Ex 1.2, 4 - Show that Modulus Function f(x) = |x| is neither - To prove injective/ surjective/ bijective (one-one & onto)

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Ex 1.2 , 4 Show that the Modulus Function f: R → R given by f(x) = 𝑥﷯ , is neither one-one nor onto, where 𝑥﷯ is x, if x is positive or 0 and 𝑥﷯ is − x, if x is negative. f(x) = 𝑥﷯ = 𝑥 , 𝑥≥0 ﷮−𝑥 , 𝑥<0﷯﷯ Check one-one For example: f (1) = 1﷯ = 1 f (– 1) = 1﷯ = 1 Check onto f: R → R f(x) = 𝑥﷯ Let f(x) = y such that y ∈ R y = 𝑥﷯ Hence value of y is defined only if y is positive, But y is a real number Hence, if y is negative, there is not corresponding element of x Hence, f is not onto

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