Theorem 3.3 If a ? b mod n then b = a + nq for some integer q and conversely. We can now show some useful algebraic properties of congruences.
Prove that a ? b (mod n) if and only if a and b leave the same remainder when Since the set of prime numbers in Z is infinite we can always find a ...
assumption these are all the primes but N is seen not to be divisible by any of If f is a polynomial with integer coefficients and a ? b (mod m)
Theorem 3.3 If a ? b mod n then b = a + nq for some integer q and conversely. We can now show some useful algebraic properties of congruences.
To show that it indeed computes the greatest common divisor of a and b we Two integers a
F. F. T. T. T. T. According to the table statements P ? Q and ? Q ?? P exercises for this chapter asks you to show that if a ? b (mod n)
look at leading coefficients to show that if f(X) = 0 and g(X) = 0 then (i) If a and b are integers that are congruent modulo n
Corollary If a and b are integers with ? 0
Our next propositions show a few properties of the solutions. Proposition 5. 1. If the linear congruence ax ? b(mod n) has a solution then there is a.
Suppose A1 = {ab
Theorem 3 2For any integers a and b and positive integer n we have: 1 a amodn 2 If a bmodn then b amodn 3 If a bmodn and b cmodn then a cmodn These results are classically called: 1 Re?exivity; 2 Symmetry; and 3 Transitivity The proofisasfollows: 1 nj(a? a) since 0 is divisible by any integer Thereforea amodn 2
Theorem 2 1For a positive integern and integersa; b; c we have a a(mod n) (congruence modnis re exive) ifa b(mod n) thenb a(mod n) (congruence modnis symmetric) and ifa b(mod n) andb c(mod n) thena c(mod n) (congruence modnistransitive) Remark Thus congruence modnis an equivalence relation onZ Proof
For a;b 2Z we have a = b mod n if and only if a and b have the same remainder when divided by n In particular for every a 2Z there is a unique r 2Z with a = r mod n and 0 r < n Proof: Let a;b 2Z Use the Division Algorithm to write a = qn + r with 0 r < n and b = pn + s with 0 s < n We need to show that a = b mod n if and only if r = s
To show that the simultaneous congruences amodm; x bmodn have a common solution inZ we give two proofs First proof: Write the rst congruence as an equation inZ sayx=a+myfor some 2Z Then the second congruence is the same as a+my bmodn: Subtractingafrom both sides we need to solve foryin (2 1) my b amodn:
Sum rule: IF a ? b(mod m) THEN a+c ? b+c(mod m) (3) Multiplication Rule: IF a ? b(mod m) and if c ? d(mod m) THEN ac ? bd(mod m) (4) De?nition An inverse to a modulo m is a integer b such that ab ? 1(mod m) (5) By de?nition (1) this means that ab ? 1 = k · m for some integer k As before there are may be many
3 22 Show that addition and multiplication mod nare well de ned operations That is show that the operations do not depend on the choice of the representative from the equivalence classes mod n Solution Suppose that a b(mod n) and c d(mod n) Then there are integers r;swith a= b+rn and c= d+ sn We nd that a+ c= b+ rn+ d+ sn = b+ d+ (r+ s)n;
( ( a + b) mod n) means that there is an integer k such that 0 ? a + b ? n k < n, and ( a mod n) + ( b mod n) means there are integers k 1 and k 2 such that 0 ? a ? n k 1 < n and 0 ? b ? n k 2 < n ? 0 ? a + b ? n ( k 1 + k 2) < 2 n which has no relation to the 0 ? a + b ? n k < n especially resulting same numbers to say they are equal.
It is because a mod b isn't simply the remainder as returned by the operator '%'. See some examples: There are some other definitions in math and other implementations in computer science according to the programming language and the computer hardware. Please see Modulo operation from Wikipedia.
Let us say ‘a’ and ‘b’ are two integers, either positive or negative. When we add the two integers, their result would always be an integer, i.e (a + b) would always be an integer.
So to find the answer to "A*B (mod 10)" all you need to do is find: "the one's place of A" and "the one's place of B" Similarly, if you instead used base N instead of base 10 to write out your numbers A and B, you'd find the exact same pattern.