Proofs without words

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

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

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

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

$[asy] unitsize(15); defaultpen(linewidth(0.7)); real r = 0.3, row1 = 3.5, row2 = 0, row3 = -3.5; void necklace(pair k, pen colors[]){ draw(shift(k)*unitcircle); for(int i = 0; i < colors.length; ++i){ pair p = k+expi(pi/2+2*pi*i/colors.length); fill(Circle(p,r),colors[i]); draw(Circle(p,r)); } } void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B); draw(A-ticklength--A+ticklength); draw(B-ticklength--B+ticklength); } /* draw necklaces */ pen BEADS1[] = {red,red,red},BEADS2[] = {blue,blue,blue},BEADS3[] = {red,red,blue},BEADS4[] = {blue,red,red},BEADS5[] = {red,blue,red},BEADS6[] = {blue,blue,red},BEADS7[] = {red,blue,blue},BEADS8[] = {blue,red,blue}; necklace((-10,(row2+row3)/2),BEADS1);necklace((-7.5,(row2+row3)/2),BEADS2); necklace((-2.5,row2),BEADS3);necklace((0,row2),BEADS4);necklace((2.5,row2),BEADS5); necklace((-2.5,row3),BEADS6);necklace((0,row3),BEADS7);necklace((2.5,row3),BEADS8); /* box them and label */ draw((-4,row2-1.3)--(4,row2-1.3)--(4,row2+1.6)--(-4,row2+1.6)--cycle,linewidth(0.9)+linetype("4 2")); draw((-4,row3-1.3)--(4,row3-1.3)--(4,row3+1.6)--(-4,row3+1.6)--cycle,linewidth(0.9)+linetype("4 2")); htick((-4,row2+2),(4,row2+2),(0,0.15)); label("$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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Proofs without words