Binary, Octal and Hexadecimal: Worked Conversions, Arithmetic and GATE Exam Traps

Decimal conversion can feel manageable until a radix point, leading padding or a carry across bases appears. Then speed and accuracy often disappear together. The single value 45.625 is (101101.101)_2, (55.5)_8 and (2D.A)_16, and adding (2D.A)_16 to (17.4)_8 lands on 61.125 by two independent checks. For complements, signed ranges, BCD and Gray code, read Number Systems and Base Conversions.

1. Binary, octal and hexadecimal begin with radix and place value

In a base, or radix, r, a digit d_i at position i contributes d_i x r^i. Positions right of the radix point use r^-1, r^-2 and so on.

One octal digit is exactly 3 bits because 8 = 2^3. One hexadecimal digit is exactly 4 bits because 16 = 2^4; here A = 10 through F = 15.

For (2D.A)_16, place value gives 2 x 16^1 + 13 x 16^0 + 10 x 16^-1 = 32 + 13 + 0.625 = 45.625. The numeral is valid. By contrast, (128)_8 contains illegal digit 8, (1G)_16 contains illegal digit G, and (101102)_2 contains illegal digit 2. The GATE CS Exam Preparation category places this topic in the wider Digital Logic learning route.

2. Binary to octal and hexadecimal conversion uses bit groups, not decimal

Group bits from the radix point outwards. Use 3 bits per octal digit and 4 per hexadecimal digit. Add zeros only at the far left of the integer part or far right of the fraction; edge padding does not change the value.

Convert (101101.101)_2 directly:

• Octal: 101 101 . 101 -> 5 5 . 5, so the result is (55.5)_8.

• Hexadecimal: 0010 1101 . 1010 -> 2 D . A, so the result is (2D.A)_16.

Now verify by place value. (101101.101)_2 = 32 + 8 + 4 + 1 + 1/2 + 1/8 = 45.625. Also, (55.5)_8 = 5 x 8 + 5 + 5/8 = 45.625, and (2D.A)_16 = 2 x 16 + 13 + 10/16 = 45.625. The .5 in (55.5)_8 means 5/8, not one half.

3. Decimal to binary conversion separates the integer and fractional algorithms

Reverse the route by treating 45 and 0.625 separately. Repeated division gives:

45/2 = 22 remainder 1
22/2 = 11 remainder 0
11/2 = 5  remainder 1
5/2  = 2  remainder 1
2/2  = 1  remainder 0
1/2  = 0  remainder 1

Read the remainders upward to obtain 101101. For the fraction, multiply repeatedly by 2:

0.625 x 2 = 1.25  -> bit 1, remainder 0.25
0.25  x 2 = 0.5   -> bit 0, remainder 0.5
0.5   x 2 = 1.0   -> bit 1, terminate

Read those bits downward to get .101. Therefore 45.625 = (101101.101)_2. Only now regroup: 101 101 . 101 -> (55.5)_8, while 0010 1101 . 1010 -> (2D.A)_16. Routing octal and hexadecimal through binary uses fixed groups, so it is shorter and less error-prone than running separate decimal algorithms.

4. Mixed-base arithmetic: add (2D.A)_16 and (17.4)_8

Never add unlike radices column by column. Convert both operands to one common base. Binary is convenient because every source digit expands into a fixed group:

(2D.A)_16 = 0010 1101.1010_2

(17.4)_8 = 001 111.100_2 = 0000 1111.1000_2 after alignment.

  0010 1101.1010
+ 0000 1111.1000
= 0011 1101.0010

The fractional sum .1010 + .1000 is 1.0010, so 1 carries into the integer part. Trimming only edge zeros gives (111101.001)_2. Regrouping produces (75.1)_8 and (3D.2)_16.

Check independently in decimal: (2D.A)_16 = 45.625, while (17.4)_8 = 8 + 7 + 4/8 = 15.5. Their sum is 61.125. The results agree because (3D.2)_16 = 3 x 16 + 13 + 2/16 = 61.125 and (75.1)_8 = 7 x 8 + 5 + 1/8 = 61.125. A disagreement points to grouping or carry handling. Combinational Circuits: MUX, Decoders, Adders shows where bit-level addition is implemented in logic.

5. Binary, octal and hexadecimal questions test five repeatable moves

The official GATE 2026 CS syllabus lists number representations and computer arithmetic under Digital Logic. Practise these five checks:

1. Direct regrouping: (735)_8 = (111011101)_2 = (1DD)_16.

2. Radix solving: (132)_b = 56_10 gives b^2 + 3b + 2 = 56, so b^2 + 3b - 54 = 0, which factors as (b + 9)(b - 6) = 0; the valid base is b = 6, because a radix must be a positive integer larger than every digit it uses, which rules out the root b = -9.

3. Legality: (128)_8 is invalid because octal has no digit 8.

4. Equality: (111111)_2 = (77)_8 = (3F)_16 = 63_10.

5. Mixed arithmetic: (2D.A)_16 + (17.4)_8 = (3D.2)_16 = 61.125_10.

Attempt them in a fixed order: validate digits, mark the radix point, use grouping when the bases are powers of two, otherwise use positional expansion or repeated division and multiplication, then verify in decimal. KnowledgeGate has over 100 practice questions on binary, octal and hexadecimal conversion.

6. Binary, octal and hexadecimal traps: padding, radix fractions, repeating bits and carries

• Grouping from the left edge: For hexadecimal, 101101.101 must become 0010 1101 . 1010. Padding belongs outside the value, never between significant bits.

• Reading a radix fraction as decimal: .5_8 = 5/8 = 0.625_10, while .5_16 = 5/16 = 0.3125_10. The same symbol has a different positional weight in each base.

• Assuming every finite decimal fraction terminates in binary: Starting with 0.1_10, repeated multiplication by 2 produces bits 0, 0, 0, 1, 1, after which the remainder returns to 0.2. Thus 0.1_10 = 0.000110011..._2. Floating Point Representation: IEEE 754 Format explains how finite precision stores such repeating values.

• Carrying as though the base were 10: In any base, a carry is one unit of the next positional weight. Run a decimal value check, as we did for 61.125.

7. Binary, octal and hexadecimal in the short version

Keep this six-line checklist beside your practice sheet:

• Validate every digit.

• Mark the radix point.

• Use 3-bit octal groups.

• Use 4-bit hexadecimal groups.

• Pad only at the outer edges.

• Verify one route by place value.

The locked results are 45.625_10 = 101101.101_2 = 55.5_8 = 2D.A_16 and (2D.A)_16 + (17.4)_8 = 111101.001_2 = 75.1_8 = 3D.2_16 = 61.125_10.

Use GATE Guidance by Sanchit Sir if you want number systems placed inside a complete subject-wise GATE sequence.

Self-test: Convert (6B.C)_16 to binary and octal, then add (13.2)_8.

Answer check: First conversion: (6B.C)_16 = (1101011.11)_2 = (153.6)_8 = 107.75_10. Final sum: (77)_16 = (1110111)_2 = (167)_8 = 119_10.