If $f(q)=g(q)$ for all rationals, prove that $f=g$ by nonstandard methods.

Question

I am trying to prove that if $f,g:\mathbb R\to\mathbb R$ are ordered ring homomorphisms, then, if $\forall q\in\mathbb Q,f(q)=g(q)$, then $f=g$.

Is it true ? Can you give a nonstandard proof of this fact ? If not, can you give a standard proof ?

Answer 1

I'm not really sure what is standard and what is not but here's an alternative approach:

Classification lemma. Let $f:\mathbb{R}\to\mathbb{R}$ be a ring homomorphism. Then either $\forall_{x\in\mathbb{R}}\ f(x)=0$ or $\forall_{x\in\mathbb{R}}\ f(x)=x$

Note that sometimes people require a ring homomorphism to map $f(1)=1$ (which I do not). In that case only $f$ being the identity is possible. Also I do not assume that $f$ preserves order. This assumption is redundant.

Proof. The proof is almost purely algebraic. The only non-algebraic properties we use are:

1. $\mathbb{Q}$ is dense in $\mathbb{R}$
2. for any $x>0$ there exists $r\in\mathbb{R}$ with $r^2=x$
3. for any $x\in\mathbb{R}$, $x^2\geq 0$.

Basic properties.

Step 1. If $f(1)=0$ then

$$f(x)=f(1\cdot x)=f(1)\cdot f(x)=0\cdot f(x)=0$$

This gives us one solution so without a loss of generality we may assume that $f(1)\neq 0$.

Step 2. $\forall_{q\in\mathbb{Q}}\ f(q)=q$. Indeed, $f(1)=f(1\cdot 1)=f(1)\cdot f(1)$ and since $f(1)\neq 0$ then $f(1)=1$. It follows that for any $n\in\mathbb{Z}$ we have $f(n)=n$ and

$$1=f(1)=f(\frac{1}{n}\cdot n)=f(\frac{1}{n})\cdot f(n)=f(\frac{1}{n})\cdot n$$

and thus $f(\frac{1}{n})=\frac{1}{n}$. Since $f$ preserves multiplication then $f(q)=q$ for $q\in\mathbb{Q}$.

Proof by contradiction.

Assume that $f(x)\neq x$ for some $x\in\mathbb{R}$.

Step 3. Since $f(-x)=-f(x)$ then we may assume that $f(x)<x$.

Step 4. We may assume that $f(x)<0<x$. Indeed, pick $q\in\mathbb{Q}$ such that $f(x)<q<x\ $ ($\mathbb{Q}$ is dense in $\mathbb{R}$). Replace $x$ by $x-q$. The desired property follows from step 2 since

$$f(x-q)=f(x)-q<0<x-q$$

Step 5. Since $x>0$ then there exists $r\in\mathbb{R}$ such that $r^2=x$. It follows that

$$f(r)^2=f(r^2)=f(x) < 0$$

Contradiction because $y^2\geq 0$ no matter what $y\in\mathbb{R}$ is. Thus $f(x)=x$ which completes the proof. $\Box$

Your statement follows trivially from the classification lemma (you just have a very limited choice for $f$ and $g$).

Answer 2

Well, I was very tired yesterday. Here is a standard proof:

Let $x\in\mathbb R$ and $n\in\mathbb N^*$ there is $q\in\mathbb Q$, such that $|x-q|<\frac 1n$

$-\frac 1n<x-q<\frac 1n$

$f(q-\frac1n)<f(x)<f(q+\frac 1n)$

$f(q)-f(\frac 1n)<f(x)<f(q)+f(\frac 1n)$

$|f(x)-f(q)|<f(\frac 1n)=\frac1n$

Similarly, $|g(x)-f(q)|<\frac1n$ and we conclude easily.

Here is a nonstandard proof:

Let $x\in\mathbb R$, there is $q\in\ ^*\mathbb Q$ such that $^*x\simeq q$ so there is an infinitesimal $dx$ such that:

$^*x=q+dx$

$^*f(^*x)=\ ^*f(q)+\ ^*f(dx)$ because $^*f$ is a ring homomorphism.

By transfer, it is true that $\forall q\in\ ^*\mathbb Q,\ ^*f(q)=\ ^*g(q)$.

We also remark that, because $^*f$ is an increasing ring homomorphism, that the image of an infinitesimal is an infinitesimal.

$^*f(^*x)=\ ^*g(q)+\ ^*f(dx)\sim\ ^*g(q)+\ ^*g(dx)=\ ^*g(^*x)$

so $f(x)=g(x)$.