# Given modulus (f(a)-f(b))=1 and a not equal b, what is f(a)?

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You should assume that `b = 0` such that:

`|f(a) - f(0)| = 1 =gt f(a) - f(0) = +- 1 =gt f(a) = f(0) +- 1`

If `f(a) = f(0) + 1` and `f(b) = f(0) - 1` , yields:

`|f(a) - f(b)| = |f(0) + 1 -f(0)+ 1| = |2| = 2!=1` contradiction

Hence, the function f(a) is constant and it could be either `f(a) = f(0) + 1` or `f(a) = f(0)- 1` such that:

`|f(a) - f(b)| = 0 != 1` contradiction

**Hence, evaluating the possible functions f(a), under given conditions, yields that there is no such a function that satisfies the given property `|f(a) - f(b)| = 1` .**