用戶:Justin545/沙盒

子頁面

• /ex
• /ex2
• /ex3
• /ex4
• /ex4/ex4a
• /ex5

模板測試

2=foo; 1=blah;

$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$

2=math_101; 1=101;

$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$

{{使用者:Justin545/沙盒/ex3|math_102|102}}

---

$a^{2}-b^{2}=(a+b)(a-b)$

${\text{16B}}$

$a^{2}-b^{2}=(a+b)(a-b)$

${\text{16C}}$

• $\underbrace {\ce {PCl5}} _{(1)}\ {\ce {->[t{\text{-}}{\ce {Bu-{\overset {\oplus }{NH3}}.{\overset {\ominus }{Cl}}}}]}}\ \underbrace {t{\text{-}}{\ce {Bu-N=PCl3}}} _{(5)}$

  175

• {{{2}}}

  [公式175]

• {{{2}}}

  [公式175]

• #not-exist

$y=ax+b$

16

$y=ax+b$

5

$y=ax+b$

49

$y=ax+b$

90

$y=ax+b$

377

$y=ax+b$

8

---

$y=ax+b$

8

$y=ax+b$

(8)

$y=ax+b$

(8)

---

block number

73

！

呀

硬

好

樂

芭

區塊(73)的特性已被指定！還有(16)、(5)、(49)、(90)、(377)！

少子化 § 生命痛苦歸零

終結 完結 殲滅 消滅 消除 斷絕 阻絕 杜絕 瓦解 解放 禁

計時

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1-o

--- 0.00194(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.5-o

--- 0.0016(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.6-o

--- 0.0018(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.7-o

--- 0.00156(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.8-o

--- 0.00224(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.9-o

--- 0.00204(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.91-o

--- 0.0022(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.92-o

--- 0.00164(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.93-o

--- 0.00172(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.931-o2

--- 0.00206(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.932-o2

--- 0.00204(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.933-o2

--- 0.0022(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.934-o2

--- 0.00204(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.935-o2

--- 0.00202(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.936-o2

--- 0.002(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.937-o2

--- 0.0039(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.938-o2

--- 0.00272(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1.939-o2

--- 0.002(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1-ex5

--- 0.00594(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.2-ex5

--- 0.00342(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.3-ex5

--- 0.00342(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.4-ex5

--- 0.00432(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.5-ex5

--- 0.00386(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.6-ex5

--- 0.00352(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.7-ex5

--- 0.00342(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.8-ex5

--- 0.00354(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.9-ex5

--- 0.00488(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.10-ex5

--- 0.00354(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.1-ex4

--- 0.00236(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.2-ex4

--- 0.0024(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.3-ex4

--- 0.0024(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.4-ex4

--- 0.00238(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.5-ex4

--- 0.00306(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.6-ex4

--- 0.00238(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.7-ex4

--- 0.00242(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.8-ex4

--- 0.00242(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.9-ex4

--- 0.0024(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Def.10-ex4

--- 0.0024(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

(Def.2-s)

--- 0.00404(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

(Def.3-s)

--- 0.0021(second)

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

(Def.4-s)

--- 0.0028(second)

$y=ax+b$

(5-s)

--- 0.00172(second)

${\ce {H+ + CO3^2- -> HCO3^-}}$

(6-s)

--- 0.00162(second)

$e^{ix}=\cos x+i\sin x$

(7-s)

--- 0.00164(second)

$\left({\frac {\partial z}{\partial x}}(x,y),{\frac {\partial z}{\partial y}}(x,y),-1\right)$

(8-s)

--- 0.0016(second)

line spacing tests

Misc.

$y=ax+b+10$

$y=ax+b+11$

$y=ax+b+1$

$y=ax+b+2$

$y=ax+b+3$

$y=ax+b+4$

$y=ax+b+5$

x	y	z
1	2	3

x	y	z
1	2	3

db6d77809cf77e2d00a078cd68e9f223

$y=ax+b$

13579

eeb14ff26fc101c7a74e492fb57c5d2e

Monolithic indent

$y=ax+b+7$

$y=ax+b+8$

$y=ax+b+9$

$y=ax+b$

41

$y=ax+b$

42

$y=ax+b$

43

$y=ax+b$

51

$y=ax+b$

52

$y=ax+b$

53

$y=ax+b$

61

$y=ax+b$

62

$y=ax+b$

63

Indentation comparisons

$y=ax+b+70$

$y=ax+b$

70.5

$y=ax+b+71$

$y=ax+b$

71.5

$y=ax+b+72$

$y=ax+b$

72.5

$y=ax+b+73$

$y=ax+b$

73.5

$y=ax+b+79$

$y=ax+b$

79.5

EN NumBlk examples

Equations may render HTML
{{NumBlk|:|<math>y=ax+b</math>|Eq. 3}}	$y=ax+b$ Eq. 3
{{NumBlk|:|<math>ax^2+bx+c=0</math>|Eq. 3}}	$ax^{2}+bx+c=0$ Eq. 3
{{NumBlk|:|<math>\Psi(x_1,x_2)=U(x_1)V(x_2)</math>|2}}	$\Psi (x_{1},x_{2})=U(x_{1})V(x_{2})$ 2
Indentation
{{NumBlk||<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3.5}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ 3.5
{{NumBlk|:|<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|1}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ 1
{{NumBlk|::|<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|13.7}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ 13.7
{{NumBlk|:::|<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|1.2}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ 1.2
Formatting of equation number
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=3.5|RawN=.}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ 3.5
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<3.5>|RawN=.}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ <3.5>
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=[3.5]|RawN=.}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ [3.5]
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3='''[3.5]'''|RawN=.}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ [3.5]
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ [3.5]
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>|RawN=.}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ [3.5]
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<math>(3.5)</math>|RawN=.}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ $(3.5)\,$
Line style
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''(3.141)'''</Big>|RawN=.|LnSty=0.2em dotted #e5e5e5}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ (3.141)
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''(3.5)'''</Big>|RawN=.|LnSty=1px dashed red}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ (3.5)
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''(3.5)'''</Big>|RawN=.|LnSty=3px dashed #0a7392}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ (3.5)
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>|RawN=.|LnSty=3px solid green}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ [3.5]
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>|RawN=.|LnSty=5px dotted blue}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ [3.5]
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>|RawN=.|LnSty=0px solid green}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ [3.5]
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>|RawN=.|LnSty=5px none green}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ [3.5]
{{NumBlk|1=:|2=<math>\mathbf{a}(t)=\frac{d}{dt}\mathbf{v}(t)</math>|3=<Big>'''[3.5]'''</Big>|RawN=.|LnSty=3px double green}}	$\mathbf {a} (t)={\frac {d}{dt}}\mathbf {v} (t)$ [3.5]
Line height and indentation (1)
The following equations :<math>3x+2y-z=1</math> :<math>2x-2y+4z=-2</math> :<math>-2x+y-2z=0</math> form a system of three equations.	The following equations $3x+2y-z=1$ $2x-2y+4z=-2$ $-2x+y-2z=0$ form a system of three equations.
The following equations {{NumBlk|:|<math>3x+2y-z=1</math>|1}} {{NumBlk|:|<math>2x-2y+4z=-2</math>|2}} {{NumBlk|:|<math>-2x+y-2z=0</math>|3}} form a system of three equations.	The following equations $3x+2y-z=1$ 1 $2x-2y+4z=-2$ 2 $-2x+y-2z=0$ 3 form a system of three equations.
The following equations <div style="line-height: 0;"> {{NumBlk|:|<math>3x+2y-z=1</math>|1}} {{NumBlk|:|<math>2x-2y+4z=-2</math>|2}} {{NumBlk|:|<math>-2x+y-2z=0</math>|3}} </div> form a system of three equations.	The following equations $3x+2y-z=1$ 1 $2x-2y+4z=-2$ 2 $-2x+y-2z=0$ 3 form a system of three equations.
