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Math Help - Number conversions

  1. #1
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    Number conversions

    9E2B.D05_{16} to binary

    1011010111001.01101_{2} to decimal

    5386.94_{10} to hexadecimal

    Thanks.
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  2. #2
    Super Member

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    Hello, Dean!

    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: . 9 = 1001_2,\;E = 1110_2,\;2 = 0010_2


    We have: . 9\quad\;\;E\quad\;\;2\quad\;\; B\quad.\quad D\quad\;\;0\quad\;\;5
    . . . . . . . \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: . 1,001, 111,0 00,10 1,011 . 110,1 00,00 0,101_2



    1011010111001.01101_{2} to decimal
    We have: . 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}

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

    . . = \;5817.40625



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

    The integer: . 5386 \:=\:150A_{16} .
    (use your favorite method)


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

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

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

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

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

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

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

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

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


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


    But someone check my work . . . please!
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