Problem A$-2$: Number base conversions II

Converting numbers with fractional parts to another base

To convert the integer part, use the same approach $($successive division$)$ as seen before: Take the integer value $(27)$ and divide by the radix $(2)...$ keep going until the quotient reaches zero.

To convert the fractional part, the approach is similar, but we use successive multiplication. Start with the fractional part $(\mathrm{.}375)$ and multiply by the radix $(2).$ You get $0\mathrm{.}75-$ the integer part of this value is the first digit of the fractional part of the binary value $(8$ in this case$).$ Now, take the fractional part $(\mathrm{.}75)$ and use that for the next step in the successive multiplication. The process continues until the fractional part reaches zero.

$($a$)$ Express ${27\mathrm{.}375}_{10}$ in binary.

$($b$)$ Express ${19\mathrm{.}2}_{10}$ in binary.

$($c$)$ You'll notice that in part $($b$),$ the fractional part in decimal $(\mathrm{.}2)$ does not convert nicely to a binary value. If we seek a particular precision, how do we know when to stop? Let's assume that there is an uncertainty of $+-1$ in the last digit of the original number; i$.$e$.,$ the value we want to convert is $(19\mathrm{.}2+-0\mathrm{.}1)($in base $10).$ Thus, we want to find out how many digits we need in the fractional part of the converted binary value so that it is at least as precise as the original value. Putting it all together, we need to satisfy the condition:

$($precision$/$ uncertainty in converted value$)($precision$/$ uncertainty in original value$)$

$2^{-n}{10}^{-m}$

$m=1$ since there was one digit in the fractional part of the original number and $n$ will be the number of digits needed in the fractional part of the converted binary answer. With all this in mind, solve for the required $n($which must be an integer$)$ to express ${19\mathrm{.}2}_{10}$ in binary, to an appropriate precision.

$($d$)$ Express the value $0E3$FED$\mathrm{.}2A9$ in decimal, to appropriate precision.

J$.$ C$.$ Sit $-$ ECE $212($Winter $2024$ LEC B$1)$ Problem set #$2$

$\frac{1}{1}$