TWO INTERSECTING LOGICAL AND·U+2A55

Character Information

Code Point
U+2A55
HEX
2A55
Unicode Plane
Basic Multilingual Plane
Category
Math Symbol

Character Representations

Click elements to copy
EncodingHexBinary
UTF8
E2 A9 95
11100010 10101001 10010101
UTF16 (big Endian)
2A 55
00101010 01010101
UTF16 (little Endian)
55 2A
01010101 00101010
UTF32 (big Endian)
00 00 2A 55
00000000 00000000 00101010 01010101
UTF32 (little Endian)
55 2A 00 00
01010101 00101010 00000000 00000000
HTML Entity
⩕
URI Encoded
%E2%A9%95

Description

U+2A55 is a specialized Unicode character that represents the logical AND operator in the form of two intersecting lines. In digital text, it's typically used in mathematics, computer programming, and engineering to denote logical conjunction operations. The character serves as a valuable tool for those working with Boolean algebra and digital logic systems, playing a critical role in defining and implementing complex algorithms, particularly in binary systems. It is especially useful in programming languages where clear visual representation of logical operators is crucial. There isn't any notable cultural, linguistic or technical context associated with this character, as its usage is primarily functional and universal. Its design, two intersecting lines, clearly conveys the concept of intersection, which aligns with the mathematical operation it represents.

How to type the symbol on Windows

Hold Alt and type 10837 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+2A55. 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+2A55 to binary: 00101010 01010101. 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 10010101