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Is (2,1), (1,-2), (-3,2), (2,-3) a function?

1 Answer

Jul 6, 2015

No - there are two distinct values for $f (2)$ $f(2)$ , so this relation does not describe a function.

Explanation:

A set of pairs $(x_{i}, y_{i})$ $(x_i, y_i)$ defines a function if

For all $i, j$ $i, j$ we have $x_{j} = x_{i} \Rightarrow y_{j} = y_{i}$ $x_j = x_i => y_j = y_i$

This is trivially satisfied if all the $x_{i}$ $x_i$ 's are distinct.

In our example, we have pairs $(2, 1)$ $(2, 1)$ and $(2, - 3)$ $(2, -3)$ which break the condition.

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