Difference between revisions of "Proofs without words"

Revision as of 23:50, 26 March 2010

The following demonstrate proofs of various identities and theorems using pictures, inspired from this gallery.

Summations

$[asy]unitsize(15); defaultpen(linewidth(0.7)); int n = 6; pair shiftR = ((n+2),0); real r = 0.3; pen colors(int i){ return rgb(i/n,0.4+i/(2n),1-i/n); } /* shading */ void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B); draw(A-ticklength--A+ticklength); draw(B-ticklength--B+ticklength); } /* triangle */ draw((-r,0)--(-r,-n+1)^^(r,-n+1)--(r,0),linetype("4 4")); for(int i = 0; i < n; ++i) draw((-i,-i)--(i,-i)); for(int i = 0; i < n; ++i) for(int j = 0; j < 2*i+1; ++j) filldraw(CR((j-i,-i),r),colors(i)); /* square */ draw(r*expi(pi/4)+shiftR--(n-1,-n+1)+r*expi(pi/4)+shiftR^^r*expi(5*pi/4)+shiftR--r*expi(5*pi/4)+(n-1,-n+1)+shiftR,linetype("4 4")); for(int i = 0; i < n; ++i) draw(shiftR+(0,-i)--shiftR+(i,-i)--shiftR+(i,0)); for(int i = 0; i < n; ++i) for(int j = 0; j < n; ++j) filldraw(CR((j,-i)+shiftR,r),colors((i>j)?i:j)); htick(shiftR+(-1,r),shiftR+(-1,-n+1-r)); label("$n$",shiftR+(-1,(-n+1)/2),W,fontsize(10)); [/asy]$

The sum of the first $n$ odd natural numbers is $n^2$ .

$[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 6; pair shiftR = ((n+2),0); real r = 0.3; pen colors(int i){ return rgb(0.4+i/(2n),i/n,1-i/n); } /* shading */ void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B); draw(A-ticklength--A+ticklength); draw(B-ticklength--B+ticklength); } /* triangle */ draw((0.5,0)--(n-0.5,-n+1),linetype("4 4")); for(int i = 0; i < n; ++i) draw((0,-i)--(i,-i)); for(int i = 0; i < n; ++i) for(int j = 0; j <= i; ++j) filldraw(CR((j,-i),r),colors(i)); /* arc arrow */ draw( arc((n,-n+1)/2, (1.5,-1.5), (n-1.5,-1.5), CW) ); fill((n-1.5,-1.5) -- (n-1.5,-1.5)+r*expi(5.2*pi/6) -- (n-1.5,-1.5)+r*expi(3.3*pi/6) -- cycle); /* manual arrowhead? avoid resizing */ /* square */ draw(shiftR+(0.5,0)--shiftR+(n-0.5,-n+1),linetype("4 4")); for(int i = 0; i < n; ++i) draw(shiftR+(0,-i)--shiftR+(i,-i)^^shiftR+(n,-n+1)-(0,-i)--shiftR+(n,-n+1)-(i,-i)); for(int i = 0; i < n; ++i) for(int j = 0; j < n+1; ++j) filldraw(CR((j,-i)+shiftR,r),colors((j <= i) ? i : n-1-i)); /* labeling and ticks */ htick(shiftR+(-1,r),shiftR+(-1,-n+1-r)); label("$n$",shiftR+(-1,(-n+1)/2),W,fontsize(10)); htick(shiftR+(-r,-n),shiftR+(n+r-1,-n),(0,0.15)); label("$n$",shiftR+((n-1)/2,-n),S,fontsize(10)); htick(shiftR+(n-r,-n),shiftR+(n+r,-n),(0,0.15)); label("$1$",shiftR+(n,-n),S,fontsize(10)); [/asy]$

The sum of the first $n$ positive integers is $n(n+1)/2$ .

