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To add two finite floating-point numbers, compare their exponents, shift the smaller significand to a common exponent, add or subtract the signed significands, normalize the result, and apply floating-point rounding. Extra low-order bits are needed to determine correct rounding when shifting or cancellation discards information. A carry beyond the leading digit increases the exponent; cancellation can decrease it. Unlike exact real addition, this operation need not be associative. Changing the child evaluation order in a fixed binary expression tree need not change its parenthesization, whereas reassociation changes which intermediate results undergo floating-point rounding.

Both signs are positive, so perform addition of the aligned significands. The true exponents are $-1$ and $3$, a difference of four. Shift the smaller significand right four places, retaining its low bits:Add at the common exponent:Renormalization raises the true exponent to $4$ and the biased exponent to $E=19=10011_2$. There are only seven fraction places to retain, so no floating-point rounding changes the result:The complete word is 0100110000001000.

The significant-digit factor $m$ in a floating-point representation $(-1{)}^{s}m{\beta }^e$, where $s$ is the sign bit, $\beta$ the radix and $e$ the exponent. For a normal positive binary significand, $1\le m<2$, and the leading one need not be stored. In a subnormal number, the leading digit is zero at the minimum normal exponent scale. The significand is a numerical factor, distinct from the stored trailing fraction bits that encode part of it.