Ratio of Dirac delta functions

My Professor tells me that

$\frac{\delta(x-a)}{\delta(x-a)} = 1$

and justifies it by considering the Dirac delta as a limit of normalized Gaussians whose width approaches zero. I don't quite buy this explanation because I've been told that this definition is only an approximation of the Dirac delta.

Is this proof fine? Or is there a more rigorous proof of this? Or is this not even true?

Thanks.

• If the equality is supposed to mean $\delta(x-a)=1\times\delta(x-a)$ then, well, it is true. But e.g. $\delta(x-a)=(1+(x-a))\times\delta(x-a)$ is also true. – user8268 May 17 '11 at 6:44
• Is this a math professor or a physics professor? :-) – joriki May 17 '11 at 7:07
• @joriki: Isn't $x/x=1$ whatever $x$ ;-) – Fabian May 17 '11 at 7:18

In particular, as user8268 pointed out in the comments, multiplication of distributions works differently. The product of two general distributions (e.g. the product of two delta distributions) cannot be defined (at least not in a straightforward sense; see here). The product of a distribution with an ordinary function can be defined, but as user8268 pointed out, this is not invertible. Division is the inverse of multiplication, so if multiplication is not invertible, division makes no sense. If you applied your professor's division rule to user8268's example, you could cancel the delta distributions and get $1=1+x-a$, which is obviously wrong.
I'd suggest that you look at what your professor was trying to use this relationship for and see whether it can be justified rigorously without resorting to division of distributions. For instance, if you have two charge distributions and you know that they're both localized on a sphere of radius $R$, you can write them as $\rho_i(\theta,\phi)\delta(r-R)$; then if by some argument you find that they're the same, $\rho_1(\theta,\phi)\delta(r-R)=\rho_2(\theta,\phi)\delta(r-R)$, you can conclude that $\rho_1(\theta,\phi)=\rho_2(\theta,\phi)$ by other means than "dividing through" by the delta distribution.