$[asy]unitsize(15); defaultpen(linewidth(0.7)); int n = 6; real r = 0.35, h = 3/4; /* radius size and horizontal spacing */ pair shiftR = (h*(n+1)+r, 0); pen colors(int i){ /* shading */ if(i == n) return red; return rgb(5/n,0.4+5/(2n),1-5/n); } void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } void makeshiftarrow(pair A, real dir, real arrowlength = r){ /* Arrow option resizes */ fill(A--A+arrowlength*expi(dir+pi/6)--A+arrowlength*expi(dir-pi/6)--cycle); } pair getCenter(int i, int j){ return ((2*j-i)*h,-i);} /* triangle */ for(int i = 0; i < n+1; ++i){ draw((-i*h,-i)--(i*h,-i)); /* horizontal lining */ for(int j = 0; j <= i; ++j) filldraw(circle(getCenter(i,j),r), colors(i)); } /* fill in circle in row 4, column 3 */ filldraw(circle(getCenter(3,2),r),blue); draw(getCenter(n,2)-- getCenter(3,2)-- getCenter(n,n+2-3)); makeshiftarrow(getCenter(n,2),pi/4,0.5); makeshiftarrow(getCenter(n,n+2-3),3*pi/4,0.5); htick(shiftR+(-1,r),shiftR+(-1,-n+1-r)); label("$n$",shiftR+(-1,(-n+1)/2),E,fontsize(10)); [/asy]$

The sum of the first $n$ positive integers is ${n+1 \choose 2}$ .^[1]

COMING

Nichomauss' Theorem: $n^3$ can be written as the sum of $n$ consecutive integers, and consequently that $1^3 + 2^3 + \cdots + n^3 = (1+2+\cdots + n)^2$ .

COMING
Another proof of the identity $1^3 + 2^3 + \cdots + n^3 = (1+2+\cdots + n)^2$ .

$[asy]defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); int n = 5, fib = 1, fib2 = 1, xsum = 1, ysum = 0; real h = 0.15; void fillsq(pair A = (0,0), real s, pen p = invisible, pen l = linewidth(1)){ filldraw(shift(A)*xscale(s)*yscale(s)*unitsquare, p, l); } void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } for(int i = 0; i < n; ++i) { fillsq((0,h*ysum),h*fib2,rgb(0.9,1,0.9)); fillsq((h*xsum,0),h*fib,rgb(1,0.9,0.9)); if(i == n-1){ label("$F_{n}^2$",h*(xsum+fib/2,fib/2),sm); label("$F_{n-1}^2$",h*(fib2/2,ysum+fib2/2),sm); } else if(i == n-2){ label("$F_{n-2}^2$",h*(xsum+fib/2,fib/2),sm); label("$F_{n-3}^2$",h*(fib2/2,ysum+fib2/2),sm); } fib = fib + fib2; fib2 = fib - fib2; xsum = fib; ysum = fib2; fib = fib + fib2; fib2 = fib - fib2; } htick(h*(xsum,0)+(1,0),h*(xsum,ysum)+(1,0)); label("$F_n$",h*(xsum,ysum/2)+(1,0), E, sm); htick(h*(0,ysum)+(0,1),h*(xsum-fib+fib2,ysum)+(0,1),(0,0.15)); label("$F_{n-1}$",h*((xsum-fib+fib2)/2,ysum)+(0,1), N, sm); htick(h*(xsum,ysum)+(0,1),h*(xsum-fib+fib2,ysum)+(0,1),(0,0.15)); label("$F_{n}$",h*((2*xsum-fib+fib2)/2,ysum)+(0,1), N, sm); [/asy]$
The identity $F_1^2 + F_2^2 + \cdots + F_n^2 = F_{n} \cdot F_{n+1}$ , where $F_i$ is the $i$ th Fibonacci number.

