Show that the function f:R→R, defined as f(x)=x2, is neither one-one nor onto.
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Step-by-Step Solution
Step 1: Check for One-One (Injectivity)
To check if a function is one-one, we assume f(x1)=f(x2) and try to prove that x1=x2. For the given function f(x)=x2, if f(x1)=f(x2), then x12=x22. This implies that x1=x2 or x1=−x2.
Step 2: Provide a Counterexample for One-One
Since x1 can be equal to −x2, the function is not one-one. For example, f(1)=1 and f(−1)=1. Here, two different elements in the domain (1 and -1) map to the same element in the codomain (1).
Step 3: Check for Onto (Surjectivity)
To check if a function is onto, we need to show that for every element y in the codomain, there exists an element x in the domain such that f(x)=y. For f(x)=x2, we set x2=y, which means x=±y.
Step 4: Provide a Counterexample for Onto
The domain and codomain are both the set of real numbers R. If we choose a negative number from the codomain, for example, y=−1, then x=±−1. This value of x is not a real number. Therefore, there is no real number x such that f(x)=−1. This means not every element in the codomain has a pre-image in the domain, so the function is not onto.