The idea that this expression is trying to relay is that the nature of the error in floating-point arithmetic is multiplicative rather than additive (which is the case for fixed-point arithmetic). This is because of the way that floating-point numbers are stored: as a mantissa multiplied by an exponent. Since the error is incurred only when rounding the mantissa, it is multiplicative (ignoring overflow and underflow).
The kind of guarantee that numerical algorithms can make is: if the input is correct up to a multiplicative error of $1 \pm \delta$, then the output will be correct up to a multiplicative error of $1 \pm \epsilon$. So $\delta$ (in your case, the machine $\delta$) measures the relative error involved when expressing real numbers as floating-point numbers, which is the information required to deduce the accuracy of numerical algorithms.