Is the braket notation of the Dirac delta function symmetric? I have a book saying,

$\int \delta(x-x')\psi(x)dx = \psi(x')$ where $\psi(x) = \langle x\lvert\psi\rangle$, so our definition of delta function would be $\langle x'\lvert x\rangle = \delta(x-x')$.

However I could find some documents (example; refer to 3. Position Space) saying,
$$\delta(x'-x'') = \langle x'\lvert x''\rangle$$
which corresponds to $\delta(x-x') = \langle x\lvert x'\rangle$.
So the result should be
$$\delta(x-x') = \langle x\lvert x'\rangle (=) \langle x'\lvert x\rangle \tag{1}$$
I think neither of them is an error, because my book uses the definition many times and I have found many documents explaining as $\delta(x-x') = \langle x\lvert x'\rangle$. Is (1) correct?
 A: As DJBunk mentions the delta function is symmetric
$$\delta(x)=\delta(-x)$$
so you certainly have
$$\delta(x-x')=\delta(x'-x).$$
But you should also know that in general we have
$$\langle a | b \rangle = \langle b | a \rangle^* $$
and since in this case the inner product is real, you will also have
$$\langle x | x' \rangle = \langle x' | x \rangle .$$
So it doesn't matter which way you write the delta function or the inner product.
A: Either way is fine: 
$\delta(x-x')= \langle x| x' \rangle= \delta(x' - x) =\langle x'| x \rangle$. 
You can see this either from the fact that in the limiting definition 
$\delta (x-x') = \lim_{\sigma \rightarrow \infty} \frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{(x-x')^2}{2 \sigma^2}}$
is symmetric in $x$ and $x'$
or that either form does the same job under an integral:
$  \int^\infty_{-\infty} dx \delta(x-x') = \int^\infty_{-\infty} dx  \delta(x' - x) = 1 $
and as Mistake Ink pointed out:
$  \int^\infty_{-\infty} dx f(x' )\delta(x-x') = \int^\infty_{-\infty} dx f(x' ) \delta(x' - x) = f(x) $
