Let the set X = (-1, 1)
Define f: X--->R(set of real numbers) such that f(x) = x/(x-1)
Check if f is onto.
Let y = x/(x-1)
Then, x = y/(y-1)
Since y belongs to R, y=1 would result in an undefined value for x.
Hence, can we conclude f is not onto?
To check if the function f is onto, we need to determine if there exists an element in the codomain (set of real numbers) for every element in the domain (set X = (-1, 1)) such that f(x) equals that element.
Let's consider the equation x = y/(y - 1), which represents the inverse of f.
If we assume y = 1, which is a valid real number, we get x = 1/(1 - 1) = 1/0. However, dividing by zero is undefined in mathematics.
This means that for y = 1, there is no real number x that satisfies the equation x = y/(y - 1), as it would result in division by zero.
Since there is no x in X that maps to y = 1, we can conclude that f is not onto.
To summarize:
1. We considered the inverse equation x = y/(y - 1).
2. We observed that when y = 1, the equation results in division by zero, which is undefined.
3. As a result, there is no x in the domain X = (-1, 1) that maps to y = 1.
4. Therefore, f is not onto.