2
=
n
2
.
Therefore, 2
m
≤
n
2

n
2
k
, yielding
k
≥
n
2
n
2

2
m
.
Cartesian product
Theorem 4
(Vizing 1963, Aberth 1964)
.
Let
G
and
H
be graphs. Then
χ
(
G
2
H
) = max
{
χ
(
G
)
, χ
(
H
)
}
.
Proof.
Since
G
2
H
contains copies of
G
and
H
as subgraphs, we have
χ
(
G
2
H
)
≥
χ
(
G
) and
χ
(
G
2
H
)
≥
χ
(
H
). Hence
χ
(
G
2
H
)
≥
max
{
χ
(
G
)
, χ
(
H
)
}
.
Proof (continued).
Denote
k
= max
{
χ
(
G
)
, χ
(
H
)
}
. Let
g
be a proper colour
ing of
G
using
χ
(
G
) colours, say 1
,
2
, . . . , χ
(
G
). Let
h
be a proper colouring
of
H
using
χ
(
H
) colours, say 1
,
2
, . . . , χ
(
H
).
For each vertex (
u, v
)
∈
V
(
G
)
×
V
(
H
), define
f
(
u, v
) =
g
(
u
) +
h
(
v
) mod
k,
which is an integer in
{
1
,
2
, . . . , k
}
. (We use
{
1
,
2
, . . . , k
}
instead of the more
commonly used
{
0
,
1
, . . . , k

1
}
for the set of reminders.)
We claim that
f
is a proper colouring of
G
2
H
.
Indeed, if (
u, v
) and
(
x, y
) are adjacent in
G
2
H
, then
g
(
u
)+
h
(
v
) and
g
(
x
)+
h
(
y
) agree in one of
the summands and differ in the other by some integer between 1 and
k
. So
g
(
u
) +
h
(
v
) and
g
(
x
) +
h
(
y
) differ by some integer between 1 and
k
. Thus
they are in different congruence classes modulo
k
.
In other words, (
u, v
)
and (
x, y
) receive different colours under
f
. Therefore,
χ
(
G
2
H
)
≤
k
.