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 non-zero 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:
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 non-zero 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 non-zero rational, and add some other rational to that. In other words:
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 .
After August 31st 2007 please use the following Gmail address: jbcosgrave at gmail.com
This page was last updated 18 February 2005 15:09:54 -0000