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Question
Show that the function f: R → R given by f(x) = x3 is injective.
Solution 1
f: R → R is given as f(x) = x3.
Suppose f(x) = f(y), where x, y ∈ R.
⇒ x3 = y3 ... (1)
Now, we need to show that x = y.
Suppose x ≠ y, their cubes will also not be equal.
⇒ x3 ≠ y3
However, this will be a contradiction to (1).
∴ x = y
Hence, f is injective.
Solution 2
Let x1, x2 ∈ R be such that
`f (x_1) = f(x_2) = x_1^3 = x_2^3`
= x1 = x2
∴ f is one-one.
Hence, f(x) = x3 is injective.
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