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You don't seem to have understood the point. Let's do it without re-using the same variable name. Solving 2=y^y^y^... lets us find that y=sqrt(2). Therefore s
by RiderOfGiraffes 16y ago
You don't seem to have understood the point. Let's do it without re-using the same variable name.
Solving 2=y^y^y^... lets us find that y=sqrt(2).
Therefore sqrt(2)^sqrt(2)^sqrt(2)^... = 2
Solving 4=z^z^z^... lets us find that z=sqrt(2).
Therefore sqrt(2)^sqrt(2)^sqrt(2)^... = 4
Therefore 2=4.
Now, could you explain your resolution of the "paradox" more clearly?
- ianferrel 16y agoYes, I did misread the proof. Thanks for the explanation.
- vessenes 16y agoTo restate; we can say: 1) Assume that an x exists such that 2=x^x^x^x^... 2) Then x = sqrt(2) 3) sqrt(2)^10 = 32 Therefore there is no such x. In very brief, we can say that there is no number so infinitesimally close to 1 that infinite exponentiation of it does not 'blow up'. Or to pull a Calc 1 term: x^x^x^x^... does not converge for all x >1. Simple.
- RiderOfGiraffes 16y agoI don't understand why your step 3 is relevant. Observe that using r=sqrt(2) we have that the sequence r, r^r, r^r^r, r^r^r^r, ... does converge to 2. We are not performing arbitrarily large exponentiation of x. Or, to pull a Calc 1 term, x^x^x^... does converge for x=sqrt(2).
- Lozzer 16y agoIt's the difference between x^(x^(x... and ((x^x)^x)...