Can anybody tell me how to set up zeno paradox, so that I can show a proof of it inductively? this is the exact question.Im not a math person, this is for a logic class, and I have no idea how to set this up.

Suppose that Achilles stands 100 yards from the finish line of a race. Suppose further that, with every step, he covers precisely half of the remaining distance to the finish line. Show that he will still be some (non-zero) distance from the finish line after a million steps.

• Zeno's paradox is based on a misunderstanding. There is nothing to be proved or disproved. In fact, it is an abuse of induction used where it is not applicable. – Somos Oct 17 at 23:57

Can anybody tell me how to set up zeno paradox, so that I can show a proof of it inductively?

Well, .... no because it's not true.

But a setup is straight forward and easy:

The proposition is: $$P(n) =$$ after $$n$$ iterations, Achilles will be $$100\times\frac 1{2^n}$$ yards from the finish line.

Pf:

Base step: $$P(0)$$. Achilles is $$100$$ yards from the finish line $$100\times \frac 1{2^0}$$ yards $$= 100$$ yards.

Induction step: $$P(k)\implies P(k+1)$$. If after $$k$$ iterations Achilles is $$100\times \frac 1{2^k}$$ yards form the finish line in the $$k+1$$th iteration, he will be half that distance $$\frac 12$$ of $$100\times \frac 1{2^k}$$ yards is $$\frac 12 \times 100\times\frac 1{2k}$$ yards = $$100\times \frac 1{2^k}\times \frac 12$$ yards $$=100\times \frac 1{2^{k+1}}$$ yards.

So the statement as stated is true.

....

But for Achilles to cross the finish does not require that a finite number of iterations occur, and a time in the space time continuum existing need not be any of the finite iterations of the proposed statement.

That after any finite iterations of Achilles will not have reached the finish line, has nothing to do with Achilles not being able to reach the finish line. An infinite number or interations does not in any way mean an infinite period of time must pass.

• hmm this makes sense. So the left side of the equation(?) in the induction case is 1/2 x 100 x 1/2k and the the right side 100 x 1/2k+1? – Moose86 Oct 19 at 2:46