Suppose that d is a positive integer which is not a perfect square and r is the (irrational) square root of d.

Question

Let f(x),g(x) be any rational functions of x (over the rational numbers).Show that f(r) = g(r) => f(-r) = g(-r). [Note that we assume that if f(x) or g(x) is p(x)/q(x) then q(r) =/=0; so, for example, q(x) = x^2 - d is not allowed. Hint: observe that since r is irrational if a,b,c,d are rational then a + br = c + dr => a=c and b=d]

Answer 1

First, we establish a lemma: if P(x) is a polynomial over ℚ of which r is a root, then -r is also a root of P

Proof: By the division algorithm, there exist polynomials Q and S over ℚ such that P(x) = Q(x)(x² - d) + S(x) and S is either zero or has degree less than 2. Substituting r, we have that 0 = P(r) = Q(r)(r² - d) + S(r) = S(r) (since r² - d = 0). Thus S(r) = 0. Now, if S is not zero, then this means that r is a root of a polynomial of degree 0 or 1, which is impossible, because degree 0 polynomials have no roots and degree 1 polynomials have only rational roots, which r is not. Therefore S is the zero function, so P(x) = Q(x)(x² - d). But then P(-r) = Q(-r)((-r)² - d) = Q(-r)(r² - d) = 0, which was to be demonstrated.

Now, moving on to the main theorem:

Suppose f(x) and g(x) are rational functions over ℚ such that f(r) = g(r). Then f and g are rational functions defined at r, and thus so is f-g. Further, (f-g)(r) = f(r) - g(r) = 0. Now, since (f-g)(x) is a rational function, it can be written as P(x)/Q(x), where P and Q are both polynomials. So P(r)/Q(r) = 0, thus P(r) = 0, so by the above lemma, P(-r) = 0. Now, since (f-g)(r) is defined, Q(r)≠0. But if Q(-r) were zero, then by the lemma, Q(r) would be zero, a contradiction. So in fact Q(-r)≠0, thus (f-g)(-r) = P(-r)/Q(-r) is defined, and since P(-r) = 0, (f-g)(-r) = 0. But this means f(-r) - g(-r) = 0, so f(-r) = g(-r). Q.E.D.