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Geometric series

$[asy]defaultpen(linewidth(0.7)); unitsize(15); int n = 10; /* # of iterations */ real s = 6; /* square size */ pair shiftR = (s+2,0); pen sm = fontsize(10); void fillrect(pair A, pair B = (0,0), pen p = invisible, pen l = linewidth(1)){ filldraw(A--(A.x,B.y)--B--(B.x,A.y)--cycle, p, l); } void htick(pair A, pair B, pair ticklength = (0,0.15)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } for(int i = 0; i < 2; ++i) /* left */ fillrect((s/2^(ceil(i/2)),s/2^(floor(i/2)))); for(int i = 0; i < n; ++i) /* right */ fillrect(shiftR,shiftR + (s/2^(ceil(i/2)),s/2^(floor(i/2)))); label("$\frac 12$",(s*3/4,s/2),sm); label("$\cdots$",(s*1/4,s/2),sm); label("$\frac 12$",shiftR+(s*3/4,s/2),sm); label("$\cdots$",shiftR+(s*1/4,s/2),sm); label("$\frac 14$",shiftR+(s*1/4,s*3/4),sm); label("$\frac 18$",shiftR+(s*3/8,s/4),sm); htick((0,-1), (s,-1)); htick(shiftR + (0,-1), shiftR + (s,-1)); label("$1$",(s/2,-1),S,sm); label("$1$",shiftR+(s/2,-1),S,sm); [/asy]$

The infinite geometric series $\frac 12 + \frac {1}{2^2} + \frac {1}{2^3} + \cdots = 1$ .

$[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 4; real h = 2; pen colors[] = {rgb(0.8,0,0),rgb(0,0.8,0)}; void drawTriGrid(real s){ for(int i = 0; i < 4; ++i){ draw( (-s*3/2,s*(3/2 - i)) -- (s*3/2,s*(3/2 - i)), linetype("2 2")); draw( (s*(3/2 - i),-s*3/2) -- (s*(3/2 - i),s*3/2), linetype("2 2")); } } void fillrect(pair A, pair B, pen p){ filldraw(A--(A.x,B.y)--B--(B.x,A.y)--cycle, p, linewidth(1)); } for(int i = 0; i < n; ++i) { fillrect( ((-1)^i*-h/3^i*(3/2),-h/3^i*(3/2)) , ((-1)^i*-h/3^i*(1/2),h/3^i*(3/2)) , colors[0]); fillrect(-((-1)^i*-h/3^i*(3/2),-h/3^i*(3/2)) ,-((-1)^i*-h/3^i*(1/2),h/3^i*(3/2)) , colors[1]); fillrect( (-h/3^i*(1/2),(-1)^i*h/3^i*(1/2)) , (h/3^i*(1/2),(-1)^i*h/3^i*(3/2)), colors[0]); fillrect(-(-h/3^i*(1/2),(-1)^i*h/3^i*(1/2)) ,-(h/3^i*(1/2),(-1)^i*h/3^i*(3/2)), colors[1]); drawTriGrid(h/3^i); } [/asy]$

The infinite geometric series $\frac 13 + \frac {1}{3^2} + \frac {1}{3^3} + \cdots = \frac 12$ .

$[asy] defaultpen(linewidth(0.7)); unitsize(15); int n = 10; real h = 6; pen colors[] = {rgb(0.9,0,0),rgb(0,0.9,0),rgb(0,0,0.9)}; pair shiftR = (h+3,0); void drawEquilaterals(pair A, real s){ filldraw(A--A+s*expi(2*pi/3)--A+(-s,0)--cycle,colors[0]); filldraw(A--A+s*expi(2*pi/3)--A+s*expi(1*pi/3)--cycle,colors[1]); filldraw(A--A+s*expi(1*pi/3)--A+(s,0)--cycle,colors[2]); } for(int i = 0; i < n; ++i) drawEquilaterals(shiftR + (0,h-h/(2^i) ), (h/(2^(i+1))) *2/3^.5); drawEquilaterals((0,0), h/3^.5); draw((-h/3^.5,0)--(h/3^.5,0)--(0,h)--cycle); label("$\vdots$",(0,3/4*h)); [/asy]$

The infinite geometric series $\frac 14 + \frac {1}{4^2} + \frac {1}{4^3} + \cdots = \frac 13$ .

