Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
What is the argument why that should somehow indicate that $\pi = 2$? – Daniel Fischer Dec 2 '13 at 10:51
The first equality is false. $x^{x^x}\neq x^{x\cdot x}$ – Callus Dec 2 '13 at 10:53
@Kartik No (hopefully). What was taught was most likely $\left(a^b\right)^c = a^{(b\cdot c)}$. In general, $a^{(b^c)} \neq \left(a^b\right)^c$. – Daniel Fischer Dec 2 '13 at 10:55
@Kartik That the exponents are equal just says $\pi \cdot \pi = \pi^2$. – Daniel Fischer Dec 2 '13 at 11:09
@Kartik It changes the meaning of the expressions in the chain of equations. $\pi^{\pi^\pi} = \pi^{\left(\pi^\pi\right)}$ is different from $\left(\pi^{\pi}\right)^\pi = \pi^{\pi\pi}$. – Daniel Fischer Dec 2 '13 at 11:15
up vote 6 down vote accepted

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!

share|cite|improve this answer

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 $b\times c=b^c$.

share|cite|improve this answer
You might add that unadorned $a^{b^c}$ means $a^{(b^c)}$ by notational convention. – 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 $\pi$?$$10^{10^{10}}=10^{10000000000}$$ $$10^{10\cdot10}=10^{100}\ne10^{10000000000}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.