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Two Equals Four - 2011/04/14
Consider the equation $2=x^{x^{x^{x^{\ldots}}}}$ and suppose we want to solve it for x. Because the exponential tower is infinite, we can also write it as $2=x^{\left({x^{x^{x^{\ldots}}}}\right)}$ But the part in brackets is the same as the whole, and hence is equal to 2. Thus we have $2=x^2$. So $x=\sqrt{2}$ and we have $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}}=2$ Now consider the equation $4=x^{x^{x^{x^{\ldots}}}}$ and again let's solve for x. As before, we can write it as $4=x^{\left({x^{x^{x^{\ldots}}}}\right)}$ and again, the part in brackets is the same as the whole, and so now we get $4=x^4$. Take the square root of each side and we get $2=x^2$ and so again $x=\sqrt{2}$. Thus $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}}=4$. So $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}}$ is 2, and it's also 4. Hence 2=4 (and halving it means 1=2).
CommentsIn an email, Iain Murray has pointed to exercise 4.20 on page 86 of the book:
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