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I am working on Spivaks calculus book and have the following question:

We have a function with properties (Chapter 3, Ex. 17, (d)): $$f(x+y)=f(x)+f(y)$$ $$f(xy)=f(x)f(y)$$ and we are trying to prove that $f(x)=x$ for $x$ a real number. In this step (d) we want to prove that $f(x)>f(y)$ if $x>y$ and here I am struggling.

My attempt was to do the following: $x>y$ is the same as $x-y>0$. Then $$f(x-y) = f(x)+f(-y) = f(x)+f(-1)f(y)=f(x)+f(y)$$ because from (a) we know that $$f(1)=1=f((-1)(-1))=f(-1)^2$$ therefore $$f(-1)=1$$ From (c) we know that $f(x)>0 $ if $x>0$ so $$f(x-y)=f(x)+f(y)>0$$ but i want $f(x)-f(y)>0$. The solution manual says instantly $f(x-y)=f(x)-f(y)$, but why is that? And what is wrong in my reasoning?

Teddy38
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Binomi
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    $f^2(-1)=1$ implies $f(-1) = \pm 1$. If $f(x)$ was indeed the identity map, $f(-1)$ would be $-1$ and not $1$. – Gary Feb 12 '21 at 15:01
  • Oh yes you are right I forgot about the $-1$ solution. But I cannot assume that it is the identity map therefore I cannot use this argument I guess. – Binomi Feb 12 '21 at 15:09
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    You can show, without knowing that $f$ is the identity, that $f(-1)=1$ would contradict the assumptions. – Gary Feb 12 '21 at 15:25
  • Oh yes I see because if $f(-1)=1=f(0-1)=f(0)-f(1)=0-1=-1$ a contradiction. I used 1. and 2. from the post below. Therefore $f(-1)=-1$ – Binomi Feb 12 '21 at 15:39
  • How did you get $f(0-1)=f(0)-f(1)$ from 1.? Shouldn't it be $f(0-1)=f(0)+f(-1)$? – Gary Feb 12 '21 at 16:08
  • Yes and then use 2. from the post below. – Binomi Feb 12 '21 at 17:09

1 Answers1

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Note that $f(x+y)=f(x)+f(y)$ implies that:

  1. $f(0)=0$ because $f(0)=f(0+0)=f(0)+f(0)$
  2. $f(-x)=-f(x)$ because $0=f(0)=f(x-x)=f(x)+f(-x)$
  3. $f(x-y)=f(x)-f(y)$ because $f(x-y)=f(x)+f(-y)=f(x)-f(y)$

These are classical properties of group homomorphisms.

There's something wrong with your reasoning though. $f(-1)^2=1$ implies that $f(-1)=\pm 1$ not $f(-1)=1$. And indeed, in reality it is $-1$, because ultimately $f(x)=x$ is supposed to be a solution, so $f(-1)=-1$. So either $f(x)=x$ is not a solution (spoiler alert: it actually is) or you've made a mistake.

Here's a different idea. First of all if $f(x_0)=0$ for some $x_0\neq 0$ then $f(x)=0$ for all $x$. That's because

$$f(x)=f(x_0x_0^{-1}x)=f(x_0)f(x_0^{-1}x)=0\cdot f(x_0^{-1}x)=0$$

Note that constant $0$ is a valid solution.

Now assume that $f$ is not constant $0$.

For any $x> 0$ we have

$$f(x)=f(\sqrt{x}\sqrt{x})=f(\sqrt{x})\cdot f(\sqrt{x})=f(\sqrt{x})^2\geq 0$$

But as we've said $f(x)$ cannot be equal $0$ because $f$ is not constant $0$. And so $f(x)>0$.

Therefore if $x> y$ then $x-y> 0$ and so $f(x-y)> 0$ meaning $f(x)-f(y)> 0$ which finally leads to $f(x)> f(y)$.

freakish
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  • Thank you. You are right, I forgot about the $-1$ solution, therefore I cannot get to the solution with my idea I think. But your step 1. and 2. made it clear although I would not have came up myself with this reasoning! – Binomi Feb 12 '21 at 15:19
  • The problem requires a bit of a cheat. Spivak's solution (which mirrors freakish's) depends on the existence of $\sqrt{x}$ for $x>0$. Fortunately, that existence is proven in the "Three Hard Theorems" chapter. – Ben Feb 16 '21 at 15:11