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I am trying to prove the real number in $(0,1)$ which has as its decimal expansion $$0.13169219728561⋯$$ obtained by concatenating the powers of $13$ is irrational. This seems harder than the analogous question for powers of $2$. Not sure how to approach this-I cannot seem to think of anything about the digits that appear in the powers of 13 to contradict a possible eventually periodic decimal expansion!

student
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  • Statement is confusing. Shouldn't there be symbols for 10, 11, 12? – herb steinberg Feb 02 '22 at 02:55
  • Is this true for any integer (other than $1$ and $0$)? It might be easiest to just prove it is never true for any integer. (I said might.....) – fleablood Feb 02 '22 at 02:58
  • How did you prove it for powers of $2$? Note $\gcd(13,10)=1$ so $13^{12}\equiv 1\pmod{13}$ That probably is key. – fleablood Feb 02 '22 at 03:00
  • The number of digits that appear in the powers of 2 can be any positive integer. So assuming that the decimal expansion obtained by concantenating them is eventually periodic can be used to get a contradiction. The number of digits in the powers of 13 can increase by 2. Thus the argument that worked for powers of 2 does not seem to work in this case. – student Feb 02 '22 at 03:10
  • How is ${13^{12}} \equiv 1(\rm{mod} 13)$? $13$ clearly divides $13^{12}$. You must mean $10^{12}\equiv 1(\rm{mod} 13)$ – student Feb 02 '22 at 03:36
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    Show that $13^n\equiv1\mod10^k$ has solutions for $n$ with $k$ arbitrarily large. The digit string will then have arbitrarily long strings of zeroes which would contradict a periodic representation. – Oscar Lanzi Feb 02 '22 at 08:30

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Following Mr. Lanczi's hint. We contend that the above decimal expansion cannot be eventually periodic. Observe that $13^m$ has $m[{\rm log}_{10} 13]+1$ digits in ites decimal expansion. We now observe that since ${\rm gcd}(13,10)=1,$ $$13^{{\phi}(n^2)} \equiv 1 ({\rm mod}(10^{n^2} ).$$ (Here $\phi$ is the Euler totient function). Thus, $$13^{{\phi}(n^2) +n} \equiv 13^{n} ({\rm mod}(10^{n^2} ).$$ And so, $$13^{{\phi}(n^2) +n}- 13^{n} \equiv 0({\rm mod}(10^{n^2} ).$$ This means that $13^{{\phi}(n^2) +n}- 13^{n}$ has at least $n^2$ consecutive terminal zeroes in its decimal expansion. And so, $$13^{{\phi}(n^2) +n} = 13^{{\phi}(n^2) +n}-13^{n} +13^{n}$$ must have at least $${n^{2}} -(n[{\rm log}_{10} 13]+1)\geq{n(n-[{\rm log}_{10} 13])-2}$$ consecutive zeroes in its decimal expansion. Since $$n(n-[{\rm log}_{10} 13])-2\rightarrow {\infty}$$ as $n\rightarrow{\infty},$ this would force the conclusion that the above number terminates in $0$, forcing the conclusion that for $m$ large enough, $13^{m} = 0$ which is absurd. Hence, the above number must be an irrational number. It may be that there is a cleaner way to write this proof-apologies if it is clumsy, but it appears to be correct.

student
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