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
asked Nov 19 at 23:12
Seji
153
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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
asked Nov 19 at 23:12
Seji
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
Nov 19 at 23:17
add a comment |
up vote
0
down vote
favorite
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
asked Nov 19 at 23:12
Seji
153
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
asked Nov 19 at 23:12
Seji
153
asked Nov 19 at 23:12
Seji
153
asked Nov 19 at 23:12
Seji
153
asked Nov 19 at 23:12
Seji
153
asked Nov 19 at 23:12
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
Nov 19 at 23:17
add a comment |
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
Nov 19 at 23:17
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
Nov 19 at 23:17
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
Nov 19 at 23:17
add a comment |
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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.
answered Nov 19 at 23:18
José Carlos Santos
146k22117217
add a comment |
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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.
answered Nov 19 at 23:18
José Carlos Santos
146k22117217
add a comment |
up vote
1
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.
answered Nov 19 at 23:18
José Carlos Santos
146k22117217
add a comment |
up vote
1
down vote
up vote
1
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.
answered Nov 19 at 23:18
José Carlos Santos
146k22117217
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.
answered Nov 19 at 23:18
José Carlos Santos
146k22117217
answered Nov 19 at 23:18
José Carlos Santos
146k22117217
answered Nov 19 at 23:18
José Carlos Santos
146k22117217
answered Nov 19 at 23:18
José Carlos Santos
146k22117217
146k22117217
add a comment |
add a comment |
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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
Nov 19 at 23:17