The meaning of 'linearly independent over the integers/rationals'
Later, if you read the Roth section, and then the Schmidt and Baker sections, there is a standard expression you will encounter:
'... linearly independent over the rationals'
(which is equivalent in meaning to
' linearly independent over the integers'
). I would like to explain what that expression means, because without understanding the
meaning
of this expression one cannot begin to appreciate the
significance
of the use of it in the Roth, Schmidt and Baker sections.
Choose any irrational number
A
. Then, infinitely many other numbers are irrational as a
consequence
. For example 2
A
, 3
A
, 4
A
,
,
, ... are
all irrational
. What numbers are being suggested there? Simply
nonzero rational multiples
of
A
. Thus, e.g.,
is (trivially) irrational, for if not, we would have
for some integers m and n (
), giving
, is rational, whereas
A
is irrational.
Now  to make a point  I choose two irrational numbers, two old friends
and
, and ask: could it be that
is irrational as a
simple
consequence
's irrationality. Meaning? Could it be that
is a
rational
multiple of
? Well it isn't (and a novice might like to wonder:
why not?
). Neither of course is
a rational multiple of
.
Whereas if I had chosen as my two irrationals (e.g.)
and
, then here each is irrational as a
simple
consequence of the other:
is
.
, and equally
is
.
.
The (standard) expression that is used to summarise what I've just pointed out is to say that:

and
are
linearly dependent
over the rationals
(or integers)

and
are
linearly
in
dependent over the rationals
(or integers)
Formal definition. Let
be
n
(real or complex) numbers (above
n
= 2), then
are said to be
linearly dependent over the rationals
(equivalently integers) if there are
rational
numbers

not all zero
 such that
. If
are
not
linearly dependent over the rationals then they are said to be linearly independent over the rationals.
Example 1.
and
are linearly dependent
over the rationals since we have
, and thus
(so
and
may be chosen to be the rationals 1 and
, but equally they could be chosen to be the integers 5 and
.
Example 2 (exercise).
and
are linearly dependent
over the rationals (and notice it doesn't matter what is the base of the logarithms).
Example 3.
and
are
linearly independent
over the rationals
. For if not, then
for some rationals
, not both zero. But then
,
, giving
, which is impossible (why?) except for
both
being zero.
Example 4 (exercise).
,
and
are
linearly dependent
over the rationals
.
I now make one last elementary point. First, return to the the earlier 'Choose any irrational number
A
...', and note that not only is any nonzero rational multiple of
A
irrational, but so also is
any such number
plus any rational number (positive, negative, or (trivially) 0). Thus, e.g.,
is irrational (why?)
Now recall that the two irrationals
and
are essentially different, in that neither's irrationality is a simple consequence of the other because they are (the clarifying) 'linearly independent over the rationals'. But note that while (the similar looking)
and
are both irrational (why?), they are linearly independent over the rationals (why?), and so neither is irrational as an
immediate
consequence of the other's irrationality. However their respective irrationalities are intimately linked to each other in that although neither is a
direct
rational multiple of the other, each however is obtainable from the other by the process I've just outlined: multiply by some nonzero rational, and add some other rational to that. In other words:

for some rational numbers
, or equivalently

for some rational numbers
These may be succinctly sumarised by noting that what we are
really saying
here is that the three numbers
are linearly dependent over the rationals; that is there are rational numbers
 not all zero  such that
.
I would hope that when you see the
proper contexts
 the relevant parts of the Roth, Schmidt and Baker sections  these definitions will become more than mere definitions, and make perfect
sense
.