Hints for proving $frac1m^k binommk < frac1n^k binomnk$

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up vote
1
down vote

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I want to prove the following exercise from a textbook:

Let $m,n in mathbbN$ and $m < n$. Prove directly that for $k = 2, dots, n$ the inequaity

$$frac1m^k binommk < frac1n^k binomnk$$

holds.

My work so far:

I know that you could probably solve this using induction, but this is not how it's intended to be done. So I want to ask for hints that doesn't involve using induction.

Since $m < n$, it follows
$$forall alpha < k: m - alpha < n - alpha$$

This implies
$$prod_i = 0^k-1 (m - i) < prod_i = 0^k-1 (n - i)$$

This implies
$$binommk = frac1k!prod_i = 0^k-1 (m - i) < frac1k!prod_i = 0^k-1 (n - i) = binomnk$$

So, what I've got is
$$binommk < binomnk$$

Now, I am pretty much stuck. If $m < n$, then $frac1n < frac1m$, so
$$frac1n^k < frac1m^k$$

which would imply that

$$frac1n^k binommk < frac1m^k binomnk$$

So I guess I've messed something up

inequality binomial-coefficients

share|cite|improve this question

asked Jul 15 at 0:23

PhysicsUndergraduateStudent

527

add a comment | 

up vote
1
down vote

favorite

I want to prove the following exercise from a textbook:

Let $m,n in mathbbN$ and $m < n$. Prove directly that for $k = 2, dots, n$ the inequaity

$$frac1m^k binommk < frac1n^k binomnk$$

holds.

My work so far:

I know that you could probably solve this using induction, but this is not how it's intended to be done. So I want to ask for hints that doesn't involve using induction.

Since $m < n$, it follows
$$forall alpha < k: m - alpha < n - alpha$$

This implies
$$prod_i = 0^k-1 (m - i) < prod_i = 0^k-1 (n - i)$$

This implies
$$binommk = frac1k!prod_i = 0^k-1 (m - i) < frac1k!prod_i = 0^k-1 (n - i) = binomnk$$

So, what I've got is
$$binommk < binomnk$$

Now, I am pretty much stuck. If $m < n$, then $frac1n < frac1m$, so
$$frac1n^k < frac1m^k$$

which would imply that

$$frac1n^k binommk < frac1m^k binomnk$$

So I guess I've messed something up

inequality binomial-coefficients

share|cite|improve this question

asked Jul 15 at 0:23

PhysicsUndergraduateStudent

527

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I want to prove the following exercise from a textbook:

Let $m,n in mathbbN$ and $m < n$. Prove directly that for $k = 2, dots, n$ the inequaity

$$frac1m^k binommk < frac1n^k binomnk$$

holds.

My work so far:

I know that you could probably solve this using induction, but this is not how it's intended to be done. So I want to ask for hints that doesn't involve using induction.

Since $m < n$, it follows
$$forall alpha < k: m - alpha < n - alpha$$

This implies
$$prod_i = 0^k-1 (m - i) < prod_i = 0^k-1 (n - i)$$

This implies
$$binommk = frac1k!prod_i = 0^k-1 (m - i) < frac1k!prod_i = 0^k-1 (n - i) = binomnk$$

So, what I've got is
$$binommk < binomnk$$

Now, I am pretty much stuck. If $m < n$, then $frac1n < frac1m$, so
$$frac1n^k < frac1m^k$$

which would imply that

$$frac1n^k binommk < frac1m^k binomnk$$

So I guess I've messed something up

inequality binomial-coefficients

share|cite|improve this question

asked Jul 15 at 0:23

PhysicsUndergraduateStudent

527

I want to prove the following exercise from a textbook:

Let $m,n in mathbbN$ and $m < n$. Prove directly that for $k = 2, dots, n$ the inequaity

$$frac1m^k binommk < frac1n^k binomnk$$

holds.

My work so far:

I know that you could probably solve this using induction, but this is not how it's intended to be done. So I want to ask for hints that doesn't involve using induction.