The following equations <div style="line-height: 0;"> {{NumBlk||<math>3x+2y-z=1</math>|1}} {{NumBlk||<math>2x-2y+4z=-2</math>|2}} {{NumBlk||<math>-2x+y-2z=0</math>|3}} </div> form a system of three equations.	The following equations $3x+2y-z=1$ 1 $2x-2y+4z=-2$ 2 $-2x+y-2z=0$ 3 form a system of three equations.
The following equations <div style="line-height: 0; margin-left: 1.6em;"> {{NumBlk||<math>3x+2y-z=1</math>|1}} {{NumBlk||<math>2x-2y+4z=-2</math>|2}} {{NumBlk||<math>-2x+y-2z=0</math>|3}} </div> form a system of three equations.	The following equations $3x+2y-z=1$ 1 $2x-2y+4z=-2$ 2 $-2x+y-2z=0$ 3 form a system of three equations.
Line height and indentation (2)
The following equations :<math>3x+2y-z=1</math> ::<math>2x-2y+4z=-2</math> :::<math>-2x+y-2z=0</math> form a system of three equations.	The following equations $3x+2y-z=1$ $2x-2y+4z=-2$ $-2x+y-2z=0$ form a system of three equations.
The following equations <div style="line-height: 0; margin-left: 1.6em;"> {{NumBlk||<math>3x+2y-z=1</math>|1}} <div style="margin-left: 1.6em;"> {{NumBlk||<math>2x-2y+4z=-2</math>|2}} <div style="margin-left: 1.6em;"> {{NumBlk||<math>-2x+y-2z=0</math>|3}} </div> </div> </div> form a system of three equations.	The following equations $3x+2y-z=1$ 1 $2x-2y+4z=-2$ 2 $-2x+y-2z=0$ 3 form a system of three equations.
The following equations <div style="line-height: 0;"> <div style="margin-left: calc(1.6em * 1);"> {{NumBlk||<math>3x+2y-z=1</math>|1}} </div> <div style="margin-left: calc(1.6em * 2);"> {{NumBlk||<math>2x-2y+4z=-2</math>|2}} </div> <div style="margin-left: calc(1.6em * 3);"> {{NumBlk||<math>-2x+y-2z=0</math>|3}} </div> </div> form a system of three equations.	The following equations $3x+2y-z=1$ 1 $2x-2y+4z=-2$ 2 $-2x+y-2z=0$ 3 form a system of three equations.
Unordered list
* <math>3x+2y-z=1</math> * <math>2x-2y+4z=-2</math> * <math>-2x+y-2z=0</math>	$3x+2y-z=1$ $2x-2y+4z=-2$ $-2x+y-2z=0$
<ul style="line-height: 0;"> <li><div style="display: inline-block; width: 100%; vertical-align: middle;">{{NumBlk||<math>3x+2y-z=1</math>|Eq. 1}}</div></li> <li><div style="display: inline-block; width: 100%; vertical-align: middle;">{{NumBlk||<math>2x-2y+4z=-2</math>|Eq. 2}}</div></li> <li><div style="display: inline-block; width: 100%; vertical-align: middle;">{{NumBlk||<math>-2x+y-2z=0</math>|Eq. 3}}</div></li> </ul>	$3x+2y-z=1$ Eq. 1 $2x-2y+4z=-2$ Eq. 2 $-2x+y-2z=0$ Eq. 3
<ul style="line-height: 0;"> <li><div style="display: inline-table; width: 100%; vertical-align: middle;">{{NumBlk||<math>3x+2y-z=1</math>|Eq. 1}}</div></li> <li><div style="display: inline-table; width: 100%; vertical-align: middle;">{{NumBlk||<math>2x-2y+4z=-2</math>|Eq. 2}}</div></li> <li><div style="display: inline-table; width: 100%; vertical-align: middle;">{{NumBlk||<math>-2x+y-2z=0</math>|Eq. 3}}</div></li> </ul>	$3x+2y-z=1$ Eq. 1 $2x-2y+4z=-2$ Eq. 2 $-2x+y-2z=0$ Eq. 3
Ordered list
# <math>3x+2y-z=1</math> # <math>2x-2y+4z=-2</math> # <math>-2x+y-2z=0</math>	$3x+2y-z=1$ $2x-2y+4z=-2$ $-2x+y-2z=0$
<ol style="line-height: 0;"> <li><div style="display: inline-block; width: 100%; vertical-align: middle;">{{NumBlk||<math>3x+2y-z=1</math>|Eq. 1}}</div></li> <li><div style="display: inline-block; width: 100%; vertical-align: middle;">{{NumBlk||<math>2x-2y+4z=-2</math>|Eq. 2}}</div></li> <li><div style="display: inline-block; width: 100%; vertical-align: middle;">{{NumBlk||<math>-2x+y-2z=0</math>|Eq. 3}}</div></li> </ol>	$3x+2y-z=1$ Eq. 1 $2x-2y+4z=-2$ Eq. 2 $-2x+y-2z=0$ Eq. 3
<ol style="line-height: 0;"> <li><div style="display: inline-table; width: 100%; vertical-align: middle;">{{NumBlk||<math>3x+2y-z=1</math>|Eq. 1}}</div></li> <li><div style="display: inline-table; width: 100%; vertical-align: middle;">{{NumBlk||<math>2x-2y+4z=-2</math>|Eq. 2}}</div></li> <li><div style="display: inline-table; width: 100%; vertical-align: middle;">{{NumBlk||<math>-2x+y-2z=0</math>|Eq. 3}}</div></li> </ol>	$3x+2y-z=1$ Eq. 1 $2x-2y+4z=-2$ Eq. 2 $-2x+y-2z=0$ Eq. 3
Border
{{NumBlk|:|<math>y=ax+b</math>|Eq. 3|Border=1}}	$y=ax+b$ Eq. 3

