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This question already has an answer here:

Which is greater, $98^{99} $ or $ 99^{98}$?

What is the easiest method to do this which can be explained to someone in junior school i.e. without using log tables.

I don't think there is an elementary way to do this. The best I could find was on Quora, in an answer by Michal Forišek on a similar question here, which is to consider $\frac{98^{99}}{99^{98}}=98.(\frac{98}{99})^{98}=98.(1-\frac{1}{99})^{98} \approx\ \frac{98}{e} >1$ and hence $98^{99} > 99^{98}$.
But the approx sign step does use definition of $e$ in terms of limits and thus cannot be considered elementary. Any other way?


Edit- I was hoping something that does not involve calculus, that is why I tagged it in number theory, but as it seems it is almost impossible to avoid calculus when exponentials are involved.

All the answers are fine, and can be explained to students in classes above seventh or eigth. My aim for this qustion was to check with you all, if I have missed some elementary trick or not,I guess I did not. I was looking for the easiest solution someone can come up with. Thanks!

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marked as duplicate by Najib Idrissi, graydad, muaddib, Tim Raczkowski, user223391 Aug 30 '15 at 0:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ 'Junior school' does not, to my knowledge, have a consistent meaning across countries. It would be better to specify which techniques are and are not permissible. (From the second-to-last sentence, perhaps you mean you want to avoid calculus altogether?) $\endgroup$ – Travis Willse Aug 23 '15 at 15:36
  • $\begingroup$ @Travis If we use the binomial formula on $(1+ \frac{1}{n})^n $ we can obtain $(1+\frac{1}{n})^n \le 1+ \sum_{i=0}^{n} 2^{-i} = 1+ \frac{ 1 - 2^{-(i+1)}}{1- 2^{-1}}< 1+ 2=3 $ $\endgroup$ – Marko Karbevski Aug 23 '15 at 15:56
  • $\begingroup$ You could introduce the $(1+\frac{1}{n})^{n}$ without giving much content about the Euler ($e$) itself. My junior high school in Indonesia (not far from your country, right?) already has taught the Euler and logarithm. I think the problem about explaining the answer is how do you explain and how your student ready to accept the material. Our efforts are useless if you can't explain properly, even there's simplest way to answer it. $\endgroup$ – làntèrn Aug 23 '15 at 15:58

13 Answers 13

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Using Bernoulli's inequality:
$$ (1+x)^n\ge 1+nx \qquad \forall \ \text{$x\ge -1$ and $n\in\mathbb N_0$ }$$ we have $$ \left(1-\frac1{99}\right)^{49}\ge 1-\frac{49}{99}=\frac{50}{99}>\frac{50}{100}=\frac12$$ Therefore $$ \frac{98^{99}}{99^{98}}=98\cdot \left(1-\frac1{99}\right)^{49}\cdot \left(1-\frac1{99}\right)^{49}>98\cdot \frac12\cdot\frac12>1.$$

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The inequality $99^{98} < 98^{99}$ is equivalent to $$\bigl( 1 + \frac{1}{98} \bigr)^{98} < 98 $$ I'll prove by induction that $$\bigl( 1 + \frac{1}{n} \bigr)^n < n \quad\text{if $n \ge 3$}\quad\text{(it's false if $n=1$ or $2$).} $$ The basis step $n=3$ is $\frac{64}{27} < 3$.

So let's assume that $$\bigl( 1 + \frac{1}{n} \bigr)^n < n \quad\text{and}\quad \bigl( 1 + \frac{1}{n+1} \bigr)^{n+1} \ge n+1 $$ and derive a contradiction. These two inequalities can be rewritten $$\left( \frac{n+1}{n} \right)^n < n \quad\text{and}\quad \bigl(\frac{n+1}{n+2}\bigr)^{n+1} \le \frac{1}{n+1} $$ Multiplying them we get $$\left( \frac{n^2 + 2n + 1}{n^2 + 2n} \right)^{n} \cdot \frac{n+1}{n+2} < \frac{n}{n+1} $$ $$\left( \frac{n^2 + 2n + 1}{n^2 + 2n} \right)^{n} \cdot \frac{n^2+2n+1}{n^2+2n} < 1 $$ $$\left( \frac{n^2 + 2n + 1}{n^2 + 2n} \right)^{n+1} < 1 $$ which is absurd.

