1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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:


    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.


    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 .
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted