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I have tried to prove this all day but couldn't come up with a convincing solution. Please i need help on how to prove this. Thanks

$\lfloor n^{1/k}\rfloor$+1 =$\lceil(n+1)^{1/k}\rceil$ where k is positive integer

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Suppose first that $n=m^k$ for some integer $m$; then $n^{1/k}=m$, and we need to show that

$$m+1=\left\lceil(m^k+1)^{1/k}\right\rceil$$

or, equivalently, that

$$m<(m^k+1)^{1/k}\le m+1\;.$$

Take $k$-th powers: this in turn is equivalent to

$$m^k<m^k+1\le(m+1)^k\;,$$

which is certainly true for all $k\in\Bbb Z^+$.

Now assume that $n^{1/k}$ is not an integer. Then

$$\left\lfloor n^{1/k}\right\rfloor+1=\left\lceil n^{1/k}\right\rceil\le\left\lceil(n+1)^{1/k}\right\rceil\;.$$

Let $m=\left\lceil n^{1/k}\right\rceil=\left\lfloor n^{1/k}\right\rfloor+1$, and suppose that $m<\left\lceil(n+1)^{1/k}\right\rceil$; then $n^{1/k}<m<(n+1)^{1/k}$, so $n<m^k<n+1$. But $m^k$ is an integer, and $n$ and $n+1$ are consecutive integers, so this is impossible. Thus,

$$\left\lfloor n^{1/k}\right\rfloor+1=\left\lceil(n+1)^{1/k}\right\rceil$$

in this case as well.

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  • $\begingroup$ Thanks Brian for the great effort...it really helps $\endgroup$ – user1999 Sep 28 '15 at 16:57
  • $\begingroup$ @user1999: You’re welcome. $\endgroup$ – Brian M. Scott Sep 28 '15 at 16:57
  • $\begingroup$ @user1999: If you think this answer is helpful you could upvote and accept it. Best regards, $\endgroup$ – Markus Scheuer Oct 2 '15 at 16:16

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