System of Numeration: ISC Class 11 Computer Science Notes

What System of Numeration is about

Weightage: System of Numeration is topic 1 of the ISC Class XI Computer Science syllabus, in Section A - Hardware. CISCE weights Section A at 25 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Computer Science syllabus, examination year 2028, Class XI unit weightage

Key concepts

Positional number systems and the base

In a positional number system, the value of a digit depends on its position. Each position has a place value that is a power of the base (radix).

For a number with digits d₃d₂d₁d₀ in base b:

Value = d₃ × b³ + d₂ × b² + d₁ × b¹ + d₀ × b⁰

Digits after the point have place values b⁻¹, b⁻², ...

System	Base	Digits
Binary	2	0, 1
Octal	8	0-7
Decimal	10	0-9
Hexadecimal	16	0-9, A (10), B (11), C (12), D (13), E (14), F (15)

A base-b system has exactly b digits, from 0 to b − 1. The base is written as a subscript: 1011₂, 37₈, 2F₁₆.

Why computers use binary: electronic circuits are reliable with two states (on/off, high/low voltage). Octal and hexadecimal are shorthands for binary: one octal digit = 3 bits, one hex digit = 4 bits.

Worked example

Find the decimal value of 2F₁₆ and 37₈. [2 marks]

Try it first, then show the solutionSolution

1. 2F₁₆ = 2 × 16 + 15 = 47.
2. 37₈ = 3 × 8 + 7 = 31.

Answer: 2F₁₆ = 47; 37₈ = 31.

Converting decimal to another base

Integer part: repeated division by the new base. Write the remainders; read them from bottom to top (last remainder is the most significant digit).

Algorithm (pseudocode):

result = ""
while n > 0
    r = n mod base
    result = digit(r) + result
    n = n div base
end while

Fraction part: repeated multiplication by the new base; the integer parts produced are the digits, read from top to bottom.

Example: 156 to binary: 156/2 = 78 r 0; 78/2 = 39 r 0; 39/2 = 19 r 1; 19/2 = 9 r 1; 9/2 = 4 r 1; 4/2 = 2 r 0; 2/2 = 1 r 0; 1/2 = 0 r 1. Reading upwards: 10011100₂. Similarly 156 = 234₈ = 9C₁₆.

Fraction: 0.625 × 2 = 1.25 (1); 0.25 × 2 = 0.5 (0); 0.5 × 2 = 1.0 (1) → 0.101₂.

The Java program below implements both directions; its output follows.

// Converts a decimal integer to any base from 2 to 16 by repeated division,
// and converts a string in any base back to decimal by positional expansion.
public class BaseConvert
{
    static final String DIGITS = "0123456789ABCDEF";

    static String toBase(int n, int base)
    {
        if (n == 0)
            return "0";
        String result = "";
        while (n > 0)
        {
            int r = n % base;                 // remainder is the next digit (from the right)
            result = DIGITS.charAt(r) + result;
            n = n / base;                     // quotient is divided again
        }
        return result;
    }

    static int toDecimal(String s, int base)
    {
        int value = 0;
        for (int i = 0; i < s.length(); i++)
        {
            int d = DIGITS.indexOf(Character.toUpperCase(s.charAt(i)));
            value = value * base + d;         // Horner's method: shift left one place, add digit
        }
        return value;
    }

    public static void main(String[] args)
    {
        int n = 156;
        System.out.println(n + " in base 2  = " + toBase(n, 2));
        System.out.println(n + " in base 8  = " + toBase(n, 8));
        System.out.println(n + " in base 16 = " + toBase(n, 16));
        System.out.println("3AF (base 16) = " + toDecimal("3AF", 16));
        System.out.println("756 (base 8)  = " + toDecimal("756", 8));
        System.out.println("101101 (base 2) = " + toDecimal("101101", 2));
    }
}

Output:

156 in base 2  = 10011100
156 in base 8  = 234
156 in base 16 = 9C
3AF (base 16) = 943
756 (base 8)  = 494
101101 (base 2) = 45

Worked example

Convert 45.625 (decimal) to binary. [2 marks]

Try it first, then show the solutionSolution

1. 45: 45/2 = 22 r 1; 22/2 = 11 r 0; 11/2 = 5 r 1; 5/2 = 2 r 1; 2/2 = 1 r 0; 1/2 = 0 r 1 → 101101.
2. 0.625: × 2 → 1 (0.25); × 2 → 0 (0.5); × 2 → 1 (0) → .101.

Answer: 45.625 = 101101.101₂.

Conversions among binary, octal and hexadecimal

Because 8 = 2³ and 16 = 2⁴, conversions use grouping of bits.

Binary → octal: group bits in threes from the binary point (leftwards for the integer part, rightwards for the fraction), padding with zeros; replace each group by its octal digit.

