Answer to Question #148110 in Discrete Mathematics for Promise Omiponle

Question

Define a function A: N x N -> N as follows: A(m, n) ={2n, if m= 0; 0, if m≥1 and n= 0; 2, if m≥1 and n= 1; A(m-1, A(m, n-1)), if m≥1 and n≥2
(a) Calculate the following:
(i)A(1,0)
(ii)A(0,1)
(iii)A(1,1)
(iv)A(2,2).

Answer 1

Consider the function $A:\mathbb N\times\mathbb N\to\mathbb N,$

$A(m,n)=\begin{cases} 2n,\ \ \ m=0\\ 0,\ \ \ \ \ m\ge 1\text{ and } n=0\\ 2, \ \ \ \ \ m\ge 1\text{ and } n =1\\ A(m-1,A(m,n-1)),\ \ \ \ m\ge 1\text{ and } n\ge 2 \end{cases}$

(i) Since $m=1$ and $n=0$ we use the formula $A(m,n)=0$, and have that $A(1,0)=0$

(ii) Since $m=0$ we use the formula $A(m,n)=2n$, and have that $A(0,1)=2$

(iii) Since $m=1$ and $n=1$ we use the formula $A(m,n)=2$, and have that $A(1,1)=2$

(iv) We use the formula $A(m,n)=2$ for $m=2$ and $n=1$ to conclude that $A(2,1)=2$.

In the following we use the formula $A(m,n)=A(m-1,A(m,n-1))$ to calculate $A(2,2)$ and $A(1,2)$, and the formula $A(m,n)=2n$ to calculate $A(0,2):$

$A(2,2)=A(1, A(2,1))=A(1,2)=A(0, A(1,1))=A(0,2)=4$