Properties of the Real Numbers 2 (G) |X + X − 5| > 0 Solve the Following Equations Over R

The Real Line Philip Pennance1 Version: – February 2019.

1. Introduction. shows that√ there is a point on the line at distance 2 from the origin. By the In the elementary school, fractions or previous claim, this point cannot corre- rational numbers are or should be de- spond to a fraction. ﬁned as points on a number line. It is a consequence of the next result that √ not all points on such a line represent 2 fractions. 1

2. Claim. √ 0 1 2 2 There do not exist natural numbers n, m such that n2 = 2m2 (i.e., there is no n fraction n whose square is 2). 3. Deﬁnition. Proof. The set of all points on the number line is called the set of real numbers and is Suppose to the contrary that such n, m denoted . exist. Since common factors between R the numerator and denominator of a 4. Claim. fraction can be cancelled it may be as- Let x ∈ . The following are equivalent. sumed that n and m have no common R factor other than 1. (a) x is a fraction (rational number). Now, n2 = 2m2 means that n2 is even. (b) x has a repeating decimal repre- Since the square of an odd number is sentation. odd, it must be that n is also even. Proof. Hence there exists a natural number k such that n = 2k. But then n2 = See, for example, Dolciani et al. Intro- 2m2 = 4k2 and so m2 = 2k2. This ductory Analysis. means that m2 and hence m are both 5. Example. even. But if n and m are both even they 1 have 2 as a common factor, contradict- The rational number 2 has repeating ing the fact that 1 is the only common decimal representation 0.500000 ··· factor. Hence the numbers n and m do 6. Example. not exist. This means that there is no fraction whose square is 2. Let x = .32456. Find natural numbers n and m such that x = n . Corollary. m Solution. Not all points on the number line core- spond to fractions. 103x = 324.56 5 Proof. 10 x = 32456.56 (105 − 103)x = 32456 − 324 This result has been know since the time of Pythagoras. Indeed, a simple Hence we may take n = 32132 and application of the Pythagorian theorem m = 105 − 103. 1https://pennance.us

1 7. Claim. 12. Claim. Let x ∈ . Let x ≥ 0. Then there exists a unique√ R non negative real number (denoted x) d(x, 0) = max{x, −x} whose square is x. ( x if x ≥ 0 , Proof. = −x if x < 0 The proof is beyond the scope of this course. Proof. √ 8. The number x is called the principal The two formulae are just two alterna- square root of x. tive ways of ensuring that distance is non negative (as required by its deﬁni- 9. Let x ∈ . The distance of x from the R tion). A human can (hopefully) evalu- origin denoted d(x, 0) is deﬁned by ate | − 5| without using these formulae. √ A machine (IQ = 0) cannot. d(x, 0) = x2 (1) 13. Claim

Justiﬁcation. Let x, x0 ∈ R. The distance between x 0 Distance is a non negative quantity. and x is given by The square root symbol is necessary to d(x, x0) = |x − x0|. ensure non-negativity. Remark. Proof. Subtraction of the same number from The absolute value of |x| is just an- 0 other notation for this distance. Thus, both x and x does not change the distance between them. Subtracting x0 |x| = d(x, 0) from both yields d(x, x0) = d(x − x0, x0 − x0.) . = d(x − x0, 0) 10. Example. = |x − x0|.

d(−5, 0) = p(−5)2 √ 14. Example. = 52 Write√ without√ the absolute value sign = 5 | 51 − 2 13|. Solution. 11. Remarks. √Notice 51√< 52 = 4×13. It follows that Equation (1) is just the 1-dimensional 51 < 2 13. Hence, version of Pythagorus’ theorem. In two √ √ √ √ dimensions, the distance P = (x, y) | 51 − 2 13| = 2 13 − 51. from the origin O = (0, 0) is given by 15. Example. p 2 2 d(P,O) = x + y Write the interval −10 < x < 12 in the form |x − a| < r. Replacing y by 0 yields the one dimensional formula. Solution.

