Real numbers in binary

ACompressed recap

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Fixed point i.f: integer ÷ 2ᶠ, step 2⁻ᶠ, unsigned range up to 2ⁱ − 2⁻ᶠ, signed −2ⁱ⁻¹ … 2ⁱ⁻¹ − 2⁻ᶠ. IEEE single: S | E (8, bias 127) | F (23); value ±1.F × 2^(E − 127). Double: 11/52, bias 1023. Special: E = 0 → zero/denormal (0.F × 2⁻¹²⁶); E = 255 → ∞/NaN. Range ≈ 3.4 × 10³⁸, min normal 2⁻¹²⁶, min denormal 2⁻¹⁴⁹; gap 2^(e − 23); integers exact to 2²⁴.

BMerged rule card

2.1.1 Fixed-point binary fractions: range, precision and rounding

• i.f fixed point: value = integer ÷ 2ᶠ; step 2⁻ᶠ (same everywhere).
• Unsigned range 0 … 2ⁱ − 2⁻ᶠ; signed 2's complement −2ⁱ⁻¹ … 2ⁱ⁻¹ − 2⁻ᶠ.
• Real → fixed: × 2ᶠ, round/truncate, write in n bits. Negatives: 2's complement of all n bits.
• Truncation error < 2⁻ᶠ; rounding error ≤ 2⁻ᶠ⁻¹.

2.1.2 Floating point and IEEE 754: sign, biased exponent, mantissa

• Normalised value: (−1)ˢ × 1.F × 2^(E − bias).
• Single: 1 + 8 + 23 bits, bias 127. Double: 1 + 11 + 52 bits, bias 1023.
• Encode: sign → binary → normalise 1.f × 2ᵉ → E = e + bias → F = bits after the point → hex in groups of 4.
• Decode: split S | E | F, e = E − bias, value = ±1.F × 2ᵉ.

2.1.3 IEEE 754 special values, denormals, range, precision and gaps

• E = 0, F = 0: ±0. E = 0, F ≠ 0: denormal (−1)ˢ × 0.F × 2^(1 − bias). E = all 1s: ±∞ (F = 0) or NaN.
• Single: max ≈ 3.4 × 10³⁸ ((2 − 2⁻²³) × 2¹²⁷), min normal 2⁻¹²⁶, min denormal 2⁻¹⁴⁹.
• Precision: 24 bits (single), 53 (double). Gap in [2ᵉ, 2ᵉ⁺¹) = 2^(e − 23); integers exact up to 2²⁴.
• Float addition aligns exponents; not associative. Bias for k exponent bits = 2ᵏ⁻¹ − 1.

CInterleaved GATE-style questions

Q1NAT1 markFixed point

How many fraction bits are needed for a step of at most 0.01?

Numerical answer: ________

Q2NAT2 marksFixed point

Signed 8.8 fixed point (16 bits, 2's complement). What is the most negative value?

Numerical answer: ________

Q3NAT1 markExponent

What is the exponent field (decimal) of 0.15625 in single precision?

Numerical answer: ________

Q4MCQ2 marksDecode

The single-precision pattern 0xBF40 0000 represents:

• (A) −0.75
• (B) −1.5
• (C) 0.75
• (D) −0.375

Q5MCQ1 markHidden bit

Why is the leading 1 of a normalised significand not stored?

• (A) It is stored in the exponent
• (B) It is the sign bit
• (C) It is only needed for denormals
• (D) It is always 1, so storing it wastes a bit

Q6NAT2 marksGap

In single precision, what is the distance between 1.0 and the next larger float? Give k where the distance is 2ᵏ.

Numerical answer: ________

Q7NAT2 marksToy format

Format: 1 sign, 4 exponent bits (bias 7), 3 fraction bits, IEEE rules. What is the smallest positive normal value? Give it as a decimal.

Numerical answer: ________

Q8MCQ1 markDenormal

A single-precision denormal number has the value:

• (A) (−1)ˢ × 1.F × 2⁻¹²⁷
• (B) (−1)ˢ × 0.F × 2⁻¹²⁷
• (C) 0
• (D) (−1)ˢ × 0.F × 2⁻¹²⁶

DAnswer Vault