$[asy] defaultpen(linewidth(1)); unitsize(15); int n = 8; /* number of layers */ real h = 3; /* square height */ pen colors[] = {rgb(0.8,0,0),rgb(0,0.8,0),rgb(0,0,0.8)}; pair shiftL = (-3*h,0); /* amount to shift second square left by */ void drawSquares(real s, pair A = (0,0)){ filldraw(shift(A)*shift(-2*s, -s)*xscale(s)*yscale(s)*unitsquare,colors[0]); filldraw(shift(A)*shift(-2*s,-2*s)*xscale(s)*yscale(s)*unitsquare,colors[1]); filldraw(shift(A)*shift(-s ,-2*s)*xscale(s)*yscale(s)*unitsquare,colors[2]); } for(int i = 0; i < n; ++i) drawSquares(h/2^i); drawSquares(h,shiftL); draw(shift(shiftL+(-2*h,-2*h))*xscale(2*h)*yscale(2*h)*unitsquare); label("$\cdots$",shiftL+(-h/2,-h/2)); [/asy]$

Another proof of the identity $\frac 14 + \frac {1}{4^2} + \frac {1}{4^3} + \cdots = \frac 13$ .

$[asy] unitsize(15); defaultpen(linewidth(1)); pen sm = fontsize(10); real r = 0.7, h = 4.5, n = 10, xsum = 0; void htick(pair A, pair B, pair ticklength = (0,0.15)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } filldraw(xscale(h)*yscale(h)*unitsquare,rgb(0.9,1,0.9)); draw((0,0)--(h/(1-r),0)--(0,h)); for(int i = 0; i < n; ++i){ xsum += r^i; draw((h*xsum,0)--(h*xsum,h*(1-(1-r)*xsum))); htick((h*(xsum-r^i),-1),(h*xsum,-1)); if(i < 6) label("$r^"+(string) i+"$",(h*(xsum-r^i/2),-1),S,sm); else if(i == 8) label("$\cdots$",(h*(xsum-r^i/2),-1.2),S,sm); } htick((-1,0),(-1,h),(.15,0)); htick((0,h+1),(h,h+1)); htick((h+1,h),(h+1,h*r),(.15,0)); label("$1$",(-1,h/2),W,sm); label("$1$",(h/2,h+1),N,sm); label("$1-r$",(h+1,h*(1+r)/2),E,sm); [/asy]$

The infinite geometric series $\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$ .

$[asy] unitsize(15); defaultpen(linewidth(1)); pen sm = fontsize(10); real r = 0.55, h = 2.5, n = 7, xsum = 0; pair shiftD = -(0,h*r/(1-r)+2.5); void htick(pair A, pair B, pair ticklength = (0,0.15)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } draw((0,h*r/(1-r))--(0,0)--(h*n,0)); for(int i = 1; i < n+1; ++i){ draw((h*i,h*(r/(1-r)-xsum-r^(i)))--(h*i,h*(r/(1-r)-xsum))--(0,h*(r/(1-r)-xsum))); if(i < 4) label("$r^"+(string) i+"$", (0,h*(r/(1-r)-xsum-r^(i)/2)), W, sm); htick((h*i,-1),(h*(i-1),-1)); if(i < n) label("$1$",(h*(i-1/2),-1),S,sm); else if(i == n) label("$\cdots$",(h*(i-1/2),-1.2),S,sm); xsum += r^i; } draw((0,h*r/(1-r))+shiftD--shiftD--(h*n,0)+shiftD); xsum = 0; for(int i = 1; i < n+1; ++i){ draw(shiftD+(h*i,0)--shiftD+(h*i,h*(r/(1-r)-xsum))--shiftD+(h*(i-1),h*(r/(1-r)-xsum))); draw(shiftD+(h*i,h*(r/(1-r)-xsum))--shiftD+(0,h*(r/(1-r)-xsum)),linetype("4 4")+linewidth(0.5)); if(i < 4) label("$r^"+(string) i+"$", shiftD+(h*i,h*(r/(1-r)-xsum-r^(i)/2)), ENE, sm); htick(shiftD+(h*i,-1),shiftD+(h*(i-1),-1)); if(i < n) label("$1$",shiftD+(h*(i-1/2),-1),S,sm); else if(i == n) label("$\cdots$",shiftD+(h*(i-1/2),-1.2),S,sm); xsum += r^i; } [/asy]$

