Number of bits required to represent $-64_{10}$ in $2s$ compliment form?

Question

I think it is $8$ bits including sign bit. But given answer is $7$ bits. Am I wrong?

Comments

2's complement range for n-bits numbers is -2^n-1 to +2^n-1 - 1

So, -64 = -2^n-1

^{$\Rightarrow n - 1 = 6$}

n = 7 is correct

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It was known @just_bhavna but is sign bit included in $7$ bits? I think no.

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7 bits

Answer 1

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Comments

Why not taking sign bit?

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range of numbers for 2's compliment is :   -  2^(n-1)  to    2^(n-1)  -1  

-2^(n-1)   is the lowest negative number we can represent in 2's compliment (iam saying lowest because in negative sign small number becomes largest and viceversa)

 

now you want to represent  -64

let me try to show you on number line

 

-2^(n-1) <----------------- -ve number--------------0---------+ve number----------------> 2^(n-1) -1   

 

now -64 will lie on negative side and   

-64 will be >=  -2^(n-1)    

=   2^(n-1) >= 64 (multiplying by minus sign , equality sign changes)

take log  on both sides

(n-1)  log2 (base2)  >=  log64(base 2)

n-1  > =  6    

n =7  bits answer :)