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I have proved that $17^{69}<10^{85}$ by using the following inequalities: $x<\exp\left(\dfrac{2(x-1)}{x+1}\right)$ for all $x\in \left]-1,1\right[$ and $x<{\mathrm e}^{x-1}$ for all $x\in \left] 1,+\infty \right[$, but I am looking for a simpler non-calculator proof.

My proof is the following: \begin{align*}\frac{17^{69}}{10^{85}}&=\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\left(\frac{5^3}{2^7}\right)^2\cdot\frac{5}{4}<\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\frac{5}{4}=\left(\frac{4913}{5000}\right)^{23}\cdot \frac{5}{4}\\&<\left(\exp\left(\frac{2\left(\frac{4913}{5000}-1\right)}{\frac{4913}{5000}+1}\right)\right)^{23}\cdot\exp\left(\frac{5}{4}-1\right)\\&=\exp\left(-\frac{174}{431}\right)\cdot\exp\left(\frac{1}{4}\right)=\exp\left(-\frac{265}{1724}\right)<1.\end{align*}

Could anyone find a simpler non-calculator proof without using big numbers?

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. $9^{89}$ is between them $\endgroup$ Apr 13, 2020 at 18:08
  • $\begingroup$ @Tanner, how can I use your tip in order to find a simpler proof? $\endgroup$
    – Angelo
    Apr 13, 2020 at 18:14
  • $\begingroup$ I've posted a new answer in which all numbers have no more than six digits (ending zeros excluded). Hope you like it. $\endgroup$
    – Ѕааԁ
    Apr 17, 2020 at 4:26
  • $\begingroup$ @Saad, thank you for your work, your proof is refined and beautiful $\endgroup$
    – Angelo
    Apr 17, 2020 at 7:10
  • $\begingroup$ It might be interesting to show $17^{243}<10^{299}$. $\endgroup$
    – Apass.Jack
    Apr 17, 2020 at 10:39

7 Answers 7

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Since $17^3 = 4913 < 492 × 10$, then$$ 17^6 < 492^2 × 10^2 = 242064 × 10^2 < 243000 × 10^2 = 3^5 × 10^5. $$ Now it suffices to prove that $(3^5 × 10^5)^{23} < (10^{85})^2$, or $3^{23} < 10^{11}$. Note that $3^9 = 27^3 = 19683 < 2 × 10^4$ and $3^5 = 243 < 25 × 10$, thus$$ 3^{23} = (3^9)^2 × 3^5 < (2 × 10^4)^2 × (25 × 10) = 10^{11}. $$

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  • $\begingroup$ Saad, thank you very much for your work, your proof is very refined and beautiful. $\endgroup$
    – Angelo
    Apr 17, 2020 at 7:01
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$$17 ^{ 13} = ((17^3)^2)^2 \cdot 17= (4913 \cdot 4913)^2\cdot 17< (242\cdot10^5)^2\cdot 17\\< 588\cdot10^{12}\cdot 17= 9996\cdot10^{12}<10^{16} $$

Hence, $$17 ^{ 69} = \left(17^{13}\right)^{\frac{69}{13}}<10^{16\cdot(5+\frac{4}{13})}= 10^{80+\frac{64}{13}} < 10^{85}.$$


Here are some minor tricks to make the computation in the first inequalities even easier.

$$4913 \cdot4913=(4910+3)(4920-7) < 4910\cdot4920$$ $$491 \cdot492=(500-9)(500-8)= 241572$$ $$242\cdot242=(240+2)(245-3)<240\cdot245=12\cdot490= 58800$$


Similar formulas:

$$\begin{array}{cl} \left.17^{4}\right/10^{5} &=0.83521\\ \left.17^{13}\right/10^{16} &=0.990458\!\cdots\\ \left.17^{69}\right/10^{85} &=0.796115\!\cdots\\ \left.17^{243}\right/10^{299} &=0.997902\!\cdots\\ \left.17^{1202}\right/10^{1479} &=0.999087\!\cdots\\ \left.17^{5524}\right/10^{6797} &=0.999636\!\cdots\\ \left.17^{7685}\right/10^{9456} &=0.999910\!\cdots\\ \vdots\\ \left.17^{302464054}\right/10^{372166569} &=0.99999999988\cdots\\ \end{array}$$

The above data is generated with, among other tools, the continued fraction of $$ \log_{17}10= 0.81271150929195899925562198972659\cdots,$$ which is, $$ [0; 1, 4, 2, 1, 17, 1, 3, 1, 1, 3, 3, 26, 1, 1, 2, 3, 2, 11, 64, 2, 3, 1, 13, 1, 8, 1, 4, \cdots].$$

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  • $\begingroup$ In fact, $17 ^{ 13}\lt10^{16}$ implies $17^{69}=17^{13\cdot5+4}\lt10^{16\cdot5}\cdot 17^4<10^{80}\cdot290^2=0.841\cdot10^{85}$. $\endgroup$
    – Apass.Jack
    Apr 20, 2020 at 20:05
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I will also say some words on this. The general procedure to easily show such inequalities without computer is to... use the computer to get "close powers" of the bases, here $17$ and $10$, then use the coarsest that still does the job, and of course not mention that this was done so! In our case, i am forgetting in this second to not mention that the "first closest powers" of $17$ and $10$ come from the convergents of the continued fraction of $a=\displaystyle\log_{10} 17=\frac {\log 17}{\log 10}$, so let us show them...

sage: c = continued_fraction( log(17)/log(10) )
sage: cvgts = [ c.convergent(k) for k in [1..7] ]
sage: cvgts
[5/4, 11/9, 16/13, 283/230, 299/243, 1180/959, 1479/1202]

So we expect that

  • $17^4=83521$ is "close" to $10^5$, yes, this is the case and $17^4\color{blue}{<}10^5$,
  • $17^9=118587876497$ is "close" to $10^{11}$, yes, and $17^{9}\color{red}{>}10^{11}$,
  • $17^{13}=9904578032905937$ is "close" to $10^{16}$, yes, and $17^{13}\color{blue}{<}10^{16}$,
  • $17^{230}=\dots$ is "close" to $10^{283}$, yes, and $17^{230}\color{red}{>}10^{283}$, and so on.

Now we "completely forget" about the above, and write some inequalities. I will use the knowledge of the "steps" $17^4$, and $17^{13}$ below (of course, without mentioning this)... It will be a "hard job" (more than four lines) to establish $17^{13}\le 10^{16}$, but then we can relax and easily show the needed inequality. The most complicated operation will be to compute $836^2$ below. So let us start now!


$$ \begin{aligned} 17^4 &= 289^2 = (290-1)^2 = 84100-580+1=83521\\ &< 83600\ , \\[3mm] % 17^{13} &=17\cdot (17^4)^3\\ &< 17\cdot 836^3\cdot 10^6 \\ &= 17\cdot 836^2\cdot 836\cdot 10^6 \\ &= 17\cdot 698896\cdot 836\cdot 10^6 \\ &< 17\cdot 700000\cdot 840\cdot 10^6 \\ &= 17\cdot 7\cdot 84\cdot 10^{12} \\ &= 9996\cdot 10 ^{12}\\ &< 10^{16}\ , \\[3mm] % 17^{69} &= (17^{13})^5\cdot 17^4\\ &<(10^{16})^5\cdot 83600\\ &< 10^{80}\cdot 10^5\\ &= 10^{85}\ . \end{aligned} $$

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Claim 1: $2.3<\ln 10.$

Claim 2: $\ln 1.7<8/15$

Both these claims can be proven easily via Taylor series, etc.