The arithmetic-geometric series $\sum_{n=1}^{\infty} nr^n = \sum_{n=1}^{\infty} \sum_{i=n}^{\infty} r^i = \sum_{n=1}^{\infty} \frac{r^{-n}}{1-r} = \frac{r}{(1-r)^2}$ , also known as Gabriel's staircase.^[2]

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Geometry

$[asy] defaultpen(linewidth(0.7)); unitsize(15); real a = 3.9, b = 5.2, c = (a^2 + b^2)^.5; pen sm = fontsize(10); void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } filldraw(xscale(a+b)*yscale(a+b)*unitsquare, rgb(1,0.9,0.8)); filldraw((b,0) --(b,a)--(0,a) --cycle, rgb(0.9,1,0.9)); filldraw((0,a) --(a,a)--(a,a+b)--cycle, rgb(0.9,1,0.9)); filldraw((a,a+b)--(a,b)--(a+b,b)--cycle, rgb(0.9,1,0.9)); filldraw((a+b,b)--(b,b)--(b,0) --cycle, rgb(0.9,1,0.9)); htick((0,-c/10),(b,-c/10),(0,0.15)); htick((-c/10,0),(-c/10,a),(0.15,0)); label("$a$",(-c/10,a/2),W,sm); label("$b$",(b/2,-c/10),S,sm); label("$c$", (a/2,a+b/2),NW,sm); label("$b-a$",((a+b)/2,b),NNE,sm); [/asy]$
First of several proofs of the Pythagorean Theorem: $c^2 = 4 \cdot \frac{ab}2 + (b-a)^2 = a^2 + b^2$ .^[3]

$[asy] pathpen = linewidth(1); unitsize(15); pen dotted = linetype("2 4"); path xaxis = (-3,0)--(3,0); pair A = (-2,2), B = (1,1.5), B3 = (-1.5,0), B2 = (B.x,-B.y), C2 = IP(xaxis, A--B2); D(xaxis,Arrows(8)); D(D(A)--D(C2)--D(B)); D(D(B2)--C2,dashed+linewidth(0.7)); D(A--D(B3)--B,dotted+linewidth(0.7)); D(B3--B2,dotted); MP("(a,b)",A,W); MP("(c,d)",B,E); MP("(c,-d)",B2,E); [/asy]$
The smallest distance necessary to travel between $(a,b)$ , the x-axis, and then $(c,d)$ for $b,d > 0$ is given by $\sqrt{(a-c)^2 + (b+d)^2}$ .

