14.4: Algebraic lemma
- Page ID
- 23671
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.