When content of the blocks and block numbers are far apart

Markup	<div style="line-height:0;"> {{NumBlk|1=:|2=<math>a^2 + b^2 = (a + b i) (a - b i)</math>|3=1}} {{NumBlk|1=:|2=<math>a^2 - b^2 = (a + b) (a - b)</math>|3=2}} {{NumBlk|1=:|2=<math>e^{i x} = \cos x + i \sin x</math>|3=3}} {{NumBlk|1=:|2=<math>\sin^2 \theta + \cos^2 \theta = 1</math>|3=4}} {{NumBlk|1=:|2=<math>\sin(2 \theta) = 2 \sin \theta \cos \theta</math>|3=5}} </div>
Renders as	$a^{2}+b^{2}=(a+bi)(a-bi)$ 1 $a^{2}-b^{2}=(a+b)(a-b)$ 2 $e^{ix}=\cos x+i\sin x$ 3 $\sin ^{2}\theta +\cos ^{2}\theta =1$ 4 $\sin(2\theta )=2\sin \theta \cos \theta$ 5

Markup

<div style="line-height:0;"> {{NumBlk|1=:|2=<math>a^2 + b^2 = (a + b i) (a - b i)</math>|3=1|LnSty=0.37ex dotted Gainsboro}} {{NumBlk|1=:|2=<math>a^2 - b^2 = (a + b) (a - b)</math>|3=2|LnSty=0.37ex dotted Gainsboro}} {{NumBlk|1=:|2=<math>e^{i x} = \cos x + i \sin x</math>|3=3|LnSty=0.37ex dotted Gainsboro}} {{NumBlk|1=:|2=<math>\sin^2 \theta + \cos^2 \theta = 1</math>|3=4|LnSty=0.37ex dotted Gainsboro}} {{NumBlk|1=:|2=<math>\sin(2 \theta) = 2 \sin \theta \cos \theta</math>|3=5|LnSty=0.37ex dotted Gainsboro}} </div>

