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During our ongoing research, we need to prove that $\pi e<\lceil \pi e\rceil$. Is $\pi e$ irrational?

How to prove it?

Thanks- mike

mike
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    Is that the ceiling function? If so what does this have to do with irrationality? – Noam Jul 15 '16 at 20:26
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    If my memory serves, it is unknown whether $\pi e$ or $\pi + e$ is irrational, but we know that at least one of them is irrational because otherwise $p(x) = (x+\pi)(x+e)$ would have a rational root. – User8128 Jul 15 '16 at 20:27
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    @Noam I assume it's because the OP wants a strict inequality. The irrationality of $\pi e$ would guarantee this. – Zestylemonzi Jul 15 '16 at 20:31
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    @Noam Maybe the strict inequality is of OP's concern? – Zhanxiong Jul 15 '16 at 20:31
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    Note: it is not difficult to estimate the product...getting $\pi\times e\sim 8.539734...$. This is enough to show that $\pi\times e < \lceil \pi \times e \rceil$, but I don't see any connection between this and irrationality. – lulu Jul 15 '16 at 20:33
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    Of course, knowing that $\pi e$ is irrational is sufficient for a strict inequality; it is far from necessary, so it does seem a bit silly to phrase the question this way. – User8128 Jul 15 '16 at 20:33
  • Yes. I need the strict inequality. Thanks! – mike Jul 15 '16 at 20:36
  • @lulu, I was trying not to use the numerical calculations to show the strict inequality. If the numerical calculation is acceptable, then it is enough for me. Thanks – mike Jul 15 '16 at 20:38
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    Even with just two significant figures for $\pi$ and $e$, we get $8.37=3.1\cdot 2.7<\pi\cdot e < 3.2\cdot 2.8 = 8.96$ implying the cieling is $9$. – JMoravitz Jul 15 '16 at 20:43

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