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Binary question

  1. Sep 18, 2013 #1
    1. The problem statement, all variables and given/known data
    What is the decimal equivalent of the largest unsigned binary that can be obtained with:
    a.8bits. b:n bits

    2. Relevant equations



    3. The attempt at a solution
    A:2^8=256. 2^8-1=255 how I can change the number to binary number?
    B.I can't get the direction of n bits
     
  2. jcsd
  3. Sep 18, 2013 #2

    UltrafastPED

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    Science Advisor
    Gold Member

    Repeated division by two - they walk you through this both ways here: http://www.purplemath.com/modules/numbbase.htm

    For n bits the maximum number is 2^n -1.

    Actually for these two examples you don't need to do any arithmetic. Just ask yourself: what is the largest number that can be displayed on a 4 digit odometer?
     
  4. Sep 22, 2013 #3
    Hope this can be some sort of a tool to help you:
    Suppose your have a decimal number X and you want to convert it into the binairy number N made of n bits.

    1st of all, in a general fashion:

    X=[itex]a_{0}[/itex]*[itex]2^{0}[/itex]+[itex]a_{1}[/itex]*[itex]2^{1}[/itex]+[itex]a_{2}[/itex]*[itex]2^{2}[/itex]......[itex]a_{n-1}[/itex]*[itex]2^{n-1}[/itex]

    If n=8, the highest number will be 255, and in binairy it is written as 11111111.
    Keep in mind that each "number 1" is the value of a specific [itex]a_{i}[/itex]

    And the reading directions of these values are opposite i.e. [itex]a_{0}[/itex] is the first "1" from the right of 11111111, [itex]a_{1}[/itex] is the 2nd one.

    For example: 11001001
    [itex]a_{0}[/itex] =1
    [itex]a_{1}[/itex] =0
    [itex]a_{2}[/itex] =0
    .
    .
    .
    [itex]a_{7}[/itex] =1

    2nd of all:
    How did I transform 255 to the number 11111111?

    Here's a general method (Bear in mind there are other methods):

    Take your n bit number (let's consider 186 to add diversity, and in this case it's 8 bits):

    Is 186 >= [itex]2^{7}[/itex]=128? Yes.
    thus: [itex]a_{7}[/itex]=1.

    Now take 186 and substract [itex]2^{7}[/itex] => 58.
    And redo the operation:

    Is 58 >= [itex]2^{6}[/itex]=64? No.
    thus: [itex]a_{6}[/itex]=0.

    Do not substract anything since 58<64

    Now go to the 3rd digit.

    Is 58 >= [itex]2^{5}[/itex]=32? Yes.
    thus [itex]a_{5}[/itex]=1.

    58-32=26.

    Is 26 >= [itex]2^{4}[/itex]=16? Yes.
    thus [itex]a_{4}[/itex]=1.

    Is 26-16=10 > [itex]2^{3}[/itex]=8? Yes.
    thus [itex]a_{3}[/itex]=1.

    Is 10-8 >= [itex]2^{2}[/itex]=4? No.
    thus [itex]a_{2}[/itex]=0.

    Is 10-8>= [itex]2^{1}[/itex]=2? Yes.
    thus [itex]a_{2}[/itex]=1.

    Is 2-2 >= [itex]2^{0}[/itex]=1? No.
    thus the last digit on the right is 0.

    Result: 10111010

    Verification: 128+0+32+16+8+0+2+0=186. The operation is correct.

    Hope this helps and clearify the idea of binairy to decimal transformation and vice-versa .
     
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