Renders as

$a^{2}+b^{2}=(a+bi)(a-bi)$ 1 $a^{2}-b^{2}=(a+b)(a-b)$ 2 $e^{ix}=\cos x+i\sin x$ 3 $\sin ^{2}\theta +\cos ^{2}\theta =1$ 4 $\sin(2\theta )=2\sin \theta \cos \theta$ 5

Markup

<div style="line-height:0;"> {{NumBlk|1=:|2=<math>a^2 + b^2 = (a + b i) (a - b i)</math>|3=1|LnSty=0.37ex dotted Gainsboro}} {{NumBlk|1=:|2=<math>a^2 - b^2 = (a + b) (a - b)</math>|3=2|LnSty=0.37ex none Gainsboro}} {{NumBlk|1=:|2=<math>e^{i x} = \cos x + i \sin x</math>|3=3|LnSty=0.37ex dotted Gainsboro}} {{NumBlk|1=:|2=<math>\sin^2 \theta + \cos^2 \theta = 1</math>|3=4|LnSty=0.37ex none Gainsboro}} {{NumBlk|1=:|2=<math>\sin(2 \theta) = 2 \sin \theta \cos \theta</math>|3=5|LnSty=0.37ex dotted Gainsboro}} </div>

Renders as

$a^{2}+b^{2}=(a+bi)(a-bi)$ 1 $a^{2}-b^{2}=(a+b)(a-b)$ 2 $e^{ix}=\cos x+i\sin x$ 3 $\sin ^{2}\theta +\cos ^{2}\theta =1$ 4 $\sin(2\theta )=2\sin \theta \cos \theta$ 5

Markup

<div style="line-height:0;"> <div style="background-color: Beige;"> {{NumBlk|1=:|2=<math>a^2 + b^2 = (a + b i) (a - b i)</math>|3=1}} </div> <div style="background-color: none;"> {{NumBlk|1=:|2=<math>a^2 - b^2 = (a + b) (a - b)</math>|3=2}} </div> <div style="background-color: Beige;"> {{NumBlk|1=:|2=<math>e^{i x} = \cos x + i \sin x</math>|3=3}} </div> <div style="background-color: none;"> {{NumBlk|1=:|2=<math>\sin^2 \theta + \cos^2 \theta = 1</math>|3=4}} </div> <div style="background-color: Beige;"> {{NumBlk|1=:|2=<math>\sin(2 \theta) = 2 \sin \theta \cos \theta</math>|3=5}} </div> </div>

Renders as

$a^{2}+b^{2}=(a+bi)(a-bi)$ 1 $a^{2}-b^{2}=(a+b)(a-b)$ 2 $e^{ix}=\cos x+i\sin x$ 3 $\sin ^{2}\theta +\cos ^{2}\theta =1$ 4 $\sin(2\theta )=2\sin \theta \cos \theta$ 5

Markup

(mouse over the row you want to highlight) {{row hover highlight}} {| class="hover-highlight" style="line-height:0; width: 100%; border-collapse: collapse; margin: 0; padding: 0;" |- | {{NumBlk|1=:|2=<math>a^2 + b^2 = (a + b i) (a - b i)</math>|3=1}} |- | {{NumBlk|1=:|2=<math>a^2 - b^2 = (a + b) (a - b)</math>|3=2}} |- | {{NumBlk|1=:|2=<math>e^{i x} = \cos x + i \sin x</math>|3=3}} |- | {{NumBlk|1=:|2=<math>\sin^2 \theta + \cos^2 \theta = 1</math>|3=4}} |- | {{NumBlk|1=:|2=<math>\sin(2 \theta) = 2 \sin \theta \cos \theta</math>|3=5}} |}

Renders as

(mouse over the row you want to highlight) $a^{2}+b^{2}=(a+bi)(a-bi)$ 1 $a^{2}-b^{2}=(a+b)(a-b)$ 2 $e^{ix}=\cos x+i\sin x$ 3 $\sin ^{2}\theta +\cos ^{2}\theta =1$ 4 $\sin(2\theta )=2\sin \theta \cos \theta$ 5

