# 平方根

（重定向自算術平方根

（算术）平方根的數學表示式

## 實數的平方根

x的平方根亦可用指數表示，如：

${\displaystyle x^{\frac {1}{2}}={\sqrt {x}}}$

${\displaystyle x}$絕對值可用${\displaystyle x^{2}}$的算數平方根表示：

${\displaystyle |x|={\sqrt {x^{2}}}\left(={\begin{cases}x&(x\geq 0)\\-x&(x<0)\end{cases}}\right)}$

### 正數的平方根

{\displaystyle {\begin{aligned}{\sqrt {x}}{\sqrt {y}}&={\sqrt {xy}}\\{\frac {\sqrt {x}}{\sqrt {y}}}&={\sqrt {\frac {x}{y}}}\end{aligned}}}

### 負數的平方根

${\displaystyle \pm {\sqrt {-x}}i}$，其中${\displaystyle {\sqrt {-x}}i={\sqrt {x}}}$

{\displaystyle {\begin{aligned}{\sqrt {x}}{\sqrt {y}}&={\sqrt {-x}}\,i\times {\sqrt {-y}}\,i={\sqrt {xy}}\,i^{2}=-{\sqrt {xy}}\\{\frac {\sqrt {x}}{\sqrt {y}}}&={\frac {{\sqrt {-x}}i}{{\sqrt {-y}}i}}={\sqrt {\frac {-x}{-y}}}={\sqrt {\frac {x}{y}}}\end{aligned}}}

## 负数与複數的平方根

${\displaystyle {\sqrt {-x}}=i{\sqrt {x}}}$

${\displaystyle (i{\sqrt {x}})^{2}=i^{2}({\sqrt {x}})^{2}=(-1)x=-x}$

### 虚数的平方根

${\displaystyle {\sqrt {i}}={\frac {\sqrt {2}}{2}}+i{\frac {\sqrt {2}}{2}}={\frac {\sqrt {2}}{2}}(1+i)}$

{\displaystyle {\begin{aligned}i&=(a+bi)^{2}\\&=a^{2}+2abi-b^{2}\end{aligned}}}

${\displaystyle {\begin{cases}2ab=1\!\\a^{2}-b^{2}=0\!\end{cases}}}$

${\displaystyle a=b=\pm {\frac {\sqrt {2}}{2}}.}$

${\displaystyle a=b={\frac {\sqrt {2}}{2}}}$

${\displaystyle i=\cos \left({\frac {\pi }{2}}\right)+i\sin \left({\frac {\pi }{2}}\right)}$

{\displaystyle {\begin{aligned}{\sqrt {i}}&=\left(\cos \left({\frac {\pi }{2}}\right)+i\sin \left({\frac {\pi }{2}}\right)\right)^{\frac {1}{2}}\\&=\cos \left({\frac {\pi }{4}}\right)+i\sin \left({\frac {\pi }{4}}\right)\\&={\frac {\sqrt {2}}{2}}+i{\frac {\sqrt {2}}{2}}\\&={\frac {\sqrt {2}}{2}}(1+i).\\\end{aligned}}}

### 复数的算术平方根

${\displaystyle z=re^{i\varphi },-\pi <\varphi \leq \pi }$

${\displaystyle {\sqrt {z}}={\sqrt {r}}\,e^{i\varphi /2}}$

${\displaystyle {\sqrt {r\left(\cos \varphi +i\,\sin \varphi \right)}}={\sqrt {r}}\left[\cos {\frac {\varphi }{2}}+i\sin {\frac {\varphi }{2}}\right]}$

### 代数公式

${\displaystyle {\sqrt {z}}={\sqrt {\frac {|z|+\operatorname {Re} (z)}{2}}}\pm i\ {\sqrt {\frac {|z|-\operatorname {Re} (z)}{2}}}}$

${\displaystyle -1=i\times i={\sqrt {-1}}\times {\sqrt {-1}}={\sqrt {-1\times -1}}={\sqrt {1}}=1}$