2 1 An open interval consists of all points Case 1. If x ≥ 2 , then whose distance from the center of the 2 2 interval is less than the radius of the |2x − 1| = x ⇐⇒ 2x − 1 = x 2 interval. In this case, the given interval ⇐⇒ x − 2x + 1 = 0 has midpoint ⇐⇒ (x − 1)2 = 0 −10 + 12 ⇐⇒ x = 1 a = = 1 2 1 Case 2. If x < 2 , then and radius |2x − 1| = x2 ⇐⇒ 1 − 2x = x2 12 − (−10) 2 r = = 11. ⇐⇒ x + 2x − 1 = 0 2 ⇐⇒ (x + 1)2 − 2 = 0 √ Hence, a point x falls in the interval ⇐⇒ x = −1 ± 2 −10 < x < 12 if and only if Both of these solutions are consistent d(x, 1) < 11. 1 with the condition x < 2 . Hence That is, if and only if √ |2x − 1| = x2 ⇐⇒ x ∈ {1, −1 ± 2}. |x − 1| < 11. 18. Example.

16. Example. Let A, B be the points shown on the following scale: √ Solve |5 − 2x| ≥ 5. 2 B A 108 2 1 Solution. Si P = 3 B + 3 A : 2 Since the absolute value of a number is Show that |P − A| = 3 |B − A|. the distance between the number and Solution. the origin. It follows that the distance 2 1 |P − A| = B + A − A between the number 5−2x and the ori- 3 3 = 2 B − 2 A gin is at least 5. This means that either 3 3 2 5 − 2x ≤ −5 or 5 − 2x ≥ 5. It follows = 3 · |B − A| that either x ≥ 5 or x ≤ 0. 2 = 3 |B − A|. 17. Example. Thus, the distance d(P,A) is two thirds Solve |2x − 1| = x2. of the distance d(A, B). Solution. Remark: It is left as an exercise to ver- Notice that ify that B < P < A. 1 2x − 1 ≥ 0 ⇐⇒ x ≥ 19. Additional Remarks. 2 Let λ1, λ2 be positive real numbers. It Hence is not diﬃcult to show that if

λ1 + λ2 = 1,

( 1 then a linear combination of points 2x − 1 if x ≥ 2 , |2x − 1| = 1 1 − 2x if x < 2 . P = λ1A + λ2B

3 does not depend on the location of the (d) Commutative law for addition: origin. Moreover the ratio of the dis- tances PA : PB is λ2 : λ1. (Note the a + b = b + a. order!). This invariance with respect to choice of origin can be seen in the for- (e) Associative law for multiplication: mulae for the midpoint 1 1 a · (b · c) = (a · b) · c. P = A + B 2 2 (f) Existence of multiplicative iden- and, in Physics, the formula for center tity: of mass. If there is a mass m1 at point A and a mass m2 at a point B then the ∃1 6= 0 : a · 1 = 1 · a = a. center of mass is given by

m1 m2 XCM = A + A (g) Existence of multiplicative inverse: m1 + m2 m1 + m2 −1 −1 Notice that the sum of the coefficients a 6= 0 =⇒ a · a = a · a = 1. of A and B is 1. These remarks extend to more points and higher dimensions. (h) Commutative law for multiplica- The formula for student grade point av- tion: erage exhibits this same invariance. a · b = b · a. 20. Remarks. (i) Distributive law It is beyond the scope of this course to rigorously extend the multiplication a · (b + c) = a · b + a · c. and division of rational numbers to real numbers. It suffices for our purposes 22. Definition. to know that this can be done, and that the resulting operations retain the Subtraction and division are defined by usual algebraic properties of the addi- a a − b = a + (−b), = a · b−1 = ab. tion and multiplication of rational frac- b tions. These properties will be taken as axioms. 23. Remark. 21. Axioms for Elementary Arithmetic. The above list is not complete. Later, There exist binary operations of addi- the axioms of order will be given and tion and multiplication, such that in the Calculus class, the important completeness axiom, without which, (a) Associative law for addition: the subtler properties of R cannot be a + (b + c) = (a + b) + c. proven. 24. Cancelation Theorems for Addition (b) Existence of additive identity: For all a, b, c ∈ R. ∃0 : 0 + a = a + 0 = a. (a) if a + c = b + c then a = b. (Right cancelation) (c) Existence of additive inverse: (b) if c + a = c + b then a = b. a + (−a) = (−a) + a = 0. (Left cancelation)

