, right? Reread the definition of continuous. If limit q_i = c, then?

 

[iceblits]

then f(q_i)=f(c) ?

 

[Dick]

then f(q_i)=f(c) ?

Almost. limit f(q_i)=f(c). That's a little different from saying f(q_i)=f(c) isn't it? Now stay with me for a little bit. Now f(q_i)=q_i-1 (we proved that, yes?), what's limit (q_i-1)? Remember we assumed limit q_i=c.

 

[iceblits]

setting f(q_i)=q_i - 1...lim (q_i - 1)=f(c)...?

 

[Dick]

setting f(q_i)=q_i - 1...lim (q_i - 1)=f(c)...?

True. But I was really hoping you'd say limit(q_i-1)=limit(q_i)-1=c-1. That would be the punch line. So f(c)=c-1. You've got all of the parts of this. Can you try and string them together in a logical order??

 

[iceblits]

True. But I was really hoping you'd say limit(q_i-1)=limit(q_i)-1=c-1. That would be the punch line. So f(c)=c-1. You've got all of the parts of this. Can you try and string them together in a logical order??

So:

a sequence of rational numbers q_i approaches any other number. That is, lim(q_i)=c By continuity, lim f(q_i)=f(c). By the formula: f(q_i)=q_i-1. So then lim (q_i-1)=lim(q_i)-1=c-1, so, since f(c)=lim(q_i-1), f(c)=c-1

 

[Dick]

So:

a sequence of rational numbers q_i approaches any other number. That is, lim(q_i)=c By continuity, lim f(q_i)=f(c). By the formula: f(q_i)=q_i-1. So then lim (q_i-1)=lim(q_i)-1=c-1, so, since f(c)=lim(q_i-1), f(c)=c-1

Change the first sentence to "Given any irrational number c there is a sequence of rational numbers q_i that approaches c." From there on, I like it. Does it make sense to you?

 

[iceblits]

Change the first sentence to "Given any irrational number c there is a sequence of rational numbers q_i that approaches c." From there on, I like it.

yes! so does this complete the proof? thank-you so much for your help by the way..you've been great..hopefully I didn't take up too much of your time and/or frustrate you too much

 

[Dick]

yes! so does this complete the proof? thank-you so much for your help by the way..you've been great..hopefully I didn't take up too much of your time and/or frustrate you too much

Welcome. Don't worry about it. You are new to proofs. It takes practice to string things together in the right order. I wouldn't have done it if you hadn't kept helping and showing interest. You might try to put together a proper inductive proof of f(nx)=nf(x)+(n-1) if you are up for it and want to put the icing on the cake. But that's good enough for me.

 

[iceblits]

Thank-you again! and yes I'll try to do that..I think I can (I've done it I think..at least shown that for n=k+1..the formula is the same..)...Yeah I don't really know much about proofs...I just recently started getting really into math and I've been trying to do as many problems as I can...and I've learned ALOT from doing this one thanks to you.