Geometric proof of sum(1 to n)

Ah, geometric proofs, so graphic, so intuitive and clear, so fitting to the way our brain works. I remember that last year I spent some time doubting Pythagoras' proof of c2 = a2 x b2, it sent me on a survey for all existing proofs, many of the more graphical ones did not 'really' convince me, specially when they involved rotating or moving shapes and then made claims about them fitting somehow, Euclid's proof on wikipedia did a better job, one of it's 'less convincing' points being the triangle area lemma. When I look at it I am pretty sure it came out of intuition (like most good proofs) and then the 'formal' proof got put into pieces after having 'devised a plan for the solution' as George Polya would put it in his classic book: HOW to solve it.

Still, it made me like geometric proofs more than I used to -- and the reason I liked algebraic ones more, in retrospect is because of the history of my Math teachers -- and so I started scribbling... I came up with a couple of tiny nice geometric proofs.

The first one is proving that and that . The key idea is (like the start of Pythagoras' theorem) is to use area when multiplying numbers by each other, and then figuring out a relationship between the resulting shapes (in this case all rectangles) and writing it down:


Another, nicer proof is the one for the sum of 1 to n being equal to n(n+1)/2 :

There are many ways to proove this one, by induction per example, or geometrically as demonstrated in one of the discrete math video lectures I am using. However, the geometric proof used there involves having to rotate shapes and comparing them, I did not really like that, so I tried to come up with a clearer proof and I found out that it is possible, I have not seen a proof that did it exactly this way, so here it is:

The idea is that we know that the square in the figure has n x n 'dots' in it, we therefore know that cutting in half by the diagonal, we have dots. This 'almost' covers the total number of dots (the answer we are looking for) and might not even be an integer.

We can see that all the last dots are cut by the diagonal in half, and those halves are all what is missing to account for the missing rest, therefore, we add them.

We have n halves since we have n rows, so adding to , we get the final answer: . We did it without having to move shapes, rotate them or compare their lengths. And that is nice!