quantum mechanics - Normalization of basis vectors with a continuous index?

I have an infinite basis which associates with each point, $x$, on the $x$-axis, a basis vector $|x\rangle$ such that the matrix of $|x\rangle$ is full of zeroes and a one by the $x^{\mathrm{th}}$ element. The book on Quantum Mechanics by Shankar says that the inner product between a basis vector and itself is not one, why not? Why can't these basis vectors be normalized to one, only to the Dirac delta function?

Answer

Any good basis should be complete. If the set of all $|x\rangle$ is complete, any other vector $|\psi\rangle$ in the Hilbert space of your system should be writeable as $|\psi\rangle=\sum_{x} |x\rangle\langle x|\psi\rangle$. This sum does not make sense for continuous variables $x$, hence the need to redefine the completeness relation with an integral (as Jan's answer demonstrates nicely). Once you employ integrals to define a meaningful completeness relation, then the relation $\langle x|y\rangle =\delta_{x,y}$ is not correct because it gives \begin{equation} |y\rangle=\int \mathrm{d}x |x\rangle\langle x|y\rangle=\int \mathrm{d}x |x\rangle\delta_{x,y}=0, \end{equation} which is inconsistent. The way out is to define $\langle x|y\rangle=\delta(x-y)$, which correctly gives \begin{equation} |y\rangle =\int \mathrm{d}x |x\rangle\langle x|y\rangle=\int \mathrm{d}x |x\rangle \delta(x-y)=|y\rangle. \end{equation} Now $\langle x|y\rangle=\delta(x-y)$ leads to $\langle x|x\rangle=\delta(0)$, that you may not like, but this is the best one can do.