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Since $\pi$ is irrational, we shall find every combination of numbers in $\pi$’s numerical form.

So, my question is equivalent to whether there exists rational number $k$ and integer $n$ such that $$\pi = k + e * 10^n$$.

Any idea?

Vinyl_cape_jawa
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Szeto
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    Every finite combination of numbers. And anyhow, is that even proven? – Ant Dec 12 '17 at 14:47
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    It was never proven that $\pi$ is a normal number. – Zubzub Dec 12 '17 at 14:48
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    The assumption that $\pi$ contains every combinations of numbers inside is not justified. One must prove that the digits of $\pi$ are normally distributed which is still an open problem. Note that $0.1001000100001...$ is transcendental and certainly does not contain every finite sequence of digits. – CyclotomicField Dec 12 '17 at 14:48
  • @CyclotomicField Does that mean the whole "every finite sequence" trope is merely a product of popular culture? Also, can normality depend on the base in use? – actinidia Dec 12 '17 at 14:51
  • According to a previous post (https://math.stackexchange.com/questions/456097/are-pi-and-e-algebraically-independent) this is an open question. – A Blumenthal Dec 12 '17 at 15:00
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    @TiwaAina Yes, it's a common misconception that has been propagated in popular media including television and film. I think that some people arrive at this result based on colloquial descriptions of $\pi$ as infinite and never repeating combined with a basic familiarity with its decimal expansion. Humans can find patterns in static so we are prone to making mistakes of this nature where additional structure is contrived out of casual resemblance. – CyclotomicField Dec 12 '17 at 15:42

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It's a commonly-believed but unproven fact that $\pi$ contains every finite sequence of numbers - there are plenty of irrationals that don't, like $0.101001000100001\ldots$. A number that contains every finite sequence of numbers is called "normal"; in a precise sense, "most" numbers are normal, but it's very hard to prove whether one number in particular is normal.

Even if $\pi$ and $e$ were both known to be normal, this would be an open question (and the answer is probably no). It's currently unknown whether such combinations as $\pi + e$ is transcendental, and this property would very easily show that it is (because then $\pi + e$ would be $1 + k + e10^n$, which is a rational combination of transcendentals).

Reese Johnston
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  • But, is the Euler’s identity e^(i*pi)=-1 a proof of pi and e’s algebraical dependence? – Szeto Dec 12 '17 at 15:10
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    @Szeto: no algebraic dependence does not allow the variables in powers. You need to find a polynomial with rational coefficients in one that equates to the other. – Ross Millikan Dec 12 '17 at 15:18