Chinese Remainder Theorem,
Rational numbers
27 October
Theorem (Chinese remainder theorem). If
{n1, . . . , nr } is a set of r natural numbers that
are pairwise relatively prime, and if {a1, . . . , ar }
are any r integers, then the system of congruences
x ≡ a1
mod n1
...
x ≡ ar
mod nr
has a unique solution modulo N =
Q
ni.
Here’s one way to find the solution.
For each i, let
N
.
ni
Let yi be a solution to the congruence
Ni =
N i yi ≡ 1
mod ni.
Set
x=
X
j
a j N j yj .
Exercise 1. Find all solutions to the system
of congruences
n≡3
mod 4
n≡1
mod 21
n≡4
mod 5
Solution.
Set
N1 = 105
N2 = 20
N3 = 84.
We must solve
105y1 ≡ 1
20y2 ≡ 1
84y3 ≡ 1
mod 4
mod 21
mod 5.
For example y1 = 1, y2 = −1, y3 = −1 will
work. Then a solution is
x = (3)(105)(1) + (1)(20)(−1) + (4)(84)(−1)
= 315 − 20 − 336
= −41.
All solutions are of the form
x = −41 + (4)(21)(5)k
for k an integer.
Rational numbers
The problem of “defining” the rational numbers is the problem of how to “define” the fractions a/b.
Let A be the set
A = {(a, b) ∈ Z × Z|b 6= 0}.
We suggestively write ab for the pair (a, b). The
set A has “too many” rational numbers, because for example
1
2
6= .
2
4
We define a relation on R ⊂ A × A on A by
a c
( , )∈R
b d
if
ad = bc.
Lemma 1. R is an equivalence relation.
Definition 1. The set of rational numbers is
the set of equivalence classes of A under the
relation R.
Proof. First, symmetry. Clearly
a a
( , )∈R
b b
since
ab = ba.
Second, reflexivity. Suppose
a c
( , )∈R
b d
i.e.
ad = bc.
Then
cb = da
so
c a
( , ) ∈ R.
d b
Finally, transitivity. Suppose
a c
( , )∈R
b d
and
c e
( , )∈R
d f
i.e.
ad = bc
cf = de.
We need to check that
af = be.
Since d 6= 0, we can check this after multipliying by d.
adf = bde.
This is
bcf = bde,
which is true.
One checks that the formulae
a
c
ad + bc
+ =
b
d
bd
ac
ac
=
bd
bd
define addition and multiplication on Q, with
the usual distributivity, associativity, and commutativity conditions.
0
0
a
c
c
a
Example 1. Show that if b = b0 and d = d0 ,
then
a
c
a0
c0
+ = 0+ 0
b
d
b
d
in Q.
Proof. By definition
c
ad + bc
a
+ =
b
d
bd
a0
c0
a0d0 + b0c0
+ 0=
.
0
0
0
b
d
bd
The two right-hand-sides are the same if
(ad + bc)(b0d0) = (a0d0 + b0c0)(bd)
or
adb0d0 + bcb0d0 = a0d0bd + b0c0bd
(2)
We have
ab0 = a0b
(3)
cd0 = c0d.
(4)
Substituting these into the left side of (2) gives
a0dbd0 + bc0b0d,
which equals the right side of (2).
0
0
a
a
c
c
Exercise 2. Show that if b = b0 and d = d0 ,
then
ac
a 0 c0
= 0 0
bd
b d
in Q.
Earlier we showed that if k has a rational square
root, then it has an integral square root. More
generally, we have
Theorem . Suppose that
f (x) = c0 + c1x + · · · + cnxn
is a polynomial with integer coefficients. If r
is a rational number such that
f (r) = 0,
and if r = pq in lowest terms, then p|c0 and q|cn.
Proof. We have
pn
p
0 = c0 + c1 + · · · + cn n .
q
q
Multiply through by q n to get
0 = c0q n + c1pq n−1 + · · · + cnpn.
or
−c0q n = c1pq n−1 + · · · + cnpn.
Since p divides the right side, it divides the left.
Since q is relatively prime to p, it follows that p
divides c0. A similar argument shows q|cn.
Example 2. Let k be an integer and n a natural number. If r is a rational number such
that
rn − k = 0
then r is an integer.
Proof. Let f (x) = −k + xn so c0 = −k and
cn = 1. If r = pq in lowest terms, then q divides
1 and so q = 1 and r is an integer.
Exercise 3 (8.20). Let f (x) = x6 + cx5 + 1.
Prove that the equation f (x) = 0 has no rational solution when c 6= ±2.
Rational points on the circle.
Another representation of rational numbers is
the following. Let L be the set of lines L starting at the origin in R2 and pasing through some
point (a, b) with a, b ∈ Z. There is a function
m
L −→ Q
defined by m(L) = slope of L.
Theorem . The function m is a bijection.
Here is another instance of that idea.
Theorem . The points (x, y) ∈ R such that
x2 + y 2 = 1 are either of the form (−1, 0) or
(x, y) =
1 − t2
2t
,
1 + t2 1 + t2
!
for some real number t. The rational points
are those for which t is rational.
Proof. First observe that such these points are
on the circle.
Next, suppose that (x, y) is on the circle, and is
not the point (−1, 0). The line through (−1, 0)
and (x, y) is given by the (point-slope) formula
Y = t(X + 1).
This line intersects the circle at (x, y) if
x2 + t2(x + 1)2 = 1
or
(1 + t)2x2 + 2t2x + t2 − 1 = 0
or
2−1
2t
t
x2 +
= 0.
x+
(1 + t2)2
(1 + t2)
One solution is x = −1, and so the other must
be
1 − t2
.
2
1+t
Note that x and y are rational if t is; and t =
y/(x + 1) is rational if x is.
Pythagorean triples
A Pythagorean triple is three integers a, b, c
such that
a2 + b2 = c2.
Theorem . The Pythagorean triples are integer multiples of triples of the form (2rs, r 2 −
s2, r 2 + s2, r 2 − s2, 2rs, r 2 + s2), where r and s
are integers.
Exercise 4 (8.25). Give an example of a Pythagore
triple which is not of the form (r 2 −s2, 2rs, r2 +
s2).