Lines, parabolas, distances and inequalities an enrichment class

Lines, parabolas, distances and inequalities an enrichment class Finbarr Holland 1. Lines in the plane A line is a particular kind of subset of the plane R 2 = R R, and can be described as the set of ordered pairs of real numbers (x, y) that satisfy a linear equation of the form ax + by + c = 0, where a, b, c are real numbers such that a 2 + b 2 > 0. If a = 0, but b 0, the solution set of this equation consists of points having the same second coordinate; such sets define horizontal lines. If b = 0, but a 0, the set consists of all points whose first coordinate is the same, which generate vertical lines. Solution sets of the latter kind are not the graphs of any functions: recall that a set is the graph of some function iff every vertical line that meets it does so in exactly one point. However, if b 0, the set {(x, y) : ax + by + c = 0} is the graph of the function f defined on R by f(x) = a b x c b. Every pair of distinct points (x 1, y 1 ), (x 2, y 2 ) determines a unique line that (passes through) contains the points. Its equation is (y y 1 )(x 2 x 1 ) = (y 2 y 1 )(x x 1 ). As long as x 2 x 1, we can write this in the form y = mx + c, by taking m = y 2 y 1 x 2 x 1, c = y 1 mx 1. This is the slope-intercept form of the corresponding line m is its slope, and (0, c) is the point where the line meets the y-axis and is usually written/spoken as the line y = mx + c, for short. Problem 1. Do the points (1, 1.828400000), (2, 2), (3, 1.000027124) lie on a line? Preprint submitted to Elsevier March 19, 2013

Solution. Since every pair of the three points determines a unique line, which can be written in slope-intercept form, all lie on the same line provided the slopes of the three lines are the same. For this to happen we require 2 1.828400000 = 1.000027124 2, i.e., 1.828400000 + 1.000027124 2 = = 1.414213562. 2 A calculator confirms this! But the confirmation is incorrect, because the number 1.414213562 is only an approximation to 2, albeit a very good one! The truth of the matter is that 2 is irrational, as we proceed to show. Theorem 1. 2 is not rational. Proof. Suppose this is false. Then there are two integers a, b, with b 0, such that 2 = a b. Since 2 stands for the unique positive real number whose square is 2, a, b are either both positive or both negative. We ll take it that they re both positive; otherwise, replace them with their negatives. Also, by cancelling their common factors, if necessary, we may assume that a, b are co-prime, i.e., they have no common divisors. So, to recap: 2 = a/b where a, b are positive integers without common factors. It follows that ( a b )2 = 2, a 2 b 2 = 2, 2b2 = a 2. This last expression means that a 2 is an even integer. Hence (?) a is even: a = 2m, for some positive integer m. Substitute back for a to obtain the equation 2b 2 = 4m 2, which means that b 2 = 2m 2. In other words, b 2 is even, whence b itself is even: b = 2n, for some positive integer n. Conclusion: a, b are both even, a [possibility that was ruled out right at the beginning. Thus we ve reached a contradiction, and so our opening assumption that 2 is rational is false. Accordingly, the statement of the theorem stands. Remark. This theorem and its proof was discovered by Pythagoras over two thousand years ago; it s a model for similar results such as the fact that 3 is also irrational. Indeed, a similar argument can be used to show that if p is a prime number, then p is irrational. Exercise 1. For what values of x are the points (1, cos x), (2, cos 2x), (3, cos 3x) on the same line? 2. New lines from old Here we look at families of lines that are graphs of functions defined on R. Such functions are of the form f(x) = mx + c, for some pair of real numbers 2

