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GATE CS 2013 · Digital Logic

Question 4

MCQPrinted as CS Q.41 markWrong answer: −1/3Easy (estimate)

Question

The smallest integer that can be represented by an 8-bit number in 2’s complement form is

Choose one option
Official answer and solution
Official answer
B
From
Official GATE 2013 answer key

Solution

Recall the formula for the range of signed integers represented in n-bit 2's complement form.

For an nn-bit binary number represented in 2's complement form, the representable range is:

[−2n−1,2n−1−1][-2^{n-1}, 2^{n-1} - 1]

For n=8n = 8 bits:

  • Minimum representable integer =−28−1=−27=−128= -2^{8-1} = -2^7 = -128 (binary: 10000000).
  • Maximum representable integer =28−1−1=27−1=127= 2^{8-1} - 1 = 2^7 - 1 = 127 (binary: 01111111).

Therefore, the smallest integer is −128-128.

Final answer

Option (B)

Common trap

Confusing 2's complement range (−128-128) with 1's complement or sign-magnitude range (−127-127).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2013 Computer Science and Information Technology (CS)
Question
4 of 65, printed as CS Q.4
Section
Computer Science and Information Technology

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