The following question (more or less) appeared in an algebra book I'm working through. It's (more or less) elementary and I found it to be a great puzzle; maybe someone will read this and enjoy it too.


The real numbers (those including pi, the square root of 2, 1/2, etc.) are plentiful. The rational numbers (fractions--4/9, 2/3, 12312451/5543161, 0, etc.) are less plentiful, though still infinite. Some real numbers are not rational--the simplest example is probably the square root of 2 (see * at end for proof). So, if a and b are rational numbers, we can only solve the equation a+b*Sqrt(2) = 0 in one way: a=b=0. Otherwise, we'd get a way to write Sqrt(2) as a rational number, but it's irrational.

One might naturally wonder if there's a pattern here: if a, b, and c are rational numbers, is it the case that we can only solve the equation a+b*Sqrt(2)+c*Sqrt(3) = 0 in one way, with a=b=c=0? Yes--if you solve that equation for Sqrt(2) and square both sides, the condition a=b=c=0 falls out from the irrationality of Sqrt(2). In particular, that means we can't solve the equation for c=-1, so that Sqrt(3) cannot be written as a+b*Sqrt(2) for some rational constants a and b.

We can generalize this notion of being able to solve an equation with some fixed constants. For fixed real numbers r1, ..., rn, can we solve the equation a1*r1+...+an*rn = 0 in more than the one trivial way of a1=a2=...=an=0? The answer is that it depends on the ri. For example, if we take r1=r2=Sqrt(2), then the answer is yes: a1=1, a2=-1 is another solution. If we take r1=1, r2=Sqrt(2), and r3=Sqrt(3), the answer is no, as I mentioned above.


The book's question was this: find an infinite set of real numbers such that for any finite subset, the resulting equation can only be solved in the one trivial way, with all the ai's set to 0. You might think {1, Sqrt(2), Sqrt(3), Sqrt(5), ...Sqrt(p prime)...} works--and it probably does--but I wasn't able to prove it. I did eventually find a set, though it took several hours. It's a really nifty question, though, especially phrased in math-speak (where it becomes very short): "find an infinite linearly independent subset of R over Q".



*If Sqrt(2) were rational, it would be a/b for some whole numbers a and b. Then (a/b)^2 = 2, so a^2 = 2*b^2. Any square has an even number of factors of a particular prime, so b^2 has an even number of factors of 2. But then a^2 has an odd number of factors of 2, while at the same time having an even number of factors of 2, which is a logical contradiction. So, our assumption must have been faulty, and Sqrt(2) must not have been rational.