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

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In summary, the conversation discusses finding all functions f: R->R such that f(x+y)=f(x)+f(y)+1 and f(1)=0. The conversation explores different approaches to solving the problem, including using differentiability and continuity. It is ultimately determined that the only function that satisfies these conditions is y=x-1. The conversation also provides hints and guidance for solving the problem.
  • #36
Dick said:
Oh, you're getting ahead of me. So using f(nx)=nf(x)+(n-1). And putting n to m and x to 1/n??

ahh yes I see that...f(m/n)=mf(1/n)+(m-1)..and then f(m/n)=m(1/n-1)+m-1...so then f(m/n)=m/n-1
 
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  • #37
iceblits said:
ahh yes I see that...f(m/n)=mf(1/n)+(m-1)..and then f(m/n)=m(1/n-1)+m-1...so then f(m/n)=m/n-1

Ok, you are almost there. So f(m/n)=m/n-1. That means f(q)=q-1 for all rational numbers q. How do you prove it's also true for irrational numbers? It's time to use the given that f is continuous.
 
  • #38
Dick said:
Ok, you are almost there. So f(m/n)=m/n-1. That means f(q)=q-1 for all rational numbers q. How do you prove it's also true for irrational numbers? It's time to use the given that f is continuous.

Oh boy..haha...maybe...since its continuous the limit as x approaches c from the right and left equals f(c)..where c is any real number?
 
  • #39
iceblits said:
Oh boy..haha...maybe...since its continuous the limit as x approaches c equals f(c)..where c is any real number?

Suppose c is irrational. Then there is a sequence of rational numbers that approach c, say q_i->c (yes?), and you've already shown f(q_i)=q_i-1. What conclusion can you draw from that?
 
  • #40
does q_i mean "q sub i" if so then, if I have shown that f(q_i)=q_i-1.. and since the function is continuous for numbers..then a sequence of rational numbers exists to form an irrational number as the number of terms in the sequence approach infinity?

edit: and this would show that..as i>>infinity...f(q_i)=q_i-1...which means the formula is true for the irrational numbers
 
  • #41
iceblits said:
does q_i mean "q sub i" if so then, if I have shown that f(q_i)=q_i-1.. and since the function is continuous for numbers..then a sequence of rational numbers exists to form an irrational number as the number of terms in the sequence approach infinity?

Yeah, q_i means "q sub i", I'm often too lazy to TeX, sorry. The ideas are more important than the format in my opinion. Do you know why you can find a sequence of rationals approaching any irrational? This may mean just pointing to a page in your book. And given q_i->c can you use continuity to show f(c)=c-1?
 
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  • #42
Dick said:
Yeah, q_i means "q sub i", I'm often to lazy to TeX, sorry. The ideas are more important than the format in my opinion. Do you know why you can find a sequence of rationals approaching any irrational? And given q_i->c can you use continuity to show f(c)=c-1?

Thats ok :) i don't even really know what TeX is..and...Im not sure why..I do know that certain irrational numbers can be written as a sequence of rational numbers..like (e=sum(1/n))..I didn't know that it extended to all irrational numbers (I do now that you've told me)..is there a name for such a theorem?

edit: is it because an irrational number is a sum of decimals?

so like pi=3+.1+.04+.001..
 
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  • #43
iceblits said:
Thats ok :) i don't even really know what TeX is..and...Im not sure why..I do know that certain irrational numbers can be written as a sequence of rational numbers..like (e=sum(1/n))..I didn't know that it extended to all irrational numbers (I do now that you've told me)..is there a name for such a theorem?

The statement of the theorem usually referred to as the statement that "the rationals are DENSE in the real numbers". Meaning given any interval (however small) around an irrational number, there is a rational number in it. So you can find a sequence of rationals converging to any irrational. Makes sense, yes?
 
  • #44
iceblits said:
edit: is it because an irrational number is a sum of decimals?

so like pi=3+.1+.04+.001..

Yes, that's basically what it is. Your book should have a theorem that tells you something very like that. But it's more like, pi is the limit of the sequence {3,3.1,3.14,3.141,3.1415,...}.
 
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  • #45
And now for extra credit :smile:

What happens if you remove the assumption that f(1) = 0?
 
  • #46
So now can I use induction to prove that the formula is true for all real?
 
  • #47
LCKurtz said:
And now for extra credit :smile:

What happens if you remove the assumption that f(1) = 0?
haha.. does it become an infinite number of functions?...like since a linear function is f(x+y)=f(x)+f(y)...then would removing the assumption give me all linear functions +1?
 
  • #48
iceblits said:
haha.. does it become an infinite number of functions?...like since a linear function is f(x+y)=f(x)+f(y)...then would removing the assumption give me all linear functions +1?

Why don't you finish the first question first?
 
  • #49
Dick said:
Why don't you finish the first question first?

right..ok so here's what I have so far (thanks to you :) ) :
I show that f(nx)=nf(x)+(n-1), where n is an integer. Then, setting x=1 and using the initial condition that f(x)=0, I show that f(n)=n-1. Then, setting x=m/n I show that f(m)=nf(m/n)+n-1..since f(m)=m-1, this becomes m-1=nf(m/n)+n-1.. so f(m/n)=m/n-1 (m and n are integers). Since the function is continuous, it exits for all points and every irrational number has a sequence of rational numbers that approach it and so f(q_i)=q_i-1, where q_i is the sequence of rational numbers approaching an irrational number. Since this formula can be derived for all real numbers i can use induction to show that f((n+1)x)=(n+1)f(x)+n equals the original formula
 
  • #50
iceblits said:
right..ok so here's what I have so far (thanks to you :) ) :
I show that f(nx)=nf(x)+(n-1), where n is an integer. Then, setting x=1 and using the initial condition that f(x)=0, I show that f(n)=n-1. Then, setting x=m/n I show that f(m)=nf(m/n)+n-1..since f(m)=m-1, this becomes m-1=nf(m/n)+n-1.. so f(m/n)=m/n-1 (m and n are integers). Since the function is continuous, it exits for all points and every irrational number has a sequence of rational numbers that approach it and so f(q_i)=q_i-1, where q_i is the sequence of rational numbers approaching an irrational number. Since this formula can be derived for all real numbers i can use induction to show that f((n+1)x)=(n+1)f(x)+n equals the original formula

It gets a little garbled towards the end. Given you have f(q)=q-1 for q a rational (which I think you get), WHY can you say f(c)=c-1 if c is irrational? Say it in your own words as clearly as you can. I'm not seeing it in that explanation.
 
  • #51
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?
 
  • #52
maybe if i add like... f(3)+f(.1)+f(.04)+f(.005)...=pi-1?
 
  • #53
iceblits said:
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?
 
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  • #54
Dick said:
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.
 
  • #55
Dick said:
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..
 
  • #56
iceblits said:
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?
 
  • #57
then f(q_i)=f(c) ?
 
  • #58
iceblits said:
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.
 
  • #59
setting f(q_i)=q_i - 1...lim (q_i - 1)=f(c)...?
 
  • #60
iceblits said:
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??
 
  • #61
Dick said:
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
 
  • #62
iceblits said:
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?
 
  • #63
Dick said:
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
 
  • #64
iceblits said:
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.
 
  • #65
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.
 

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