Proofs without words

Revision as of 23:39, 17 March 2010 by Azjps (talk | contribs) (start, will add more later (1 per day?))
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $n(n+1)/2$.