m, c. Varying these two parameters we get a collection of linear functions whose graphs are non-vertical lines. Given also, g(x) = m x + c, we can determine new lines in many different ways: eg, by forming the functions mm x + cc, (m ± m )x + (c ± c. Another way is to form the compositions f g, g f. Thus f g(x) = f(g(x)) = mg(x) + c = m(m x + c ) + c = mm x + mc + c; g f(x) = g(f(x)) = m (f(x) + c = m (mx + c) + c = m mx + mc + c. Thus, the graphs of f g and g f are parallel lines which are identical iff mc + c = m c + c. In particular, this shows that the operation of composition is not commutative, i.e., general, f g g f. A particularly interesting sub-family of lines are those having non-zero slope, i.e., those given by functions of the form f(x) = mx+c, with m 0. Not only is R the domain of such functions, but R is also their range, so that, given y R, we can solve y = mx + c for x. This gives rise to the function g(x) = m x + c, where m = 1 m, c = c m. Check that f g = g f = id, where id is the identity function given by id(x) = x, whose graph is the diagonal line y = x. This means that this pair of functions are each other s inverses. Note also, that if, for some pair of subsets A, B of R, S = {(x, y) : x A, y B}, its inverse set, denoted by S 1, is defined by S 1 = {(x, y) : (y, x) S} = {(x, y) : x B, y A}. In particular, the inverse of the y-axis is the x-axis, and vice-versa; the inverse of a circle is a circle, though not necessarily the same one; the inverse of the graph of the quadratic y = x 2 is the parabola {(x, y) : y 2 = x, x 0}, whose axis is the positive real axis. As this example shows, the inverse of the graph of a function need not be the graph of a function. Note too, that the inverse of a set is obtained by reflecting each of its elements in the diagonal line. Exercise 2. Prove that the reflection of (a, b) in the line y = x is given by (b, a). Exercise 3. Prove that the inverse of (the graph of) the line y = mx + c, with m 0, is (equal to the graph of) the line y = 1 m x c m. Exercise 4. Describe the compositions f g and g f for the following pairs of functions, and define their domains and ranges. 1. 2. f(x) = 1 x 2, 1 x 1; g(x) = sin x, x R. f(x) = (1 x) 4, x R; g(x) = x 5, x R. Exercise 5. Prove that the composition of two polynomials is a polynomial. 3

3. The modulus function defined on R The domain of definition of this important function is R. Its value at x is given by x, which is short for any of the following expressions that define it: x = { x, if x 0, x 2 = max(x, x) = x, if x 0. Note that: (i) x 0, with equality iff x = 0; (ii) xy = x y for all real numbers x, y; (iii) x = x, for all real numbers x; (iv) the graph of y = x coincides with that of y = x, when x 0, and with that of y = x, when x 0; (v) the region {(x, y) : y x } is a convex set, i.e., the line segment joining any two of its points is a subset of it. Exercise 6. Sketch the graph of the function x x 1 + x + 1. Exercise 7. Sketch the graph of the function x ( x 1 + x + 1 ) / ( x 2 + x + 2 ). Definition 1. We define the (usual) distance between two points a, b R by a b. In particular, x is the distance between the origin 0 and x, Theorem 2 (The triangle inequality). If a, b R, then This inequality is strict unless ab 0. a b a + b. Proof. It s enough (?) to prove that a b 2 ( a + b ) 2. Now a b 2 = (a b) 2 (because x 2 = x 2 x) = a 2 2ab + b 2 = a 2 2ab + b 2 a 2 + 2 ab + b 2 (because x x x) = a 2 + 2 a b + b 2 = ( a + b ) 2, as desired. Moreover, there is equality iff ab = ab, i.e., iff ab 0. Exercise 8. Prove that xy = x y for all real x, y. Exercise 9. Prove that a b a b, a, b. Exercise 10. Let ρ(a, b) = a b, a, b R. 1 + a b Prove that ρ(a, b) ρ(a, 0) + ρ(b, 0). [Hint: If 0 t s, then t/(1 + t) s/(1 + s).] 4

4. Distance in the plane How do we measure distance between two points P = (p 1, p 2 ), Q = (q 1, q 2 ) in the plane? The usual way is to define this to be given by (p1 q 1 ) 2 + (p 2 q 2 ) 2, which we label d 2 (P, Q), the subscript 2 indicating that other definitions are possible. It is known as the crow flies metric. Note that d 2 (P, Q) 0, with equality iff P and Q coincide. Also, d 2 (P, Q) = d 2 (Q, P ), and d 2 (λp, λq) = λ d 2 (P, Q). In addition, d 2 (P + R, Q + R) = d 2 (P, Q). The last two properties describe how d 2 behaves with respect to scaling and translation. All of these properties hold for d 1 (P, Q) = p 1 q 1 + p 2 q 2, which is a little less awkward to work with than d 2. This is known as the taxi-cab metric. In everyday life, it is this metric that we use most often. In addition, the following triangle inequality holds for d 1. Theorem 3. If P, G, R are three points, then Proof. We re required to prove that d 1 (P, Q) d 1 (P, R) + d 1 (R, Q). p 1 q 1 + p 2 q 2 ( p 1 r 1 + p 2 r 2 ) + ( r 1 q 1 + r 2 q 2 ). But by Theorem 2, and p 1 q 1 = (p 1 r 1 ) (q 1 r 1 ) p 1 r 1 + q 1 r 1, p 2 q 2 = (p 2 r 2 ) (q 2 r 2 ) p 2 r 2 + q 2 r 2. The result follows by adding these inequalities, and then rearranging the terms on the right side. A similar result holds for d 2, but the proof is more cumbersome. It is made easier if we first establish an intermediate result. Lemma 1. Suppose a, b, c, d are real numbers. Then ac + bd a 2 + b 2 c 3 + d 2. Moreover, there is equality here iff ad = bc, i.e., iff the points (a, b) and (c, d) lie on a line through (0, 0). Proof. The stated inequality is equivalent (?) to this one: (ac + bd) 2 (a 2 + b 2 )(c 2 + d 2 ), 5

which in turn is equivalent to (ac) 2 + 2acbd + (bd) 2 a 2 c 2 + a 2 d 2 + b 2 c 2 + b 2 d 2 and so to 2abcd (ad) 2 + (bc) 2 i.e., to 0 (ad bc) 2. But the square of every real number is non-negative. Hence the previous result is true, and working backwards through the chain of inequalities, we see that our claim is true. Furthermore, it is plain that equality holds right the way through iff ad = bc. This ends the proof. Corollary 1. If a, b, c, d are real numbers, then (ac + bd) a 2 + b 2 c 3 + d 2. Moreover, there is equality here iff ad = bc, i.e., iff the points (a, b) and (c, d) lie on a line through (0, 0). Exercise 11. Prove that, if a, b are real numbers, then, for every real number x, a 2 + b 2 a cos x + b sin x a 2 + b 2. For what values of x is there equality in both of these inequalities? Before using lemma 1 to establish the triangle inequality for d 2, we prove the well-known distance formula described in the next theorem. 5. Distance from a point to a line The result that follows is due to Euclid, and has had a profound affect on the development of Mathematics. Euclid based his proof of it on pure geometric arguments, which you may encounter in Strand 2 of Project Maths. Theorem 4. Given a line L with equation ax + by + c = 0, where a 2 + b 2 > 0, and a point P 0 = (x 0, y 0 ), there is a unique point Q L such that d 2 (Q, P 0 ) < d 2 (P, P 0 ) for all other points P L. In addition, d 2 (Q, P 0 ) = ax 0 + by 0 + c a2 + b 2. Proof. Let P = (x, y) L, so that c = ax by, in which case, ax 0 + by 0 + c = a(x 0 x) + b(y 0 y) a(x 0 x) + b(y 0 y) (because x x, x R) a 2 + b 2 (x x 0 ) 2 + (y y 0 ) 2 (by Lemma 1) = a 2 + b 2 d 2 (P, P 0 ). Thus ax 0 + by 0 + c a2 + b 2 d 2 (P, P 0 ), P L; 6

and the equality sign holds here, again by Lemma 1, iff ax + by + c = 0, and a(y 0 y) = b(x 0 x). Since a 2 + b 2 > 0 these equations have a unique solution pair (x, y), which lies on L, and on the line perpendicular to L that passes through P 0. This is the point Q. It is the foot of the perpendicular from P 0 on to L. Exercise 12. Solve the equations ax + by + c = 0, a(y y 0 ) = b(x x 0 ) for x, y under the assumption that a 2 + b 2 > 0. Exercise 13. Redo Theorem 4 with d 1 in place of d 2. 6. The triangle inequality for d 2 We ll prove the analogue of Theorem 2 by showing that the d 2 -distance between two points P, Q doesn t exceed the sum of the d 2 -distances from the origin O to P and Q. Theorem 5. d 2 (P, Q) d 2 (P, O)) + d 2 (Q, O). The inequality is strict unless the points P, Q and O = (0, 0) are collinear. Proof. Because the numbers are positive, the stated inequality is equivalent to that obtained by squaring both sides of the one on display. In other words, it suffices to prove that 2. (p 1 q 1 ) 2 + (p 2 q 2 ) 2 (p 2 1 + p2 2 + q1 2) 2 + q2 Expanding out both sides of this, and cancelling common terms, we see that this is equivalent to the following statement: (p 1 q 1 + p 2 q 2 ) p 2 1 + p2 2 q1 2 + q2 2. But this is now a consequence of Corollary 1. The case of equality is left as an exercise, as is the following corollary. Corollary 2. Suppose P, Q, R are three points. Prove that with equality iff P, Q, R are collinear. d 2 (P, Q) d 2 (P, R)) + d 2 (R, Q), 7

7. The square function Lemma 1 is a special case of a more general inequality known as the Cauchy- Schwarz inequality. To deal with this, it is convenient to recall a few facts about quadratic polynomials. These are functions defined on R whose values at x are expressions of the form ax 2 + bx + c, where a, b, c are real numbers, and a 0. The simplest of these occurs when b = c = 0, and a 0, resulting in the generic form of a quadratic: y = ax 2, whose graph is symmetric about the y- axis. If a > 0, it is a subset of the upper half-plane, and the lowest point on it is the origin. Thus, if a > 0, the function x ax 2 assumes its minimum value at x = 0. If a < 0, the corresponding graph is a subset of the lower half-plane, and the highest point on it is the origin. Thus, if a < 0, the function x ax 2 assumes its maximum value at x = 0. These features persist for a general quadratic with equation f(x) = ax 2 + bx+c. The sign of a determines whether its graph is cup-shaped or cap-shaped. This is seen by completing the square on x. Doing this results in a new form for f: y = f(x) = a(x 2 + b b x) + c = a(x + a 2a )2 b2 4a + c = a(x + b 2a )2 + 4ac b2. 4a Changing axes by setting X = x + b/(2a) and Y = y (4ac b 2 )/(4a) we obtain a simplified expression: Y = ax 2, and we re back to the generic form of a quadratic discussed earlier. Hence, the graph of f is symmetric about the line x = b/(2a), and its minimum value is (4ac b 2 )/(4a) if a > 0, while its maximum value is (4ac b 2 )/(4a) if a < 0. The sign of the discriminant b 2 4ac determines whether the equation f(x) = ax 2 + bx + c = 0 has real or complex roots. Theorem 6. If a 0, the equation f(x) = ax 2 + bx + c = 0 has real roots iff b 2 4ac. Proof. Suppose 4ac > b 2 and x is a real root of f. Complete the square on x, as above, thereby getting 0 = f(x) = a(x + b 4ac [ 2a )2 b2 + = a (x + b 4ac ] 4a 2a )2 b2 + 4a 2. Since x is real, the number in square brackets is positive, and a 0. Hence, the displayed statement can t hold. In other words, if the equation has real roots, then b 2 4ac. Conversely, if the latter condition holds, then we can extract the unique positive square root of (b 2 4ac)/(4a 2 ) and form the real numbers α = b 2a + b2 4ac 4a 2, β = b 2a b2 4ac 4a 2. It can now be verified by direct substitution that f(α) = f(β) = 0, which means that f has real (possibly coincident) roots. 8

