something which cannot be done (if this last point doesn’t make much sense to you then don’t worry, it is not crucial to the argument).
Now for the punch-line – the answer to whether such basis exists is yes! It is called a Hamel basis after Georg Hamel (1877 to 1954), a German mathematician. One way to prove this, used in the reference above, uses Zorn’s lemma. The details of this soon get complicated, but essential idea is that to find a basis for a set of vectors you first choose one vector. If that is a basis then we’re done. If not then there is at least one other vector which is independent of the first vector, so we include the latter with the first. If this is a basis then we are done, but if not then there is a third vector independent of the first two so we choose that and add it to the basis. We keep repeating this, with Zorn’s lemma showing that eventually we will form a basis (again, the details of this get complicated). We are near our aim. Suppose we have a Hamel basis, i.e. a basis for the real numbers over the rationals, i.e. a set of real numbers such that all real numbers can be formed by a finite combination of rational multiples of the basis numbers. Suppose we call the set of these basis numbers { r i } (the notation is slightly misleading as it makes us think of r 1 , r 2 , r 3 , … when in fact the set is not countable, but as above this is not important to the argument). Let T( x ) be the coefficient of one particular vector, r m , in the expansion of x .
So, if x 1
= … + a 1
r m
+ … and x 2
= … + a 2
r m
+ … then x 1
+ x 2
= … + ( a 1
+ a 2
) r m
+ …
Therefore T( x 1
) = a 1
and T( x 2
) = a 2
and T( x 1
+ x 2
) = a 1
+ a 2
, i.e. (2) is satisfied.
r
r
n
n
However T( r m
) = 1 as r m
= 1 r m
, but if we let x =
r m,
, for n ≠ m then T( x ) = T( r n
) = 0, not
as
r
r
m
m
would be implied by (1).
It has taken a while, but we have found a transformation which satisfies (2) but not (1).
The axiom of choice
It is perhaps making one final note, namely that the transformation found may seem quite strange – in fact, we cannot even list the basis which was used so don’t know for sure what it looks like. Worse still, Zorn’s lemma, and indeed the outline proof above that a Hamel basis exists, relies on what is called the axiom of choice, a rather strange axioms, essentially simply saying that if you have a set of objects then you can always pick one from the set. This seems obvious for a simple finite set of numbers, e.g. {1, 5, 7, 9} as we could state a rule to always pick the smallest number, or the largest. However, things quickly become less obvious. For example these rules would not pick a number for the set { x : 0 < x < 1} where there is no smallest or largest number. We might be tempted to change our rule to ‘pick the number half way between the upper and lower bounds of the set’. However this would fail for a set such as { x : 0 < x < 1 or 2 < x < 3} where there is no half way number, 1½ not being in the set. Similar problems occur no matter how we try to define the function. However, it seems to make sense to still say ‘pick an object, I don’t care how’. Bertrand Russell gave the analogy of it being easy to state a rule to pick one
58
Made with FlippingBook - professional solution for displaying marketing and sales documents online