2
$\begingroup$

there

Is the following statement is true?

$\forall k \in \mathbf{N},\exists{m}\in\mathbf{N}, k!\pi+\frac{\pi}{6}\le{m!}\le{k!\pi}+\frac{5\pi}{6}$

I tried by descendant proof but was not satisfied by the arguement.

$\endgroup$
  • $\begingroup$ How can it be true ? $\frac{m!}{k!}$, a product of integers, must be very close to $\pi$ ?! $\endgroup$ – Yves Daoust Aug 8 '14 at 9:32
  • $\begingroup$ Did you try picking some $k$, say, $k=4$, and seeing whether there's an $m$? $\endgroup$ – Gerry Myerson Aug 8 '14 at 9:48
  • $\begingroup$ every irrational can be arbitarary approximate by rationals. so may be it can be happan. $\endgroup$ – Bumblebee Aug 8 '14 at 9:58
  • 1
    $\begingroup$ @Nilan, get back to me after you have found an integer $m$ such that $$24\pi+(\pi/6)\le m!\le24\pi+(5\pi/6)$$ $\endgroup$ – Gerry Myerson Aug 8 '14 at 23:33
1
$\begingroup$

If such $m$ exists, $k<m$ obviously, so $(k+1)!\le m!$. But if $k$ is sufficiently large, $(k+1)!>k!\pi+\frac56\pi$, which contradicts with the condition. Thus, the statement is false.

$\endgroup$
1
$\begingroup$

It's not true. There is no $m\in\mathbb N$ such that $$1!\cdot \pi+\frac{\pi}{6}\le m!\le 1! \cdot \pi+\frac{5}{6}\pi.$$ Here, note that $$1!\cdot \pi+\frac{\pi}{6}=\frac{7}{6}\pi\approx 3.7,\ \ 1!\cdot\pi+\frac{5}{6}\pi=\frac{11}{6}\pi\approx 5.8.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.