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Which is greater between $$\sqrt[8]{8!}$$ and $$\sqrt[9]{9!}$$?

I want to know if my proof is correct...

\begin{align} \sqrt[8]{8!} &< \sqrt[9]{9!} \\ (8!)^{(1/8)} &< (9!)^{(1/9)} \\ (8!)^{(1/8)} - (9!)^{(1/9)} &< 0 \\ (8!)^{(9/72)} - (9!)^{8/72} &< 0 \\ (9!)^{8/72} \left(\left( \frac{8!}{9!} \right)^{(1/72)} - 1\right) &< 0 \\ \left(\frac{8!}{9!}\right)^{(1/72)} - 1 &< 0 \\ \left(\frac{8!}{9!}\right)^{(1/72)} &< 1 \\ \left(\left(\frac{8!}{9!}\right)^{(1/72)}\right)^{72} &< 1^{72} \\ \frac{8!}{9!} < 1 \\ \frac{1}{9} < 1 \\ \end{align}

if it is not correct how it would be?

Michael Lugo
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Sonia f
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  • Some good general strategies for determining $a ≷ b$ are calculating the sign of $a - b$, or if you know they have the same sign, whether $\frac{a}{b}$ is greater than or less than 1.. – Davislor May 01 '18 at 22:46
  • Comments nuked. A civil tone is required on this site. If you wish to continue this discussion, please do so in a group chat. – Alexander Gruber May 03 '18 at 15:50
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    $\sqrt[n]{n!}$ is the geometric average of the integers from $1$ to $n$. This is obviously a growing function. –  May 03 '18 at 19:59
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    @YvesDaoust Your comment is the best answer here. – Andrey Portnoy May 25 '18 at 03:14

10 Answers10

64

$$(\sqrt[8]{8!})^ {72}= (8!)^9 = (8!) (8!)^8 $$

$$(\sqrt[9]{9!})^ {72} = (9!)^8 = (9\times 8!)^8 = 9^8 (8!)^8$$

The second one, wins hands down.

Mohammad Riazi-Kermani
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    This relies on the fact that $$(9^8 > 8!)$$. Which is true, but perhaps not obvious? – Ross Presser May 02 '18 at 03:34
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    @RossPresser Could you explain what you mean by not obvious? – Alex D May 02 '18 at 03:54
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    Both products have eight factors, and all of the factors in one product are larger than all the factors in the other, clearly the larger product has the larger factors. – Nij May 02 '18 at 04:21
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    @RossPresser, One side is $99999999$, the other is $87654321$. The first one of these is clearly far larger – Rhys Hughes May 02 '18 at 10:38
23

You are implicitly writing that $$\frac{(8!)^{(9/72)}}{(9!)^{(8/72)}} = \left( \frac{8!}{9!} \right)^{(1/72)}$$ which is wrong.

Arnaud Mortier
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16

Another way to see this is to convert both sides into two "averages".

  • "Uniform distribution on $\{\log1,\dots,\log8\}$": $$\frac{\log1 + \dots + \log8}{8}$$
  • "Uniform distribution on $\{\log1,\dots,\log9\}$": $$\frac{\log1 + \dots + \log9}{9}$$

Intuitively, the later has a greater "expectation", so the result follows. If you're not satisfied with this probabilistic interpretation, break the later into a sum of two terms and regroup them as

\begin{align} & \frac{\log1 + \dots + \log8}{8} < \frac{\log1 + \dots + \log8}{9} + \frac{\log9}{9} \\ &\iff (\log1 + \dots + \log8) \left(\frac18-\frac19\right) < \frac{\log9}{9} \\ &\iff \frac{\log1 + \dots + \log8}{72} < \frac{\log9}{9} \\ &\iff \log1 + \dots + \log8 < 8 \log9 \end{align}

The last inequality is true since $\log$ is strictly increasing.

