if a function is onetoone and onto then it is invertible. Hi, For each ordered pair (a,b) of real numbers with $a \ne 0$, let $\alpha_{a,b} : \mathbb{R} \rightarrow \mathbb{R}$ be defined by $\alpha_{a,b}(x) = ax+b$. Let $A$ denote the set of all such mappings. Then composition is an operation on A, $\alpha_{a,b} \circ \alpha_{c,d} = \alpha_{ac, ad+b}$. Consider the operation $\circ$ on the set A of the above exercise Prove that each $\alpha_{a,b} \in A$ is invertible by verifying that it is onetoone and onto. (with $a \ne 0$). So, this is my attempt: Assuming $\alpha_{a,b}$ both onetoone and onto, we'll go to show that $\alpha_{a,b}$ is invertible describing an inverse. Assuming $y = ax+b \in \mathbb{R}$ in the codomain, since it is onto, there is at least an element $x \in \mathbb{R}$ in the domain such that $\alpha_{a,b}(x) = y$ But, since $\alpha_{a,b}$ onetoone, then that element must be unique. We consider another function $\beta(y) = x$, this can be done for each element $y \in \mathbb{R}$ in the codomain, therefore we obtain a function $\beta : \mathbb{R} \rightarrow \mathbb{R}$ inverse of $\alpha_{a,b}$. Can you help me please? What do you think about it? Many thanks! 
Apply the definition of bijection. Than try to figure out if it's possible to do the same with $Y^2=R^2X^2$ 
Quote:
Quote:

Quote:
To verify that the function is injective I have also tried this: $\alpha_{a,b} = ax+b \mbox{ is injective if } \\ x_1 \ne x_2 \Rightarrow \alpha_{a,b}(x_1) \ne \alpha_{a,b}(x_2), (x_1, x_2 \in \mathbb{R}) \\ ax_1 + b \ne ax_2 + b \\ ax_1  ax_2 \ne 0 \\ x_1  x_2 \ne 0 \, (\forall a \ne 0) \\ x_1 \ne x_2$ 
All times are GMT 8. The time now is 01:50 AM. 
Copyright © 2018 My Math Forum. All rights reserved.