# How does $a^2 + b^2 = c^2$ work with ‘steps’? [duplicate]

We all know that $$a^2+b^2=c^2$$ in a right-angled triangle, and therefore, that $$c, so that walking along the red line would be shorter than using the two black lines to get from top left to bottom right in the following graphic: Now, let's assume that the direct way using the red line is blocked, but instead, we can use the green way in the following picture: Obviously, the green way isn't any shorter than the black one, it's just $$a/2+b/2+a/2+b/2 = a+b$$. Now, we can divide the green path again, just like the black path, and get to the purple path. Dividing this one in two halfs again, we get the yellow path: Now obviously, the yellow path is still as long as the black path from the beginning, it's just $$8*a/8+8*b/8=a+b$$. But if we do this segmentation again and again, we approximate the red line - without making the way any shorter. Why is this so?

• math.stackexchange.com/questions/12906/is-value-of-pi-4 Jul 4, 2013 at 17:59
• A simple reason: From basic Euclidean geometry we can prove Pythagoras Theorem. So we know a^2+b^2=c^2. Your question is equivalent to asking, why doesn't a+b=c, given a^2+b^2=c^2? Well, they are not equal because, for example, 3^2+4^2=5^2, yet 3+4 does not =5 (so we found a counterexample). In reality, you are asking why a sub i + b sub i does not equal c sub i. If any did, then adding all the as and bs and cs together on respective side of equations would result in a+b=c (which we know is false), since i is arbitrary, ai+bi=ci can't hold for some but not others, so it holds for none. Oct 30, 2016 at 12:27

An example where the limit is properly found is dividing a circle into $n$ equal parts and computing the sum of the line segments connecting the endpoints of the arcs. This $does$ converge to the length of the circle because the height of each arc gets arbitrarily small compared to the length of each arc as $n$ gets large.