- #36
iceblits
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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
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??
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
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.
iceblits said:Oh boy..haha...maybe...since its continuous the limit as x approaches c equals f(c)..where c is any real number?
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?
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?
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?
iceblits said:edit: is it because an irrational number is a sum of decimals?
so like pi=3+.1+.04+.001..
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?LCKurtz said:And now for extra credit
What happens if you remove the assumption that f(1) = 0?
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?
Dick said:Why don't you finish the first question first?
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
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?
Dick said:Why don't you finish the first question first?
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?
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..
iceblits said:then f(q_i)=f(c) ?
iceblits said:setting f(q_i)=q_i - 1...lim (q_i - 1)=f(c)...?
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??
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
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.
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