# Prove that $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without Calculus.

Prove that $$2018^{2019}> 2019^{2018}$$ without induction, without Newton's binomial formula and without Calculus. This inequality is equivalent to $$2018^{1/2018}>2019^{1/2019}$$

One of my 'High school' student asked why the inequality is true. The whole class became interested in the problem.The demonstration that such inequality is true, using calculus, can be found here. But my students are not familiar with calculus.

I can also show by induction and Newton's binomial formula that $$n^{(n + 1)}> (n + 1)^n$$, to $$n> 3$$, but my students are not familiar with mathematical induction. Another limitation of my students is that they have not yet learned the Newton's binomial formula.

How to prove the inequality $$2018^{2019}> 2019^{2018}$$ without induction, without Newton's binomial formula and without calculus? That is how to prove this inequality for High school students without using Newton's binomial formula?

• Your students sure do know how to multiply. The result is big, but - given enough time and paper - the task is within their capabilities ;) – Hagen von Eitzen Feb 25 at 20:41
• I imagine somebody stumbled upon this and a calculator confirmed the claim---thus, convincing them. However, that calculator so taken for granted most likely uses an algorithm which was proven to work via calculus. It sounds like a good opportunity to tell them why they should continue in math. Most work in the world is accomplished by elbow grease or by making better tools. Some work is achieved by extreme cleverness. – Robert Wolfe Feb 25 at 22:39
• @Robertwolfe Actually, calculus is not necessary to prove that the inequality of interest holds. In fact, I posted two ways forward that rely on precalculus tools only. – Mark Viola Feb 26 at 1:16
• just curious, are logarthims already taught? May be you can intuitively argue $\frac{1}{x}$ decays faster than $\log (x)$ that the product becomes smaller as $x$ increases. For instance, $x = 10^6$ implies $\frac{1}{x} = 10^{-6}$ where $\log_{10}(10^6)$ is only 6. – dineshdileep Feb 26 at 2:30
• @MathOverview : I have added a late answer which uses only fractions and the GM-HM inequality. – trancelocation Feb 28 at 18:14

Note that $$2019^{2048}<2018^{2049}$$ implies that $$2019^{2018}2019^{30}=2019^{2048}<2018^{2049}<2018^{2019}2019^{30},$$ which implies that $$2019^{2018}<2018^{2019}$$. We look at $$2048$$ in the exponent because it is a power of $$2$$.

Claim: $$2019^{2048}<2018^{2049}$$.

Proof: $$2019^{2048}-2018^{2048}=$$ $$(2019^{1024}+2018^{1024})(2019^{512}+2018^{512})...(2019^{2}+2018^{2})(2019+2018)(2019-2018)$$ by repeatedly factoring differences of squares. Each term of the form $$2019^i+2018^i<2 \cdot 2019^i$$, so taking each of these inequalities into account, we get that $$2019^{2048}-2018^{2048}<2^{10}\cdot 2019^{2047}$$ since $$1+2+4+8+...+512+1024=2^{11}-1$$. Then we combine terms with like bases to get that $$2019^{2048}-1024\cdot 2019^{2047}=995\cdot 2019^{2047}<2018^{2048}.$$ We now multiply both sides by $$2018$$ to get $$2019^{2048}<995\cdot 2018\cdot 2019^{2047}<2018^{2049},$$ which is the desired result.

• Excelent idea! +1 – MathOverview Mar 5 at 20:36

I would try to motivate this by showing that $$f(x) = x^{1/x}$$ is a monotone function for reasonably small $$x$$.

Another approach is to note that $$\frac{2019^{2018}}{2018^{2018}} = \left(1 + \frac{1}{2018}\right)^{2018}$$ and so your inequality is equivalent to showing $$\left(1 + \frac{1}{2018}\right)^{2018} < 2018,$$ which does not sound very far-fetched, since LHS is close to $$e$$...

UPDATE

Please see saulspatz's answer for how to prove this last claim with a hand computation only.

