Achilles will actually meet up with the Tortoise at an exact distance of 556 meters. This you may think doesn’t sound logical. If we apply mathematics to the specified distances using the given set of parameters, we can actually find the mathematical distance at which the Achilles and the Tortoise meetup.

First off we must enter in the equation. 500 + 50 + 5 + 1/2 + 1/4 + 1/8 + 1/16 + ….. + 1/(2^n) = ? Now one can easily sum up the first few numbers 500 + 50 + 5 = 555. The next step is to sum up the infinite series being described here which is typically described in mathematics as Sn = [Σ] 1/(2^n) . One then proceeds to multiply both sides by 2 making it because of a special  2Sn = 2/2 + 2/4 + 4/8+…2/((2)^(n-1))+ 2/(2^n). This reveals an interesting relationship so the equation now becomes 2Sn = 1 + [1/2 + 1/4 + 1/8 + 1/16 + ….. + 1/((2)^(n-1)] = 1 + [Sn – 1/(2^n)]. One then subtracts both sides by Sn making the equation now Sn = 1 –  1/(2^n). Now by applying calculus and taking the limit of 1 – 1/(2^n) this then gives us the value of Sn. Sn = 1

So now we have our final term which is simply 1 So adding all the numbers together gives us 500 + 50 + 5 + 1 = 556 meters is the point at which Achilles will finally meet up with the Tortoise. Now this case only solves this particular paradox because of the poorly worded statement about the last few changes to the distance being a geometric sum 1/2 + 1/4 + 1/8 + 1/16 + ….. + 1/(2^n). Also this was not a solvable equation back in ancient Greece since the invention of calculus did not take place until the 1700s by Isaac Newton. But there you have it, the solution to an age old paradox.

[Σ]

m = my age in year
s  = son in years
G  = grandson in years

G + s + g = 140 years

substitute G for into each

G + 12G + 7G  = 140

20G = 140

G  = 140/20

G  = 7 years

s = 49 years

m = 84 years

Romance

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