Since $m < n$, it follows
$$forall alpha < k: m - alpha < n - alpha$$

This implies
$$prod_i = 0^k-1 (m - i) < prod_i = 0^k-1 (n - i)$$

This implies
$$binommk = frac1k!prod_i = 0^k-1 (m - i) < frac1k!prod_i = 0^k-1 (n - i) = binomnk$$

So, what I've got is
$$binommk < binomnk$$

Now, I am pretty much stuck. If $m < n$, then $frac1n < frac1m$, so
$$frac1n^k < frac1m^k$$

which would imply that

$$frac1n^k binommk < frac1m^k binomnk$$

So I guess I've messed something up

inequality binomial-coefficients

share|cite|improve this question

asked Jul 15 at 0:23

PhysicsUndergraduateStudent

527

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 0:23

PhysicsUndergraduateStudent

527

asked Jul 15 at 0:23

PhysicsUndergraduateStudent

527

asked Jul 15 at 0:23

PhysicsUndergraduateStudent

527

527

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

Hint:   write it as $;
displaystyle frac1k!,prod_i=0^k-1 ,left(1-dfracimright) ;lt; frac1k!,prod_i=0^k-1 ,left(1-dfracinright);$ and compare the terms.

share|cite|improve this answer

answered Jul 15 at 0:32

dxiv

54.2k64797

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But why is it that $1 - fracim < 1 - fracin$?
  – PhysicsUndergraduateStudent
  Jul 15 at 0:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PhysicsUndergraduateStudent Because $frac1n lt frac1m$ (as you wrote in the post) $implies fracin lt fracim$ $implies 1-fracin gt 1-fracim$.
  – dxiv
  Jul 15 at 0:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ohh I see. But that leaves me wondering if my reasoning was correct as well, that is: Does $$frac1n^k binommk < frac1m^k binomnk$$ hold as well then?
  – PhysicsUndergraduateStudent
  Jul 15 at 0:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @PhysicsUndergraduateStudent That holds as well, but it's a weaker inequality than asked, since: $$ frac1n^k binommk < colorredfrac1m^k binommk < frac1n^k binomnk < frac1m^k binomnk $$
  – dxiv
  Jul 15 at 0:45
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

$1-frac j m < 1- frac j n$ for $j=0,1,2,...,k-1$. Just multiply these inequalities and simplify.

share|cite|improve this answer

answered Jul 15 at 0:33

Kavi Rama Murthy

21k2830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry, but I really don't see why $1-frac j m < 1- frac j n$ holds if $m < n$
  – PhysicsUndergraduateStudent
  Jul 15 at 0:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $1-frac j x$ is an increasing function of $x$. An even simpler argument is $m <n$ implies $frac 1 m > frac 1 n$ which implies $frac j m > frac j n$ which in turn implies $1-frac j m <1- frac jn $
  – Kavi Rama Murthy
  Jul 15 at 5:00
  
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Hint:   write it as $;
displaystyle frac1k!,prod_i=0^k-1 ,left(1-dfracimright) ;lt; frac1k!,prod_i=0^k-1 ,left(1-dfracinright);$ and compare the terms.

share|cite|improve this answer

answered Jul 15 at 0:32

dxiv

54.2k64797

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But why is it that $1 - fracim < 1 - fracin$?
  – PhysicsUndergraduateStudent
  Jul 15 at 0:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PhysicsUndergraduateStudent Because $frac1n lt frac1m$ (as you wrote in the post) $implies fracin lt fracim$ $implies 1-fracin gt 1-fracim$.
  – dxiv
  Jul 15 at 0:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ohh I see. But that leaves me wondering if my reasoning was correct as well, that is: Does $$frac1n^k binommk < frac1m^k binomnk$$ hold as well then?
  – PhysicsUndergraduateStudent
  Jul 15 at 0:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @PhysicsUndergraduateStudent That holds as well, but it's a weaker inequality than asked, since: $$ frac1n^k binommk < colorredfrac1m^k binommk < frac1n^k binomnk < frac1m^k binomnk $$
  – dxiv
  Jul 15 at 0:45
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Hint:   write it as $;
displaystyle frac1k!,prod_i=0^k-1 ,left(1-dfracimright) ;lt; frac1k!,prod_i=0^k-1 ,left(1-dfracinright);$ and compare the terms.