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Miscellaneous

$[asy]unitsize(15); defaultpen(linewidth(0.7)); real a=2.5,b=5,s=a+b; pen colors[] = {rgb(0.9,0.2,0.2), rgb(0.2,0.9,0.2), rgb(0.2,0.2,0.9)}; pen sm = fontsize(8); void fillrect(pair A, pair B, pen p = invisible, pen l = linewidth(1)){ filldraw(A--(A.x,B.y)--B--(B.x,A.y)--cycle, p, l); } void htick(pair A, pair B, pair ticklength = (0.2,0)){ draw(A--B); draw(A-ticklength--A+ticklength); draw(B-ticklength--B+ticklength); } fillrect((0,0),(s,s)); fillrect((a,b),(s,s),colors[0]); filldraw((0,a)--(a,a)--(s/2,s/2)--(a,b)--(a,s)--(0,s)--cycle,colors[1],linewidth(1)); filldraw((0,0)--(b,0)--(b,b)--(a,a)--(0,a)--cycle,colors[2],linewidth(1)); draw((0,0)--(a,a),linewidth(1)); draw((s/2,s/2)--(b,a)--(a,a)--(a,b),linewidth(0.7)+linetype("4 2")); htick((s+1,0),(s+1,b)); htick((s+1,b),(s+1,s)); /* in labels, a,b swapped */ label("$a$",(s+1,b/2),E);label("$b$",(s+1,(s+b)/2),E); label("$ab$",(a+s,b+s)/2,sm); label("$\frac{(a+b)^2}{4}$",(a,a+s)/2,sm); label("$\frac{a^2}2$",(s/2,a*2/3),sm); label("$\frac{b^2}2$",(a/4,a*2/3),sm); [/asy]$
The Root-Mean Square-Arithmetic Mean-Geometric Mean inequality, $\color{red}{ab} \color{black} \le \color{green} \frac{(a+b)^2}{4} \color{black} \le \color{blue} \frac{a^2 + b^2}{2}$ .

$[asy] unitsize(15); defaultpen(linewidth(0.7)); void htick(pair A, pair B,pair ticklength = (0,0.15)){ draw(A--B); draw(A-ticklength--A+ticklength); draw(B-ticklength--B+ticklength); } real a=10,b=3,r=(a+b)/2; pen sm = fontsize(8), dark = linewidth(1); pen colors[] = {rgb(0.9,0.2,0.2) + dark, /* GM */ rgb(0.2,0.9,0.2) + dark, /* AM */ rgb(0.2,0.2,0.9) + dark, /* QM */ rgb(0.2,0.9,0.9) + dark }; /* HM */ pair A = (r-b,(r^2-(r-a)^2)^.5),B=foot((A.x,0),(0,0),A); draw(arc((0,0),r,0,180)--cycle); dot(A); dot((0,r)); dot((A.x,0)); dot((0,0)); draw(B--A,colors[3]); label("HM",(A+B)/2, E, sm+colors[3]); draw((0,0)--(0,r),colors[1]); label("AM",(0,r*2/3), NW, sm+colors[1]); draw((A.x,0)--A,colors[0]); label("GM",(A.x,A.y/2), SE, sm+colors[0]); draw((A.x,0)--(0,r),colors[2]); label("RMS",(A.x/5,r*4/5), NE, sm+colors[2]); draw((-r,0)--A--(r,0), linetype("4 2")); draw((0,0)--B--(A.x,0), linetype("4 2")); draw(rightanglemark((-r,0),A,(r,0))); draw(rightanglemark((0,0),B,(A.x,0))); htick((-r,-1),(A.x,-1)); htick((A.x,-1),(r,-1)); label("$a$",((-r+A.x)/2,-1),S); label("$b$",((r+A.x)/2,-1),S); [/asy]$

The Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality.^[5]

