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Prove that A(m+1,n)A(m,n)whenever m and n are nonnegative integers.

Short Answer

Expert verified

It has been proved.

Step by step solution

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01

Introduction

A non negative integer is an integer that that is either positive or zero. It's the union of the natural numbers and the number zero. Sometimes it is referred to as , and it can be defined as the as the set {0,1,2,3,.....}.

02

Solution

Consider m and n be non-negative integers.

The definition of Ackermann’s function is as follows

A(m,n)=2n    ifm=00    ifm1andn=02    ifm1andn=1A(m1,A(m,n1))    ifm1andn2

Individuals shall prove that A(m+1,n)A(m,n)using mathematical induction

This will be proved by double induction, first on m and n then

03

Step 3: Basis step

First consider m = 0

For all n, individuals have A (m,n) = 2n , where 2n > 2n

For n = 0 and n = 1 , individuals have A (m + 1, n) = 2n , where 2n > 2n

For n > 1 , individuals have

A (m + 1,n) = 2n , where 2n > 2n

Consider be a non-negative integer individuals assume that

A (m + 1,t) > A (m,t) ................(1)

04

Inductive step

Individuals prove that result holds for n + 1 , that is,

Greater than equal to,

A(m+2,n + 1) > A (m+1,n+1)

For n = 0 , individuals get 0 > 0

For n = 1, individuals get 2 > 2

Individuals assume that

A(m+2,n ) > A (m+1,n) .......(2)

Consider,

A(m+2,n + 1) = A (m+1,A (m + 2 , n))

Using definition,

A(m+1,A(m+1,n))Use(1)A(m+A(m+1,n))Use(2)=A(m+1,n+1)

Using definition backward

Hence, it is proved by principle of mathematical induction.

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