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Oksi-84 [34.3K]
3 years ago
6

How do I solve this? Please include clear instructions as I don't have much experience with functions and relations :(

Mathematics
1 answer:
mojhsa [17]3 years ago
8 0

Answer:

9

Step-by-step explanation:

From 3),

A(2,3)=A(2-1,A(2,3-1))=A(1,A(2,2))

Consider A(2,2). From 3),

A(2,2)=A(2-1,A(2,2-1))=A(1,A(2,1))

Consider A(2,1). From 3),

A(2,1)=A(2-1,A(2,1-1))=A(1,A(2,0))

Form 2),

A(2,0)=A(2-1,1)=A(1,1)=\{{\text{Use 3)}}\}=A(1-1,A(1,1-1))=A(0,A(1,0))=\{\text{Use 2)}\}=A(0,A(1-1,1))=A(0,A(0,1))=\{\text{Use 1)}\}=A(0,1+1)=A(0,2)=\{\text{Use 1)}\}=2+1=3

Then

A(2,1)=A(1,3)=\{{\text{Use 3)}}\}=A(1-1,A(1,3-1))=A(0,A(1,2))=\{{\text{Use 3)}}\}=A(0,A(1-1,A(1,2-1)))=A(0,A(0,A(1,1)))=\{\text{A(1,1)}=3 \}=A(0,A(0,3))=\{{\text{Use 1)}}\}=A(0,3+1)=A(0,4)=\{{\text{Use 1)}}\}=4+1=5

Then

A(2,2)=A(1,5)=\{{\text{Use 3)}}\}=A(1-1,A(1,5-1))=A(0,A(1,4))=\{{\text{Use 1)}}\}=A(1,4)+1=\{{\text{Use 3)}}\}=A(1-1,A(1,4-1))+1=A(0,A(1,3))+1=\{{\text{Use 1)}}\}=A(1,3)+1+1=A(1,3)+2=\{{\text{Use 3)}}\}=A(1-1,A(1,3-1))+2=A(0,A(1,2))+2=\{{\text{Use 1)}}\}=A(1,2)+1+2=A(1,2)+3=\{{\text{Use 3)}}\}=A(1-1,A(1,2-1))+3=A(0,A(1,1))+3=\{{\text{Use 3)}}\}=A(1,1)+1+3=\{A(1,1)=3\}=3+4=7

Hence,

A(2,3)=A(1,7)=\{{\text{Use 3)}}\}=A(1-1,A(1,7-1))=A(0,A(1,6))=\{{\text{Use 1)}}\}=A(1,6)+1=...=1+1+7=9

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