How to prove that $G(x)=ax+b$ is a one-to-one correspondence where $a\neq0$ and $a,b\in\mathbb{R}$. $G(x) = ax + b$ where $a$ is not equal to $0$ and $b$ are real numbers.
Prove $G$ is a one-to-one correspondence.
I understand that for every $a$ there is a corresponding $b$-value that does not repeat; however, I do not understand how I can prove this.  
 A: With $G(x)=ax+b$, where $a\neq 0$ and $a,b\in\mathbb{R}$, we have the following:
Onto: Suppose $y\in\mathbb{R}$. Then let $x=\frac{y-b}{a}$. We then have that
\begin{align}
f\left(\frac{y-b}{a}\right) &= a\left(\frac{y-b}{a}\right)+b\\[0.75em]
&= (y-b)+b\\[0.5em]
&= y.
\end{align}
Thus, $G(x)$ is onto.
One-to-one: Suppose that $G(x_1)=G(x_2)$. We then have that
\begin{array}{rcl}
G(x_1) &= & G(x_2)\\
ax_1+b &= & ax_2+b\\
ax_1 &= &ax_2\\
x_1 &= & x_2.
\end{array}
Thus, $G(x)$ is one-to-one.

Thus, we have that $G(x)$ is onto and one-to-one. That is, we have shown that $G(x)$ is a bijection or one-to-one correspondence, thus concluding the proof. $\blacksquare$
A: There are four components to one to one correspondence. I'm assuming you mean between x and G(x).


*

*For each x in the domain set (real numbers, in this case), there
must be an element G(x).

*G(x) must be a unique element of the range. There can't be two results, like the positive and negative square roots of a number.

*And for each element y of the range of G
(which is not explicitly stated, so you can probably assume real
numbers, depending on who your proof is for) there must be an
element x such that G(x) = y. 

*That element x must be unique, so that G(x)=G(x') implies that x=x'.


You need to prove all four of those. #4 is probably easiest to prove by negation.
A: A function $f:X\to Y$ is a one-to-one correspondence between $X$ and $Y$ if every element in $Y$ has a unique preimage in $X$. (Since $f$ is a function, we already know that every $x\in X$ has a unique image $y\in Y$).
In other words, $\forall y \in Y \,\,\exists ! x\in X\,\, f(x) = y$.
Usually we break this up into two parts by proving that $f$ is a surjection and an injection. To show that $f$ is surjective, we prove that $\forall y \in Y \,\,\exists x\in X\,\, f(x) = y$. To show that $f$ is injective, we prove that this $x\in X$ is unique.
Let $g: \mathbb R \to \mathbb R$ defined by $g(x) = ax+b$, where $a\not=0$ and $b$ are real numbers.
Fix a $y_0\in \mathbb R$. Notice that if we evaluate $g$ at $x=\frac{y_0-b}{a}$ (a valid choice since $a\not = 0$), we'll have
$$g\left(\frac{y_0-b}{a}\right) = a\left(\frac{y_0-b}{a}\right) + b = y_0.$$
Now since $y_0$ was arbitrary, indeed, $\forall y \in \mathbb R\,\, \exists x\in\mathbb R\,\, g(x) = y.$ That is, $g$ is surjective.
To prove that this choice for $x$ is unique for any $y_0$, we suppose that there are two such choices $x_1, x_2$ such that $g(x_1) = y_0$ and $g(x_2) = y_0$. We will then show that, in fact, these two such choices are equal, and hence unique. That is, does $g(x_1) = g(x_2) \implies x_1 = x_2$ ? So we have
$$g(x_1) = g(x_2) \underset{\text{definition of }\ g}{\implies} ax_1 + b = ax_2 + b \underset{\text{subtract}\  b}{\implies} ax_1 = ax_2 \underset{\text{divide by}\ a\not =0}{\implies} x_1=x_2.$$
That is, $g$ is injective.
Now we have $\forall y \in \mathbb R \,\,\exists ! x\in \mathbb R\,\, g(x) = y$, and so $g$ is a one-to-one correspondence.
