How to show that $\pi^\pi$ is not a integer?

Question

This goes without saying, but, I can't use a calculator to evaluate $\pi^\pi$. I think we need to find a integer $x$ such that $$x<\pi^\pi < x+1. \tag{1}$$

However, since I have no ideia what $\pi^\pi$ looks like, probably I will not find $x$, but if I can prove that such integer $x$ exists, will be enough. But this seems like a difficult problem.

I can use a calculator for other things, for example: evaluating $\pi,\pi^2,e^{27}$ or $\log, \sin$ etc. I tried taking the $\log$ base $\pi$ in $(1)$ to simplify $\pi^\pi$ to just $\pi$.

Maybe this approach is a wrong one.

This is not a "homework problem", is just something that I found it interesting to do and learn more.

I'll appreciate any insight and improved tags. Thanks!

Also, I don't think using a power series for $\pi^x$ is fair, because that's how calculators find the number in the first place.

What about $\pi^{\pi^{{\pi}^{\pi}}}$ ? This is an open problem, so I wanted to see if there's a way to prove that $\pi^\pi$ is not integer in a more "analytical way" but also using mathematical softwares if needed, but thanks for the answers.

Answer 1

$$3.141 < π < 3.142$$

$${3.141}^{3.141} \approx 36.146 < 36.147 < π^{3.141} < π^π < π^{3.142} < {3.142}^{3.142} \approx 36.461$$

$$36.147 < π^π < 36.461$$

Therefore, $π^π$ is not an integer.

Answer 2

The main question to answer is "How do you define $\pi$?"

Once we know that, we can start to find upper and lower bounds for $\pi$ that show what your calculator showed you: That

$$ 36 < \pi^{\pi} < 37. $$

Answer 3

We have $\pi<\frac{22}7$. To see that $\pi^\pi<37$, we check that $(22/7)^{22/7}<37$, or equivalently $$22^{22}<37^7\cdot 7^{22}. $$ This is something that can be done without calculator (though I won't): $$ 341427877364219557396646723584<94931877133\cdot 3909821048582988049$$ It is possible to make a similar lower estimate of the same form, but that would require much higher powers - good luck. So in reality, a pocket calculator throws out $36.4621596$ and even if we allow an incredible 1% error in the calculation, this is safely between $36$ and $37$.

Answer 4

Archimedes proved directly from geometry (by inscribing and circumscribing regular 96-gons to a circle) that $$ 3+10/71<\pi<3+1/7. $$ It is fairly straightforward to manipulate the inequalities to show that $$ 36<(3+10/71)^{3+10/71}\textrm{ and } (3+1/7)^{3+1/7}<37. $$ It reduces to comparing large integers, e.g. $31^{31}<7^{31}37^7$ for the second one. Since the function $x^x$ is monotone increasing for $x>1$ (by taking the derivative, or arguing from monotonicity of addition and multiplication for positive integers), $36 < π^π < 37$ is not an integer.

Answer 5

First of all, notice that $(x^x)' = x^x(1+\ln(x))$ meaning that we need the precision of about at least $0.01$ for discussing the integer evaluation around $30$ where we expect the possible integer to land.

Take that

$$\sqrt{163} - \sqrt{67} \approx 4\ln(\pi)$$

(The constant is actually quite precise $4.0025$.)

Now again with quite some precision

$$e^{\pi \sqrt{163}} \approx 640320^3+744$$

$$e^{\pi \sqrt{67}} \approx 5280^3+744$$

both very close to an integer due to Heegner numbers involved.

Now

$$\frac{e^{\pi \sqrt{163}}}{e^{\pi \sqrt{67}}}=e^{\pi (\sqrt{163}-\sqrt{67})} \approx e^{4 \pi \ln(\pi)}$$

Meaning

$$\pi^{4\pi} \approx \frac{640320^3+744}{5280^3+744} \approx \frac{640320^3}{5280^3} = \left ( \frac{1334}{11} \right)^3 \approx \left ( \frac{1331}{11} \right)^3 = 121^3=11^6$$

If $\pi^{\pi}$ is an integer so is $\pi^{2\pi}$ and we find it being close to $1331$.

However $\pi^{\pi} \approx \sqrt{11^3}=\sqrt{1331}$ cannot be an integer as there is no close square to $1331$, at least not within the error margin of the estimations involved.

(Notice that the method completely avoids dealing with big numbers.)

On top of that, this is how we've derived one of those magical close to integer expressions

$$\frac{11\sqrt{11}}{\pi^{\pi}}=1.000568$$

which is possibly the shortest possible expression connected directly to the proof that $\pi^{\pi}$ is not an integer, for the constant $1.000568$ is too small and the fractional part of $11\sqrt{11}$ too close to $0.5$ to be anywhere near the integer after division for the numbers in that range.