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

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.

• Not sure but think this is an open problem. – Karl Jun 22 '19 at 7:18
• @Karl I think transcendental is an open problem, not integral. – Juan Sebastian Lozano Jun 22 '19 at 7:22
• So, if we don't know if the number is irrational or transcedental, we can never prove if it's an integer without expliciting calculating said number? This sounds terrible. – Pinteco Jun 22 '19 at 7:23
• @Pinteco You prove it isn't an integer by bounding it away from integers through good enough estimates. To prove it is an integer might be much harder. Since the value appears to be about 36.5 that should be possible. Getting good estimates which can be computed by hand is tougher. It is quite easy with the help of a calculator, though. – Mark Bennet Jun 22 '19 at 7:27
• This is an extremely artificial problem. The strategy is very clear, and it will require some calculation, no matter what. It is just a matter of taste whether the computation $36.33< 3.14^{3.14} < \pi^\pi < 3.142^{3.142}<36.5$ is within the acceptable range of "calculation without using a calculator". After all, you can do this using a paper and a pen in a relatively short period of time... – A. Pongrácz Jun 22 '19 at 8:35

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.

• I think you have a typo: 96-gons, not 91. – DanielWainfleet Jun 22 '19 at 21:48
• +1. My edit was for $310/71$ and $31/7.$ I.e. $310/71$ looks like $\frac {310}{71}.$ – DanielWainfleet Jun 22 '19 at 21:54

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

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

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 $$\pi^{\pi} = \sqrt{11^3}=\sqrt{1331}$$ 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.)

$$3.141 < π < 3.142$$

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

$$36.147 < π^π < 36.461$$

Therefore, $$π^π$$ is not an integer.

• I think, there is a typos in Line 2, "36,147" shoud be "36.147". – Zongxiang Yi Jun 22 '19 at 7:44
• If you know $3.141 < \pi< 3.142$ then you would know $36.4621<\pi^\pi<36.4622$. Note that $f(x)=x^x$ is an increasing function. So the problem turns to an estimation of $\pi$. Thus I agree to lngix's answer. – Zongxiang Yi Jun 22 '19 at 7:53

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 $$37$$.

• I think you can perhaps estimate the derivative to show that $22/7$ is good enough and $355/113$ is unnecessary. – Mark Bennet Jun 22 '19 at 7:39