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Question 157

For all non-negative integers x and y, f(x, y) is defined as below:

f(0, y) = y + 1

f(x + 1, 0) = f(x, 1)

f(x+ 1, y+ 1)= f(x, f(x+ 1, y))

Then, what is the value of f(1,2)?

Solution

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For f(1,2). First consider x=0 and y=1 and use 3rd given equation, we get f(0,f(1,1)) now for f(1,1) take x=0 and y=0 we get f(0,f(1,0)), for f(1,0) which we use 2nd equation we get f(0,1) whose value is 2. So we have f(0,f(1,0))= f(0,2) whose value is 3 then put this in f(0,f(1,1)) we get f(0,3) we get as 4

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