How does it follow from the Pascal's Triangle that binomial coefficient are integers











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So I was reading this lemma which states:
Let $m,n$ be natural numbers such that $1 leq m leq n$. Then
begin{equation*}
{nchoose m-1} + {nchoose m} = {n+1choose m}.
end{equation*}
It follows from this lemma using induction that the binomial coefficients are integers, rather than just rational numbers.



Using induction I could only prove that this equation is true for all natural numbers. How would I show that these coefficients are integers?






binomial-coefficients binomial-theorem







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  • Note that $binom{n}{0}=1$ for all $ninBbb N$ as well as $binom{n}{k}=0$ for $k<0$. Applying double induction then along with Pascal's identity and the well known property that the sum of two natural numbers is again a natural number will imply the result.
    – JMoravitz
    2 days ago

















up vote
0
down vote

favorite












So I was reading this lemma which states:
Let $m,n$ be natural numbers such that $1 leq m leq n$. Then
begin{equation*}
{nchoose m-1} + {nchoose m} = {n+1choose m}.
end{equation*}
It follows from this lemma using induction that the binomial coefficients are integers, rather than just rational numbers.



Using induction I could only prove that this equation is true for all natural numbers. How would I show that these coefficients are integers?






binomial-coefficients binomial-theorem







share|cite|improve this question






















  • Note that $binom{n}{0}=1$ for all $ninBbb N$ as well as $binom{n}{k}=0$ for $k<0$. Applying double induction then along with Pascal's identity and the well known property that the sum of two natural numbers is again a natural number will imply the result.
    – JMoravitz
    2 days ago















up vote
0
down vote

favorite









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0
down vote

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So I was reading this lemma which states:
Let $m,n$ be natural numbers such that $1 leq m leq n$. Then
begin{equation*}
{nchoose m-1} + {nchoose m} = {n+1choose m}.
end{equation*}
It follows from this lemma using induction that the binomial coefficients are integers, rather than just rational numbers.



Using induction I could only prove that this equation is true for all natural numbers. How would I show that these coefficients are integers?






binomial-coefficients binomial-theorem







share|cite|improve this question













So I was reading this lemma which states:
Let $m,n$ be natural numbers such that $1 leq m leq n$. Then
begin{equation*}
{nchoose m-1} + {nchoose m} = {n+1choose m}.
end{equation*}
It follows from this lemma using induction that the binomial coefficients are integers, rather than just rational numbers.



Using induction I could only prove that this equation is true for all natural numbers. How would I show that these coefficients are integers?






binomial-coefficients binomial-theorem




binomial-coefficients binomial-theorem






share|cite|improve this question













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asked 2 days ago









Seji

153




153












  • Note that $binom{n}{0}=1$ for all $ninBbb N$ as well as $binom{n}{k}=0$ for $k<0$. Applying double induction then along with Pascal's identity and the well known property that the sum of two natural numbers is again a natural number will imply the result.
    – JMoravitz
    2 days ago




















  • Note that $binom{n}{0}=1$ for all $ninBbb N$ as well as $binom{n}{k}=0$ for $k<0$. Applying double induction then along with Pascal's identity and the well known property that the sum of two natural numbers is again a natural number will imply the result.
    – JMoravitz
    2 days ago


















Note that $binom{n}{0}=1$ for all $ninBbb N$ as well as $binom{n}{k}=0$ for $k<0$. Applying double induction then along with Pascal's identity and the well known property that the sum of two natural numbers is again a natural number will imply the result.
– JMoravitz
2 days ago






Note that $binom{n}{0}=1$ for all $ninBbb N$ as well as $binom{n}{k}=0$ for $k<0$. Applying double induction then along with Pascal's identity and the well known property that the sum of two natural numbers is again a natural number will imply the result.
– JMoravitz
2 days ago












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I suppose that, in this context, $binom nm$ is defined as$$binom nm=frac{n!}{m!(n-m)!}.$$With this definition, it is clear that $binom nminmathbb Q$, but it is not clear that it is an integer.



However, it is clear that $binom n0=binom nn=1$, which is an integer. And, since$$binom n{m-1}+binom nm=binom{n+1}m,$$it follows (by induction on $n$), that each $binom nm$ is an integer.






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    I suppose that, in this context, $binom nm$ is defined as$$binom nm=frac{n!}{m!(n-m)!}.$$With this definition, it is clear that $binom nminmathbb Q$, but it is not clear that it is an integer.



    However, it is clear that $binom n0=binom nn=1$, which is an integer. And, since$$binom n{m-1}+binom nm=binom{n+1}m,$$it follows (by induction on $n$), that each $binom nm$ is an integer.






    share|cite|improve this answer

























      up vote
      2
      down vote













      I suppose that, in this context, $binom nm$ is defined as$$binom nm=frac{n!}{m!(n-m)!}.$$With this definition, it is clear that $binom nminmathbb Q$, but it is not clear that it is an integer.



      However, it is clear that $binom n0=binom nn=1$, which is an integer. And, since$$binom n{m-1}+binom nm=binom{n+1}m,$$it follows (by induction on $n$), that each $binom nm$ is an integer.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        I suppose that, in this context, $binom nm$ is defined as$$binom nm=frac{n!}{m!(n-m)!}.$$With this definition, it is clear that $binom nminmathbb Q$, but it is not clear that it is an integer.



        However, it is clear that $binom n0=binom nn=1$, which is an integer. And, since$$binom n{m-1}+binom nm=binom{n+1}m,$$it follows (by induction on $n$), that each $binom nm$ is an integer.






        share|cite|improve this answer












        I suppose that, in this context, $binom nm$ is defined as$$binom nm=frac{n!}{m!(n-m)!}.$$With this definition, it is clear that $binom nminmathbb Q$, but it is not clear that it is an integer.



        However, it is clear that $binom n0=binom nn=1$, which is an integer. And, since$$binom n{m-1}+binom nm=binom{n+1}m,$$it follows (by induction on $n$), that each $binom nm$ is an integer.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        José Carlos Santos

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