# Show that the Mod

It is given that f : R → R, given by f (x) = | x|

We can see that f(-1) = |-1| = 1, f(1) = |1| = 1

⇒ f(-1) = f(1), but -1 ≠ 1.

⇒ f is not one-one.

Now, we consider -1 ϵ R.

We know that f(x) = |x| is always positive

Therefore, there doesn't exist any element x in domain R such that f(x) = |x| = -1

⇒ f is not onto.

Therefore, modulus function is neither one-one nor onto.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

Fill in theMathematics - Exemplar

State True Mathematics - Exemplar

State True Mathematics - Exemplar

State True Mathematics - Exemplar

Let A = {1, 2, 3}Mathematics - Exemplar

Show that the relMathematics - Board Papers

Let N denote the Mathematics - Board Papers