## 2的算术平方根

${\displaystyle {\sqrt {2}}}$是無理數，可由歸謬法證明：

1. ${\displaystyle {\sqrt {2}}}$有理數，可表示為${\displaystyle {\frac {p}{q}}}$，其中pq互質之正整數。
2. 因為${\displaystyle \left({\sqrt {2}}\right)^{2}={\frac {p^{2}}{q^{2}}}=2}$，故${\displaystyle p^{2}}$是2的倍數，p也是2的倍數，記為2k，其中k為正整數。
3. 但是${\displaystyle 2q^{2}=p^{2}=4k^{2}}$，故${\displaystyle q^{2}=2k^{2}}$${\displaystyle q^{2}}$是2的倍數，q也是2的倍數。
4. 依上兩式，pq都是2的倍數，和pq為互質之正整數的前題矛盾。依歸謬法，得證${\displaystyle {\sqrt {2}}}$不是有理數，即${\displaystyle {\sqrt {2}}}$是無理數。

## 計算方法

### 長除式算法

1. 首先將要開平方根的數從小數點分別向右及向左每兩個位一組分開，如98765.432內小數點前的65是一組，87是一組，9是一組，小數點後的43是一組，之後是單獨一個2，要補一個0而得20是一組。如1 04.85 73得四組，順序為1' 04. 85' 73'。
2. 將最左的一組的數減去最接近又少於它的平方數，並將該平方數的開方（應該是個位數）記下。
3. 將上一步所得之差乘100，和下一組數加起來。
4. 將記下的數乘20，然後將它加上某個個位數，再乘以該個個位數，令這個積不大於又最接近上一步所得之差，並將該個個位數記下，且將上一步所得之差減去所得之積。
5. 記下的數一次隔兩位記下。
6. 重覆第3步，直到找到答案。
7. 可以在數字的最右補上多組的00'以求得理想的精確度為止。

${\displaystyle {\begin{array}{ll}\quad {\color {Red}1}~~{\color {Green}4}.~~{\color {Blue}1}~~{\color {Purple}4}~~{\color {Orange}2}\\{\sqrt {2|00.00|00|00}}\\\quad {\underline {1\quad ~}}&\quad {\color {Red}1}\times {\color {Red}1}\leq 2\\\quad 1~00&a={\color {Red}1}0,b={\color {Green}4}\\\quad {\underline {~~\,96\quad ~}}&\quad \Rightarrow (2a+b)b=2{\color {Green}4}\times {\color {Green}4}=96\leq 100\\\qquad ~4~00&a={\color {Red}1}{\color {Green}4}0,b={\color {Blue}1}\\\qquad ~{\underline {2~81\quad ~}}&\quad \Rightarrow (2a+b)b=28{\color {Blue}1}\times {\color {Blue}1}=281\leq 400\\\qquad ~1~19~00&a={\color {Red}1}{\color {Green}4}{\color {Blue}1}0,b={\color {Purple}4}\\\qquad ~{\underline {1~12~96\quad ~}}&\quad \Rightarrow (2a+b)b=282{\color {Purple}4}\times {\color {Purple}4}=11296\leq 11900\\\qquad \quad ~~6~04~00&a={\color {Red}1}{\color {Green}4}{\color {Blue}1}{\color {Purple}4}0,b={\color {Orange}2}\\\qquad \quad ~~{\underline {5~65~64}}&\quad \Rightarrow (2a+b)b=2828{\color {Orange}2}\times {\color {Orange}2}=56564\leq 60400\\\qquad \quad \quad ~\,38~36\\\end{array}}}$

