Proof on Functions /Set Theory Let $S$ be the set of all numbers of the form $a + b\sqrt 2$ where $a$ and $b$ are rational. Let $f : S \to R$ be a function such that $f(x+y)=f(x)+f(y)$ for all $x$ and $y$ in $S$. Then $f(x)=f(1)x$ for all $x$ in $S$.
How to prove disprove this? And can you go really slow with explaining?
Here are some steps done in the book (tho I don't understand the steps at all):
On trying to prove the statement, you may succeed in proving that $f(a) =
f(1)a$ for all rational $a$ ("I didn't succeed"), and perhaps go
on to prove that $f(a+b \sqrt 2) = f(1)a+f( \sqrt 2)b$ for all rational $a$ and $b$ ("Nah, didn't manage to do that either"). At this
point, you may notice that it would suffice to prove that $f( \sqrt 2) = f(1) \sqrt 2$ ("WHY?").
 A: Not true.
Let $a,b$ arbitrary real numbers. Set
$$
f(p+q\sqrt{2})=pa+qb.
$$
Then $f$ satisfies $f(x+y)=f(x)+f(y)$, and $f(x)\ne xf(0)$, in general, unless
$b=a\sqrt{2}$.
A: Let's go step by step, and see exactly where the proof fails.
You may notice that the set S forms a vector space with basis $(1,\sqrt 2)$. If this does not meant anything to you, do not worry.
The point is that every element of S can be expressed in terms of the base, so it would be natural to assume that the image of every such element is somehow determined by the images of the basis.
Let's try showing first that the simplest members of S, the natural numbers, satisfy the property. Notice that every natural number n is the sum of n 1's. Thus:
$$
f(n)=f(1+\dots+1)=f(1)+\dots+f(1)=f(1)n
$$
With this accomplished, we want to proceed to show the property for the rational numbers. As before, a rational number $\frac{a}{b}$ can be expressed as the sum of a numbers of the form $\frac{1}{b}$. So we need to show that the numbers of this form have the property. This is easy:
$$
f(1)=f(\frac{b}{b})=bf(\frac{1}{b})\implies f(\frac{1}{b})=\frac{1}{b}f(1)
$$
Therefore:
$$
f(\frac{a}{b})=f(\frac{1}{b}+\dots+\frac{1}{b})=f(\frac{1}{b})+\dots+f(\frac{1}{b})=a\cdot f(\frac{1}{b})=\frac{a}{b} f(1)
$$
So we only have to deal with the numbers with $\sqrt 2$ present; that is, of the form $\frac{a}{b}+\frac{c}{d}\sqrt 2$. Trying a similar analysis:
$$
f(\frac{c}{d}\sqrt 2)=f(\frac{1}{d}\sqrt 2+\dots+\frac{1}{d}\sqrt 2)=
f(\frac{1}{d}\sqrt 2)+\dots+f(\frac{1}{d}\sqrt 2)=c\dot f(\frac{1}{d}\sqrt 2)=\frac{c}{d}(f(\frac{1}{d}\sqrt 2)+\dots+f(\frac{1}{d}\sqrt 2))=\frac{c}{d}f(\frac{d}{d}\sqrt 2)=\frac{c}{d}f(\sqrt 2)
$$
So:
$$
f(\frac{a}{b}+\frac{c}{d}\sqrt 2)=f(\frac{a}{b})+f(\frac{c}{d}\sqrt 2)=
\frac{a}{b}f(1)+\frac{c}{d}f(\sqrt 2)
$$
And here we ran into a roadblock, since the property cannot be shown to be satisfied by $\sqrt 2$.
A valid counterexample is conjugation: $f(p+q\sqrt 2)=p-q\sqrt 2)$. Then $f(\sqrt 2) = -\sqrt 2\ne \sqrt 2 f(1) = \sqrt 2$.