GNUSupporter 8964民主女神 地下教會
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  • Same idea as mine, so +1 :) – Oldboy May 01 '18 at 19:16
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    You can also say that $\sqrt[8]{8!}$ is the ___geometric___ mean of the numbers ${1,2,3,\ldots,8}$ while $\sqrt[9]{9!}$ is the geometric mean of ${1,2,3,\ldots,9}$. And you expect the this mean to increase when you throw in one more number which is greater than all the old ones. – Jeppe Stig Nielsen May 01 '18 at 22:09
16

We have obviously $8!<9^8$. Hence, it follows that $(9!)^8=(8!\cdot9)^8=(8!)^8\cdot 9^8>(8!)^8\cdot(8!)=(8!)^9$. This implies that $(9!)^{1/9}>(8!)^{1/8}$.

Gerhard S.
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  • Upvoted this as the best example of starting from a known fact and deriving the theorem, rather than assuming what we are trying to prove, which was the first and most important mistake the OP made. – Davislor May 01 '18 at 22:41
14

This step doesn't look right to me: \begin{gather} (8!)^{(9/72)} - (9!)^{8/72} < 0 \\[6px] (9!)^{8/72} \left(\left( \frac{8!}{9!} \right)^{(1/72)} - 1\right) < 0 \end{gather} When you divide $(8!)^{9/72} $ by $(9!)^{8/72}$, you should get $$ \frac{(8!)^{9/72}}{(9!)^{8/72}} = \frac{(8!)^{9/72}}{(9\cdot8!)^{8/72}} = \frac{(8!)^{1/72}}{(9)^{8/72}} = \frac{(8!)^{1/72}}{(9)^{1/9}} = \left( \frac{(8!)^{(1/8)}}{9} \right)^{1/9} $$

Also, note that your proof is basically of this form: \begin{gather} a < b \\[6px] a - b < 0 \\[6px] \left(\frac{a}{b} -1\right) < 0\\[6px] \frac{a}{b} < 1 \end{gather} You can skip several steps and just do \begin{gather} a < b \\[6px] \frac{a}{b} < 1 \end{gather}

egreg
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Acccumulation
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11

$$(8!)^9=(8!)^8\cdot 8! < (8!)^8\cdot 9^8= (9!)^8 $$

user
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5

I would compare it by using logarithms:

$$\ln a = \frac{1}{8}(\ln1 + ...+\ln8)$$

$$\ln b = \frac{1}{9}(\ln1 + ...+\ln8+\ln9)=\frac{8}{9} \ln a + \frac{1}{9}\ln9$$

$$\ln b-\ln a=\frac{1}{9}(\ln9-\ln a)=\frac{1}{9} \ln\frac{9}{a}$$

$a$ is obviously less than 8 so:

$$\frac{9}{a} \gt 1$$

$$\ln\frac{9}{a}\gt0$$

$$\ln b - \ln a \gt 0$$

$$b \gt a$$

Oldboy
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5

Instead of comparing the diference, comparing the ratio should be easy in such cases:

$$\left(\frac{8!^{1/8}}{9!^{1/9}}\right)^{8}=\frac{8!}{9!^{8/9}}=\frac{8!}{9!}(9!)^{1/9}=\left(\frac{9!}{9^9}\right)^{1/9}<1$$

this quantity is inferior than $1$ because $9!<9^9$

if $a$ is strictly positive real and $n$ is positive rational and also $a^n<1$ then $a<1$

which says:

$$\frac{8!^{1/8}}{9!^{1/9}}<1$$ QED.

TheSimpliFire
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Abr001am
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3

For any growing sequence, the arithmetic mean of the first terms is a growing function.$^*$

By taking the logarithm, this extends to the geometric mean.

$$^*t_1,t_2,\cdots t_n<t_{n+1}\implies s_n<nt_{n+1}\implies\frac{s_n}n<\frac{s_n+t_{n+1}}{n+1}=\frac{s_{n+1}}{n+1}.$$

0

You need only note that $9^8>8!,$ then we'd be done. And $9^8>8!$ since all the eight factors on the left exceed all the eight factors on the right.

Then multiplying both sides by $(8!)^8$ gives $(8!)^89^8>(8!)^88!,$ or $$(9!)^8>(8!)^8,$$ and by taking $1/72$nd powers, we obtain the result $$(9!)^{1/9}>(8!)^{1/8}.$$

Allawonder
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