The guide book asks me for
A real number $x$ in range of a machine with $\beta=2$ (binary) and $n=24$ (24 mantissa positions), such that it satisfy $fl(x)=x(1+\delta)$, with $|\delta|$ as big as possible
in addition it ask me if
Can $|\delta|$ reach the bound $\epsilon$?.
My process: Let $x=q\cdot 2^m$.
If $fl(x)=Tr(q)\cdot 2^m$ then $|\delta|\leq\beta^{1-n}=2^{-23}=\epsilon_{tr}$
If $fl(x)=Ex(q)\cdot 2^m$ then $|\delta|\leq\frac{1}{2}\beta^{1-n}=2^{-24}=\epsilon_{ex}$.
So since $2^{-23}>2^{-24}$ i take $fl(x)=Tr(q)\cdot2^m=(q\cdot2^m)+\delta(q\cdot2^m)$.
But i don't know if i'm correct, and if so, what's next?
Can you help me?.
Your floating point numbers are consistent with IEEE single precision. Here the unit roundoff is $u=2^{-24}$. Now consider real numbers $x \in [1,2)$. The spacing between the floating numbers in this interval is $2u$, i.e., the numbers are $$1, 1+2u, 1+4u, \dotsc, 2-2u.$$ It follows that the error is bounded by $u$. This bound is achieved at the midpoints between successive floating point numbers, i.e., at $$1+u, 1+3u, 1+5u, \dotsc, 2-u.$$ The largest relative error occurs at the left endpoint, i.e., at the real number $x=1+u$. The floating point representation is either $\text{fl}(x)=1$ or $\text{fl}(x)=1+2u$ depending on the rounding mode. In either case, the relative error $\delta = (x -\text{fl}(x))/x$ satisfies $$|\delta| = u/x = u/(1+u) < u.$$ We conclude that $|\delta|$ cannot achieve the value $u$ for any $x \in [1,2)$.
The extension to general $x$ in the representational range is straightforward.