The "mod" operator is the remainder from a division. Since you can't divide by zero, "mod zero" doesn't make sense, either (i.e., it is undefined).
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The modulus of a number relative to a base is the positive remainder when itis divided by that base. So -24 mod 7 = (-28 + 4) mod 7 = -28 mod 7 + 4 mod 7 = 4 mod 7 = 4while 24 mod 7 = (21 + 3) mod 7 = 21 mod 7 + 3 mod 7 = 3 mod 7 = 3.
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No because it is impossible. Let mod(x.3) denote the remainder when x is divided by 3. Let n be any integer. Then mod(n,3) = 0,1 or 2. When mod(n,3) = 0, mod(n2,3) = 0 When mod(n,3) = 1, mod(n2,3) = 1 When mod(n,3) = 2, mod(n2,3) = 4 and, equivalently mod(n2,3) = 1. So, there are no integers whose squar leaves a remainder of 2 when divided by 3.
This is very easy to prove using modulo arithmetic. Basically, what you do is to look at the remainder when a number (n) is divided by 3. Let k(mod 3) represent the remainder when a number is divided by 3. Since the divisor is 3, there are only 3 possible values for k, that is: n = 0(mod 3), 1(mod3) or 2(mod3). Suppose n = 0(mod 3) then n2 + 1 = 0 + 1(mod 3) = 1(mod 3) so that n2 + 1 leaves a remainder of 1 when divided by 3 and so is not divisible by 3. Suppose n = 1(mod 3) then n2 + 1 = 12 + 1(mod 3) = 2(mod 3) so that n2 + 1 leaves a remainder of 2 when divided by 3 and so is not divisible by 3. Suppose n = 2(mod 3) then n2 + 1 = 22 + 1(mod 3) = 5(mod 3) = 2(mod 3) so that n2 + 1 leaves a remainder of 2 when divided by 3 and so is not divisible by 3. Thus, for all possible values of n, division by 3 leaves a positive remainder. And so the result follows.
03 = 0