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The only "elementary" way I can think of is to write $99^{98} = (98 + 1)^{98}$ and then expand using the binomial expansion formula, and then show you get a sum of $99$ terms where each term is less than or equal to $98^{98}$, and the sum of the last two terms $98 + 1$ is strictly less than $98^{98}$. Then your sum is strictly less than $98 \cdot 98^{98} = 98^{99}$.

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    $\begingroup$ Expanding $(a+b)^n$ gives $n+1$ terms. So you need to absorb the last $1^{98}$ into one of the other terms. $\endgroup$ – Daniel Fischer Aug 23 '15 at 15:58
  • $\begingroup$ Binomial expansion is derived from "Probability" material and I'm sure junior student has already known some questions involving probability. $\endgroup$ – làntèrn Aug 23 '15 at 16:01
  • $\begingroup$ @DanielFischer Yes you're right, thank you for catching that. I'll update with the correction. $\endgroup$ – user2566092 Aug 23 '15 at 16:03
  • $\begingroup$ @lantern I hoped as much, I honestly don't know if the binomial expansion is covered in sufficiently advanced "junior" curricula on basic probability or algebra, or at least algebra 2, but if it is I think this is basically one of the only "elementary" ways to get the answer without some sort of calculus. But one has to realize that the terms in the binomial expansion are all essentially $98^{98}$ divided by factorials (actually less that that usually, but anyways...) $\endgroup$ – user2566092 Aug 23 '15 at 16:19
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Let $$\displaystyle f(x) = x^{\frac{1}{x}}\;,$$ where $x>0$

Now $$\displaystyle f'(x) = x^{\frac{1}{x}}\cdot \left[\frac{1-\ln (x)}{x^2}\right]$$

So here $f'(x)>0$ for $x<e$ and $f'(x)<0$ for $x>e\approx 2.714$

So function $f(x)$ is an Strictly Decreasing function for $x>e\approx 2.714$

So $$f(98)>f(99)\Rightarrow 98^{\frac{1}{98}}>99^{\frac{1}{99}}\Rightarrow 98^{99}>99^{98}$$

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    $\begingroup$ Junior school. $\endgroup$ – Vincenzo Oliva Aug 23 '15 at 15:31
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    $\begingroup$ @VincenzoOliva Junior school does not, to my knowledge, have a universal meaning across countries. $\endgroup$ – Travis Willse Aug 23 '15 at 15:34
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    $\begingroup$ @Travis: Maybe, but the OP explicitly stated the specific meaning we're interested in. $\endgroup$ – Vincenzo Oliva Aug 23 '15 at 15:36
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    $\begingroup$ OP specified that they didn't want a solution that uses log tables, and made a vague comment about limits (which one may or may not construe as excluding calculus altogether), but this does not nearly "state a specific meaning". $\endgroup$ – Travis Willse Aug 23 '15 at 15:40
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    $\begingroup$ @Travis: As far as I'm concerned, it had been made clear that calculus and log tables were not taken in consideration. Anyway, OP's edit clarifies it all. $\endgroup$ – Vincenzo Oliva Aug 23 '15 at 15:45
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$$\frac{98^{99}}{99^{98}}=\frac{98}{\left(1+\frac{1}{98}\right)^{98}}>\frac{98}{97}>1$$

