Character Information

Code Point
U+2A53
HEX
2A53
Unicode Plane
Basic Multilingual Plane
Category
Math Symbol

Character Representations

Click elements to copy
EncodingHexBinary
UTF8
E2 A9 93
11100010 10101001 10010011
UTF16 (big Endian)
2A 53
00101010 01010011
UTF16 (little Endian)
53 2A
01010011 00101010
UTF32 (big Endian)
00 00 2A 53
00000000 00000000 00101010 01010011
UTF32 (little Endian)
53 2A 00 00
01010011 00101010 00000000 00000000
HTML Entity
⩓
URI Encoded
%E2%A9%93

Description

The Unicode character U+2A53 represents the DOUBLE LOGICAL AND symbol. This typographical element is used primarily in digital text for its specific role as a logical operator. It functions to represent a conjunction that requires both operands to be true for the overall expression to evaluate to true. In computer science and programming, it plays a crucial part in boolean expressions and algorithms, ensuring accurate and reliable outcomes. Despite its technical purpose, the DOUBLE LOGICAL AND symbol doesn't have any significant cultural or linguistic context. It is purely a utility character within the realm of digital text processing and computational logic.

How to type the symbol on Windows

Hold Alt and type 10835 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+2A53. 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+2A53 to binary: 00101010 01010011. 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:
    11100010 10101001 10010011