$[asy] unitsize(15); defaultpen(linewidth(0.7)); real r = 0.3, row1 = 3.5, row2 = 0, row3 = -3.5; void necklace(pair k, pen colors[]){ draw(shift(k)*unitcircle); for(int i = 0; i < colors.length; ++i){ pair p = k+expi(pi/2+2*pi*i/colors.length); fill(Circle(p,r),colors[i]); draw(Circle(p,r)); } } void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B); draw(A-ticklength--A+ticklength); draw(B-ticklength--B+ticklength); } /* draw necklaces */ pen BEADS1[] = {red,red,red},BEADS2[] = {blue,blue,blue},BEADS3[] = {red,red,blue},BEADS4[] = {blue,red,red},BEADS5[] = {red,blue,red},BEADS6[] = {blue,blue,red},BEADS7[] = {red,blue,blue},BEADS8[] = {blue,red,blue}; necklace((-10,(row2+row3)/2),BEADS1);necklace((-7.5,(row2+row3)/2),BEADS2); necklace((-2.5,row2),BEADS3);necklace((0,row2),BEADS4);necklace((2.5,row2),BEADS5); necklace((-2.5,row3),BEADS6);necklace((0,row3),BEADS7);necklace((2.5,row3),BEADS8); /* box them and label */ draw((-4,row2-1.3)--(4,row2-1.3)--(4,row2+1.6)--(-4,row2+1.6)--cycle,linewidth(0.9)+linetype("4 2")); draw((-4,row3-1.3)--(4,row3-1.3)--(4,row3+1.6)--(-4,row3+1.6)--cycle,linewidth(0.9)+linetype("4 2")); htick((-4,row2+2),(4,row2+2),(0,0.15)); label("$p$",(0,row2+2),N,fontsize(10)); htick((-11.5,(row2+row3)/2+2),(-6,(row2+row3)/2+2),(0,0.15)); label("$a$",(-17.5/2,(row2+row3)/2+2),N,fontsize(10)); [/asy]$

Fermat's Little Theorem: $a^p \equiv a \pmod{p}$ for $\text{gcd}\,(a,p) = 1$ (above $a=2,p=3$ ).

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References

^ MathOverflow
^ Wolfram MathWorld
^ Attributed to the Chinese text Zhou Bi Suan Jing.
^ This is more of a proof without words of the AM-GM inequality $\frac{a+b}{2} \ge \sqrt{ab}$ ; though the lengths of the segments labeled RMS and HM can easily be verified to have values of $\sqrt{\frac{a^2+b^2}{2}}, \frac{2}{\frac 1a + \frac 1b}$ , respectively, it might not be obvious from the diagram. It still serves as a useful graphical demonstration of the inequality.