Now, using the above inequalities, we have $1.7^{69}<e^{69\cdot \frac{8}{15}}<10^{16},$ or, multiplying $10^{69}$ on both sides, $17^{69}<10^{85}.$

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    $\begingroup$ Since it is easy, could you show how to prove $2.3\lt \ln10$? $\endgroup$
    – Apass.Jack
    Apr 17, 2020 at 10:43
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    $\begingroup$ Decompose as: $\ln(10)=2\ln(1+1/2)-2\ln(1-1/2)+\ln(1+1/9),$ and use taylor series. $\endgroup$
    – Kenta S
    Apr 17, 2020 at 11:01
  • $\begingroup$ many terms of the first two series will cancel out, making calculations simpler. $\endgroup$
    – Kenta S
    Apr 17, 2020 at 11:01
  • $\begingroup$ How about $\ln 1.7 < 8/15$? Why not update your answer with detailed explanations? $\endgroup$
    – Apass.Jack
    Apr 22, 2020 at 11:44
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You already got $$\frac{17^{69}}{10^{85}}\lt\left(\frac{4913}{5000}\right)^{23}\cdot \frac{5}{4}$$ from which we have $$\begin{align}\frac{17^{69}}{10^{85}}&\lt\left(\frac{4913}{5000}\right)^{23}\cdot \frac{5}{4} \\\\&\lt \left(\frac{4950}{5000}\right)^{23}\cdot \frac{5}{4} \\\\&=\left(\frac{99}{100}\right)^{23}\cdot \frac{5}{4} \\\\&=\bigg(1-\frac{1}{100}\bigg)^{23}\cdot \frac{5}{4} \\\\&=\frac 54\sum_{k=0}^{23}\underbrace{\binom{23}{k}\bigg(-\frac{1}{100}\bigg)^k}_{f(k)}\ \ \ \ \ \text{(binomial theorem)} \\\\&=\frac 54(f(0)+f(1)+\cdots +f(22)+\underbrace{f(23)}_{\lt 0}) \\\\&\lt \frac 54(f(0)+f(1)+\cdots +f(22)) \\\\&=\frac 54\bigg(f(0)+f(1)+f(2)+\sum_{k=1}^{10}(f(2k+1)+f(2k+2))\bigg) \\\\&=\frac 54\bigg(f(0)+f(1)+f(2) \\&\qquad+\sum_{k=1}^{10}\bigg(\binom{23}{2k+1}\bigg(-\frac{1}{100}\bigg)^{2k+1}+\binom{23}{2k+2}\bigg(-\frac{1}{100}\bigg)^{2k+2}\bigg)\bigg) \\\\&=\frac 54\bigg(f(0)+f(1)+f(2) \\&\qquad+\sum_{k=1}^{10}\bigg(\frac{-23!(\frac{1}{100})^{2k+1}}{(2k+1)!(23-2k-1)!}+\frac{23!(\frac{1}{100})^{2k+2}}{(2k+2)!(23-2k-2)!}\bigg)\bigg) \\\\&=\frac 54\bigg(f(0)+f(1)+f(2) \\&\qquad+\sum_{k=1}^{10}\frac{23!(\frac{1}{100})^{2k+2}}{(2k+2)!(22-2k)!}\bigg(-100(2k+2)+(22-2k)\bigg)\bigg) \\\\&=\frac 54\bigg(f(0)+f(1)+f(2)+\underbrace{\sum_{k=1}^{10}\frac{23!(\frac{1}{100})^{2k+2}(-202k-178)}{(2k+2)!(22-2k)!}}_{\lt 0}\bigg) \\\\&\lt\frac 54\bigg(f(0)+f(1)+f(2)\bigg) \\\\&=\frac 54\bigg(1-\frac{23}{100}+\frac{253}{10000}\bigg) \\\\&=\frac 54\cdot\frac{10000-2300+253}{10000} \\\\&=\frac{39765}{40000} \\\\&\lt 1\qquad\blacksquare\end{align}$$

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Firstly, can be obtained the next numerical inequalities.

  • $$5\cdot17^3 = 24565 < 24576 = 6\cdot 16^3,$$ $$\mathbf{\left(\dfrac{17}{16}\right)^3 <\dfrac65}.\tag1$$
  • $$3^5\cdot2^6 =15552 < 15625 = 5^6,$$ $$\mathbf{3^5<\left(\dfrac52\right)^6}.\tag2$$
  • $$(1.024)^4 < 1.0486^2 < 1 + 0.0972 + 0.0025 < 1.1,$$ $$1.1^7 = 1 + 0.7 + 0.21 + 0.035 + 0.0035 + 0.00021 + 0.000007 + 0.0000001 < 2,$$ $$2^{280} = (1.024)^{28}\cdot10^{84} < 2\cdot10^{84},$$ $$\mathbf{2^{279}<10^{84}}.\tag3$$

Then, taking in account $(1)-(3),$ one can get: $$\left(\dfrac{17}{16}\right)^{69} < \left(\dfrac65\right)^{23} = 27\cdot(3^5)^4\cdot\left(\dfrac25\right)^{23} < 27\left(\dfrac52\right)^{24}\left(\dfrac25\right)^{23} =\dfrac{135}2,$$ $$17^{69} < \dfrac{135}{2}\,\dfrac {2\cdot2^{279}}{16}<\dfrac{135}{16}\cdot10^{84},$$

$$\color{brown}{\mathbf{17^{69}< \dfrac{135}{16}\cdot10^{84}}},$$ $$\color{brown}{\mathbf{17^{69}<10^{85}.}}$$ Thus, there is a simple proof of a more rigorous inequality.

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Proof that $17^{69} < 10^{85}$ without using calculator, there are many ways this can be done you'll need to understand number theory, so I'll highlight a few examples $$17^{69} < 10^{85}$$ $$17^{69} < 10^{17×5}$$ $$17^{1/17} < 10^{5/69}$$ Remember $$\lim_{n→∞} \sqrt[n]{n} = 1$$ Meaning the value of 17^{1/17} is between $1$ to $1.4$ $$17^{1/17} < 10^{5/69}$$ Using long division $\frac{5}{69} = 0.07246$ $$17^{1/17} < 10^{x}$$ But $10^x$ can also be between $1$ to $1.4$ if $x$ is from $0$ to $≈0.1$ So the the value $\frac{5}{69}$ is too small, therefore $$17^{69} \mathbb{<} 10^{85}$$

Another way $$17^{69} < 10^{85}$$ $$17^{69} < 10^{17×5}$$ $$17^{69/17} < 10^{5}$$ Using long division $\frac{69}{17} = 4.0588$ $$17^{4.0588} < 10^{5}$$ If we assume that $17^{4.0588} = 20^x$, since the base became bigger, the power would be smaller for them to be equal

$17 → 20$$, $4.0588 → ≈3$

$$17^{4.0588} ≈ 20^3$$ $$≈20^3 < 10^5$$ But we know clearly that $8000 < 10000$ Therefore $$17^{69} \mathbb{<} 10^{85}$$

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    $\begingroup$ These are not sufficient to constitute proofs - the inequality is much more fine-grained than your estimates. For example, $20^x$ does not dominate $17^{69/17}$ until roughly $x = 3.84$, while $20^x$ dominates $10^5$ starting roughly around $x=3.85$, so the margin of error in estimation is small. The situation for $17^{1/17}$ vs $10^{5/69}$ is even worse. $\endgroup$
    – akkapi
    Apr 21, 2020 at 15:03
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    $\begingroup$ The proof is wrong. $\sqrt2\approx1.414,\ \sqrt[3]3\approx1.442.$ Why $\sqrt[17]{17}<1.4?$ $\endgroup$ Apr 21, 2020 at 16:02
  • $\begingroup$ The proof isn't wrong, I took rough estimate to create an estimation within range, I know that $\sqrt[17]{17}$ is lesser than 1.4, that's why I put it in a range of $1$ to $1.4$, I want to write an answer of a very simple understanding, if not I could have made my estimation more precised, with bigger proof $\endgroup$ Apr 21, 2020 at 18:08
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    $\begingroup$ 1) How does $17^{1/17} < 10^{5/69}$ follow from the statements preceding this: even if we grant that $17^{1/17}$ is in the interval $[1,1.4]$, as you say, $10^x$ is in the interval $[1,1.4]$ for $x \in [0,1/10]$, but so is $5/69 \in [0,1/10]$. How can you compare the two inequalities just based on this? When you "expand" your proof, you must either have larger computations or expose some numerical relationships, none of which this does. 2) For the second part: $17^{69/17} > 20^3$ so knowing $20^3 < 10^5$ doesn't buy you anything. Your estimates are not strong enough for this problem. $\endgroup$
    – akkapi
    Apr 21, 2020 at 18:21

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