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    $\begingroup$ You still need to justify the first greater-than sign. $\endgroup$ – celtschk Aug 23 '15 at 15:34
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    $\begingroup$ @celtschk Something much stronger also holds. Namely $(1+ \frac{1}{n})^n<3$ for $n \in \mathbb{N}$ $\endgroup$ – Marko Karbevski Aug 23 '15 at 15:45
  • $\begingroup$ $e<3$ is the limit as $n\rightarrow\infty$ $\endgroup$ – mathreadler Aug 23 '15 at 15:52
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    $\begingroup$ All you really need is $(1+{1\over n})^n\le n-1$ for $n$ greater than something. There's an easy induction proof for that: $$\left(1+{1\over n+1} \right)^{n+1}=\left(n+2\over n+1 \right)\left(1+{1\over n+1} \right)^n\lt\left(n+2\over n+1 \right)\left(1+{1\over n} \right)^n\lt\left(n+2\over n+1 \right)(n-1)\lt n$$ since $n^2+n-2\lt n^2+n$. $\endgroup$ – Barry Cipra Aug 23 '15 at 15:53
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    $\begingroup$ @celtschk If we use the binomial formula on $(1+ \frac{1}{n})^n $ we can obtain $(1+\frac{1}{n})^n \le 1+ \sum_{i=0}^{n} 2^{-i} = 1+ \frac{ 1 - 2^{-(i+1)}}{1- 2^{-1}}< 1+ 2=3 $ $\endgroup$ – Marko Karbevski Aug 23 '15 at 15:57
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We could try and generalize a little bit and ask ourselves if $n ^ {n+1}$ is always bigger than $(n+1) ^ n$ where n is a natural number.

This turns into verifying if the inequality $n^{n+1} > (n+1)^n$ holds.

The left hand side can be rewritten as $n \cdot n^n$ and the inequality becomes $n \cdot n^n > (n+1) ^ n$

Being $n^n$ always positive, the inequality holds dividing both sides by it so we have $n > (\frac{n+1}{n})^n$ which can be written also as $n > (1+\frac{1}{n})^n$.

The term $(1+\frac{1}{n})^n$ tends to $e$ when n tends to infinity, therefore if $n \geq 3$, the latter inequality holds as the right hand side cannot be more than 2.71...

It does not hold for n < 3, in fact 1^2 (=1) < 2^1 (=2) and 2^3 (=8) < 3^2 (=9). For 3 we have 3^4 (=81) > 4^3 (=64) and so on.

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  • $\begingroup$ This is by far the best answer in terms of how elementary it is. $\endgroup$ – theREALyumdub Aug 26 '15 at 15:38
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Hint We can write $$\frac{99^{98}}{98^{99}} = \frac{1}{98}\left(1 + \frac{1}{98}\right)^{98}.$$

Now, one can show via an easy induction (if you find yourself stuck, see this one-line proof) that $\left(1 + \frac{1}{n}\right)^n < n$ for $n \geq 3$, so the above quantity is less than $$\frac{1}{98} (98) = 1 ,$$ and rearranging gives $$99^{98} < 98^{99}.$$

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I hope at least we can use the Binomial Theorem with integer exponents. I will also assume knowledge of the usual $\binom mn$ notation for the coefficients, but that is not really necessary; you can rewrite the argument to do without that notation.

Observe that for $m \geq 1$, $$\binom{98}{m + 1} = \frac{98 \cdot 97 \cdot 96 \cdot \cdots \cdot (98 - m)} {1 \cdot 2 \cdot 3 \cdot \cdots \cdot (m + 1)} = \frac{98 - m}{m+1} \binom{98}{m} \leq \frac{98}{2} \binom{98}{m}. $$ from which we can conclude that for $m \geq 2$, $$\binom{98}{m} \leq \left(\frac{98}{2}\right)^{m-1} \binom{98}{1} \leq \frac{1}{2^{m-1}}\cdot 98^m \leq \frac12\cdot 98^m.$$

Therefore

\begin{align} 99^{98} &= (98+1)^{98} \\ &= 98^{98} + \binom{98}{1} 98^{97} + \underbrace{ \binom{98}{2} 98^{96} + \binom{98}{3} 98^{95} + \cdots + \binom{98}{97} 98 + 1}_{97 \text{ terms}} \\ & \leq 98^{98} + 98^{98} + \underbrace{ \frac12\cdot 98^{98} + \frac12\cdot 98^{98} + \cdots + \frac12\cdot 98^{98} + \frac12\cdot 98^{98}}_{97 \text{ terms}} \\ & \leq \left(2 + 97 \cdot \frac12\right)98^{98} \\ & < 98 \cdot 98^{98} = 98^{99}. \end{align}

So $98^{99}$ is greater.

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    $\begingroup$ This is essentially my answer, just with a different way of manipulating the binomial expansion to show each term is less than or equal to $98^{98}$ with some terms strictly less than. I didn't have the patience to write down my argument explicitly in terms of formulas, and I didn't explain explicitly why each term in the binomial expansion is less than or equal to $98^{98}$ (I just figured that would be fairly obvious if someone knows and understands the formula for binomial expansion). So, +1, at least for going the extra mile. =) $\endgroup$ – user2566092 Aug 23 '15 at 16:36
  • $\begingroup$ @user2566092 I agree, if one follows the advice in your answer it comes out to the same thing, except for some unimportant details (mainly the choices of what terms to compare to what when adding them up). I might not have written this up if your answer had been posted already when I started writing; in any case I hope your answer stays higher on the page, since it's better to have people read the nice summary before they see the gory details. $\endgroup$ – David K Aug 23 '15 at 23:43
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$98^{99}>99^{98}\Leftrightarrow 98^{\frac {1}{98}}>99^{\frac {1}{99}}$ and $(n^{\frac {1}{n}})$ is strictly decreasing.

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  • $\begingroup$ How do you show that $n^{1/n}$ is decreasing without using calculus? $\endgroup$ – Barry Cipra Aug 23 '15 at 16:07
  • $\begingroup$ @BarryCipra $n^{n+1}>(n+1)^n \Leftrightarrow n>(1+1/n)^n$ and this is the induction step... $\endgroup$ – Haha Aug 23 '15 at 16:10
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you should know that the maxima of the function $$f(x)=x^\frac 1x$$ attains maxima at $x=e^\frac 1e$

So, as 98 is closer to $e$, hence we have $98^\frac 1{98}\gt 99^\frac 1{99}$

So, $98^{99}\gt99^{98}$

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    $\begingroup$ Junior school. $\endgroup$ – Vincenzo Oliva Aug 23 '15 at 15:32
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This is the most elementary proof I can muster, and assumes students are familiar with inequalities and the interest formula.

Assume $ 98^{99} > 99^{98} $ .

The following algebra holds:

$$ 98^{99} > 99^{98} $$ $$ 98 * 98^{98} > 99^{98} $$ $$ 98 > (\frac{99}{98})^{98} $$ $$ 98 > (1 + \frac{1}{98})^{98} $$

You could ask your students to compute $ (1 + \frac{1}{98}) ^{98} $ on their calculators, which will do so because the number is not excessively large. Or you could explain that this mysterious, 2.71 number actually appears with every $ (1 + \frac{1}{n})^{n} $, n greater than a certain value. Either way the assumed statement is proven true with the calculation.

In either case, the formula $ (1 + \frac{1}{n})^{n} $ is elementary since it computes basic interest over time. I learned this in grade school around the age of 16 I would say...

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I have generalized a bit more this result finding when $$n^{n+m} > (n+m)^n$$
Of course $n^{n+m} = n^n \cdot n^m$ and being $n^m > 0$ it is possible to divide both sides of the inequality to find that $$n^m > (\frac{n+m}{n})^n$$ that becomes $$n^m > (1+\frac{m}{n})^n$$

Now $lim_{n \to \infty} (1+\frac{m}{n})^n = e^m$ (https://en.wikipedia.org/wiki/List_of_limits) so eventually we have $$n^m > e^m$$ therefore for $m>1$ the same condition as before holds: from $n>=3$ the condition is verified.

For instance, suppose $n=3$ and $m=2$ : we can state that $3^{3+2} > (2+3)^3 \rightarrow 3^5 > 5^3 \rightarrow 243 > 125$.

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@ Bhaskar Vashishth: If you want a best answer as you comment to Lee Mosher take a graphic of $f(x)= x^{\frac{1}{x}}-1-\frac1{x}$and you can see that your inequality can be generalized to all real greater than $2.293$ for which this function is positive. Obviously, after to see that $(x+1)^x<x^{x+1}$ is easily equivalent to $x^{\frac{1}{x}}>1+\frac1{x}$. I feel all generalization of your inequality has been started from this graphic or some other equivalent.On the other hand, thinking a little about it seems to me that this should be something obvious for everybody.

enter image description here

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