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.

Ah, an infinity of connecting roads. Therein lies your problem. If there were an infinite number of such roads, then it could be argued you would take an infinite length of time to negotiate them, and consequently travel an infinite distance doing so. This is similar to Zeno’s paradox, with which I’m sure you are familiar.If you travel down perpendicular roads between the two points, then you will always travel (effectively) the distance A to B, and then B to C, even if there are an infinitude of such roads. Think about it – you are simply sub-dividing the hypotenuse distance, but still travelling the good old A-B +B-C distance and NOT A-C.Try this. Imagine there are now 10 sub-divisions. You cannot travel diagonally. Blow up one of the "triangles" to the original size and you’ll see what I mean. Keep on sub-dividing, and enlarging the picture. You still cannot travel diagonally, but must follow a rectiliear path.I once had a related idea regarding rational thought. The probability of soemthing happening is the number of desired outcomes/the number of possible outcomes. So, if you have one thought, then the probability of the next one being connected is the number of possible connected thoughts divided by the number of possible thoughts. This latter may be argued to be infinite in number and anything divided by infinity is zero. This suggests that the probability of two connected thoughts in series is zero. It may, of course, also be argued that the number of connected thoughts is also infinite, but it is a smaller infinity that that of the number of unconnected thoughts. For more on infinities and their "sizes", check out Cantor or Godel.Have fun!

I’m not sure I see your reasoning there, Shuggie.I am familiar with some of Zeno’s paradoxes, and also with the various explanations of them. I guess this is, sort of, like perhaps the tortoise and the hare, but not quite.My explanation however, is that as the number of roads becomes larger, the distance between each parallel road becomes less. As linear distance is quantized ( at the Planck length), then at some point, the roads cannot get closer together than the Planck distance. At this point, you can now travel directly between the current A and C rather than having to go via B.From there, it’s just a matter of multiplying up again.

Yeah, yeah, but your diagonal will still equate to sqr root of 2*(Planck length squared). i.e. longer than the Planck length. <br/> <br/> <br/> <br/>From: