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Let f = {(1,1), (2, 3), (0, 1), (1, 3)}b...

Let `f = {(1,1), (2, 3), (0, 1), (1, 3)}`be a function from `Z to Z` defined by `f(x) = a x + b`, for some integers `a, b.`
Determine `a, b`.

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The correct Answer is:
N/a

f = {(1,1), (2,3),(0,-1),(-1,-3)} is a function from Z to Z such that
f (x) = ax + b (a, b `in` integers)
`because" "f(1)=a(1)+b=a+b`
but it is clear from f(1,1) that f(1) = 1.
`because" "a+b=1" "...(1)`
similarly, f(2) = a (2) + b =2a + b but from (2,3) it is clear that
`2a+b=3" "...(2)`
Subtracting equation (1) from equation (2), a = 2 Now, put a=2 in equation (1), 2 + b or b= - 1
Therefore, a= 2, b = -1
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