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Provide a counterexample to the following statement: If fog is one-to-one, then f is one-to-one.

User Wins
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Final answer:

A counterexample to the statement 'If fog is one-to-one, then f is one-to-one' can be shown using functions. Let's consider two functions: f(x) = x^2 and g(x) = √ (x). If we compose these functions, fog(x) = f(g(x)), we get (√ (x))^2 = x, which is the identity function. Here, fog is one-to-one because every input has a unique output. However, f(x) = x^2 is not one-to-one since different inputs can have the same output. Therefore, this counterexample disproves the given statement.

Step-by-step explanation:

A counterexample to the statement 'If fog is one-to-one, then f is one-to-one' can be shown using functions. Let's consider two functions: f(x) = x^2 and g(x) =√ (x). If we compose these functions, fog(x) = f(g(x)), we get (√ (x))^2 = x, which is the identity function. Here, fog is one-to-one because every input has a unique output. However, f(x) = x^2 is not one-to-one since different inputs can have the same output. Therefore, this counterexample disproves the given statement.

User Lilit
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