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Modulo 2 arithmetic is performed digit by digit on binary numbers. Each digit is considered independently from its neighbours. Numbers are not carried or borrowed. Addition/Subtraction. Modulo 2 addition/subtraction is performed using an exclusive OR (xor) operation on the corresponding binary digits of each operand.
When your modulus n is a power of 2, then x mod n will count up in binary from 0 to n-1, back to 0, to n-1, etc; for modulus n that looks like binary 01xxxxx, x mod n will cycle through every of those low-order bits xxxxx.
25 Ιαν 2020 · Modulo-2 operators. In our modulo-2 arithmetic system, we define new operators. These operators are frequently very similar to boolean logical operators, so we will discuss those here too. Add. To add two numbers, we take the modulo-2 of the result. Here is a truth table for an add operation: +. 0. 1.
24 Νοε 2017 · Empirically, I can know that (a+b+c) mod 2 = (a-b-c) mod 2. e.g.,) 1+2+3 = 6, 6 mod 2 = 0 1-2-3 = -4, -4 mod 2 = 0. 1+2+4 = 7, 7 mod 2 = 1 1-2-4 = -5, -5 mod 2 = 1. It seems that it is only possible when we use binary modulo (mod 2). Is there any formal proof for this?
19 Ιουν 2010 · There is only a simple way to find modulo of 2^i numbers using bitwise. There is an ingenious way to solve Mersenne cases as per the link such as n % 3, n % 7... There are special cases for n % 5, n % 255, and composite cases such as n % 6.
The xor of two one-bit numbers is their sum modulo 2. But the xor of two $n$-bit numbers can't possibly be their sum modulo 2: any value modulo 2 is either zero or one but the xor of two $n$-bit numbers could be anything between 0 and $2^n-1$, inclusive.
24 Νοε 2017 · Empirically, I can know that (a+b+c) mod 2 = (a-b-c) mod 2. e.g.,) 1+2+3 = 6, 6 mod 2 = 0. 1-2-3 = -4, -4 mod 2 = 0. 1+2+4 = 7, 7 mod 2 = 1. 1-2-4 = -5, -5 mod 2 = 1. It seems that it is only possible when we use binary modulo (mod 2). Is there any formal proof for this?