4 Proof of (a). Proof. Suppose a + c = b + c. Since c has Using respectivly the property of the an additive inverse −c it follows that multiplicative inverse, the distributive (a+c)+(−c) =( b+c)+(−c). Associa- law, and the property of the additive tivity gives a+(c+(−c))= b+(c+(−c)). inverse, we obtain x + (−1) · x = 1 · By the property of the additive inverse x + (−1) · x = (1 + (−1)) · x = 0.x. the latter implies a + 0 = b + 0. Fi- It follows (using the previous theorem) nally by the property of the additive that x + (−1) · x = 0. Since the addi- identity 0 we have a = b, which was to tive inverse is unique we conclude that be proven. The proof of (b) is similar. −1 · x = −x. 25. Claim. 29. Claim.

The additive inverse of any real number For all a, b ∈ R,(−a) · b = −(a · b). x is unique. Proof. Proof. Using the previous theorem and the as- Suppose x has two additive inverses, a sociative law and b. We claim that a = b. Since −(a · b) = −1 · (a · b) = (−1 · a) · b = a and b are additive inverses of x we (−a) · (b). have a + x = 0 and x + b = 0. Hence a + x = b + x. By the cancellation the- 30. Claim. orem a = b. For all a, b ∈ , 26. Claim. R ab = 0 ⇒ (a = 0 or b = 0). (2) For all a ∈ R, −(−a) = a. Proof. Proof. By the property of additive inverse, −a Suppose that ab = 0. If a = 0 the con- has an inverse −(−a) such that −a + clusion is true. If a 6= 0 then a has [−(−a)] = 0. Since −a is the inverse a unique multiplicative inverse a−1. It of a, it satisﬁes (−a) + a = 0. There- follows that a−1(ab) = a−10. By [4] fore (−a)+[−(−a)] = (−a)+a. By the above the right hand side of this equa- cancellation theorem −(−a) = a. tion is zero. By associativity and the deﬁnition of multiplicative inverse, the 27. Claim. left hand side a−1(ab) = (a−1a)b = 1b = For all x ∈ R, 0 · x = 0. b. We conclude that in this case b = 0. In both possible cases we have proven Proof. that (a = 0 or b = 0) as required. Since 0 is the additive inverse. 0x+0 = 31. Example. 0x = (0 + 0)x. By the distributive law (0 + 0)x = 0x + 0x. By transitivity Solve x2 = 4. of equality, 0x + 0 = 0x + 0x. Using Solution. the cancellation theorem it follows that 0x = 0. x2 = 4 ⇒ x2 − 22 = 0 and so (x − 2)(x + 2) = 0. By (2) either 28. Claim. x − 2 = 0 or x + 2 = 0. Hence ei- For all x ∈ R, −x = −1 · x ther x = 2 or x = −2. Substitution

5 verifies that both values are indeed so- 35. Example. lutions. Use completion of the square to express x2+3x−8 as a difference of two squares. 32. (Difference of Squares Identity). Solution. a2 − b2 = (a − b)(a + b). 32 32 Proof. x2 + 3x − 8 = x + − − 8 2 2 Using the distributive law twice yields 32 9 = x + − + 8 (a − b)(a + b) = a(a + b) − b(a + b) 2 4 2 2 = a + ab − ba − b 32 41 = x + − a2 − b2. 2 4 !2 32 r41 33. Claim. (Completing the Square). = x + − . 2 2 2 4 b b x2 + bx = x + − . 2 2