@@ Line 157: / Line 157: @@
 </center>
-<!--<center><asy>defaultpen(linewidth(0.7)); unitsize(15);
+<center><!--<asy>defaultpen(linewidth(0.7)); unitsize(15);
-</asy><br>
+</asy><br>--> COMING
+[[Nichomauss' Theorem]]: <math>n^3</math> can be written as the sum of <math>n</math> consecutive integers, and consequently that <math>1^3 + 2^3 + \cdots + n^3 = (1+2+\cdots + n)^2</math>. <br><br>
+</center>
+<center>COMING<br>
-[[Nichomauss' Theorem]]: <math>n^3</math> can be written as the sum of <math>n</math> consecutive integers, and <math>1^3 + 2^3 + \cdots + n^3 = (1+2+\cdots + n)^2</math>.
+Another proof of the identity <math>1^3 + 2^3 + \cdots + n^3 = (1+2+\cdots + n)^2</math>.<br><br></center>
-</center>-->
 <center><asy>defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10);
@@ Line 316: / Line 320: @@
 for(int i = 1; i < n+1; ++i){
   draw(shiftD+(h*i,0)--shiftD+(h*i,h*(r/(1-r)-xsum))--shiftD+(h*(i-1),h*(r/(1-r)-xsum)));
+ draw(shiftD+(h*i,h*(r/(1-r)-xsum))--shiftD+(0,h*(r/(1-r)-xsum)),linetype("4 4")+linewidth(0.5));
   if(i < 4)
-   label("$r^"+(string) i+"$", shiftD+(0,h*(r/(1-r)-xsum-r^(i)/2)), W, sm);
+   label("$r^"+(string) i+"$", shiftD+(h*i,h*(r/(1-r)-xsum-r^(i)/2)), ENE, sm);
   htick(shiftD+(h*i,-1),shiftD+(h*(i-1),-1));
   if(i < n)
@@ Line 326: / Line 331: @@
 }
 </asy><br><br>
-The [[arithmetic-geometric series]] <math>\sum_{n=0}^{\infty} nr^n</math>, also known as Gabriel's staircase.{{ref|2}}<br><br></center>
+The [[arithmetic-geometric series]] <math>\sum_{n=1}^{\infty} nr^n = \sum_{n=1}^{\infty} \sum_{i=n}^{\infty} r^i = \sum_{n=1}^{\infty} \frac{r^{-n}}{1-r} = \frac{r}{(1-r)^2}</math>, also known as Gabriel's staircase.{{ref|2}}<br><br></center>
 <center>[[#toc|Back to Top]]</center>
 == Geometry ==
+<center><asy>
+defaultpen(linewidth(0.7)); unitsize(15); real a = 3.9, b = 5.2, c = (a^2 + b^2)^.5; pen sm = fontsize(10);
+void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); }
+filldraw(xscale(a+b)*yscale(a+b)*unitsquare, rgb(1,0.9,0.8));
+filldraw((b,0)  --(b,a)--(0,a)  --cycle, rgb(0.9,1,0.9));
+filldraw((0,a)  --(a,a)--(a,a+b)--cycle, rgb(0.9,1,0.9));
+filldraw((a,a+b)--(a,b)--(a+b,b)--cycle, rgb(0.9,1,0.9));
+filldraw((a+b,b)--(b,b)--(b,0)  --cycle, rgb(0.9,1,0.9));
+htick((0,-c/10),(b,-c/10),(0,0.15)); htick((-c/10,0),(-c/10,a),(0.15,0)); label("$a$",(-c/10,a/2),W,sm); label("$b$",(b/2,-c/10),S,sm); label("$c$", (a/2,a+b/2),NW,sm); label("$b-a$",((a+b)/2,b),NNE,sm);
+</asy><br>
+First of several proofs of the [[Pythagorean Theorem]]: <math>c^2 = 4 \cdot \frac{ab}2 + (b-a)^2 = a^2 + b^2</math>.{{ref|3}}<br><br></center>
 <center><asy>
 pathpen = linewidth(1); unitsize(15); pen dotted = linetype("2 4");
@@ Line 340: / Line 360: @@
 </asy><br>
 The smallest distance necessary to travel between <math>(a,b)</math>, the x-axis, and then <math>(c,d)</math> for <math>b,d > 0</math> is given by <math>\sqrt{(a-c)^2 + (b+d)^2}</math>. <br><br></center>
 <center>[[#toc|Back to Top]]</center>
@@ Line 395: / Line 415: @@
 label("$a$",((-r+A.x)/2,-1),S); label("$b$",((r+A.x)/2,-1),S);
 </asy><!--[[Image:RMS-AM-GM-HM.gif]]--><br>
-The [[Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality]].{{ref|3}}<br><br>
+The [[Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality]].{{ref|5}}<br><br>
 </center>
@@ Line 436: / Line 456: @@
 #{{note|1}} MathOverflow
 #{{note|2}} Wolfram MathWorld
-#{{note|3}} This is more of a proof without words of the [[AM-GM]] inequality <math>\frac{a+b}{2} \ge \sqrt{ab}</math>; though the lengths of the segments labeled RMS and HM can easily be verified to have values of <math>\sqrt{\frac{a^2+b^2}{2}}, \frac{2}{\frac 1a + \frac 1b}</math>, respectively, it might not be obvious from the diagram. It still serves as a useful graphical demonstration of the inequality.
+#{{note|3}} Attributed to the Chinese text [http://en.wikipedia.org/wiki/Zhou_Bi_Suan_Jing Zhou Bi Suan Jing].
+#{{note|5}} This is more of a proof without words of the [[AM-GM]] inequality <math>\frac{a+b}{2} \ge \sqrt{ab}</math>; though the lengths of the segments labeled RMS and HM can easily be verified to have values of <math>\sqrt{\frac{a^2+b^2}{2}}, \frac{2}{\frac 1a + \frac 1b}</math>, respectively, it might not be obvious from the diagram. It still serves as a useful graphical demonstration of the inequality.
 [[Category:Proofs]]

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