Question

Ordering problem

Answer 1

Starting from two positive integers a and b, and this inequality, how well can we order all possible multiples of them?
\[a<b\]

We can pretty easily see these relations have to be true:
\[a^2<ab<b^2\]
\[a^3<a^2b<ab^2<b^3\]
\[a^4<a^3b<a^2b^2<ab^3<b^4\]
etc...

But it doesn't seem like there's much we can say or do about statements like this, where the sum of the exponents are different:

\(b^2<a^3\) or \(a^3 < b^2\) ?

Is there some way to get some way to order them without knowing a or b?

Answer 2

no u gotta know a and b

Answer 3

i see well,
a<b
a^2<b^2
a.a^2<a.b^2
a^3<a.b^2 that is kinda not generalization, i havn't seen away to compare before. hmm

Answer 4

Since we know a<b if we start at \(a^n\) and replace one of them with b, then the new number has to be larger like this: \(a^n < a^{n-1}b\) and you can keep doing that until you get all b's.

@imqwerty I agree with you we have to know a and b, but is there some sorta temporary structure we can work with? I think there's more, for instance:

\[ab < a^2b^2\]

but I'm not sure how we can generalize this. I want to have the most knowledge as possible even if it involves some temporary guessing or something haha.

Answer 5

a and b are integers and if a<b
then the least integer b can be is a+1

so I think we only have to look at the inequalities you mentioned above only comparing a and a+1
\[b^2<a^3? \\ \text{so this one is } \\ (a+1)^2 < a^3 \text{ which is true for integer } a \ge 3\]

Answer 6

\[a^3<b^2? \\ \text{ this one is } \\ a^3<(a+1)^2 \text{ which is true for positive integers } a=1,2\]

Answer 7

I like that, so we can kinda like try to find the range of where it holds true ok cool, that seems exactly the sorta thing I am looking for.

Answer 8

\[a^3<(a+k)^2 \text{ where} k \in \mathbb{Z}^+\]
general thing might be harder

Answer 9

http://www.wolframalpha.com/input/?i=%28a%2Bk%29%5E2%3Ca%5E3
nevermind wolfram did it
so we should be able to

Answer 10

a^3-(a+1)^2=0

interesting! its positive function for a>=3

Answer 11

Yeah I guess the more general form would be something scarier like:

\[a^wb^x < a^yb^z\] and then throwing in some kind of stuff like:
\[b=a+k\]
\[a^w(a+k)^x < a^y(a+k)^z\]

hahaha... gross... hmmm it might not be completely hopeless though.

Answer 12

start from
\(a^{n-1}>(a+1)^n\)
if this could be generalized for all n then the rest ur asking for is solved!

Answer 13

well it need not to be like this but see cases and conditions.
"above i've just made conjecture i'm not sure it works when n is odd"

Answer 14

Oooh ok I like this thinking, actually when I looked at it I thought you got the sign backwards cause:

\[a^n < a(a+1)^n\]

Ok going to go downstairs and get some snack and come back... :D

Answer 15

oooh one thing, does it matter it \(\gcd(a,b)=1\) or not? I think just as long as \(\gcd(a,b) \ne a\) we are fine here.

Answer 16

\(a^{n-1}-(a+1)^n<0\) like this u mean ?
i said i'm not sure about what we need lol

Answer 17

Hahaha I like this it's giving me ideas, I keep thinking of stuff then realizing it won't work so I am stuck haha.

Answer 18

I really like how we have these little 'islands' of ordering though with all the same sum of exponents, I want to stitch these together in some consistent way where knowing a and b turns it from a range to a single ordering or somethin... hmmm

Answer 19

Can we say this is true?

\[\frac{a+b}{2} < \frac{a^2+ab+b^2}{3} < \frac{a^3+a^2b+ab^2+b^3}{4} < \cdots \]

Answer 20

its something like ...
https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

Answer 21

ok like I was thinking about what you said ikram, and I got this sorta thing:

\[a^n<b^n\]
plug in \(b=a+k\)
\[a^n< (a+k)^n\]
\[0 < \sum_{r=1}^n \binom{n}{r} a^{n-r} k^r\]
But I don't know if that helps us, I feel like there' s a bunch of different ways to try to take it it's hard to decide haha

Answer 22

but u made this only

(a+b)/2 <(a+b)^2/3<(a+b)^3/4<....<(a+b)^n/n+1

my problem with u kai is u don't stick to one idea :P please focus on one bhahaha jk

Answer 23

so we need to show that sequence is monotone increasing , we haven't show that yet. how would we know if for some n that function won't look like this ?

Answer 24

even thought i feel disappointed the moment i see induction prove lol but i think we need it xD

Answer 25

Haha ok I did focu on it it's just too annoying to write on paper. That idea is the average of the islands is ordered so that gives us some stem to grow off of:

In general:

$$\frac{1}{n} \sum_{k=0}^n a^kb^{n-k} < \frac{1}{n+1} \sum_{k=0}^{n+1} a^kb^{n+1-k}$$

Answer 26

ok that helps!

Answer 27

OS is trying to die on me here it takes an hour to type anything lol

Answer 28

the lag has reduced for me \(\Huge\ddot\smile\)

Answer 29

i know a better smile
\(\Huge \color{blue}{\text{ت}} \)

Answer 30

mine is better

Answer 31

nah here a one eyed smile :P
\(\Huge \color{blue}{\text{ن}} \)
and three eyed smile xD
\(\Huge \color{blue}{\text{ث}} \)

Answer 32

ikram's an alien here's her real smile:


\(\Huge \color{green}{\text{ث}} \)

also here's a duck
\(\Huge \color{orange}{\text{ي}} \)

Answer 33

cool :P

Answer 34

Ok I was able to figure something out by taking the sum and shifting around some terms:

\[T_n = \frac{1}{n} \sum_{k=0}^n a^kb^{n-k}\]

\[T_{n+1} = \frac{a^{n+1}}{n+1} + \frac{bn}{n+1} T_n\]

Now we can look at this a little easier:

\[T_n < T_{n+1}\] by pluggin that in:

\[T_n < \frac{a^{n+1}}{n(1-b)+1}\]

Unless I messed up along the way, it would appear that the average decreases...? Since b is positive that's saying T_N is less than 0... So I should just rewrite it with the opposite sign #_#

Answer 35

hmmm