Upload
milos
View
51
Download
0
Embed Size (px)
DESCRIPTION
Modular Arithmetic. Lecture 9: Oct 5. Modular Arithmetic. Def : a b (mod n) iff n|( a - b) iff a mod n = b mod n. e.g. 12 2 (mod 10) 107 207 (mod 10) 7 1 (mod 2) 1 -1 (mod 2) 13 -1 (mod 7) -15 0 (mod 5). Modular Addition. - PowerPoint PPT Presentation
Citation preview
Modular Arithmetic
Lecture 9: Oct 5
Def: a b (mod n) iff n|(a - b) iff a mod n = b mod n.
Modular Arithmetic
e.g. 12 2 (mod 10)
107 207 (mod 10)
7 1 (mod 2)
1 -1 (mod 2)
13 -1 (mod 7)
-15 0 (mod 5)
Lemma: If a c (mod n), and b d (mod n) then
a+b c+d (mod n).
Modular Addition
Example 1 12 2 (mod 10), 25 5 (mod 10)
=> 12 + 25 (mod 10)
2 + 5 (mod 10) 7 (mod 10)
Example 2 87 2 (mod 17), 222 1 (mod 17)
=> 87 + 222 (mod 17)
2 + 1 (mod 17)
3 (mod 17)
Example 3 101 2 (mod 11), 141 -2 (mod 11)
=> 101 + 141 (mod 11) 0 (mod 11)
Lemma: If a c (mod n), and b d (mod n) then
a+b c+d (mod n).
Modular Addition
a c (mod n) => a = c + nx for some integer x
b d (mod n) => b = d + ny for some integer y
To show a+b c+d (mod n), it is equivalent to showing that n | (a+b-c-d).
Consider a+b-c-d.
a+b-c-d = (c+nx) + (d+ny) – c –d = nx + ny.
It is clear that n | nx + ny.
Therefore, n | a+b-c-d.
We conclude that a+b c+d (mod n).
Proof
Lemma: If a c (mod n), and b d (mod n) then
ab cd (mod n).
Modular Multiplication
Example 1 9876 6 (mod 10), 17642 2 (mod 10)
=> 9876 * 17642 (mod 10)
6 * 2 (mod 10)
2 (mod 10)
Example 2 10987 1 (mod 2), 28663 1 (mod 2)
=> 10987 * 28663 (mod 2) 1 (mod 2)
Example 3 999 5 (mod 7), 674 2 (mod 7)
=> 999 * 674 (mod 7) 5 * 2 (mod 7) 3 (mod 7)
Lemma: If a c (mod n), and b d (mod n) then
ab cd (mod n).
Modular Multiplication
a c (mod n) => a = c + nx for some integer x
b d (mod n) => b = d + ny for some integer y
To show ab cd (mod n), it is equivalent to showing that n | (ab-cd).
Consider ab-cd.
ab-cd = (c+nx) (d+ny) – cd
= cd + dnx + cny + n2xy – cd = n(dx + cy + nxy).
It is clear that n | n(dx + cy + nxy). Therefore, n | ab-cd.
We conclude that ab cd (mod n).
Proof
Exercise
1444 mod 713
= 144 * 144 * 144 * 144 mod 713
= 20736 * 144 * 144 mod 713
= 59 * 144 * 144 mod 713
= 8496 * 144 mod 713
= 653 * 144 mod 713
= 94032 mod 713
= 629 mod 713
20736 * 20736 mod 713
= 59 * 59 mod 713
= 3481 mod 713
= 629 mod 713
Application
Is a number written in decimal evenly divisible by 9 if and only
if the sum of its digits is a multiple of 9?
Example 1. 9333234513171 is divisible by 9.
9+3+3+3+2+3+4+5+1+3+1+7+1 = 45 is divisible by 9.
Example 2. 128573649683 is not divisible by 9.
1+2+8+5+7+3+6+4+9+6+8+3 = 62 is not divisible by 9.
A coincidence?
This can be proved easily using modular arithmetic.
NO
Application
Claim. A number written in decimal is divisible by 9 if and only
if the sum of its digits is a multiple of 9?
Hint: 10 1 (mod 9).
Let the decimal representation of n be dkdk-1dk-2…d1d0.
This means that n = dk10k + dk-110k-1 + … + d110 + d0
Note that di10i mod 9
= (di mod 9) (10i mod 9) mod 9
= (di mod 9) (10 mod 9) (10 mod 9) … (10 mod 9) mod 9
= (di mod 9) (1 mod 9) (1 mod 9) … (1 mod 9) mod 9
= di mod 9
i terms
Application
Claim. A number written in decimal is divisible by 9 if and only
if the sum of its digits is a multiple of 9?
Let the decimal representation of n be dkdk-1dk-2…d1d0.
This means that n = dk10k + dk-110k-1 + … + d110 + d0
Note that di10i mod 9 = di mod 9.
Hence n mod 9 = (dk10k + dk-110k-1 + … + d110 + d0) mod 9
= (dk10k mod 9 + dk-110k-1 mod 9 + … + d110 mod 9 + d0 mod 9) mod 9
= (dk mod 9 + dk-1 mod 9 + … + d1 mod 9 + d0 mod 9) mod 9
= (dk + dk-1 + … + d1 + d0) mod 9
Hint: 10 1 (mod 9).
The multiplicative inverse of a number a is another number a’ such that:
a · a’ 1 (mod n)
Multiplication Inverse
Does every number has a multiplicative inverse in modular arithmetic?
For real numbers, every nonzero number has a multiplicative inverse.
For integers, only 1 has a multiplicative inverse.
An interesting property of modular arithmetic is that there are multiplicative inverse for intgers.
For example, 2 * 5 = 1 mod 3, so 5 is a multiplicative inverse for 2 under modulo 3 (and vice versa).
