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December 27th, 2016, 03:50 AM  #1 
Senior Member Joined: Jan 2013 From: Italy Posts: 152 Thanks: 7  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! 
December 27th, 2016, 09:04 AM  #2 
Senior Member Joined: Dec 2012 Posts: 917 Thanks: 23 
Apply the definition of bijection. Than try to figure out if it's possible to do the same with $Y^2=R^2X^2$ 
December 27th, 2016, 10:29 AM  #3 
Senior Member Joined: Aug 2012 Posts: 956 Thanks: 189  
December 27th, 2016, 11:03 AM  #4  
Senior Member Joined: Jan 2013 From: Italy Posts: 152 Thanks: 7  Quote:
To verify that the function is injective I have also tried this: $\alpha_{a,b} = ax+b \mbox{ is injective if } \\ x_1 \ne x_2 \Rightarrow \alpha_{a,b}(x_1) \ne \alpha_{a,b}(x_2), (x_1, x_2 \in \mathbb{R}) \\ ax_1 + b \ne ax_2 + b \\ ax_1  ax_2 \ne 0 \\ x_1  x_2 \ne 0 \, (\forall a \ne 0) \\ x_1 \ne x_2$  

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function, invertible, onetoone 
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