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Spivak Calculus, 4th ed., problem 3-17:

If $f(x)=0$ for all $x$, then $f$ satisfies $f(x+y)=f(x)+f(y)$ for all $x$ and $y$, and also $f(x\cdot y ) =f(x)\cdot f(y)$ for all $x$ and $y$. Now suppose that $f$ satisfies these two properties, but that $f(x)$ is not always $0$. Prove that $f(x)=x $ for all $x$, as follows:

  1. Prove that $f(1)=1.$
  2. Prove that $f(x)=x$ if $x$ is rational.
  3. Prove that $f(x)>0$ if $x>0$. (This part is tricky, but if you have been paying attention to the philosophical remarks in accompanying the problems in the last two chapters, you will know what to do.)
  4. Prove that $f(x)>f(y)$ if $x>y$.
  5. Prove that $f(x)=x$ for all $x$. Hint: Use the fact that between any two numbers there is a rational number.

Parts 1 and 2 are fine with me, but with 3, I'm not exactly sure what he's looking for. Some kind of a pre-continuity continuity argument...anyone have any ideas? I'm also not sure about part 5.

Note that the completeness of the real numbers has not yet been defined, nor has continuity been defined.

Eric Auld
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    He said last two chapters, not next :) 3. is suspect, to be sure, because we don't know that every positive number is a square (although he's made such "philosophical remarks"). 5. follows by density of the rationals, applying 2. and 4. – Ted Shifrin Apr 13 '14 at 22:54
  • see http://math.stackexchange.com/questions/449404/is-an-automorphism-of-the-field-of-real-numbers-the-identity-map – Will Jagy Apr 13 '14 at 23:06

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For part $3$: If $x>0$ then $x= y^2$ for some $y$ , then $f(x)=f(y^2)=(f(y))^2 \ge 0$. Now to get rid of the equal under the greater than if $f(x)=0$ then $f(q)=f(q \frac{x}{x})=f(x)f(\frac{q}{x})=0$ for any $q$ which is absurd.

For $5$ suppose not, i.e. $f(x)>x$ (or $f(x)<x$) for some $x$ ,use the density of the rationals to find some $q\in \Bbb{Q}$ such that $x<q<f(x)$. Now apply 2 and 4.

user10444
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