- #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
The general form of the function f(x) that satisfies f(x+y)=f(x)+f(y)+1 is f(x) = x - 1. This means that for any value of x, the function will output x-1.
We can prove that f(x) = x - 1 satisfies the given conditions by plugging in the values of x and y into the equation f(x+y)=f(x)+f(y)+1 and showing that the left side is equal to the right side. We can also use mathematical induction to prove that the function holds for all values of x and y.
No, there cannot be other functions that satisfy the given conditions. The given conditions uniquely determine the function f(x) = x - 1. This is known as the "uniqueness theorem" in mathematics.
The initial condition f(1)=0 means that when x=1, the function f(x) outputs 0. This is important because it allows us to solve for the constant term in the function f(x) = x - 1. Without this initial condition, the function would have multiple possible solutions.
This type of function is commonly used in mathematical modeling and physics to describe linear relationships between variables. It can also be used in financial calculations, such as calculating compound interest. Additionally, this type of function can be used to model the behavior of certain chemical reactions or physical processes.