10
$\begingroup$

Question:

Find the greater number: $1000^{1000}$ or $1001^{999}$

My Attempt:

I know that: $(a+b)^n \geq a^n + a^{n-1}bn$.

Thus, $(1+999)^{1000} \geq 999001$

And $(1+1000)^{999} \geq 999001$

But that doesn't make much sense.

I want some hints regarding how to solve this problem.

Thanks.

$\endgroup$
5
  • $\begingroup$ Also, I don't think $999^{1000} = 999,000 = 999 \cdot 1000$... $\endgroup$
    – grantfgates
    Commented May 14, 2014 at 6:22
  • $\begingroup$ mathforum.org/kb/… $\endgroup$
    – lab bhattacharjee
    Commented May 14, 2014 at 6:24
  • 1
    $\begingroup$ The way you're getting your bounds isn't a useful way to do things. You've picked the two very smallest terms of the expression to add together; on the other end of the binomial expansion, you have terms like $999^{1000}$, which swamp your bound by about 3000 orders of magnitude. $\endgroup$
    – user2357112
    Commented May 14, 2014 at 6:26
  • $\begingroup$ You can simply take log and compare, since log is a monotone function. $\endgroup$
    – Y.H. Chan
    Commented May 14, 2014 at 7:47
  • 1
    $\begingroup$ possible duplicate of Given $a>b>2$ both positive integers, which of $a^b$ and $b^a$ is larger? $\endgroup$
    – Najib Idrissi
    Commented Sep 3, 2015 at 11:31

3 Answers 3

Reset to default
34
$\begingroup$

Look at the quotient $$ \frac{1001^{999}}{1000^{1000}}=\frac1{1001}\underbrace{\left(1+\frac1{1000}\right)^{1000}}_{\approx e}$$

$\endgroup$
8
  • 2
    $\begingroup$ Wow, striking. +1 $\endgroup$
    – Alex Wertheim
    Commented May 14, 2014 at 6:25
  • 2
    $\begingroup$ @Hagen The approach looks nice. But how do I use it to get the answer? I am a bit newbie so please help me out :) $\endgroup$
    – Gaurang Tandon
    Commented May 14, 2014 at 6:28
  • $\begingroup$ This is the ratio of the numbers you are comparing. The top is apparently much smaller than the bottom, around $e/1001$ times smaller. $\endgroup$
    – orion
    Commented May 14, 2014 at 7:53
  • $\begingroup$ @GaurangTandon This pretty much is the answer... the thing on the right is less than one. Let's call it $c$. Then $1001^{999} = c1000^{1000}$, where $c$ is around $e/1001$. $\endgroup$
    – user98602
    Commented May 14, 2014 at 7:53
  • 2
    $\begingroup$ @GaurangTandon The approximation is certainly $<1001$, which is all we need here. More rigorously (but awfully wasteful), $(1-\frac1{1001})^{1000}>1-\frac{1000}{1001}$ by Bernoulli's inequality, hence the reciprocal $(1+\frac1{1000})^{1000}$ is $<1001$ $\endgroup$
    – Hagen von Eitzen
    Commented Jan 25, 2016 at 13:32
3
$\begingroup$

$\forall r \in \Bbb N-\{1\}$, we have by applying the AM-GM inequality to the $r$ numbers $r-1$ of which equal $r+1$ and one $1$, we have, $$\frac {1+(r-1)(r+1)}{r} \gt (1 \times (r+1)^{r-1})^\frac {1}{r}$$ wherefrom we have, $r^r \gt (r+1)^{r-1}.$

$\endgroup$
4
  • $\begingroup$ What is the AM-GM inequality ? $\endgroup$
    – Gaurang Tandon
    Commented May 14, 2014 at 6:41
  • $\begingroup$ See this page, it is a very famous inequality... en.wikipedia.org/wiki/… $\endgroup$
    – Indrayudh Roy
    Commented May 14, 2014 at 6:44
  • $\begingroup$ 2 questions: 1) What is the ∀ symbol ? 2) How did you actually use the AM-GM inequality ? $\endgroup$
    – Gaurang Tandon
    Commented May 14, 2014 at 6:50
  • $\begingroup$ $ \forall$ mean for all, i.e. for every element $r\in \Bbb N-\{1\}$. $\endgroup$
    – Indrayudh Roy
    Commented May 14, 2014 at 6:52
0
$\begingroup$

Divide the two numbers and prove the result is grater or less than 1. Then, respectively, the first number or the second number is bigger than the other one.

This is my first post here, so I hope somebody will help me with the proper formatting, but generally its like this:

$$ \frac{1000^{1000}}{1001^{999}} = \frac{1}{1001}\cdot \frac{1000^{1000}}{1001^{1000}} = \frac{1}{1001}\cdot \left( \frac{1000}{1001}\right)^{1000} $$ As you can see here: $\frac{1}{1001} < 1$ and $(1000/1001)^{1000} < 1$, so $1000^{1000}/1001^{999} < 1$ So from here we can conclude that

$1000^{1000} < 1001^{999}$

$\endgroup$
5
  • $\begingroup$ I formatted the equations for you, which makes it easier to see that you've made an algebraic mistake when you pull out the $1/1001$ factor. $\endgroup$
    – mrf
    Commented May 14, 2014 at 6:52
  • $\begingroup$ Hey, thank you, now I see that i made a mistake :D I'll try to correct it, but until then its wrong. Sorry, maybe I need a cup of coffee :D ;) $\endgroup$
    – LGalabov
    Commented May 14, 2014 at 6:55
  • $\begingroup$ You can delete your post while it's wrong, then undelete it once you've fixed it. That way you avoid confusing people and you also avoid downvotes due to the mistake. $\endgroup$
    – MvG
    Commented May 14, 2014 at 7:18
  • 2
    $\begingroup$ @MvG: Unregistered users cannot delete posts. $\endgroup$
    – Asaf Karagila ♦
    Commented May 14, 2014 at 8:14
  • $\begingroup$ This is wrong, the first step should multiply by $1001$ instead of $1/1001$ $\endgroup$
    – Guido
    Commented Aug 9, 2014 at 15:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .