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Proving $prod_pin pi(m+1,2m)p leq 2m choose m $ [duplicate]

Product of all prime numbers on the interval [m+1, 2m] is $le left(beginmatrix 2m \mendmatrixright)$Proving $n choose p equiv Bigl[fracnpBigr] (textmod p)$An estimate for relatively prime numbersUpper bound for $prod_ 5 leq p <n p^fracnp-1$Is the set of all prime numbers bounded?Proving claim regargind upper bound of primorialShow that $prod_1le ple nple4^n$ using the fact that $prod_m+1le ple 2mple 2mchoose m$ where $p$ are prime.Second degree polynomials in one variable (with integer coefficients) and limiting behavior of the number of prime values they takeprove that the product of primes in a given interval is less than or equal to binomasymptotic bound on number of primes from 1 to nprime numbers in an interval

2

$begingroup$

This question already has an answer here:

• 
  Product of all prime numbers on the interval [m+1, 2m] is $le left(beginmatrix 2m \mendmatrixright)$
  
  2 answers
  
  

Let $pi (m,n)$ denote the set of prime numbers in the interval $[m,n]$

Show that
$prod_pin pi(m+1,2m)p leq 2m choose m $.

---

My attempt: $$2m choose m =fracm!(prod_pin pi(m+1,2m)p)(prod_qin [m+1,2m]-pi(m+1,2m)q)(m!)^2=frac(prod_pin pi(m+1,2m)p)(prod_qin [m+1,2m]-pi(m+1,2m)q)m!$$

All I have to do now is explain why: $$fracprod_qin [m+1,2m]-pi(m+1,2m)qm! geq 1$$

but I'm struggling with a formal proof.

combinatorics number-theory prime-numbers

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asked Mar 29 at 11:31

user401516user401516

974311

$endgroup$

marked as duplicate by Mike Earnest, Jyrki Lahtonen, José Carlos Santos, Santana Afton, Leucippus Mar 31 at 0:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

2

$begingroup$

This question already has an answer here:

• 
  Product of all prime numbers on the interval [m+1, 2m] is $le left(beginmatrix 2m \mendmatrixright)$
  
  2 answers
  
  

Let $pi (m,n)$ denote the set of prime numbers in the interval $[m,n]$

Show that
$prod_pin pi(m+1,2m)p leq 2m choose m $.

---

My attempt: $$2m choose m =fracm!(prod_pin pi(m+1,2m)p)(prod_qin [m+1,2m]-pi(m+1,2m)q)(m!)^2=frac(prod_pin pi(m+1,2m)p)(prod_qin [m+1,2m]-pi(m+1,2m)q)m!$$

All I have to do now is explain why: $$fracprod_qin [m+1,2m]-pi(m+1,2m)qm! geq 1$$

but I'm struggling with a formal proof.

combinatorics number-theory prime-numbers

share|cite|improve this question

asked Mar 29 at 11:31

user401516user401516

974311

$endgroup$

marked as duplicate by Mike Earnest, Jyrki Lahtonen, José Carlos Santos, Santana Afton, Leucippus Mar 31 at 0:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Product of all prime numbers on the interval [m+1, 2m] is $le left(beginmatrix 2m \mendmatrixright)$
  
  2 answers
  
  

Let $pi (m,n)$ denote the set of prime numbers in the interval $[m,n]$

Show that
$prod_pin pi(m+1,2m)p leq 2m choose m $.

---

My attempt: $$2m choose m =fracm!(prod_pin pi(m+1,2m)p)(prod_qin [m+1,2m]-pi(m+1,2m)q)(m!)^2=frac(prod_pin pi(m+1,2m)p)(prod_qin [m+1,2m]-pi(m+1,2m)q)m!$$

All I have to do now is explain why: $$fracprod_qin [m+1,2m]-pi(m+1,2m)qm! geq 1$$

but I'm struggling with a formal proof.

combinatorics number-theory prime-numbers

share|cite|improve this question

asked Mar 29 at 11:31

user401516user401516

974311

$endgroup$

share|cite|improve this question

asked Mar 29 at 11:31

user401516user401516

974311

share|cite|improve this question

asked Mar 29 at 11:31

user401516user401516

974311

asked Mar 29 at 11:31

user401516user401516

974311

asked Mar 29 at 11:31

user401516user401516

974311

1 Answer
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3

$begingroup$

Hint:

Any prime $pin(m+1,2m)$ divides $(2m)!$ and does not divide $m!cdot m!$. It follows that $p$ must divide $binom2mm$.

Now consider the prime factorization of $binom2mm$.

share|cite|improve this answer

answered Mar 29 at 11:34

YankoYanko

8,3492830

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add a comment | 

3

$begingroup$

Hint:

Any prime $pin(m+1,2m)$ divides $(2m)!$ and does not divide $m!cdot m!$. It follows that $p$ must divide $binom2mm$.

Now consider the prime factorization of $binom2mm$.

share|cite|improve this answer

answered Mar 29 at 11:34

YankoYanko

8,3492830

$endgroup$

add a comment | 

3

$begingroup$

Hint:

Any prime $pin(m+1,2m)$ divides $(2m)!$ and does not divide $m!cdot m!$. It follows that $p$ must divide $binom2mm$.

Now consider the prime factorization of $binom2mm$.

share|cite|improve this answer

answered Mar 29 at 11:34

YankoYanko

8,3492830

$endgroup$

add a comment | 

$begingroup$

Hint:

Any prime $pin(m+1,2m)$ divides $(2m)!$ and does not divide $m!cdot m!$. It follows that $p$ must divide $binom2mm$.

Now consider the prime factorization of $binom2mm$.

share|cite|improve this answer

answered Mar 29 at 11:34

YankoYanko

8,3492830

$endgroup$

share|cite|improve this answer

answered Mar 29 at 11:34

YankoYanko

8,3492830

answered Mar 29 at 11:34

YankoYanko

8,3492830

answered Mar 29 at 11:34

YankoYanko

8,3492830