Proposition 1. Suppose a 0 and ax 2 + bx + c 0, x R. Then a > 0 and b 2 4ac. Proof. By hypothesis, if x 0, then a + b x + c x 2 0. By letting x here, we deduce that a 0. But a 0. Hence a > 0. But 0 f( b 4ac b2 ) =, 2a 4a whence we deduce that 4ac b 2 is non-negative, as stated. Example 1. The region S = {(x, y) : y x 2 } is convex. Proof. The claim is that S contains the line segment joining any two of its points. So, suppose P = (x 1, x 2 ), Q = (y 1, y 2 ) both belong to S, so that y 1 x 2 1, y 2 x 2 2. Let R = (z 1, z 2 ) be on the line segment [P, Q]. Then there is a number t [0, 1] such that z 1 = tx 1 + (1 t)x 2, z 2 = ty 1 + (1 t)y 2. Hence z 2 1 = (tx 1 + (1 t)x 2 ) 2 = t 2 x 2 1 + 2t(1 t)x 1 x 2 + (1 t) 2 x 2 2 t 2 x 2 1 + t(1 t) ( x 2 1 + x 2 2) + (1 t) 2 x 2 2 = ( t 2 + t(1 t) ) x 2 1 + ( t(1 t) + (1 t) 2) x 2 2 = tx 2 1 + (1 t)x 2 2 ty 1 + (1 t)y 2 (because 0 t 1) = z 2, which means that R S. Hence, [P, Q] S, as claimed. As a final result, we derive the Cauchy-Schwarz inequality, a spec8ial case of which is Lemma 1. Theorem 7. Suppose a 1, a 2,..., a n, b 1, b 2,..., b n are 2n real numbers. Let A = a 2 1 + a2 2 + + a2 n, and B = b 2 1 + b2 2 + + b2 n. Then a 1 b 1 + a 2 b 2 + + a n b n AB. Equality holds here iff either (i) AB = 0 or (ii) AB > 0 and there is some real number t such that ta i = b i, i = 1, 2,..., n. Proof. Let C = a 1 b 1 + a 2 b 2 + + a n b n. The claim is that C AB. In the first place, if AB = 0, then either all the a s are zero or all the b s are, and, in whichever of these cases holds, C = 0 = AB. So, it s enough to prove the result when AB > 0. With this assumption in place, consider the function p(x) = (xa 1 b 1 ) 2 + (xa 2 b 2 ) 2 + (xa n b n ) 2, x R. 9

Expanding out each term, and gathering the coefficients of x 2, x and 1, we see that p(x) = A 2 x 2 2Cx + B 2. Thus, p is a quadratic polynomial which takes only non-negative values. By Proposition 1, 4C 2 4A 2 B 2, i.e., C AB, which is the desired inequality. To deal with the case of equality, note that (?) C = AB iff p(x) = (Ax C/A) 2, x R. In which case, p(c/a 2 ) = 0. Equivalently, Ca i = A 2 b i, i = 1, 2,..., n. Thus, if AB > 0 and C = AB, part (ii) follows. Conversely, if AB > 0, and ta i b i = 0, i = 1, 2,..., n, for some real t, then p(t) = 0. But p(x 0, x R. Hence, t is a double (?) root of p, whence C 2 = A 2 B 2, i.e., C = AB. This completes the proof. Corollary 3. In the same notation, if the b s are a rearrangement of the a s then a 1 b 1 + a 2 b 2 + + a n b n a 2 1 + a 2 2 + + a 2 n. For instance, ab + bc + ca a 2 + b 2 + c 2, and the inequality is strict unless a = b = c. Exercise 14. Prove that if a, b, c, x, y are real numbers, then a 2 + b 2 + c 2 a cos x cos y + b cos x sin y + c sin x a 2 + b 2 + c 2.. The following exercises depend on Theorem 7, and can be done in the same way that Theorems 2 and 5 were treated. Exercise 15. Suppose a 1, a 2,..., a n, b 1, b 2,..., b n are 2n real numbers. Prove that n (a i b i ) 2 n a 2 i + n b 2 i. Exercise 16. Suppose a 1, a 2,..., a n, b 1, b 2,..., b n, c 1, c 2,..., c n are 3n real numbers. Prove that n (a i b i ) 2 n (a i c i ) 2 + n (c i b i ) 2. 10