Find a Bijection between the two intervals Find a bijection from one to the other:
$R$ and $(\sqrt(2), \infty)$
I was attempting this proof with $f(x) = \sqrt{2} + 2^x$

Proof:
  Consider the function $f(x) = \sqrt{2} + 2^x$
  
  Let $\sqrt{2} + 2^a = \sqrt{2} + 2^b$
  
  $2^a = 2^b => a = b$
  
  Thus f is injective


This is where I got stuck. My text says I should start out with the following:
Now let $ c > 0$
Then $\sqrt{2}+2^c = c$ 
I haven't been able to figure out how to solve this, which makes me think that I chose the wrong bijection? 
Can anyone offer any advice or other possible bijections?
 A: It sounds like you want to show surjectivity now.  You don't want this:
$$ \sqrt2 + 2^c = c$$
You want this:
$$ \sqrt2 + 2^x = c$$
and you want to solve for $x$.  Also, you want $c > \sqrt2$, not $c > 0$.  The reason is that you want to show the following:

$ f : \Bbb{R} \to (\sqrt2, +\infty)$, defined to be $f(x) = \sqrt2 + 2^x$, is surjective.  That is, for all $c \in (\sqrt2, +\infty)$ there exists $x \in \Bbb{R}$ such that $f(x) = c$.

So you just want to explicitly find $x \in \Bbb{R}$ such that $f(x) = c$.  I claim that $x = \log_2(c - \sqrt2)$ works.  Here's why.  First note that $c > \sqrt2$ means $c-\sqrt2 > 0$, so $\log_2(c-\sqrt2)$ is a valid expression.  Furthermore,
\begin{align*}
  f(x) &= f(\log_2(c-\sqrt2))\\
    &= \sqrt2 + 2^{\log_2(c-\sqrt2)}\\
    &= \sqrt2 + c - \sqrt2\\
    &= c.
\end{align*}
Therefore $f(x) = c$, so $f$ is surjective.
A: Your bijection is correct. 
Since $g(x)=2^x$ has the image $(0,\infty)$ the image of $f(x)=\sqrt{2}+2^x$ is $(\sqrt{2},\infty)$.
$f$ is injective:
$f(a)=f(b)\Rightarrow \sqrt{2}+2^a=\sqrt{2}+2^b \Rightarrow 2^a=2^b \Rightarrow a=b$
$f$ is onto:
Given $c \in (\sqrt{2},\infty)$ take $x_0=\log_{2}(c-\sqrt{2})$ and then $f(x_0)=\sqrt{2}+2^{\log_{2}(c-\sqrt{2})}=\sqrt{2}+c-\sqrt{2}=c$
So $f$ is bijective.