Multiplication Inverse
Does every number has a multiplicative inverse in modular arithmetic?
Multiplication Inverse
What is the pattern?
Multiplication Inverse
Why 2 does not have a multiplicative inverse under modulo 6?
Suppose it has a multiplicative inverse y. 2y 1 (mod 6)
=> 2y = 1 + 6x for some integer x=> y = ½ + 3x This is a contradiction since both x and y are integers.
Claim. An integer k does not have an multiplicative inverse under modulo n, if k and n have a common factor >= 2 (gcd(k,n) >= 2).
Proof. Same as above. Leave it as an exercise.
Theorem. If gcd(k,n)=1, then have k’
k·k’ 1 (mod n).
k’ is an inverse mod n of k
Proof: Since gcd(k,n)=1, there exist s and t so that sk + tn = 1.
So tn = 1 - sk
This means n | 1 – sk.
This means that 1 – sk 0 (mod n).
This means that 1 = sk (mod n).
So k’ = s is an multiplicative inverse for k.
Multiplication Inverse
What about if gcd(k,n)=1?
Would k always have an multiplicative inverse under modulo n?
gcd(k,n)=spc(k,n)
Cancellation
There is no general cancellation in modular arithmetic.
Note that (mod n) a lot like =.
If a b (mod n), then a+c b+c (mod n).
If a b (mod n), then ac bc (mod n)
However, if ac bc (mod n),
it is not necessarily true that a b (mod n).
For example, 4·2 1·2 (mod 6), but 4 1 (mod 6)
Cancellation
Why a·k b·k (mod n) when a ≠ b?
This means that ak = bk + nx.This means that (a-b)k = nx, which means a-b=(nx)/k.Since 0 < a < n and 0 < b < n, it implies that –n < a-b < n.Therefore, nx/k must be < n. For this to happen, n and k must have a common divisor >= 2!
Without loss of generality, assume 0 < a < n and 0 < b < n.Because if a·k b·k (mod n), then also (a mod n)·k (b mod n)·k (mod n).
smaller than n.
Okay, so, can we say something when gcd(n,k)=1?
Cancellation
Claim: If i·k j·k (mod n), and gcd(k,n) = 1,
then i j (mod n)
For example, multiplicative inverse always exists if n is a prime!
Proof. Since gcd(k,n) = 1, there exists k’ such that kk’ 1 (mod n).
i·k j·k (mod n).
=> i·k·k’ j·k·k’ (mod n).
=> i j (mod n)
This makes arithmetic modulo prime a field,
a structure that “behaves like” real numbers.
Arithmetic modulo prime is very useful in coding theory.
If p is prime & k not a multiple of p, then we can cancel k. So
k mod p, 2k mod p, …, (p-1)k mod p
are all different.
Fermat’s Little Theorem
This means that
k mod p, 2k mod p,…,(p-1)k mod p
must be a permutation of
1, 2, ···, (p-1)
(each number appears exactly once)
Fermat’s Little Theorem
1 kp-1 (mod p)
Theorem: If p is prime & k not a multiple of p
Proof.
1·2···(p-1) (k mod p · 2k mod p··· (p-1)k mod p) mod p
(k·2k ··· (p-1)k) mod p
(kp-1)·1·2 ··· (p-1) (mod p)
So, by cancelling 1·2 ··· (p-1) on both sides, we have
1 kp-1 (mod p)
A permutation
Wilson’s Theorem
Theorem: p is a prime if and only if
(p-1)! -1 (mod p)
First we consider the easy direction.
If p is not a prime, assume p >= 5, (for p=4, 3! 2 (mod 4) )
Then p=qr for some 2 <= q < p and 2 <= r < p.
If q ≠ r, then both q and r appear in (p-1)!, and so (p-1)! 0 (mod p).
If q = r, then p = q2 > 2q (since we assume p > 5 and thus q > 2).then both q and 2q are in (p-1)!,and so again (p-1)! 0 (mod p).
Wilson’s Theorem
Theorem: p is a prime if and only if
(p-1)! -1 (mod p)
To prove the more interesting direction, first we need a lemma.
Lemma. If p is a prime number, x2 1 (mod p) if and only if x 1 (mod p) or x -1 (mod p)
Proof. x2 1 (mod p)
iff p | x2 - 1
iff p | (x – 1)(x + 1)
iff p | (x – 1) or p | (x+1)
iff x 1 (mod p) or x -1 (mod p)
Lemma: p prime and p|a·b iff p|a or p|b.
Wilson’s Theorem
Theorem: p is a prime if and only if
(p-1)! -1 (mod p)
Let’s get the proof idea by considering a concrete example.
10! 1·2·3·4·5·6·7·8·9·10 mod 11
1·10·(2·6)·(3·4)·(5·9)·(7·8) mod 11
1·-1·(1)·(1)·(1)·(1) mod 11
-1 mod 11
Besides 1 and 10, the remaining numbers are paired up into multiplicative inverse!
Wilson’s Theorem
Theorem: p is a prime if and only if
(p-1)! -1 (mod p)
Proof.Since p is a prime, every number from 1 to p-1 has a multiplicative inverse.
By the Lemma, every number 2 <= k <= p-2 has an inverse k’ with k≠k’.
Since p is odd, the numbers from 2 to p-2 can be grouped into pairs
(a1,b1),(a2,b2),…,(a(p-3)/2,b(p-3)/2) so that aibi 1 (mod p)
Therefore, (p-1)! 1·(p-1)·2·3·····(p-3)·(p-2) (mod p)
1·(p-1)·(a1b1)·(a2b2)·····(a(p-3)/2b(p-3)/2) (mod p)
1·(-1)·(1)·(1)·····(1) (mod p)
-1 (mod p)