share|cite|improve this answer

answered Jul 15 at 0:32

dxiv

54.2k64797

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But why is it that $1 - fracim < 1 - fracin$?
  – PhysicsUndergraduateStudent
  Jul 15 at 0:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PhysicsUndergraduateStudent Because $frac1n lt frac1m$ (as you wrote in the post) $implies fracin lt fracim$ $implies 1-fracin gt 1-fracim$.
  – dxiv
  Jul 15 at 0:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ohh I see. But that leaves me wondering if my reasoning was correct as well, that is: Does $$frac1n^k binommk < frac1m^k binomnk$$ hold as well then?
  – PhysicsUndergraduateStudent
  Jul 15 at 0:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @PhysicsUndergraduateStudent That holds as well, but it's a weaker inequality than asked, since: $$ frac1n^k binommk < colorredfrac1m^k binommk < frac1n^k binomnk < frac1m^k binomnk $$
  – dxiv
  Jul 15 at 0:45
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Hint:   write it as $;
displaystyle frac1k!,prod_i=0^k-1 ,left(1-dfracimright) ;lt; frac1k!,prod_i=0^k-1 ,left(1-dfracinright);$ and compare the terms.

share|cite|improve this answer

answered Jul 15 at 0:32

dxiv

54.2k64797

Hint:   write it as $;
displaystyle frac1k!,prod_i=0^k-1 ,left(1-dfracimright) ;lt; frac1k!,prod_i=0^k-1 ,left(1-dfracinright);$ and compare the terms.

share|cite|improve this answer

answered Jul 15 at 0:32

dxiv

54.2k64797

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 0:32

dxiv

54.2k64797

answered Jul 15 at 0:32

dxiv

54.2k64797

answered Jul 15 at 0:32

dxiv

54.2k64797

54.2k64797

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But why is it that $1 - fracim < 1 - fracin$?
  – PhysicsUndergraduateStudent
  Jul 15 at 0:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PhysicsUndergraduateStudent Because $frac1n lt frac1m$ (as you wrote in the post) $implies fracin lt fracim$ $implies 1-fracin gt 1-fracim$.
  – dxiv
  Jul 15 at 0:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ohh I see. But that leaves me wondering if my reasoning was correct as well, that is: Does $$frac1n^k binommk < frac1m^k binomnk$$ hold as well then?
  – PhysicsUndergraduateStudent
  Jul 15 at 0:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @PhysicsUndergraduateStudent That holds as well, but it's a weaker inequality than asked, since: $$ frac1n^k binommk < colorredfrac1m^k binommk < frac1n^k binomnk < frac1m^k binomnk $$
  – dxiv
  Jul 15 at 0:45
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But why is it that $1 - fracim < 1 - fracin$?
  – PhysicsUndergraduateStudent
  Jul 15 at 0:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PhysicsUndergraduateStudent Because $frac1n lt frac1m$ (as you wrote in the post) $implies fracin lt fracim$ $implies 1-fracin gt 1-fracim$.
  – dxiv
  Jul 15 at 0:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ohh I see. But that leaves me wondering if my reasoning was correct as well, that is: Does $$frac1n^k binommk < frac1m^k binomnk$$ hold as well then?
  – PhysicsUndergraduateStudent
  Jul 15 at 0:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @PhysicsUndergraduateStudent That holds as well, but it's a weaker inequality than asked, since: $$ frac1n^k binommk < colorredfrac1m^k binommk < frac1n^k binomnk < frac1m^k binomnk $$
  – dxiv
  Jul 15 at 0:45
  
  

  
  
  
  

But why is it that $1 - fracim < 1 - fracin$?
– PhysicsUndergraduateStudent
Jul 15 at 0:40

But why is it that $1 - fracim < 1 - fracin$?
– PhysicsUndergraduateStudent
Jul 15 at 0:40

@PhysicsUndergraduateStudent Because $frac1n lt frac1m$ (as you wrote in the post) $implies fracin lt fracim$ $implies 1-fracin gt 1-fracim$.
– dxiv
Jul 15 at 0:41

@PhysicsUndergraduateStudent Because $frac1n lt frac1m$ (as you wrote in the post) $implies fracin lt fracim$ $implies 1-fracin gt 1-fracim$.
– dxiv
Jul 15 at 0:41

Ohh I see. But that leaves me wondering if my reasoning was correct as well, that is: Does $$frac1n^k binommk < frac1m^k binomnk$$ hold as well then?
– PhysicsUndergraduateStudent
Jul 15 at 0:44

Ohh I see. But that leaves me wondering if my reasoning was correct as well, that is: Does $$frac1n^k binommk < frac1m^k binomnk$$ hold as well then?
– PhysicsUndergraduateStudent
Jul 15 at 0:44

2

2

@PhysicsUndergraduateStudent That holds as well, but it's a weaker inequality than asked, since: $$ frac1n^k binommk < colorredfrac1m^k binommk < frac1n^k binomnk < frac1m^k binomnk $$
– dxiv
Jul 15 at 0:45

