Mathematics for Computer Science
MIT 6.042J/18.062J
Remainder
Arithmetic;
Euler’s Theorem
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8F.2
Congruence mod n
If a  b (mod n), then
ac  bc (mod n)
pf: n | (a - b) implies
n | (a-b)c, and so
n | ((ac) – (bc))
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.3
remainder arithmetic
a  rem(a,n) (mod n)
pf: 0  ra,n  n, so
rem(ra,n,n) = ra,n
so  by Remainder Lemma
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.4
remainder arithmetic
important:
a  rem(a,n) (mod n)
keeps (mod n) arithmetic
in the remainder range
0 to n-1
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.5
remainder arithmetic
example: 2879  ? (mod 4)
9
9
287  3 since r287,4 = 3
2
2
2
= ((3 ) ) 3
((1 )2)23 since r9,4 =1
= 3 (mod 4)
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.6
Congruence mod n
So arithmetic (mod n) a lot
like ordinary arithmetic
the main difference:
8·2  3·2 (mod 10)
83
(mod 10)
no general cancellation
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.7
cancellation (mod n)
When can you cancel k?
--when k has no common
factors with n
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.8
cancellation (mod n)
a·k  b·k (mod n) and
gcd(k,n)=1
implies a  b (mod n)
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.9
inverses (mod n)
If gcd(k,n)=1, then have k’
k·k’  1 (mod n).
k’ is an inverse mod n of k
pf: sk + tn = 1, so
just let k’ ::= s
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.10
cancellation (mod n)
If a·k  b·k (mod n)
and gcd(k,n) = 1, then
multiply by k’:
(a·k)·k’  (b·k)·k’ (mod n)
a·(k·k’)
 b·(k·k’)
1
1
so a  b (mod n)
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.11
cancellation (mod n)
Conversely, suppose can cancel k.
So
0k, 1k, 2k, …, (n-1)k
are all different (mod n).
So their remainders on division
by n are all different.
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.12
cancellation (mod n)
This means that
0,r1k,n, r2k,n, …, r(n-1)k,n
must be a permutation of
0,1, 2, …, (n-1)
example on the board:
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.13
permuting (mod 9)
0 1 2 3 4 5 6 7 8
2 02 4 6 8 1 3 5 7
7 07 5 3 1 8 6 4 2
5 is
-1
2
Copyright © Albert R. Meyer, 2007. All rights reserved.
4 is
October 26, 2007
-1
7
lec 8W.14
cancellation (mod n)
This means that
0, r1k,n, r2k,n, …, r(n-1)k,n
must be a permutation of
0, 1, 2, …, (n-1)
so 1 appears, say jth in list:
rem(jk,n)
=1
so
j is k inverse
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.15
cancellation (mod n)
summary:
k is cancellable (mod n) iff
k has an inverse (mod n) iff
k is relatively prime to n
gcd(k,n)=1
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.16
Euler
 function
(n) ::= # k  0,1,…,n-1 s.t.
(7) = 6
(12) = 4
k rel. prime to n
1,2,3,4,5,6
0,1,2,3,4,5,6,7,8,9,10,11
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8F.22
Calculating

If p prime, everything from
1 to p-1 is rel. prime to p, so
(p) = p – 1
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8F.23
Euler
 function
(49)?
0,1,2,3,4,5,6,7,8,9,…,13,14,15,…,21,…
every 7th number is divisible by 7
so, (49) = 49-7
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8F.24
Calculating

For 0,pk-1 every pth element is
not rel. prime to pk:
k-2
k
0,1,...,p-1,p,...,2p,...,(p )p,...,p -1
k
(1/p)p elements
not rel. prime to
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
k
p
lec 8F.25
Calculating
k
(p )
Copyright © Albert R. Meyer, 2007. All rights reserved.
=

k-1
k
k
p
p ––(1/p)p
October 26, 2007
lec 8F.26
Calculating

Lemma :
For a,b relatively prime,
(ab) = (a)(b)
pf: Pset 7 now;
another way in 3 weeks
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8F.28
Euler’s Theorem
For k relatively
prime to n,
(n)
k
´ 1 (mod n)
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8F.29
inverses (mod n)
so when gcd(k,n)=1,
k
(n)-1
(k
)
Copyright © Albert R. Meyer, 2007. All rights reserved.
=
(n)
k
October 26, 2007
1
lec 8W.31
inverses (mod n)
so when gcd(k,n)=1,
(n)-1
k
is a (mod n)
inverse of k
--an alternative to finding
inverses with the pulverizer
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.32
remainder arithmetic
99885
28
?
(mod 5)
99885
99886
28
3
[r28,5 = 3]
rem(99886,4)
=3
[(5)=4]
1
 3 = 2 (mod 5)
Copyright © Albert R. Meyer, 2007. All rights reserved.
October 26, 2007
lec 8W.33