Binary → hexadecimal: group in fours the same way.

Octal or hex → binary: replace each digit by its 3-bit or 4-bit binary equivalent.

Octal ↔ hexadecimal: go through binary.

Examples:

• 10011100₂ → (010)(011)(100) = 234₈; (1001)(1100) = 9C₁₆.
• 3AF₁₆ → 0011 1010 1111 = 1110101111₂ (= 943).
• 756₈ → 111 101 110 = 111101110₂ → (0001)(1110)(1110) = 1EE₁₆ (= 494).
• 1101.11₂ → octal: (001)(101).(110) = 15.6₈; hex: (1101).(1100) = D.C₁₆.

Worked example

Convert 756₈ to hexadecimal through binary. [2 marks]

Try it first, then show the solutionSolution

1. 756₈ → 111 101 110 → 111101110₂.
2. Group in fours from the right: 0001 1110 1110 → 1EE₁₆.

Answer: 756₈ = 1EE₁₆ (both equal 494 in decimal).

Addition in different bases

Addition works exactly as in decimal: add column by column from the right; when a column sum reaches the base, write (sum − base) and carry 1 to the next column.

Binary rules: 0 + 0 = 0; 0 + 1 = 1; 1 + 1 = 10 (write 0, carry 1); 1 + 1 + 1 = 11 (write 1, carry 1).

Examples:

• Binary: 101101 + 11011 = 1001000₂ (45 + 27 = 72).
• Octal: 567₈ + 354₈: 7 + 4 = 11 = 8 + 3 → write 3, carry 1; 6 + 5 + 1 = 12 = 8 + 4 → write 4, carry 1; 5 + 3 + 1 = 9 = 8 + 1 → write 1, carry 1 → 1143₈ (375 + 236 = 611).
• Hexadecimal: 9AF₁₆ + 4C8₁₆: F + 8 = 23 = 16 + 7 → 7 carry 1; A + C + 1 = 23 → 7 carry 1; 9 + 4 + 1 = 14 = E → E77₁₆ (2479 + 1224 = 3703).

Check every answer by converting to decimal.

Worked example

Add 110111₂ and 101011₂, and check in decimal. [2 marks]

Try it first, then show the solutionSolution

1. Column sums from the right: 1+1 = 10 (0, c1); 1+1+1 = 11 (1, c1); 1+0+1 = 10 (0, c1); 0+1+1 = 10 (0, c1); 1+0+1 = 10 (0, c1); 1+1+1 = 11 (1, c1); final carry 1.
2. Result 1100010₂; check: 55 + 43 = 98 = 1100010₂.

Answer: 1100010₂ (= 98).

Subtraction using 1's and 2's complement

Computers subtract by adding a complement, just as in decimal 9's/10's complement.

1's complement of a binary number: invert every bit (0 → 1, 1 → 0). 2's complement = 1's complement + 1.

A − B using 1's complement (same number of bits):

1. Add A and the 1's complement of B.
2. If there is an end carry, add it back to the result (end-around carry): the answer is positive.
3. If there is no carry, the answer is negative: take the 1's complement of the result and put a minus sign.

A − B using 2's complement:

1. Add A and the 2's complement of B.
2. If there is an end carry, discard it: the answer is positive.
3. If there is no carry, the answer is negative: take the 2's complement of the result.

Example (1's): 11010 − 01101 (26 − 13): 1's comp of 01101 = 10010; 11010 + 10010 = 1 01100; add carry: 01100 + 1 = 01101 = 13.

Example (2's): 01101 − 11010 (13 − 26): 2's comp of 11010 = 00110; 01101 + 00110 = 10011, no carry → negative; 2's comp of 10011 = 01101 → −13.

Signed numbers: in an n-bit 2's complement system the leftmost bit is the sign bit (1 = negative); the range for 8 bits is −128 to +127. The program below does 13 − 26 with 8-bit patterns.

// Subtracts two 8-bit numbers by adding the 2's complement, and shows the bit patterns.
public class TwosComplement
{
    static String bits8(int v)
    {
        String s = Integer.toBinaryString(v & 0xFF);   // keep only the lowest 8 bits
        while (s.length() < 8)
            s = "0" + s;
        return s;
    }

    public static void main(String[] args)
    {
        int a = 13, b = 26;
        int onesB = (~b) & 0xFF;                        // invert every bit
        int twosB = (onesB + 1) & 0xFF;                 // add 1
        int sum = (a + twosB) & 0x1FF;                  // keep a possible 9th (carry) bit
        System.out.println("a          = " + bits8(a));
        System.out.println("b          = " + bits8(b));
        System.out.println("1's comp b = " + bits8(onesB));
        System.out.println("2's comp b = " + bits8(twosB));
        System.out.println("a + 2's b  = " + bits8(sum) + (sum > 0xFF ? " with carry" : " no carry"));
        System.out.println("as signed  = " + (byte) sum);
    }
}

Output:

a          = 00001101
b          = 00011010
1's comp b = 11100101
2's comp b = 11100110
a + 2's b  = 11110011 no carry
as signed  = -13

The last line confirms that 11110011 read as a signed 8-bit number is −13.

