Proof for pascal's triangle consisting only of natural numbers While going through Spivak, i encountered the problem of proving that every number in pascal's triangle is positive via induction. Another property that was proven before this was $\left( {\begin{array}{*{20}c} n+1 \\ k \\ \end{array}} \right)=\left( {\begin{array}{*{20}c} n \\ k-1 \\ \end{array}} \right)+\left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right)$
I figured that i can do this by proving that if the nth row consists of natural numbers, so must the (n+1)th row. I also proved that the first row consists of natural numbers through simple evaluation of $({\begin{array}{*{20}c} 1 \\ 1 \\ \end{array}})$
The problem is, how do i prove the part about how the nth row being natural implies that the (n+1)th row is also natural? I can deduce that every element in the (n+1)th row is a sum of two elements of the nth row, and hence, should be the sum of two natural numbers, i.e a natural number of their own.
But this is just an english statement and does not sound like a proper proof to me. How do i state this properly?
 A: Base:
$$\binom00=1\in\mathbb N.$$
Induction hypothesis:
$$\forall k: 0\le k\le n:\binom nk\in\mathbb N.$$
Induction step: (mind the strict inequalities)
$$\forall k: 0< k<n+1:\binom {n+1}k=\binom n{k-1}+\binom nk\land\binom n{k-1}\in\mathbb N\land \binom nk\in\mathbb N\\\implies\\
\forall k: 0< k<n+1:\binom {n+1}k\in\mathbb N.$$
As in addition
$$\binom{n+1}0=\binom{n+1}{n+1}=1\in\mathbb N,$$ we have
$$\forall k: 0\le k\le n+1:\binom{n+1}k\in\mathbb N.$$
Then by induction
$$\forall n,k:n\ge0,0\le k\le n:\binom nk\in\mathbb N.$$

Note that the Base is actually never used in the induction step, as the endpoints are handled differently. We just added it for the property to hold even with $n=0$.
A: The first row consist of $R_1(1) = 1$;
Each row $R_n$ consists of terms $R_n(i); i = 1...n$ where $R_n(1) = R_{n-1}(1)$, $R_n(i) = R_{n-1}(i) + R_{n-1}(i -1);i>1$ and $i < n$ and $R_n(n) = R_{n-1}(n-1)$.
Initial step:  $R_1 = \{1\}$ consist of only natural numbers.
Induction step: If $R_n$ consists of only natural numbers then $R_{n+1}$ consists of only natural numbers.
Proof:  $R_{n+1}(1) = R_n(1) \in \mathbb N$.  $R_{n+1}(n+1) = R_n(n) \in \mathbb N$.  And for all others $R_{n+1}(i)= R_{n}(i) + R_{n}(i -1)$ which is the sum of two natural numbers and so a natural number.
So $R_{n+1}$ consists of natural numbers.
Hence via induction we can conclude all rows consists of only natural number.
