Author |
Topic: power ranges (Read 1474 times) |
|
Noke Lieu
Uberpuzzler
pen... paper... let's go! (and bit of plastic)
Gender: 
Posts: 1884
|
 |
power ranges
« on: Mar 20th, 2012, 3:06pm » |
 Quote  Modify
|
If A and B are positive integers, and a<b, does it follow that ab < ba ? When?
|
|
 IP Logged |
a shade of wit and the art of farce.
|
|
|
pex
Uberpuzzler
Gender:
Posts: 880
|
|
Re: power ranges
« Reply #2 on: Mar 21st, 2012, 1:46am » |
Quote Modify
|
on Mar 20th, 2012, 3:06pm, Noke Lieu wrote:| If A and B are positive integers, and a<b, does it follow that ab < ba ? When? |
|
In fact, it is "usually" the other way around. Your statement holds if a=1, and if (a,b)=(2,3), but not in any other cases. It is not a coincidence that things change around the number e 
|
|
IP Logged |
|
|
|
Noke Lieu
Uberpuzzler
pen... paper... let's go! (and bit of plastic)
Gender:
Posts: 1884
|
|
Re: power ranges
« Reply #3 on: Mar 21st, 2012, 10:09pm » |
Quote Modify
|
Yeah- I wound up at when a/b > logba
I was hoping that it'd be more enlightening than that. I guess pex, you make a nice point in that vein.
I supose I shouldn't have restricted it to integers, but even then...
|
|
IP Logged |
a shade of wit and the art of farce.
|
|
|
pex
Uberpuzzler
Gender:
Posts: 880
|
|
Re: power ranges
« Reply #4 on: Mar 22nd, 2012, 12:38am » |
Quote Modify
|
on Mar 21st, 2012, 10:09pm, Noke Lieu wrote:| Yeah- I wound up at when a/b > logba |
|
... which is equivalent to (log a)/a < (log b)/b. And (log x)/x is maximized at x=e.
Over the integers, (log 3)/3 is the maximum. Further, (log 2)/2 = (log 4)/4. Finally, (log 1)/1 = 0 and (log n)/n is positive for all other integers. My answer was derived from these properties.
So your puzzle was more interesting than you think 
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print