Two Equals Four

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Two Equals Four - 2011/04/14

I'm amazed that I need to say this, but it appears that I do ...

I know this is fallacious, and I do know what's wrong with it. Please don't assume that I really do think that 2=4, or that 1=2.


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Here's a cool puzzle.

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).

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In an email, Iain Murray has pointed to exercise 4.20 on page 86 of the book:

  • Information Theory, Inference, and Learning Algorithms
    • by David J.C. MacKay FRS

A copy of the book can be found online here: The exercise itself is in this PDF: and the solution/discussion is on page 89:



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