How to do floating point numbers in binary, Summaries of Computer science

How to do floating point numbers in binary

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Floating Point Binary Numbers Computer Science — Student Guide
Floating Point Binary Numbers
How computers store decimal numbers using just 0s and 1s
Computers can only store 1s and 0s (binary digits). That's fine for whole numbers like 42 or 255 —
but what about 6.75 or −0.001? This document explains the trick computers use: floating point
representation.
1. Quick Revision — Binary & Powers of Two
You already know that binary whole numbers work like this:
Binary Calculation Decimal value
1101 1×8 + 1×4 + 0×2 + 1×1 13
1010 1×8 + 0×4 + 1×2 + 0×1 10
11111111 128+64+32+16+8+4+2+1 255
The same idea extends to the right of the binary point (like a decimal point, but for binary). Each
position to the right is half the previous one:
Position Value
1s place (2⁰) 1
½s place (2 ¹)0.5
¼s place (2 ²)0.25
⅛s place (2 ³)0.125
1/16s place (2 ⁴)0.0625
So 101.11 in binary means: 4 + 0 + 1 + 0.5 + 0.25 = 5.75 in decimal. Easy! The challenge is
storing this efficiently using a fixed number of bits.
2. The Three Parts of a Floating Point Number
The most common floating point standard is IEEE 754 single precision, which uses exactly 32
bits split into three fields:
Sign Exponent Mantissa (Fraction)
1 bit 8 bits 23 bits
Sign (1 bit): 0 = positive, 1 = negative. Simple!
Exponent (8 bits): Controls how big or small the number is — like the power in scientific notation.
A bias of 127 is added before storing, so the stored value is always positive.
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Floating Point Binary Numbers

How computers store decimal numbers using just 0s and 1s

Computers can only store 1s and 0s (binary digits). That's fine for whole numbers like 42 or 255 — but what about 6.75 or −0.001? This document explains the trick computers use: floating point representation.

1. Quick Revision — Binary & Powers of Two You already know that binary whole numbers work like this: Binary Calculation Decimal value (^1101) 1×8 + 1×4 + 0×2 + 1×1 13 (^1010) 1×8 + 0×4 + 1×2 + 0×1 10 (^11111111) 128+64+32+16+8+4+2+1 255 The same idea extends to the right of the binary point (like a decimal point, but for binary). Each position to the right is half the previous one: Position Value 1s place (2⁰) 1 ½s place (2 ¹)⁻ 0. ¼s place (2 ²)⁻ 0. ⅛s place (2 ³)⁻ 0. 1/16s place (2 ⁴)⁻ 0. So 101.11 in binary means: 4 + 0 + 1 + 0.5 + 0.25 = 5.75 in decimal. Easy! The challenge is storing this efficiently using a fixed number of bits. 2. The Three Parts of a Floating Point Number The most common floating point standard is IEEE 754 single precision , which uses exactly 32 bits split into three fields: Sign Exponent Mantissa (Fraction) 1 bit 8 bits 23 bits Sign (1 bit): 0 = positive, 1 = negative. Simple! Exponent (8 bits): Controls how big or small the number is — like the power in scientific notation. A bias of 127 is added before storing, so the stored value is always positive.

Mantissa / Fraction (23 bits): Stores the significant digits of the number. There is always a hidden leading 1 that isn't stored (called the implicit leading bit ), so we effectively get 24 bits of precision. 💡 Scientific Notation Analogy Think of it like scientific notation in decimal: 6.022 × 10²³. Here 6.022 is the mantissa and 23 is the exponent. Floating point does exactly the same thing — but in binary: 1.mantissa × 2^exponent.

3. Step-by-Step: Converting 6.75 to IEEE 754 Let's convert 6.75 into its 32-bit IEEE 754 binary representation, one step at a time. 1 Convert the whole part to binary Divide 6 by 2 repeatedly: 6 → 3 → 1 → 0, reading remainders upward: 110 2 Convert the fractional part to binary Multiply 0.75 by 2: 0.75 × 2 = 1 .5 → first bit: 1 Multiply 0.5 by 2: 0.5 × 2 = 1 .0 → second bit: 1 (remainder 0, stop) So 0.75 in binary =. 3 Combine and write in binary scientific notation Full binary: 110.11 → normalise to 1.1011 × 2² (shift point left 2 places) 4 Fill in the three fields Sign: 6.75 is positive → 0 Exponent: Power is 2, add bias 127: 2 + 127 = 129 → 10000001 Mantissa: Digits after the leading 1 in 1.1011 are 1011 , padded to 23 bits: 10110000000000000000000 Putting it all together: Sign Exponent (129) Mantissa 0 10000001 10110000000000000000000 ✓ Check your answer Full 32-bit pattern: 0 10000001 10110000000000000000000. To verify: sign = +, exponent = 129−127 = 2, mantissa = 1.1011 ₂= 1 + ½ + ¼ + 0 + ⅛ ... wait — 1.1011 = 1 + 0.

  • 0.0625... actually: 1 + 0.5 + 0.125 + 0.0625 = 1.6875, × 2² = 6.75 ✓