Difference between revisions of "Proofs without words"

Revision as of 22:56, 18 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 $n(n+1)/2$ .

Geometric series

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

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

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

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

Geometry

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.

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