MODULAR ARITHMETIC I

Transcription

1 MODULAR ARITHMETIC I. Modular Arithmetic and Check Digit Schemes The check digit schemes that we have been studying have some features in common. Certain information (book title publisher etc., personal data, an details about an item for purchase) are encoded in a string of numbers, and veri cation consists of a peculiar mathematical computation involving "vertical" multiplication and "horizontal" addition. They di er in several ways though, () The number of digits used to encode information (0 for the Zip Code, 2 for UPC, etc.). Let s call that the length of the scheme, (2) The multipliers ((; ; : : : ; ) for the zip code, (3; ; 3 : : : ; ) for UPC, (2; 7; 6 : : : ; ) for the NM Driver s License, etc.). These multipliers are called the weight of the scheme. (3) The number that the computation needs to be a multiple of. Call that the modulus of the scheme. So UPC is a length 2 scheme with weight (3; ; 3 : : : ; ) and modulus 0. Exercise. Describe ISBN-0 and ISBN-3 in terms of length, weight, and modulus. Exercise 2. Construct a check digit scheme of length 3 with modulus 0. Will it detect every single digit error? Will it detect every transposition error? Exercise 3. Construct a check digit scheme of length 3 with modulus. Will it detect every single digit error? Will it detect every transposition error? The shortcuts we used to determine whether a given number was a valid for a particular code relied on what is known as modular arithmetic. Consider the following well known use of mod 2 arithmetic. When we tell time in the U.S., the hour value is a whole number between and 2. Three hours after 0 o clock will be o clock. In general, to see what time it will be n hours after 0 o clock, you add n to 0, and then remove enough multiples of 2 until you have a value between and 2. For instance, 37 hours past 0 o clock, the time will

2 2 MODULAR ARITHMETIC I be o clock since = 47 = In telling time, we identify 3 o clock with o clock, 4 o clock with 2 o clock, and so on. In this clock arithmetic, if we add 2 hours to any time, we get the same time (but changing AM to PM and vice-versa). Therefore, 2 acts in clock arithmetic like 0 acts in ordinary arithmetic. Exercise 4. Suppose you begin work at 8 AM and work two and one half 8 hour shifts. What time is it when you nally leave work? There is nothing special about 2 with respect to obtaining a new type of arithmetic. For UPC and ISBN-3 we use mod 0 arithmetic and for ISBN-0 and the NM Driver s license number we use mod arithmetic. Later when we discuss cryptography we will use a modulus that is roughly 200 digits long! Exercise 5. The inhabitants of the planet Nonus tell time on a clock with 9 hours. Suppose that a Nonusian arrives at work at 4 and works 2 ve hour shifts. At what time does it leave work? These types of calculations are made just on the numbers from to the modulus. But notice that the modulus acts just like 0 when we add. So let s formalize this and de ne modular addition and multiplication on the symbols f0; g for arithmetic mod 2, f0; ; 2g for arithmetic mod 3, etc. by imitating clock arithmetic. The addition and multiplication tables for mod 2 arithmetic are for mod 3: + mod mod and the mod 6 tables are + mod mod mod mod :

3 MODULAR ARITHMETIC I 3 Exercise 6. Fill in the missing entries in the addition and multiplication tables for mod 5 and mod 7 arithmetic (leaving out the 0 rows and columns for multiplication. + mod mod mod : Notice that in all cases, 0 serves as an additive identity element and serves as the multiplicative identity. Consider some multiplication tables for mod 0 multiplication. For example, the multiplication table for 3 is 3; 6; 9; 2 = 2; 5 = 5; 8 = 8; 2 = ; 24 = 4; 27 = 7; 3; 6; 9; 2; 5; 8; ; 4; 7: Since 3 7 = we say that 3 and 7 are multiplicative inverses mod 0. Exercise 7. Do 2 and 5 have multiplicative inverses mod 0? Sometimes we can solve algebraic equations using modular arithmetic. For example consider the equation 3x = 8 in mod 0 arithmetic. Even though it doesn t make sense to divide by 3 we can use the fact that 7 is the multiplicative inverse of 3: Multiply both sides of the equation by 7 to get 7 3x = 7 8 x = 56 x = 6: Reality check: Notice that 3 6 = 8 = 8 + 0: So our result is consistent with ordinary arithmetic.

4 4 MODULAR ARITHMETIC I We can t solve 2x = 7 in mod 0 arithmetic. Just go through the multiplication table of 2 : 2; 4; 6; 8; 0; 2; ::: to see that 7 never appears. We can solve 2x = 7 in mod arithmetic: Since 6 2 = 2 = we have 2x = 7 6 2x = 6 7 x = 42 = = 9: So it becomes interesting to ask when exactly can we solve equations ax = b in modular arithmetic. It makes sense to guess that we can solve such an equation whenever a has a multiplicative inverse for the modulus in question. That is true and we can also determine exactly the relationship between a and the modulus m that guarantees that a has such an inverse. For the case of m = 2 we only have 0 and to work with. Equations 0x = b are never very interesting, so for m = 2 the equations x = 0 x = have the solutions x = 0 and x = respectively. For m = 3; we have equations x = b where b can be 0,, or 2: The rst equation is easy to handle, and for the second we have 2 2 = 4 = ; so 2 2x = 2 b x = 2 b: For m = 4 things get a bit more complicated. We have x = b 3x = b:

5 MODULAR ARITHMETIC I 5 where b can be 0,,2, or 3. The rst and third equations are easy to handle because 3 3 = 9 = : But if b = or b = 3 the second equation has no solution: 2 0 = 0 2 = = 4 = = 6 = 2: For m = 0 the equations 2x = and 5x = have no solutions because the multiplication tables for 2 and 5 are 2; 4; 6; 8; 0; 2; 4; 6; 8 5; 0; 5; 0; 5; 0; 4; 0; 5: For m = 5; 7; (primes!) however, every equation ax = b with a 6= 0 can be solved because every such a has a multiplicative inverse. It turns out that for any modulus m those a s having a multiplicative inverse modulo m are exactly the same as those for which ax = 0 have no solution other than x = 0: They are the same as the a s only divisor with m is :