# In Desperate Need of Help! Analysis Homework

## Homework Statement

Let f: R→R have the property that for every u and v in R.
f(u+v)=f(u)+f(v)
(a) Prove: If f(1)=m, then f(x)=mx for all rational numbers x.
(b) Prove: If f is continuous, then f(x)=mx for all x ∈ R.

## The Attempt at a Solution

I am really lost on this problem. We have kind of rushed through the topic of continuity, but I believe I should use the MVT on part (a) and the definition of continuity on part (b).

## Answers and Replies

For part a, here is a hint:

Notice that f(1) = f(1/2 + 1/2) so... Also, f(2) = f(1 + 1) so...

For part b, here is a hint: once you have shown part a, you will know that f(x) = mx at all values of x that are rational. Now, let w be irrational. Then, you can find rationals "near" w, in fact on any open interval around w there are rationals in that interval. Use part a and continuity to conclude that f(w) = mw.

How do you know that f(1) = f( 1/2 + 1/2) and f(2) = f( 1+1)? I am really bad at writing proofs.

tiny-tim
Science Advisor
Homework Helper
How do you know that f(1) = f( 1/2 + 1/2) and f(2) = f( 1+1)? I am really bad at writing proofs.

erm … because 1 = 1/2 + 1/2 and 2 = 1 + 1 …

that's all Russell Berty is saying! Thanks! I feel so special now. erm … because 1 = 1/2 + 1/2 and 2 = 1 + 1 …

that's all Russell Berty is saying! My professor suggested using f(3/2) = f(1+1/2) = f(1) + f(1/2) = f(1/2 + 1/2) + f(1/2) = f(1/2) + f(1/2) + f(1/2) = 3f(1/2).
I guess I am having so much difficulty with this problem because I do not understand what it is really asking me to prove. I really need this broken down. It takes me a little longer than others I guess. You want to show that f(x) = mx for rational x, and then from the continuity of f, show that this is true for irrational x too.

If x is rational, we can write it in the form p/q where both p and q are integers. Now we have:

f(p/q) = f(1/q) + f(1/q) + ... + f(1/q) (p times) = pf(1/q)

Prove that f(1/q) = m/q and then you're done with rational x. Use f's continuity to extend it to irrational x too.

tiny-tim
Science Advisor
Homework Helper
My professor suggested using f(3/2) = f(1+1/2) = f(1) + f(1/2) = f(1/2 + 1/2) + f(1/2) = f(1/2) + f(1/2) + f(1/2) = 3f(1/2).

I think your professor is being rather obscure. Let's start again …

f(1) = m …

ok: then what's f(1 + 1 + … 1) = f(n) for a whole number n ?

and what's f(1/n + 1/n + … 1/n) ? Is the first part mf(n)? Is the second part mf(1/n)?

tiny-tim
Science Advisor
Homework Helper
Is the first part mf(n)? …

oooh, that doesn't even make sense Be systematic …

write it out step by step

f(1) = m,

so f(n) = f(1 + 1 + … 1) = f(1) + f(1) + … + f(1) = … ? And f(1/n) = … ?

Ok, the first part is f(1)=pm, and the second part is f(1/n)=p(m/n). I guess?

I am starting to feel about this problem the way my Algebra II students feel about their homework assignments. It is really giving me grief. I know the proof is probably very simple, but the more I think about the problem the more confused I become No, f(1) = m.

Now if n is a positive integer, then f(n) = f(1 + 1 + 1 + ... + 1) n amount of times = f(1) + f(1) + ... + f(1) n amount of times = ??

Also, m = f(1) = f(1/n + 1/n + ... + 1/n) n amount of times = f(1/n) + f(1/n) + ... + f(1/n) n amount of times = ??

Ok, does the first part = mn, and the second part = m/n?

My professor also mentioned that for part (a) we should also include the proof that f(a/b)=af(1/b).
For that part of (a) I have f(a/b)=f(a+1/b)=f(a)+f(1/b)=? This is where I got lost.

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There you go. You've showed that f(1/q) = m/q. Can you show what f(p/q) is equal to? And show your work next time please :)

Ok, f(p/q)=f(p+1/q)=f(p)+f(1/q)=f(1/q+1/q)+f(1/q)=?

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f(p/q) is NOT equal to f(p + 1/q). W

e'll use a/b instead of p/q. To show that f(a/b) = af(1/b):

f(a/b) = f(1/b + 1/b + ... + 1/b) a amount of times = f(1/b) + ... + f(1/b) a amount of times = af(1/b). Can you now conclude what f(a/b) is equal to?

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So, f(a/b)=ma/b. I hope this is right, because this assignment is due at 6:35p.m. I will try to finish on my own. Thanks for everyone's help!