In IEEE arithmetic, assume that a and b are normalized floating point numbers. Then $($i$)$ fl$($a $-$ b$)=-$ f $*$ l$($b $-$ a$)($ii$)$ fl$($a $+$ a$)=$ fl$(2$a$)($iii$)$ fl$(0\mathrm{.}5$a$)=$ fl$($a $/2)(2)($iv$)$ fl$(($a $+$ b$)+$ c$)=$ fl$($a $+($b $+$ c$))$