It's my understanding that when you XOR something, the result is the sum of the two numbers mod $2$.

Why then does $4 \oplus 2 = 6$ and not $0$? $4+2=6$, $6%2$ doesn't equal $6$. I must be missing something about what "addition modulo 2" means, but what?

100 // 4

010 // XOR against 2

110 = 6 // why not zero if xor = sum mod 2?

up vote 6 down vote accepted

The confusion here stems from a missing word. A correct statement is "The result of XORing two bits is the same as that of adding those two bits mod 2."

For example, $(0+1)\bmod 2 = 1\bmod 2 = 1=(0\text{ XOR }1)$


$(1+1) \bmod 2= 2\bmod 2 = 0 =(1\text{ XOR }1)$

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.

You are confusing operations on a single bit with operations on a byte,or word.(Multiple bits) A single bit represents either 0 or 1 depending on its value. If you add two bits, and ignore the carry, you are adding "mod2". 0+0 = 0 0+1 = 1 1+0 = 1 1+1 overflows, or carries, and you have 0

this is exactly the same as XOR

However it is NOT true for words. Consider a nibble (4 bits) 0110 + 0111 = 6 + 7 = 13 = 1101 0110 xor 0111 = 0001 = 1

I think you're misunderstanding the property.

I believe that property is more concerned with the boolean result of the XOR function, not the numeric value. What I mean is that each "number" you're adding is actually just a boolean value of True (1) or False (0) and adding those together, mod 2, is the same as XOR of those values.

This page may clear that up further: The Magic of XOR

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.