Let us consider the number $$\Large\pi^{\pi^\pi}=\pi^{\pi\cdot\pi}=\pi^{\pi^2}$$

As the bases are equal, the exponents must be equal, So $$\pi=2$$

You can take any $x$ instead of $\pi$.

What is wrong in this proof?

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    $\begingroup$ What is the argument why that should somehow indicate that $\pi = 2$? $\endgroup$ – Daniel Fischer Dec 2 '13 at 10:51
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    $\begingroup$ The first equality is false. $x^{x^x}\neq x^{x\cdot x}$ $\endgroup$ – Callus - Reinstate Monica Dec 2 '13 at 10:53
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    $\begingroup$ @Kartik No (hopefully). What was taught was most likely $\left(a^b\right)^c = a^{(b\cdot c)}$. In general, $a^{(b^c)} \neq \left(a^b\right)^c$. $\endgroup$ – Daniel Fischer Dec 2 '13 at 10:55
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    $\begingroup$ @Kartik That the exponents are equal just says $\pi \cdot \pi = \pi^2$. $\endgroup$ – Daniel Fischer Dec 2 '13 at 11:09
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    $\begingroup$ @Kartik It changes the meaning of the expressions in the chain of equations. $\pi^{\pi^\pi} = \pi^{\left(\pi^\pi\right)}$ is different from $\left(\pi^{\pi}\right)^\pi = \pi^{\pi\pi}$. $\endgroup$ – Daniel Fischer Dec 2 '13 at 11:15

Lets write $a \uparrow b$ to mean $a^b$.

Then the following reasoning is correct: $$(\pi \uparrow \pi)\uparrow \pi = \pi \uparrow (\pi \cdot \pi) = \pi \uparrow (\pi \uparrow 2)$$

However, we cannot necessarily deduce that the RHS equals

$$(\pi \uparrow \pi) \uparrow 2$$

because exponentiation isn't associative. Indeed, Google calculator tells me that:

  • $\pi \uparrow (\pi \uparrow 2) \approx 80662.6659386$

  • $(\pi \uparrow \pi) \uparrow 2 \approx 1329.48908322$

so if the calculator is correct to even the first decimal place, then

$$(\pi \uparrow \pi) \uparrow 2 \neq \pi \uparrow (\pi \uparrow 2).$$

Moral of the story: if in doubt, find better notation!


I think you are mixing up $$ \left(\pi^\pi\right)^\pi=\pi^{\pi^2} $$ with $$ \pi^{\left(\pi^\pi\right)}\neq \pi^{\pi^2}. $$

In general, $$ \left(a^b\right)^c\neq a^{\left(b^c\right)} $$ but if $a=b=c=2$ it is true since then $b\times c=b^c$.

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    $\begingroup$ You might add that unadorned $a^{b^c}$ means $a^{(b^c)}$ by notational convention. $\endgroup$ – Marc van Leeuwen Dec 2 '13 at 11:33

If you can take any $x$ instead of $\pi$, why didn't you take $3$ or $10$ instead of $\pi$?$$10^{10^{10}}=10^{10000000000}$$ $$10^{10\cdot10}=10^{100}\ne10^{10000000000}$$


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