Computing $x_+ - x$ where $x_+$ is the rounded binary number of $x$ This question stems directly from Converting $\frac{2}{7}$ to a binary number in a $32$ bit computer
I want to check if $x_+$ (increase the value of the last bit by one unit and discard all the bits after the last bit) is closer to $x$ than $x_-$ (discard all bits after the last bit).
$x = (0.\underbrace{100100100100100100100100}_{24 \text{ bits}}10\overline{010} )_{2} \times 2^{-1}$
$x_{-} = 0.\underbrace{100100100100100100100100}_{24 \text{ bits}} \times 2^{-1}$
$ x_+ = \left ((0.\underbrace{10010010010010010010010}_{23\text{ bits}}\mathbf{0})_{2} + 2^{-24} \right )\times 2^{-1} = (0.100100100100100100100101)_{2} \times 2^{-1}$
$$x-x_{-} = 0.\underbrace{0 \cdots 0 }_{24 \text{ zeros}}10\overline{010} \times 2^{-1} = 0.10\overline{010} \times 2^{-25} = \frac{2}{7} \times 2^{-25}$$
$$x_+ - x = 0.\underbrace{1}_{24\text{ th bit in }x_+}00\overline{0} \times 2^{-24} - 0.10010010\overline{010}\times 2^{-24} = 0.1101\overline{101}\times 2^{-25}$$
I don't think I did the $x_+ - x$ calculuation correctly because if I "borrow" from the the 24th bit in $x_+$, then that $24$th bit becomes a $0$. Then how we subtract that $24$th bit in $x_+$ from the $24$th bit in $x$?
 A: There are still at least two errors in the calculation.
For one thing,
$$0.10\overline{010}_2 =
0.\overline{100}_2 = 100_2 \times 0.\overline{001}_2 =  \frac 47.$$
Therefore the correct value of $x - x_-$ is
$$ x - x_- = \frac 47 \times 2^{-25}. $$
For another thing, the discarded bits should consistently have the same value,
$0.10\overline{010} \times 2^{-25} = 0.\overline{100} \times 2^{-25},$ 
both when computing $x - x_-$ and when computing $x_+ - x.$
You have written $0.10\overline{010} \times 2^{-25}$
in one place and $0.10\overline{010} \times 2^{-\mathbf{24}}$ in the other.
The correct value of $x_+ - x$ is
$$ x_+ - x = 1 \times 2^{-25} - 0.\overline{100} \times 2^{-25}
  = 0.\overline{011} \times 2^{-25} = \frac 37 \times 2^{-25}.
$$
Checking,
$$ x_+ - x_- = (x_+ - x) + (x - x_-)
 = \frac 47 \times 2^{-25} + \frac 37 \times 2^{-25}
 = 2^{-25} = 2^{-24} \times 2^{-1},
$$
which is precisely the value of the least significant bit of the IEEE single-precision approximation of $\frac27,$
namely the bit you set to $1$ in $x_+$ and to $0$ in $x_-.$
