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Binary question

  • Thread starter ccky
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  • #1
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Homework Statement


What is the decimal equivalent of the largest unsigned binary that can be obtained with:
a.8bits. b:n bits

Homework Equations





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
 

Answers and Replies

  • #2
UltrafastPED
Science Advisor
Gold Member
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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?
 
  • #3
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
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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