• +1 In fact I can show that $2<\left(1 + \frac{1}{2018}\right)^{2018} <3$ and that $\left(1 + \frac{1}{2018}\right)^{2018}$ always approaches a fixed number (which by definition is $e=2,718...$). But I do not know any other way to show that $2<\left(1 + \frac{1}{2018}\right)^{2018} <3$ without induction. I need a simpler argument to convince my students. – MathOverview Feb 25 at 21:01
• I have given a proof that $(1 + 1/n)^n < 3$ for $n \ge 3$ in another answer. (To show it's at least 2 is trivial - take the first two terms of the binomial expansion.) – Michael Lugo Feb 25 at 22:06
• This answer is incomplete since you are stating something about the number $e$. But what is $e$ here? How does one know that $a_n=\left(1+\frac1n\right)^n<e$ when you have not yet defined $e$. And if you define it as the limit of $a_n$, how do you know that $a_n$ converges? – Mark Viola Feb 25 at 22:41
• @MarkViola I did not intend to finish the problem, just to give a direction. There are plenty of others here who demonstrated what you are asking for. – gt6989b Feb 25 at 22:49
• @gt6989b I am not asking for myself and in fact posted two ways forward that rely on elementary methodologies (i.e., precalculus) only. But your answer is simply incomplete in that it relies on circular logic. So, the issue isn't whether you finished; you did not. It is that your answer makes a claim the proof of which is at the very heart of the OP's question. – Mark Viola Feb 25 at 22:57

$$1+{1\over2018}<1+{1\over2000}=1.0005$$ $$\left(1+{1\over2018}\right)^{2018}<(1.0005)^{2048}$$

The right hand side can be evaluated by repeated squaring, and, since there is plenty of leeway, you can simplify the calculations by rounding up as you go. Since $$(1.0005)^{2048}<3,$$ this calculation should be well within you students' abilities.

• Do you mean $2018$ in the exponent? – mfl Feb 25 at 21:16
• @mfl No, I chose $2048$ because $2048=2^{11}$ and it makes the remaining calculations easier. $x^{2048}= \left(x^{1024}\right)^2$ and so on. – saulspatz Feb 25 at 21:20
• How did you find $1.000...$ without Calculus. – hamam_Abdallah Feb 25 at 22:23
• @hamam_Abdallah $$\frac{1}{2000} = 0.0005$$ – gt6989b Feb 25 at 22:31
• You used Calculus. – hamam_Abdallah Feb 25 at 22:31

Borrowing from a recent answer of mine (it proves more than is needed here, and it avoids using the Binomial Theorem, but it does use induction, and it is arguably a bit too complicated, even for its original purpose, so I've simplified it considerably for the present application):

Define $$a_n = \left(1+\frac{1}{n}\right)^n \quad (n \geqslant 1).$$

If $$x \geqslant y > 0$$, and $$n$$ is a positive integer, then $$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \cdots + y^{n-1}) \geqslant n(x - y)y^{n-1}.$$ This can be convincingly justified without an explicit use of induction, just a plausible use of ellipsis.

Therefore, for $$n > 1$$, \begin{align*} a_n - a_{n-1} & = \left(1+\frac{1}{n}\right)^n \! - \left(1+\frac{1}{n-1}\right)^{n-1} \\ & = \frac{1}{n}\left(1+\frac{1}{n}\right)^{n-1} \!\! - \left[ \left(1+\frac{1}{n-1}\right)^{n-1} \!\! - \left(1+\frac{1}{n}\right)^{n-1}\right] \\ & \leqslant \frac{1}{n}\left(1+\frac{1}{n}\right)^{n-1} \!\! - \frac{1}{n}\left(1+\frac{1}{n}\right)^{n-2} \\ & = \frac{1}{n^2}\left(1+\frac{1}{n}\right)^{n-2} \\ & = \frac{a_n}{(n+1)^2}, \end{align*} whence $$a_n \leqslant a_{n-1}\left(1 - \frac{1}{(n+1)^2}\right)^{-1} \quad (n > 1).$$ But, for $$n > 1$$, $$\left(1 - \frac{1}{(n+1)^2}\right)^{-1} \!\! = 1 + \frac{1}{(n+1)^2 - 1} < 1 + \frac{1}{n-1} = \frac{n}{n-1},$$ therefore $$\frac{a_n}{n} < \frac{a_{n-1}}{n-1} \quad (n > 1),$$ and - again without explicit use of induction - one can deduce: $$\frac{a_n}{n} \leqslant \frac{a_3}{3} = \frac{64}{81} < 1 \quad (n \geqslant 3),$$ whence: $$(n + 1)^n < n^{n+1} \quad (n \geqslant 3).$$

We wish to show that

$$\left(2018\right)^{2019}>\left(2019\right)^{2018}$$

METHODOLOGY $$1$$:

This is equivalent to showing

$$2018>\left(1+\frac1{2018}\right)^{2018}\tag1$$

Let us generalize $$(1)$$ and analyze the sequence $$\left(1+\frac1n\right)^n$$.

We first note that for $$n\ge 2$$

$$\left(1+\frac1n\right)^n<\frac1{\left(1-\frac1n\right)^n}\tag2$$

Furthermore, since the sequence $$\left(1-\frac1n\right)^n$$ is monotonically increasing, which we show using only Bernoulli's Inequality in the Appendix to this solution, we have

$$\left(1-\frac1n\right)^n\ge \left(1-\frac12\right)^2=\frac14\tag 3$$

Using $$(3)$$ in $$(2)$$ reveals that

$$\left(1+\frac1n\right)^n<4$$

whence we obtain the result $$n>\left(1+\frac1n\right)^n$$ for $$n>4$$. Setting $$n=2018$$ yields the inequality in $$(1)$$.

And we are done.

METHODOLOGY $$2$$

This is equivalent to showing

$$\log(2018)>2018\log\left(1+\frac1{2018}\right)$$

From THIS ANSWER, clearly we have $$2018\log\left(1+\frac1{2018}\right)<1$$. In addition, it is trivial to see that $$\log(2018)>1$$.

And we are done!

APPENDIX: Proof that $$\left(1-\frac1n\right)^n$$ monotonically decreases

Let $$a_n=\left(1-\frac1n\right)^n$$. Then, for $$n\ge 2$$, the ratio $$\frac{a_{n+1}}{a_n}$$ is given by

\begin{align} \frac{a_{n+1}}{a_n}&=\frac{\left(1-\frac1{n+1}\right)^{n+1}}{\left(1-\frac1n\right)^n}\\\\ &=\left(1-\frac1n\right)\left(\frac{1-\frac1{n+1}}{1-\frac1n}\right)^{n+1}\\\\ &=\left(1-\frac1n\right)\left(1+\frac{1}{n^2-1}\right)^{n+1}\tag {A1}\\\\ &\ge \left(1-\frac1n\right)\left(1+\frac{n+1}{n^2-1}\right)\tag{A2}\\\\ &=1 \end{align}

Hence, we see that $$a_{n+1}\ge a_n$$ for $$n\ge 2$$. And clearly $$a_2\ge a_1=0$$ so that $$a_n$$ is monotonically increasing for $$n\ge 1$$, which completes the proof.

Note in going from $$(A1)$$ to $$(A2)$$ we used Bernoulli's Inequality.

• Great answer +1 – user370967 Mar 4 at 22:38
• @mathwolf_qed Thank you. Much appreciated. – Mark Viola Mar 5 at 0:49
• @mathoverview Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Mar 14 at 4:00

It is the same as $$2018>(1+\frac{1}{2018})^{2018}$$, so we only need to show that $$a_n=\left(1+\frac{1}{n}\right)^n$$ is bounded. There are many proofs of this. Here is one possibility.

Let $$b_n=(1+\frac{1}{n})^{n+1}$$. We observe that $$b_n$$ is decreasing since \begin{aligned}b_{n-1}=\left(\frac n{n-1}\right)^n&=\left[\text{Geometric mean of }\underbrace{\frac n{n-1},\ldots,\frac{n}{n-1}}_n,1\right]^{n+1}\\ &\ge \left[\text{Harmonic mean of }\underbrace{\frac n{n-1},\ldots,\frac{n}{n-1}}_n,1\right]^{n+1}\\ &=\left(\frac{n+1}n\right)^{n+1}=b_n.\end{aligned} Hence, $$a_n\le b_n\le b_1=4$$.

However, you may need to make them believe that geometric mean is larger than harmonic mean.

• @hamam_Abdallah $GM\le HM$ is equivalent to $GM\le AM$ by taking reciprocals. To prove the latter, I believe Google will tell you many many elementary proofs. – Eclipse Sun Feb 25 at 22:41
• @CalumGilhooley Is it good now? – Eclipse Sun Mar 5 at 4:25
• Yes, very neat. – Calum Gilhooley Mar 5 at 20:25

Notice that $$2018^x > x+1$$ for any $$x>0$$ (plot each function if need be to convince your high school students).

Then $$x = \frac{2019}{2018} -1 > 0$$ gives

$$\frac{2018^{\frac{2019}{2018}}}{2018} > \frac{2019}{2018}$$

which gives the desired result after clearing denominators and taking each side to the $$2018$$ power.

Note: This argument is more commonly used on comparing $$e^2$$ to $$2^e$$.

• How would a high schooler believe that the plot is accurate? How would they know that what is drawn is, in fact, the function you write down algebraically? – John B Feb 26 at 3:19
• @JohnB Is it too much to assume that a high school student is somewhat familiar with the plot of both $a^x$ (where $a > 1$) and $x+1$ when $x \geq 0$? – user328442 Feb 26 at 3:27
• I would imagine that they would not have much information about the behavior of tangent lines on $y=a^x$ near $x=0$ to be able to compare them to the slope of $y=x+1$. But I only speculate about their perspective. I personally would not be entirely convinced. – John B Feb 26 at 3:42

Compared to my first answer, this answer makes a simpler use of the same familiar identity, which is proved without explicit use of induction, thus: \begin{align*} x^m - y^m & = x^m - (x^{m-1}y - x^{m-1}y) - (x^{m-2}y^2 - x^{m-2}y^2) - \cdots - (xy^{m-1} - xy^{m-1}) - y^m \\ & = (x^m - x^{m-1}y) + (x^{m-1}y - x^{m-2}y^2) + (x^{m-2}y^2 - x^{m-3}y^3) + \cdots + (xy^{m-1} - y^m) \\ & = (x - y)x^{m-1} + (x - y)x^{m-2}y + (x - y)x^{m-3}y^2 + \cdots + (x - y)y^{m-1} \\ & = (x - y)(x^{m-1} + x^{m-2}y + x^{m-3}y^2 + \cdots + y^{m-1}). \end{align*} The rest of the proof is a straightforward calculation. (It is based on a trick which was inspired loosely by nbarto's answer, but I am to blame for it!)

Suppose $$n \geqslant 3$$.

In the above identity, take $$x = m = n+1$$, $$y = n$$, and group together all but the first two terms in the brackets, obtaining: $$(n+1)^{n+1} - n^{n+1} = (n+1)^n + (n+1)^{n-1}n + [(n+1)^{n-2}n^2 + \cdots + n^n].$$ There are $$n-1$$ terms in the square brackets, they are in strictly decreasing order, and the largest of them is $$(n+1)^{n-2}n^2$$. Therefore: \begin{align*} (n+1)^{n+1} - n^{n+1} & < (n+1)^n + (n+1)^{n-1}n + (n-1)(n+1)^{n-2}n^2 \\ & = (n+1)^n + (n+1)^{n-2}[n(n+1) + n^2(n-1)] \\ & = (n+1)^n + (n^3+n)(n+1)^{n-2}. \end{align*} On the other hand: \begin{align*} (n+1)^{n+1} - (n+1)^n & = n(n+1)^n \\ & = (n+1)^n + (n-1)(n+1)^n \\ & = (n+1)^n + (n-1)(n+1)^2(n+1)^{n-2} \\ & = (n+1)^n + (n^2-1)(n+1)(n+1)^{n-2} \\ & = (n+1)^n + (n^3 + n^2 - n - 1)(n+1)^{n-2}. \end{align*} But: $$(n^3 + n^2 - n - 1) - (n^3 + n) = n^2 - 2n - 1 = (n - 1)^2 - 2 > 0,$$ therefore: $$(n+1)^{n+1} - n^{n+1} < (n+1)^{n+1} - (n+1)^n,$$ therefore $$n^{n+1} > (n+1)^n$$. $$\square$$