Worked example

Find the 8-bit 2's complement representation of −45. [2 marks]

Try it first, then show the solutionSolution

1. +45 = 00101101; 1's complement = 11010010.
2. Add 1: 11010011.

Answer: −45 = 11010011 (8-bit 2's complement).

Multiplication in different bases

Multiply digit by digit as in decimal, but carry in the base: if a product is p, write p mod base and carry p div base. Then add the partial products in that base. (The syllabus asks for multiplication without fractional parts.)

Binary rules: 0 × 0 = 0; 0 × 1 = 0; 1 × 1 = 1. Multiplication is shift and add: for each 1 in the multiplier, write the multiplicand shifted left.

Examples:

• Binary: 1011 × 101: partial products 1011 and 101100 (shifted two places); sum 110111₂ (11 × 5 = 55).
• Octal: 47₈ × 6₈: 7 × 6 = 42 = 5 × 8 + 2 → write 2, carry 5; 4 × 6 + 5 = 29 = 3 × 8 + 5 → write 5, carry 3 → 352₈ (check: 47₈ = 39, and 39 × 6 = 234 = 3 × 64 + 5 × 8 + 2).
• Hexadecimal: 2B₁₆ × 1A₁₆ = 45E₁₆ (43 × 26 = 1118).

Worked example

Multiply 1101₂ by 110₂ and check in decimal. [2 marks]

Try it first, then show the solutionSolution

1. Multiplier bits 0, 1, 1 (right to left): partial products 0, 11010, 110100.
2. Sum 11010 + 110100 = 1001110₂; check: 13 × 6 = 78 = 1001110₂.

Answer: 1001110₂ (= 78).

System of Numeration in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• Arithmetic in Other Bases: Addition, Subtraction by 1's and 2's Complement, Multiplication
• Number Systems: Positional Notation and Conversions among Decimal, Binary, Octal and Hexadecimal

System of Numeration: syntax and key facts

Positional value

Value = Σ dᵢ × bⁱ (i = position from the right, starting at 0; negative for fractional places)

Grouping

1 octal digit = 3 bits; 1 hexadecimal digit = 4 bits

Complements

1's complement: invert all bits; 2's complement = 1's complement + 1

Range of n-bit numbers

Unsigned: 0 to 2ⁿ − 1; 2's complement signed: −2ⁿ⁻¹ to 2ⁿ⁻¹ − 1

8 bits: 0 to 255 unsigned, −128 to 127 signed.

Carry rule

If a column sum ≥ base: write sum − base, carry 1

System of Numeration: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks System of Numeration

The question patterns to expect, and the method that works for each.

Tricks to remember System of Numeration

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Divide down, read up

  Integers: divide repeatedly, read remainders upwards. Fractions: multiply repeatedly, read integer parts downwards.

  Use it when: Decimal to any base.

• Octal 3, hex 4

  8 = 2³ and 16 = 2⁴.

  Use it when: Grouping bits.

• 1's: flip; 2's: flip and add one

  Shortcut for 2's: keep bits up to and including the first 1 from the right, flip the rest.

  Use it when: Complements.

• Carry in 1's, drop in 2's

  End-around carry for 1's complement; discard the carry for 2's complement.

  Use it when: Complement subtraction.

System of Numeration: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 2 + 2 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in System of Numeration

The slips examiners see most often, and what to write instead.

• ✗Mistake: Reading remainders from top to bottom

  The first remainder is the least significant digit.

  ✓Do this: Divide down, read up.

  Usually costs: 1-2 marks.

• ✗Mistake: Grouping bits from the left end of the integer part

  Groups must start at the binary point; otherwise digits shift.

  ✓Do this: Pad zeros at the far left (integer) and far right (fraction).

  Usually costs: 1 mark.

• ✗Mistake: Keeping the end carry in 2's complement subtraction

  In 2's complement, the end carry is discarded; only 1's complement adds it back.

  ✓Do this: Carry in 1's, drop in 2's.

  Usually costs: 1 mark.

• ✗Mistake: Using different bit lengths for A and B

  Complements must be taken with the same number of bits.

  ✓Do this: Pad the shorter number with leading zeros first.

  Usually costs: 1 mark.

Practice questions on System of Numeration

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The 2's complement of 01011000 is:
   (a) 10100111
   (b) 10101000
   (c) 10100110
   (d) 01011001

   Hint

   Flip all bits, then add 1.

   Answer

   (b) 10101000.

2. Q2 · 1 mark · Very short answerExam level

   Convert 11010110₂ to hexadecimal.

   Hint

   Groups of four.

   Answer

   1101 0110 = D6₁₆.

3. Q3 · 2 marks · Short answerExam level

   Convert 0.8125 (decimal) to binary.

   Hint

   Multiply by 2 repeatedly.

   Answer

   0.8125 × 2 = 1.625 (1); 0.625 × 2 = 1.25 (1); 0.25 × 2 = 0.5 (0); 0.5 × 2 = 1.0 (1) → 0.1101₂.

4. Q4 · 2 marks · Short answerHarder exam level

   Add 7A6₁₆ and 1F9₁₆.

   Hint

   Carry when a column reaches 16.

   Answer

   6 + 9 = 15 = F; A + F = 25 = 16 + 9 → 9, carry 1; 7 + 1 + 1 = 9 → 99F₁₆ (1958 + 505 = 2463).

5. Q5 · 5 marks · Short answerHarder exam level

   Subtract 1011₂ from 1101₂ using 2's complement (4-bit), and 1101₂ from 1011₂ using 1's complement.

   Hint

   Discard carry in 2's; no carry in 1's means negative.

   Answer

   1101 − 1011: 2's comp of 1011 = 0101; 1101 + 0101 = 1 0010; discard carry → 0010 = 2 (13 − 11). 1011 − 1101: 1's comp of 1101 = 0010; 1011 + 0010 = 1101, no carry → negative; 1's comp of 1101 = 0010 → −0010 = −2 (11 − 13).

6. Q6 · 2 marks · Short answerHarder exam level

   Multiply 365₈ by 4₈.

   Hint

   Carry in eights.

   Answer

   5 × 4 = 20 = 2 × 8 + 4 → 4, carry 2; 6 × 4 + 2 = 26 = 3 × 8 + 2 → 2, carry 3; 3 × 4 + 3 = 15 = 8 + 7 → 7, carry 1 → 1724₈ (245 × 4 = 980).

7. Q7 · 2 marks · Short answerHOTS / competency

   What decimal values do the 8-bit patterns 10000000 and 11111111 represent (a) as unsigned numbers, (b) in 2's complement?

   Hint

   Sign bit has weight −128 in 2's complement.

   Answer

   (a) Unsigned: 128 and 255. (b) 2's complement: −128 and −1.

8. Q8 · 2 marks · Short answerHOTS / competency

   Convert 2D3.A₁₆ to octal.

   Hint

   Through binary; regroup from the point.

   Answer

   2D3.A₁₆ = 0010 1101 0011 . 1010 = 1011010011.101₂; groups of three: 001 011 010 011 . 101 = 1323.5₈ (= 723.625).

9. Q9 · 5 marks · Long answerHOTS / competency

   Write an algorithm (in English or pseudocode) to add two binary numbers given as strings of equal length, and trace it for 1011 + 0110.

   Hint

   Right to left with a carry variable.

   Answer

   Set carry = 0 and result = "". For i from the last index down to 0: s = bit(a, i) + bit(b, i) + carry; prepend (s mod 2) to result; carry = s div 2. After the loop, if carry = 1, prepend 1. Trace: i=3: 1+0+0 = 1 → '1', c0; i=2: 1+1+0 = 2 → '0', c1; i=1: 0+1+1 = 2 → '0', c1; i=0: 1+0+1 = 2 → '0', c1; final carry → '10001'. Check: 11 + 6 = 17 = 10001₂.

10. Q10 · 2 marks · Short answerHOTS / competency

    Explain why computers prefer 2's complement over 1's complement for negative numbers.

    Hint

    Zero and carries.

    Answer

    1's complement has two zeros (+0 = 0000, −0 = 1111) and needs an end-around carry. 2's complement has a single zero, the end carry is simply discarded, and the same adder circuit handles signed and unsigned addition, so hardware is simpler.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Digits of base 16?Show answer

   0-9 and A-F.

2. 2.Bits per octal digit?Show answer

   3.

3. 3.Bits per hex digit?Show answer

   4.

4. 4.2's complement?Show answer

   1's complement + 1.

5. 5.8-bit signed range?Show answer

   −128 to 127.

6. 6.1's complement end carry?Show answer

   Added back (end-around carry).

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Value = Σ digit × baseᵖᵒˢⁱᵗⁱᵒⁿ.
• ☐Decimal → base: divide, read remainders upwards; fractions: multiply, read downwards.
• ☐Octal = 3 bits, hex = 4 bits; group from the point.
• ☐Carry when the column sum reaches the base.
• ☐1's: invert, end-around carry; 2's: invert + 1, discard carry; no carry = negative.
• ☐Multiply digit by digit; carry p div base; binary = shift and add.

System of Numeration: frequently asked questions