ALL functions such that:

iceblits

and then...setting x to m/n I get..f(m)=nf(m/n)+n-1..and f(m)=m-1..so then
m-1=nf(m/n)+n-1..so then f(m/n)=m/n-1?

iceblits

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

Dick

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.

iceblits

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?

Dick

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?

iceblits

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

Dick

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?

iceblits

Thats ok :) i dont 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..

Dick

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?

Dick

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,...}.

LCKurtz

And now for extra credit

What happens if you remove the assumption that f(1) = 0?

iceblits

So now can I use induction to prove that the formula is true for all real?

iceblits

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?

Dick

Why don't you finish the first question first?

iceblits

right..ok so heres 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

Dick

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.

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 Im 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?