To understand it we need to know first of all what a ‘basis’ for a vector field is. The details can be found in any linear algebra book, but the main idea is that with any set of vectors, each vector can be formed by multiples of a smaller subset of the vectors. For example, all two dimensional vectors
y x = x
0 1 and
1 0 , e.g.
4 3 = 3
0 1 + 4
1 0 , or more generally
0 1 + y 0 1 and
can be formed by multiples of
1 0 . It is worth noting that whilst
0 1 and
1 0 may be the simplest basis, we could also use
2 0 , where the multiple of the second vector is just half what is was in the simple case, or we could
1 1 and
1 1 . The key thing turns out to be that there is no way to form
0 0 as multiples of
even use
the basis vectors without the multiples both being zero, for example the only way to have 0 0 = a 0 1 + b 1 0 is for both a and b to be zero – this is what we mean by 0 1 and
1 0 being ‘linearly
independent’.
Now we ask the question – can we find a basis for all real numbers such that all real numbers can be expressed as a finite combination of rational multiples of the basis numbers?
It is worth noting that we need to require this to be finite combination of numbers/vectors as there is no obvious way to define an infinite sum of vectors while keep the vector space closed, i.e. such that any combination of vectors is itself a vector. It is also worth noting that if we were to allow real multiples then this would be trivial as any real number can be expressed as that number multiplied by 1. It may seem arbitrary to restrict ourselves to rational multiples, but it is perhaps worth remembering that we are simply trying to find a transformation which satisfies (2) but not (1) for real numbers. Finally one might say that we have been looking at vectors but have now switched to simply numbers. This does not matter as real numbers satisfy all of the conditions for what is called a vector space, i.e. behave as though they are vectors, e.g. the sum of two reals is a real, and a + b = b + a etc. If this this still sounds strange then we can simply form all vectors of the form a a where a For a more concreate example, we might suppose that our basis ‘vectors’ are the numbers 2 1 and 4 1 . We can then form other numbers such as 4 3 = 1 2 1 + 1 4 1 , or 1 = 1 2 1 + 2 4 1 . The astute reader may have noticed that in fact this is not a proper basis as we can create zero using non zero multiples of the vectors, namely 0 = 1 2 1 – 2 4 1 . In fact, to form a basis we only need one rational number and we can get all of the others. However, we cannot get 2 as a finite combination of rational numbers so we need to add 2 to the basis. It can be shown that we also need 3 as this cannot be formed is real. then 3 – a 2 = b ( 3 – a 2 ) 2 = b 2 3 + 2 a 2 – 2 a 6 = b 2 which would imply that 6 is rational which we know it isn’t. In a similar way we can see that we need all of the other square roots of non-square numbers, and then all cube roots of non-cube numbers, and finally , 2 , 3 , … e , e 2 , e 3 etc. (though the latter are a bit harder to show). In fact, we can see that ultimately no countable set of vectors could form a basis as otherwise we could list the reals in a countable way, by a combination of a rational number and a rational multiple of 2 . If it could, e.g. 3 = a 2 + b
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