@PhysicsUndergraduateStudent That holds as well, but it's a weaker inequality than asked, since: $$ frac1n^k binommk < colorredfrac1m^k binommk < frac1n^k binomnk < frac1m^k binomnk $$
– dxiv
Jul 15 at 0:45

add a comment | 

up vote
0
down vote

$1-frac j m < 1- frac j n$ for $j=0,1,2,...,k-1$. Just multiply these inequalities and simplify.

share|cite|improve this answer

answered Jul 15 at 0:33

Kavi Rama Murthy

21k2830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry, but I really don't see why $1-frac j m < 1- frac j n$ holds if $m < n$
  – PhysicsUndergraduateStudent
  Jul 15 at 0:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $1-frac j x$ is an increasing function of $x$. An even simpler argument is $m <n$ implies $frac 1 m > frac 1 n$ which implies $frac j m > frac j n$ which in turn implies $1-frac j m <1- frac jn $
  – Kavi Rama Murthy
  Jul 15 at 5:00
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

$1-frac j m < 1- frac j n$ for $j=0,1,2,...,k-1$. Just multiply these inequalities and simplify.

share|cite|improve this answer

answered Jul 15 at 0:33

Kavi Rama Murthy

21k2830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry, but I really don't see why $1-frac j m < 1- frac j n$ holds if $m < n$
  – PhysicsUndergraduateStudent
  Jul 15 at 0:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $1-frac j x$ is an increasing function of $x$. An even simpler argument is $m <n$ implies $frac 1 m > frac 1 n$ which implies $frac j m > frac j n$ which in turn implies $1-frac j m <1- frac jn $
  – Kavi Rama Murthy
  Jul 15 at 5:00
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

$1-frac j m < 1- frac j n$ for $j=0,1,2,...,k-1$. Just multiply these inequalities and simplify.

share|cite|improve this answer

answered Jul 15 at 0:33

Kavi Rama Murthy

21k2830

$1-frac j m < 1- frac j n$ for $j=0,1,2,...,k-1$. Just multiply these inequalities and simplify.

share|cite|improve this answer

answered Jul 15 at 0:33

Kavi Rama Murthy

21k2830

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 0:33

Kavi Rama Murthy

21k2830

answered Jul 15 at 0:33

Kavi Rama Murthy

21k2830

answered Jul 15 at 0:33

Kavi Rama Murthy

21k2830

21k2830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry, but I really don't see why $1-frac j m < 1- frac j n$ holds if $m < n$
  – PhysicsUndergraduateStudent
  Jul 15 at 0:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $1-frac j x$ is an increasing function of $x$. An even simpler argument is $m <n$ implies $frac 1 m > frac 1 n$ which implies $frac j m > frac j n$ which in turn implies $1-frac j m <1- frac jn $
  – Kavi Rama Murthy
  Jul 15 at 5:00
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry, but I really don't see why $1-frac j m < 1- frac j n$ holds if $m < n$
  – PhysicsUndergraduateStudent
  Jul 15 at 0:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $1-frac j x$ is an increasing function of $x$. An even simpler argument is $m <n$ implies $frac 1 m > frac 1 n$ which implies $frac j m > frac j n$ which in turn implies $1-frac j m <1- frac jn $
  – Kavi Rama Murthy
  Jul 15 at 5:00
  
  

  
  
  
  

I am sorry, but I really don't see why $1-frac j m < 1- frac j n$ holds if $m < n$
– PhysicsUndergraduateStudent
Jul 15 at 0:38

I am sorry, but I really don't see why $1-frac j m < 1- frac j n$ holds if $m < n$
– PhysicsUndergraduateStudent
Jul 15 at 0:38

$1-frac j x$ is an increasing function of $x$. An even simpler argument is $m <n$ implies $frac 1 m > frac 1 n$ which implies $frac j m > frac j n$ which in turn implies $1-frac j m <1- frac jn $
– Kavi Rama Murthy
Jul 15 at 5:00

$1-frac j x$ is an increasing function of $x$. An even simpler argument is $m <n$ implies $frac 1 m > frac 1 n$ which implies $frac j m > frac j n$ which in turn implies $1-frac j m <1- frac jn $
– Kavi Rama Murthy
Jul 15 at 5:00

add a comment | 

 

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