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Mathematics LibreTexts

14.4: Algebraic lemma

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    23671
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    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.