Seen that $$\,x\mapsto x\,$$ grows considerably faster than $$\,x\mapsto\ln x\,$$ and
when accepting more or less (without calculus at any rate) that $$\,x\,/\ln x\,$$ is strictly increasing, then $$\frac{2018}{\ln 2018} \:<\: \frac{2019}{\ln 2019}\,.$$

• What does considerably means Mathematically ? – hamam_Abdallah Feb 25 at 22:26
• This answer lacks rigor. – Mark Viola Feb 25 at 22:43

Hint:

From

$$n^{1/n}>(n+1)^{1/(n+1)}$$

we draw

$$n^{1/n-1/(n+1)}=\left(\frac{n+1}n\right)^{1/(n+1)},$$

$$n>\left(1+\frac1n\right)^n.$$

If we can show that $$\left(1+\dfrac1n\right)^n$$ is bounded above, say by $$3$$, we are done.

Late answer but I think worth noting:

A possible way is using the inequality between the geometric and harmonic mean:

• $$\sqrt[n]{a_1 \cdots a_n} \geq \frac{n}{\frac{1}{a_1} + \cdots + \frac{1}{a_n}}$$ for $$a_1, \ldots , a_n > 0$$.

Now, we use $$2018^{2019}> 2019^{2018} \Leftrightarrow \sqrt[2018]{2018} > 1+\frac{1}{2018}$$

and show the inequality on the RHS:

$$\begin{eqnarray*} \sqrt[2018]{2018} & = & \sqrt[2018]{2 \cdot 1009 \cdot 1^{2016}} \\ & \stackrel{GM-HM}{\geq} & \frac{2018}{\frac{1}{2} + \frac{1}{1009} + 2016}\\ & = & 1+ \frac{2-\frac{1}{2} - \frac{1}{1009}}{\frac{1}{2} + \frac{1}{1009} + 2016} \\ & > & 1+ \frac{1}{2018} \\\end{eqnarray*}$$

gt6989b's answer has shown that it suffices to show that $$(1 + 1/n)^n < n$$ when $$n = 2018$$. In fact this is true for $$n \ge 3$$. If you're allowing the "usual" binomial theorem (with positive integer exponents), then you can show this as follows:

$$\left( {1 + {1 \over n}} \right)^n = \sum_{k=0}^n {n \choose k} \left( {1 \over n}\right)^k = 2 + \sum_{k=2}^n {{n \choose k} \over n^k}.$$

Now I claim that, for $$k \ge 3$$, $${{n \choose k} \over n^k} < \left( {1 \over 2} \right) ^{k-1}.$$

First observe that $${n \choose k} = {n(n-1) \cdots (n-k+1) \over k!} < {n^k \over k!}$$ and so $${{n \choose k} \over n^k} < {1 \over k!}$$ But the denominator $$k!$$ has a factor of 2 and $$k-2$$ factors which are greater than 2, so that denominator is at least $$2^{k-1}$$.

Applying this inequality to the first displayed equation you get $$\left( {1 + {1 \over n}} \right)^n = 2 + \sum_{k=2}^n {1 \over 2^{k-1}}$$ and summing the geometric series gives $$(1 + 1/n)^n < 3$$.

Partial answer: we can show using combinatorics that $$(n+1)^n < n^n \cdot (n+1)$$ for $$n > 1$$.

If you have a string of $$n$$ beads that can have $$n+1$$ colors $$c_1, \ldots c_{n+1}$$, there must be at least $$1$$ color that does not appear in the string. Shifting the colors to fill in the gap, we obtain a string of length $$n$$ where each bead is colored $$c_1$$ to $$c_n$$. The string we obtain is not unique; there are $$n+1$$ ways we could have shifted the colors to obtain that string. Hence the weak inequality. The strict inequality holds because the LHS is not divisible by $$n$$.

