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I would like to know why if it is not valid. Thanks in advance.

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Mardia
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    Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – Martin R Aug 31 '22 at 18:33
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    $f(x)=0$ is a solution. – Arthur Aug 31 '22 at 18:34
  • @Martin R I'm sorry, I wrote in picture since I didn't have a labtop. Should I delete this post? – Mardia Aug 31 '22 at 18:35
  • @Arthur Can you elaborate more? I'm don't see why that is significant. – Mardia Aug 31 '22 at 18:38
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    Mostly it looks good. To be careful, you should show that $f(1/b) \neq 0$. Because when you say "this implies $f(a)=a$", you are implicitly dividing both sides by $f(1/b)$ – NazimJ Aug 31 '22 at 18:44
  • @NazimJ Do I still need to show that when I assumed a/b to be rational? – Mardia Aug 31 '22 at 18:45
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    If $x\geq0$, then there is $y$ such that $y^2=x$. Then $f(x)=f(y^2)=f(y)^2\geq0$. So, $f$ is non-negative on the non-negative numbers. Then if $x\geq y$, then $f(x)=f(x-y+y)=f(x-y)+f(y)\geq f(y)$, since $f(x-y)\geq0$ for $x-y$ being non-negative. Use that $f$ is monotonic to finish. Observe that the value $f(a)$ at an irrational value $a$ will have to be sandwiched between the values at all rationals $r,s$ wish $r<a<s$. So, $r=f(r)\leq f(a)\leq f(s)=s$. This forces $f(a)=a$ by taking limits with $r,s\to a$. – plop Aug 31 '22 at 18:47
  • See https://math.stackexchange.com/q/423492/42969 for a comprehensive overview of the Cauchy functional equation. – Martin R Aug 31 '22 at 18:48
  • Yes you can write that $r =\sum \frac {a_i}{10^i}$ but we can not assume that $f(\sum\frac{a_i}{10^i}) = \sum f(\frac {a_i}{10^i})$ for infinite sums. – fleablood Aug 31 '22 at 18:48
  • The first sentence on your piece of paper (assuming you mean $x$ when you say $r$) is directly contradicted by the solution $f(x)=0$. – Arthur Aug 31 '22 at 18:51
  • @fleablood Why is that? Since I showed that f(rational) = rational, can't I just sum all the rationals again after applying the function? – Mardia Aug 31 '22 at 18:53
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    @Mardia Because that is not a sum. It is a sum plus a limit, a series. In this comment you can see how to finish the argument. – plop Aug 31 '22 at 18:55
  • @Arthur You are right. I must have missed an important information when I wrote down the question. This is a question 17 on chapter 2 of spivaks. – Mardia Aug 31 '22 at 18:57
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    "can't I just sum all the rationals again after applying the function" Absolutely not. You only know it for 2 term sums and by induction you can show it for any $n$ terms sums but you can't ever get to infinity by doing things one step at a time. Infinite sums are tricky and the common sense rules of rearranging and "adding just one more" simply don't apply. This is a consequence of the key difference between rational and irrational numbers. There's too much to go into in a comment but no, you just can't do that. – fleablood Aug 31 '22 at 19:00
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    And what @user85667 said.... $\sum_{k=1}^{\infty} b_i$ although it is written to look like an infinite sum is actually $\lim_{n\to \infty} \sum_{k=1}^n b_i$. There is no such thing as "infinite sums" only limits of infinite sequences of values (which in turn may be finite sums). Now it is true that $f(\sum^n b_i) = \sum^n f(b_i)$ for all $n$ and all $b_i$ but it does not follow that $f(\lim_{n\to \infty} \sum^nb_i)=\lim_{n\to \infty}\sum^n f(b_i)$. – fleablood Aug 31 '22 at 19:06
  • @user85667 "In this comment you can see how to finish the argument." But that is assuming $f$ is continuous. We can show it is true for all reals if we assume $f$ is continuous. But if $f$ is not continuous, we can use the multiplicative property to show it is true for all algebraic numbers, but I don't think we can every prove it for trancendental numbers. In fact I think we can prove $f(x)=x$ if $x$ is algebraic and $f(x)=0$ otherwise will qualify. – fleablood Aug 31 '22 at 19:14
  • @fleablood No, in the comment it is proven that $f$ is monotonic and that is enough to finish. – plop Aug 31 '22 at 19:27

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You can always write an irrational number as a sum of rationals. Then the last step in your argument depends on knowing that the function $f$ is continuous.

If you know that, then knowing that addition is preserved you can show $f(x) = cx$ for some constant $c$.

This is a well studied problem: see https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation

I don't know whether assuming that the function preserves multiplication too suffices to get the continuity.

Ethan Bolker
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    I think a reasonable confusion is that if $f(x+y) = f(x) + f(y)$ and by induction that would mean $f(\sum_{k=1}^n x_i) = \sum_{k=1}^n f(x_i)$ then we wouldn't it be valid so say $f(\sum_{k=1}^\infty x_i) = \sum_{k=1}^\infty f(x_i)$. That is the OPs error. And the general "infinite sums are tricky! always be careful" applies. But in more detail. Induction implies that the property holds for all finite sums. But you can not get to infinity by any number of inductive steps so does not hold for infinite sums. Details and counter examples .... well, we can go on forever. – fleablood Aug 31 '22 at 18:54
  • @fleablood Haven't started analysis yet so I don't see why it is wrong here since to me it just seems like a tautology in this case but I appreciate it very much. Thank you! – Mardia Aug 31 '22 at 18:59
  • As a simple example of where it can fail, consider $f(x) = 0$ if $x$ is rational, and $f(x) = 1$ if $x$ is irrational. For all of the finite sums $\sum_{i=1}^N f(x_i) = \sum_{i = 0}^N 0 = 0 = f(\sum_{i=1}^N x_i)$, but clearly in the limit you get $f(x) = 1 \neq 0 = \sum_{i=0}^\infty f(x_i)$. – ConMan Sep 01 '22 at 00:34
  • @ConMan But this function doesn't satisfy the f(x+y) = f(x)+f(y). – Mardia Sep 02 '22 at 12:33
  • It doesn't, I was just using it as an example of how "true for all finite $N$" and "true in the limit as $N$ goes to infinity" aren't always consistent. – ConMan Sep 03 '22 at 03:54