LATIN SUBSCRIPT SMALL LETTER I·U+1D62

Character Information

Code Point
U+1D62
HEX
1D62
Unicode Plane
Basic Multilingual Plane
Category
Modifier Letter

Character Representations

Click elements to copy
EncodingHexBinary
UTF8
E1 B5 A2
11100001 10110101 10100010
UTF16 (big Endian)
1D 62
00011101 01100010
UTF16 (little Endian)
62 1D
01100010 00011101
UTF32 (big Endian)
00 00 1D 62
00000000 00000000 00011101 01100010
UTF32 (little Endian)
62 1D 00 00
01100010 00011101 00000000 00000000
HTML Entity
ᵢ
URI Encoded
%E1%B5%A2

Description

U+1D62, the Latin Subscript Small Letter I, is a typographical character that primarily serves as an element within digital text. Its role is typically found in mathematical expressions or scientific notations where it represents a lowercase 'i' subscripted below another character or symbol to signify a specific variable or element within a series. In certain cases, the Latin Subscript Small Letter I can be employed in linguistic contexts for representing dialects or languages that use subscript letters as part of their orthography. However, its usage is relatively rare compared to other characters from the Basic Multilingual Plane of Unicode. Despite its limited application, the U+1D62 character remains an essential tool for those working in fields that require precise notation and clear communication of complex concepts.

How to type the symbol on Windows

Hold Alt and type 7522 on the numpad. Or use Character Map.

  1. Step 1: Determine the UTF-8 encoding bit layout

    The character has the Unicode code point U+1D62. In UTF-8, it is encoded using 3 bytes because its codepoint is in the range of 0x0800 to 0xffff.

    Therefore we know that the UTF-8 encoding will be done over 16 bits within the final 24 bits and that it will have the format: 1110xxxx 10xxxxxx 10xxxxxx
    Where the x are the payload bits.

    UTF-8 Encoding bit layout by codepoint range
    Codepoint RangeBytesBit patternPayload length
    U+0000 - U+007F10xxxxxxx7 bits
    U+0080 - U+07FF2110xxxxx 10xxxxxx11 bits
    U+0800 - U+FFFF31110xxxx 10xxxxxx 10xxxxxx16 bits
    U+10000 - U+10FFFF411110xxx 10xxxxxx 10xxxxxx 10xxxxxx21 bits
  2. Step 2: Obtain the payload bits:

    Convert the hexadecimal code point U+1D62 to binary: 00011101 01100010. Those are the payload bits.

  3. Step 3: Fill in the bits to match the bit pattern:

    Obtain the final bytes by arranging the paylod bits to match the bit layout:
    11100001 10110101 10100010