Proof. Exercise. (Use the diﬀerence of two squares identity). 34. Remark. The geometric interpretation of completing the square should be clear from the following diagram. x

+ = + = b

Exercises

1. Explain carefully with reference to the 2. Find the exact value of |3.99 − 4|. appropriate deﬁnitions why each of the following formulae are not true for all 3. Write the following intervals in the form real numbers. |x − a| < r” √ (a) 4 = ±2 (a) −7 < x < −1 √ (b) x2 = ±x √ (b) e < x < π (c) x2 = x √ 4. Solve the following. Write each solution (d) x2 = −x as an interval or union of intervals.

6 (a) |x| ≤ 2 (a) x = −x 1 1 (b) | − 2x| < −2 (b) = − x x (c) |x − 5| > 2 (c)0 .5x − 4 = (1/3)x + 6 (d)2 < |x| < 3 (d) x − 4 = 2x − 7 (e) |5 − 2x| ≥ 5 (e) x2 − 3x + 2 = 0 (f) |8x − 7| < 8 10. Use the properties of the real numbers 2 (g) |x + x − 5| > 0 solve the following equations over R. (h) |x − 1| + |x − 2| = 12 (a) ax + b = c (i)3 < |2 − 4x| < 6. (b) x2 − 2x − 63 = 0 √ (j)0 < x2 < 1. (c) x2 − 5 = 0. 5. Disprove each of the following by a 11. Show, using the axioms, that the fol- counter example x ∈ R for which the lowing are identities. formula is false. (a) If c 6= 0 and ac = bc, then a = b. (a) |x| = ±x. (b) a − b = −(b − a) (b) |x| = x. (c)( a−1)−1 = a, a 6= 0 (c) |x| = −x. (d) −(ab) = (−a)b = a(−b) (d) | − x| = x ab a (e) = c, where b, c 6= 0. 6. Disprove each of the following by ﬁnd- c b ing counter examples: 12. Use the properties of the real numbers (a) If x < y then |x| < |y|. to prove the following identities: (b) |x| < 3 ⇐⇒ x < 3 or − x < 3. (a)( ab)2 = a2b2 (c) |x| > 3 ⇐⇒ x > 3 and − x > 3. (b) a4 −b4 = (a−b)(a3 +a2b+ab2 +b3) (d) x2 > 16 ⇒ (x > 4 or x > −4) (c) a3 + b3 = (a + b)(a2 − ab + b2) 3 3 2 2 3 7. Let x ∈ R. Show that the following are (d)( a + b) = a + 3a b + 3ab + b equivalent: 13. Factorize (a) x2 ≥ |x| (a)2 x2 + 3x − 8 (b) x = 0 or x ≤ −1 or x ≥ 1. √ (b) x4 + x2 − 11 8. Let (c) x3 + x2 + x + 1 S = {x ∈ R : |x−1| < 5 ⇒ |3x−1| < 7}. 14. Solve by factorization: Write S as an interval or union of inter- (a) x2 + 5x − 14 = 0 vals. (b)2 x2 + 3x − 2 = 0 9. Solve the following equation. Mention (c) 12x2 + x − 6 = 0 the axiom(s) which justify each step in your solution. 15. Solve by completing the square:

7 (a) x2 + 6x + 6 = 0 16. (a) Show that (b)2 x2 + 3x − 2 = 0 √ −b ± b2 − 4ac 2c 2 3 = √ (c) x − πx − e = 0 2a −b ± b2 − 4ac √ (d) πx2 + ex + 17 = 5 (b) Solve (e) x4 − 4x2 + 4 = 0 (f)( x − 5)2 = 4 x2 + (1 + )x + 1 = 0 √ √ (g) x2 − (1 + 17)x + 17 = 0 where = 10−10 by using each of (h) ax2 + bx + c = 0 where a > 0. the above formulae.