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The number $43$ in $2's$ complement representation is

1. $01010101$
2. $11010101$
3. $00101011$
4. $10101011$
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it is really very confusing question. but now i got it.
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In 2's complement representation positive integers are represented in its normal binary form while negative numbers are represented in its 2's complement form

$2's$ complement representation is not same as $2's$ complement of a number. In $2's$ complement representation positive integers are represented in its normal binary form while negative numbers are represented in its $2's$ complement form. So, (c) is correct here.

http://www.ele.uri.edu/courses/ele447/proj_pages/divid/twos.html

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nice point arjun sir

MSB of 2's complement number has a weight of -2n-1

This is exactly the sort of question, which come under "silly mistakes " Category After exam.

Reading the question carefully is very important , answer will be 1's complement of 43 , i.e. C)

00101011.

As we know 2's complement of a number vary from

-( 2

(n-1) ) to 2(n-1) - 1

43 is between -64 to 63, so n= 7. Therefore minimum 7 bits is needed.

In 2's complement representation, for positive number the number can be written as 0 followed by its binary representation.

for negative number 1 followed by 2's complement of it's binary representation.

As 43 is a positive number so 0 followed by binary representation of 43

43 = 32+8+2+1 = 0101011

But options are in 8 bit representation so we can make 7bit number to 8bit by sign extension in 2's complement representation i.e if first bit is 1 the new leftmost bit will be 1 and if it is zero then new leftmost bit will be 0.
so we can rewrite 43 as 00101011 which is option C.

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