${\displaystyle {\sqrt {200}}\approx 14.14213562373095048801668872421}$

 ${\displaystyle {\sqrt {1}}}$ ${\displaystyle =\,}$ 1 ${\displaystyle {\sqrt {2}}}$ ${\displaystyle \approx }$ 1.4142135623 7309504880 1688724209 6980785696 7187537694 8073176679 7379907324 78462 ${\displaystyle {\sqrt {3}}}$ ${\displaystyle \approx }$ 1.7320508075 6887729352 7446341505 8723669428 0525381038 0628055806 9794519330 16909 ${\displaystyle {\sqrt {4}}}$ ${\displaystyle =\,}$ 2 ${\displaystyle {\sqrt {5}}}$ ${\displaystyle \approx }$ 2.2360679774 9978969640 9173668731 2762354406 1835961152 5724270897 2454105209 25638 ${\displaystyle {\sqrt {6}}}$ ${\displaystyle \approx }$ 2.4494897427 8317809819 7284074705 8913919659 4748065667 0128432692 5672509603 77457 ${\displaystyle {\sqrt {7}}}$ ${\displaystyle \approx }$ 2.6457513110 6459059050 1615753639 2604257102 5918308245 0180368334 4592010688 23230 ${\displaystyle {\sqrt {8}}}$ ${\displaystyle \approx }$ 2.8284271247 4619009760 3377448419 3961571393 4375075389 6146353359 4759814649 56924 ${\displaystyle {\sqrt {9}}}$ ${\displaystyle =\,}$ 3 ${\displaystyle {\sqrt {10}}}$ ${\displaystyle \approx }$ 3.1622776601 6837933199 8893544432 7185337195 5513932521 6826857504 8527925944 38639 ${\displaystyle {\sqrt {11}}}$ ${\displaystyle \approx }$ 3.3166247903 5539984911 4932736670 6866839270 8854558935 3597058682 1461164846 42609 ${\displaystyle {\sqrt {12}}}$ ${\displaystyle \approx }$ 3.4641016151 3775458705 4892683011 7447338856 1050762076 1256111613 9589038660 33818 ${\displaystyle {\sqrt {13}}}$ ${\displaystyle \approx }$ 3.6055512754 6398929311 9221267470 4959462512 9657384524 6212710453 0562271669 48293 ${\displaystyle {\sqrt {14}}}$ ${\displaystyle \approx }$ 3.7416573867 7394138558 3748732316 5493017560 1980777872 6946303745 4673200351 56307 ${\displaystyle {\sqrt {15}}}$ ${\displaystyle \approx }$ 3.8729833462 0741688517 9265399782 3996108329 2170529159 0826587573 7661134830 91937 ${\displaystyle {\sqrt {16}}}$ ${\displaystyle =\,}$ 4 ${\displaystyle {\sqrt {17}}}$ ${\displaystyle \approx }$ 4.1231056256 1766054982 1409855974 0770251471 9922537362 0434398633 5730949543 46338 ${\displaystyle {\sqrt {18}}}$ ${\displaystyle \approx }$ 4.2426406871 1928514640 5066172629 0942357090 1562613084 4219530039 2139721974 35386 ${\displaystyle {\sqrt {19}}}$ ${\displaystyle \approx }$ 4.3588989435 4067355223 6981983859 6156591370 0392523244 4936890344 1381595573 28203 ${\displaystyle {\sqrt {20}}}$ ${\displaystyle \approx }$ 4.4721359549 9957939281 8347337462 5524708812 3671922305 1448541794 4908210418 51276

### 牛頓法

${\displaystyle x_{n+1}={\frac {1}{2}}\left(x_{n}+{\frac {S}{x_{n}}}\right)}$

${\displaystyle x_{0}=3^{6}=729.000\,\!}$
${\displaystyle x_{1}={\frac {1}{2}}\left(x_{0}+{\frac {S}{x_{0}}}\right)={\frac {1}{2}}\left(729.000+{\frac {125348}{729.000}}\right)=450.472}$
${\displaystyle x_{2}={\frac {1}{2}}\left(x_{1}+{\frac {S}{x_{1}}}\right)={\frac {1}{2}}\left(450.472+{\frac {125348}{450.472}}\right)=364.365}$
${\displaystyle x_{3}={\frac {1}{2}}\left(x_{2}+{\frac {S}{x_{2}}}\right)={\frac {1}{2}}\left(364.365+{\frac {125348}{364.365}}\right)=354.191}$
${\displaystyle x_{4}={\frac {1}{2}}\left(x_{3}+{\frac {S}{x_{3}}}\right)={\frac {1}{2}}\left(354.191+{\frac {125348}{354.191}}\right)=354.045}$
${\displaystyle x_{5}={\frac {1}{2}}\left(x_{4}+{\frac {S}{x_{4}}}\right)={\frac {1}{2}}\left(354.045+{\frac {125348}{354.045}}\right)=354.045}$

### 連分數

${\displaystyle {\sqrt {1}}=1}$
${\displaystyle {\sqrt {2}}=[1;2,2,2,2...]}$
${\displaystyle {\sqrt {3}}=[1;1,2,1,2...]}$
${\displaystyle {\sqrt {4}}=2}$
${\displaystyle {\sqrt {5}}=[2;4,4,4,4...]}$
${\displaystyle {\sqrt {6}}=[2;2,4,2,4...]}$
${\displaystyle {\sqrt {7}}=[2;1,1,1,4,1,1,1,4...]}$
${\displaystyle {\sqrt {8}}=[2;1,4,1,4...]}$
${\displaystyle {\sqrt {9}}=3}$
${\displaystyle {\sqrt {1}}0=[3;6,6,6,6...]}$
${\displaystyle {\sqrt {1}}1=[3;3,6,3,6...]}$
${\displaystyle {\sqrt {1}}2=[3;2,6,2,6...]}$
${\displaystyle {\sqrt {1}}3=[3;1,1,1,1,6,1,1,1,1,6...]}$
${\displaystyle {\sqrt {1}}4=[3;1,2,1,6,1,2,1,6...]}$
${\displaystyle {\sqrt {1}}5=[3;1,6,1,6...]}$
${\displaystyle {\sqrt {1}}6=4}$
${\displaystyle {\sqrt {1}}7=[4;8,8,8,8...]}$
${\displaystyle {\sqrt {1}}8=[4;4,8,4,8...]}$
${\displaystyle {\sqrt {1}}9=[4;2,1,3,1,2,8,2,1,3,1,2,8...]}$
${\displaystyle {\sqrt {2}}0=[4;2,8,2,8...]}$

### 尺规作图

#### 解法

1. 畫線AB，延長BAC使AC=1
2. BC的中點為圓心，OC為半徑畫圓

#### 證明

• O=（0,0）
• AB=n
1. 直徑為BC的圓就是${\displaystyle x^{2}+y^{2}=\left({\frac {n+1}{2}}\right)^{2}}$（圓的方程式：${\displaystyle x^{2}+y^{2}=}$r2）（其中,r表示半径。）
2. ${\displaystyle \left({\frac {n+1}{2}}-1\right)}$A,D所在的x座標）代入上面的方程式
3. ${\displaystyle \left({\frac {n+1}{2}}-1\right)^{2}+y^{2}=\left({\frac {n+1}{2}}\right)^{2}}$
4. 解方程，得y=n

## 參考資料

1. ^ Analysis of YBC 7289. ubc.ca. [19 January 2015].
2. ^ Anglin, W.S. (1994). Mathematics: A Concise History and Philosophy. New York: Springer-Verlag.
3. ^ Heath, Sir Thomas L. The Thirteen Books of The Elements, Vol. 3. Cambridge University Press. 1908: 3.
4. ^ Abramowitz, Milton; Stegun, Irene A. Handbook of mathematical functions with formulas, graphs, and mathematical tables. Courier Dover Publications. 1964: 17. ISBN 0-486-61272-4. （原始内容存档于2016-04-23）., Section 3.7.27, p. 17 互联网档案馆存檔，存档日期2009-09-10.
5. ^ Cooke, Roger. Classical algebra: its nature, origins, and uses. John Wiley and Sons. 2008: 59. ISBN 0-470-25952-3. （原始内容存档于2016-04-23）.
6. ^ 劳汉生《珠算与实用算术》ISBN 7-5375-1891-2/O