1. ## Number conversions

$\displaystyle 9E2B.D05_{16}$ to binary

$\displaystyle 1011010111001.01101_{2}$ to decimal

$\displaystyle 5386.94_{10}$ to hexadecimal

Thanks.

2. Hello, Dean!

$\displaystyle 9E2B.D05_{16}$ to binary
Since 16 is a power of 2, there is a special trick we can use.

Convert each hexidecimal digit into its (four-digit) binary equivalent.
. . For example: .$\displaystyle 9 = 1001_2,\;E = 1110_2,\;2 = 0010_2$

We have: . $\displaystyle 9\quad\;\;E\quad\;\;2\quad\;\; B\quad.\quad D\quad\;\;0\quad\;\;5$
. . . . . . . $\displaystyle \overbrace{1001}^{\downarrow}\overbrace{1110}^{\do wnarrow}\overbrace{0010}^{\downarrow}\overbrace{10 11}^{\downarrow}\;\cdot\;\overbrace{1101}^{\downar row}\overbrace{0000}^{\downarrow}\overbrace{0101}^ {\downarrow}$

Answer: .$\displaystyle 1,001, 111,0 00,10 1,011 . 110,1 00,00 0,101_2$

$\displaystyle 1011010111001.01101_{2}$ to decimal
We have: .$\displaystyle 2^{12} + 2^{10} + 2^9 + 2^7 + 2^5 + 2^4 + 2^3 + 1 + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^5}$

. . $\displaystyle = \;4096 + 1024 + 512 + 128 + 32 + 16 + 8 + 1 + \frac{1}{4} + \frac{1}{8} + \frac{1}{32}$

. . $\displaystyle = \;5817.40625$

$\displaystyle 5386.94_{10}$ to hexadecimal
I did this one in two stages.

The integer: .$\displaystyle 5386 \:=\:150A_{16}$ .

The decimal: .$\displaystyle 0.94$ .
(This will take a while)

. . Let $\displaystyle 0.94 \:=\:0.abc{d}e\!f\!gh..._{16}$

Multiply by 16: .$\displaystyle 15.04 \:=\:a.bc{d}e\!f\!gh..._{16}$
. . Hence: .$\displaystyle a \:=\:15\:=\:F_{16}$

We have: .$\displaystyle 0.04 \:=\:0.bc{d}e\!f\!gh...$ .[1]
Multiply by 16: .$\displaystyle 0.64 \:=\:b.c{d}e\!f\!gh...$
. . Hence: .$\displaystyle b = 0$

We have: .$\displaystyle 0.64 \:=\:0.c{d}e\!f\!gh...$
Multiply by 16: .$\displaystyle 10.24 \:=\:c.{d}e\!f\!gh...$
. . Hence: .$\displaystyle c \:=\:10 \:=\:A_{16}$

We have: .$\displaystyle 0.24 \:=\:0.{d}e\!f\!gh...$
Multiply by 16: .$\displaystyle 3.84 \:=\:d.e\!f\!gh...$
. . Hence: .$\displaystyle d \:=\:3$

We have: .$\displaystyle 0.84 \:=\:0.e\!f\!gh...$
Multiply by 16: .$\displaystyle 13.44 \:=\:e.f\!gh...$
. . Hence: .$\displaystyle e \:=\:13 \:=\:D_{16}$

We have: .$\displaystyle 0.44 \:=\:0.f\!gh...$
Multiply by 16: .$\displaystyle 7.04 \:=\:f.gh...$
. . Hence: .$\displaystyle f \:=\:7$

We have: .$\displaystyle 0.04 \:=\:g.h...$
. . But this brings us back to [1].
Hence, the decimal has a repeating cycle: .$\displaystyle \overline{0A3D7}$

Answer: .$\displaystyle 5386.94 \;=\;150A.F\overline{0A3D7}_{16}$

But someone check my work . . . please!