14.4: Algebraic lemma
The following lemma was used in the proof of Proposition 14.3.1 .
Assume \(f:\mathbb{R}\to\mathbb{R}\) is a function such that for any \(x,y\in\mathbb{R}\) we have
- \(f(1)=1\) ,
- \(f(x+y)=f(x)+f(y)\) ,
- \(f(x\cdot y)=f(x)\cdot f(y)\) .
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
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)
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
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.