
# 14.4: Algebraic lemma


The following lemma was used in the proof of Proposition 14.3.1.

Lemma $$\PageIndex{1}$$

Assume $$f:\mathbb{R}\to\mathbb{R}$$ is a function such that for any $$x,y\in\mathbb{R}$$ we have

1. $$f(1)=1$$,

2. $$f(x+y)=f(x)+f(y)$$,

3. $$f(x\cdot y)=f(x)\cdot f(y)$$.

Then $$f$$ is the identity function; that is, $$f(x)=x$$ for any $$x\in \mathbb{R}$$.

Note that we do not assume that $$f$$ is continuous.

A function $$f$$ satisfying these three conditions is called a field automorphism. Therefore, the lemma states that the identity function is the only automorphism of the field of real numbers. For the field of complex numbers, the conjugation $$z\mapsto \bar{z}$$ (see Section 14.3) gives an example of a nontrivial automorphism.

Proof

By (b) we have

$$f(0)+f(1)=f(0+1).$$

By (a),

$$f(0)+1=1;$$

whence

$f(0)=0.$

Applying (b) again, we get that

$$0=f(0)=f(x)+f(-x).$$

Therefore,

$f(-x)=-f(x) \ \ \ \ \text{for any} \ \ \ \ x\in \mathbb{R}.$

Applying (b) recurrently, we get that

$$\begin{array} {l} {f(2) = f(1) + f(1) = 1 + 1 = 2;} \\ {f(3) = f(2) + f(1) = 2 + 1 = 3;} \\ {\ \ \ \ ...} \end{array}$$

Together with 14.4.2, the latter implies that

$$f(n)=n \ \ \ \text{for any integer}\ \ \ n.$$

By (c)

$$f(m)=f(\dfrac{m}{n})\cdot f(n).$$

Therefore

$$f(\dfrac{m}{n})=\dfrac{m}{n}$$

for any rational number $$\dfrac{m}{n}$$.

Assume $$a\ge 0$$. Then the equation $$x\cdot x=a$$ has a real solution $$x = \sqrt{a}$$. Therefore, $$[f(\sqrt{a})]^2=f(\sqrt{a})\cdot f(\sqrt{a})=f(a)$$. Hence $$f(a)\ge 0$$. That is,

$a\ge 0 \ \ \ \Longrightarrow\ \ \ f(a)\ge 0.$

Applying 14.3.2, we also get

$a\le 0 \ \ \ \Longrightarrow \ \ \ f(a) \le 0.$

Now assume $$f(a)\ne a$$ for some $$a\in\mathbb{R}$$. Then there is a rational number $$\dfrac{m}{n}$$ that lies between $$a$$ and $$f(a)$$; that is, the numbers

$$x=a-\dfrac{m}{n}\ \ \ \text{and}\ \ \ y=f(a)-\dfrac{m}{n}$$

have opposite signs.

By 14.4.3

$$\begin{array} {rcl} {y + \dfrac{m}{n}} & = & {f(a) =} \\ {} & = & {f(x + \dfrac{m}{n}) =} \\ {} & = & {f(x) + f(\dfrac{m}{n}) =} \\ {} & = & {f(x) + \dfrac{m}{n};} \end{array}$$

that is, $$f(x)=y$$. By 14.4.4 and 14.4.5 the values $$x$$ and $$y$$ cannot have opposite signs — a contradiction.