Imagine two streets at right angles to one another and imagine that a pedestrian wants to get from point A on one street to point C on the other. One can construct a right-angled triangle ABC with hypotenuse AC. Imagine two streets at right angles…

Imagine two streets at right angles to one another and imagine that a pedestrian wants to get from point A on one street to point C on the other. One can construct a right-angled triangle ABC with hypotenuse AC.

Imagine two streets at right angles to one another and imagine that a pedestrian wants to get from point A on one street to point C on the other. One can construct a right-angled triangle ABC with hypotenuse AC.

Now consider that there is no direct route from A to C and no other roads. The route to be travelled will be A – B – C and the distance travelled will be AB + BC, rather than the straight line distance AC = SQRT( AB*AB + BC*BC).

Now consider that there is actually an intermediate road parallel to BC between A and B, intersecting AB at point D. Further consider that there is a similar road parallel to AB intersecting BC at point E, and that these two new roads intersect at point F.

The pedestrian now has the alternative route A – D – F – E – C with distance AD + DF + FE + EC. However AD + DE = AC, and DF + EC = BC, thus the distance walked is still AC + BC.

One can extend this arrangement, introducing extra roads into the matrix, but at all times the distance travelled is the same as the route A – B – C. As the number of roads increases, so the inter-road distance diminishes, but the total route remains the same; so how is it that at the limit, where the number of extra roads is infinite, and the inter-road distance approaches zero, and thus the route from A to C approximates a straight line, the total distance becomes SQRT( AB*AB + BC*BC)?

I know I’m obviously missing something here, but for the life of me I can’t see what.