Proof of hypothetical syllogism by constructive dilemma

$(p\rightarrow q)\leftrightarrow (\neg q\rightarrow \neg p)$

/ / Lemma: Logical equivalences involving conditional statements B

$p\land \top \leftrightarrow p$

/ / Lemma: Identity laws A

$p\lor \neg p\leftrightarrow \top$

/ / Lemma: Negation laws A

$[(P\rightarrow Q)\land (R\rightarrow S)\land (P\lor R)]\rightarrow (Q\lor S)$

/ / Lemma: Constructive dilemma

$(p\rightarrow q)\leftrightarrow (\neg p\lor q)$

/ / Lemma: Logical equivalences involving conditional statements A

$(p\rightarrow q)\land (q\rightarrow r)$

.1 / / premise

$(p\rightarrow q)\leftrightarrow (\neg q\rightarrow \neg p)\;\left|\;{\begin{array}{l}p:p\\q:q\end{array}}\right.$

.11 / .1 / Logical equivalences involving conditional statements B

$(p\rightarrow q)\leftrightarrow (\neg q\rightarrow \neg p)$

.12 / .11

$(\neg q\rightarrow \neg p)\land (q\rightarrow r)$

.13 / .1 .12

$p\land \top \leftrightarrow p\;|\;p:(\neg q\rightarrow \neg p)\land (q\rightarrow r)$

.14 / .13 / Identity laws A

$[(\neg q\rightarrow \neg p)\land (q\rightarrow r)]\land \top \leftrightarrow (\neg q\rightarrow \neg p)\land (q\rightarrow r)$

.15 / .14

$(\neg q\rightarrow \neg p)\land (q\rightarrow r)\land \top \leftrightarrow (\neg q\rightarrow \neg p)\land (q\rightarrow r)$

.16 / .15

$(\neg q\rightarrow \neg p)\land (q\rightarrow r)\land \top$

.17 / .13 .16

$(q\rightarrow r)\land (\neg q\rightarrow \neg p)\land \top$

.18 / .17

$p\lor \neg p\leftrightarrow \top \;|\;p:q$

.19 / .18 / Negation laws A

$q\lor \neg q\leftrightarrow \top$

.2 / .19

$(q\rightarrow r)\land (\neg q\rightarrow \neg p)\land (q\lor \neg q)$

.21 / .18 .2

$[(P\rightarrow Q)\land (R\rightarrow S)\land (P\lor R)]\rightarrow (Q\lor S)\;\left|\;{\begin{array}{l}P:q\\Q:r\\R:\neg q\\S:\neg p\end{array}}\right.$

.22 / .21 / Constructive dilemma

$[(q\rightarrow r)\land (\neg q\rightarrow \neg p)\land (q\lor \neg q)]\rightarrow (r\lor \neg p)$

.23 / .22

$[(q\rightarrow r)\land (\neg q\rightarrow \neg p)\land (q\lor \neg q)]\rightarrow (\neg p\lor r)$

.24 / .23

$(p\rightarrow q)\leftrightarrow (\neg p\lor q)\;\left|\;{\begin{array}{l}p:p\\q:r\end{array}}\right.$

.25 / .24 / Logical equivalences involving conditional statements A

$(p\rightarrow r)\leftrightarrow (\neg p\lor r)$

.26 / .25

$[(q\rightarrow r)\land (\neg q\rightarrow \neg p)\land (q\lor \neg q)]\rightarrow (p\rightarrow r)$

.27 / .24 .26

$[(q\rightarrow r)\land (\neg q\rightarrow \neg p)\land \top ]\rightarrow (p\rightarrow r)$

.28 / .2 .27

$[(\neg q\rightarrow \neg p)\land (q\rightarrow r)\land \top ]\rightarrow (p\rightarrow r)$

.29 / .28

$[(\neg q\rightarrow \neg p)\land (q\rightarrow r)]\rightarrow (p\rightarrow r)$

.3 / .16 .29

$[(p\rightarrow q)\land (q\rightarrow r)]\rightarrow (p\rightarrow r)$

.31 / .12 .3 / conclusion