This section will give a proof of a very interesting result. The result is remarkable, but I
think its proof is even more interesting than the result obtained. It is based on
ideas used by Jacobi in the mid 1830’s. It involves the sum of integer multiples of
numbers whose ratio is irrational. If a∕b is irrational, then it is not possible that
ma + nb = 0 because if this were so, you would have

−-m
n

=

b
a

and so the ratio of
a,b is rational after all. Even though you cannot get 0 (which you can get if the
ratio of a and b is rational) you can get such an integer combination arbitrarily
small.

Theorem 2.15.1If a,b are real numbers and a∕b is not rational, then for everyε > 0 there exist integers m,n such that

|ma + nb|

< ε.

Proof:Let P_{N} denote all combinations of the form ma + nb where m,n are integers and

|m |

,

|n|

≤ N. Thus there are

(2N + 1)

^{2} of these integer combinations and all of them are
contained in the interval I ≡

[− N (|a|+ |b|),N (|a|+ |b|)]

. Now pick an integer M such
that

2 2
(2N ) < M < (2N +1)

I know such an integer exists because

(2N + 1)

^{2} = 4N^{2} + 4N + 1 and so

(2N + 1)

^{2}−

(2N )

^{2} = 4N + 1 > 2. Now partition the interval I into M equal intervals. If l is
the length of one of these intervals, then

Now as mentioned, all of the points of P_{N} are contained in I and there are more of these
points than there are intervals in the partition of I which has only M intervals. Therefore,
some interval in M contains two points of P_{N}. But each interval has length no more than
C∕N and so there exist k,

ˆk

,l,

ˆl

integers such that

|| ( )|| C
|ka + lb− ˆka + ˆlb | ≡ |ma + nb| < N

Now let ε > 0 be given. Choose N large enough that C∕N < ε. Then the above equation holds
for some integers m,n. ■