I hope this will inspire someone to prove that $$(n+1)^n < n^n \cdot n$$.

Rewrite $$\begin{array} {rll} 2018 \cdot (2018)^{2018} &> 2019^{2018}\\ 2018 &\gt (1+1/2018)^{2018} \\ \ln(2018) &\gt 2018 \cdot \ln(1+1/2018) \\ & \qquad= 2018\cdot(1/2018-1/2018^2/2 + ...)\\ & \qquad \qquad \text{by Mercatorseries expansion of the } \ln() \end{array}$$ leading to $$\ln(2018) \gt 1-1/2018/2 + 1/2018^2/3 - ... + ...$$ which is obviously true.

However, I don't know whether we can assume a highschooler would know the Mercator-series for the natural logarithm?

Without Calculus, Newton's binomial theorem, or induction: Using Maple one gets:

is(2018^2019>2019^2018); true

This answer is a fusion of answers given by "@gt6989b" and "@saulspatz". The answers from "@user328442" is intuitively plausible. Thank you all for your responses that I found pertinent with "+1". I would especially like to thank the three of you mentioned above.

I'll write the answer in a friendly way for a 'High school' student. So it's a bit pedantic and long. I insist on putting this answer because there are subtleties in the numerical approximation. Subtleties that are circumvented by the "@saulspatz" user approach.

Note that $$\frac{2019^{2018}}{{2018}^{2019}} = \frac{2019^{2018}}{2018\cdot {2018}^{2018}} = \frac{1}{2018}\cdot \frac{2019^{2018}}{ {2018}^{2018}} = \frac{1}{2018}\cdot \left(\frac{2019}{2018}\right)^{2018} \\ \quad \\ = \frac{1}{2018}\cdot \left(\frac{2018+1}{2018}\right)^{2018} = \frac{1}{2018}\cdot \left(1+\frac{1}{2018}\right)^{2018}$$

We will prove that $$\frac{1}{2018}\cdot \left(1+\frac{1}{2018}\right)^{2018}<1$$ In fact, \begin{align*} \frac{1}{2018}\cdot \left(1+\frac{1}{2018}\right)^{2018} \leq & \frac{1}{2018}\cdot \left(1+\frac{1}{2000}\right)^{2018} \\ \leq & \frac{1}{2018}\cdot \left(1+\frac{1}{2000}\right)^{2048} \\ = & \frac{1}{2018}\cdot (1,0005)^{2^{11}} \\ = & \frac{1}{2018}\cdot (1.00100025)^{2^{10}} \\ \leq & \frac{1}{2018}\cdot (1.0015)^{2^{10}} \\ = & \frac{1}{2018}\cdot (1.00300225)^{2^{9}} \\ \leq & \frac{1}{2018}\cdot (1.0035)^{2^{9}} \\ = & \frac{1}{2018}\cdot (1.00701225)^{2^{8}} \\ \leq & \frac{1}{2018}\cdot (1.0075)^{2^{8}} \\ = & \frac{1}{2018}\cdot (1.01505625)^{2^{7}} \\ \leq & \frac{1}{2018}\cdot (1.0155)^{2^{7}} \\ = & \frac{1}{2018}\cdot (1.03124025)^{2^{6}} \\ \leq & \frac{1}{2018}\cdot (1.035)^{2^{6}} \\ = & \frac{1}{2018}\cdot (1.071225)^{2^{5}} \\ \leq & \frac{1}{2018}\cdot (1.075)^{2^{5}} \\ = & \frac{1}{2018}\cdot (1.155625)^{2^4} \\ \leq & \frac{1}{2018}\cdot (1.175)^{2^4} \\ = & \frac{1}{2018}\cdot (1.380625)^{2^3} \\ \leq & \frac{1}{2018}\cdot 2^{2^3} \\ =& \frac{1}{2018}\cdot 256 <1 \end{align*}