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362 Answers to Exercises and Solutions to Problems

See Problem 358, Crux Muthemuticorum 4 (1978); 161; 5 (1979), 84.

4.14. First solution. The inequality clearly holds when y = 0, so we need

consider only nonnegative values of y. The difference

---=^1 y-x 421: - y12 + (8 - y2)1

1 +x2 Y+x 4(1+ X2>(Y + z>

has the numerator (y/2)(8 - y2) w h en x = y/2, and so its minimum is

nonnegative if and only if y 5 22/2. The answer is 2fi.

Second solution. As above, we suppose that y > 0. Since y + x > 0, the

inequality is equivalent to

y _< min{x + 2/x : 2 > 0).

By the AGM inequality, x + 2/x > 24, with equality iff x = 2/z = a,

so that the largest y is 24.

4.15. f(t) = (t - r)g(t) where

g(t) = b”-lt”-l +... + bit + bo.

Since f(t) is irreducible, it has no double zeros and so g(r) # 0. Let M =

clbkl(lrj + l)k, so that [g(t)1 5 M for It - r( _< 1. Since f(t) is irreducible of

degree exceeding 1, f(t) h as no rational zero, so that for any rational p/q,

If(p/q)I = s/q” 2 l/q” for some integer s.

If Ip/q - rl 5 1, then

l/9” L lf(d9>l = Id9 - 4 Idda) I I Wpl9 - d.

If Ip/q-rl 2 1, then Ip/q-rl 2 l/q”. Let k = min(l,l/M). Then Ip/q-rl 2

k/q” as required.

Remark. This result asserts that a rational approximation of a zero of

a polynomial over Z is, in some sense, not very close to the zero. To look

at the matter in another way, note that from this result follows that, if for

each m, a nonrational number w satisfies Iw - p/q1 < l/q’” for infinitely

many distinct rational numbers p/q, then w is not the solution of a polyno-

mial equation with integer coefficients. (For suppose it were the solution of

an irreducible equation of degree n; then we would have k/q” 5 [p/q - rl <

119 “+l for infinitely many rationals p/q. Since for any denominator q,

IPl9 - 4 < 119 “+l <^1 can occur at most finitely often, we must have

that q < l/k for infinitely many positive integers q - an impossibility.)

Other results on approximation of nonrationals by rationals can be found in

Chapter 11 of G.H. Hardy & E.M. Wright, An Introduction to the Theory

of Numbers (4th ed., Oxford, 1960).

4.16. Multiply the numerator and denominator of the second (resp. third)
member by x (resp. xy) and simplify to obtain the constant value 2.