An inequality is a statement that quantities are not equal. If two quantities are unequal, the first is either greater than or less than the other. The inequality symbols are: ≠, meaning unequal to; >, meaning greater than; and <, meaning less than. Thus, 4 ≠ 3 is read “four is unequal to three”; 7 > 2 is read “seven is greater than two”; and 1 < 5 is read “one is less than five.”
Two inequalities may be of the same order or of opposite order. In inequalities of the same order, the same inequality symbol is used; in inequalities of the opposite order, opposite inequality symbols are used. Thus, 5 > 3 and 10 > 7 are inequalities of the same order; 5 > 3 and 7 < 10 are inequalities of opposite order.
Inequalities of the same order may be combined, as follows. The inequalities x < y and y < z may be combined into x < y < z, which states that y is greater than x and less than z. The inequalities a > b and b > c may be combined into a > b > c, which states b is less than a and greater than c.
Axioms are statements that are accepted as true without proof and are used in the same way as theorems.
AXIOM 1: A quantity may be substituted for its equal in any inequality.
Thus if x > y and y = 10, then x > 10.
AXIOM 2: If the first of three quantities is greater than the second, and the second is greater than the third, then the first is greater than the third.
Thus if x > y and y > z, then x > z.
AXIOM 3: The whole is greater than any of its parts.
Thus, AB > AM and m/BAD > m∠BAC in Fig. 13-1.
Fig. 13-1
AXIOM 4: If equals are added to unequals, the sums are unequal in the same order.
Since 5 > 4 and 4 = 4, we know that 5 + 4 > 4 + 4 (or 9 > 8). If x–4 < 5, then x–4 + 4 < 5 + 4 or x < 9.
AXIOM 5: If unequals are added to unequals of the same order, the sums are unequal in the same order.
Since 5 > 3 and 4 > 1, we have 5 + 4 > 3 + 1 (or 9 > 4). If 2x–4 < 5 and x + 4 < 8, then 2x–4 + x + 4 < 5 + 8 or 3x < 13.
AXIOM 6: If equals are subtracted from unequals, the differences are unequal in the same order.
Since 10 > 5 and 3 = 3, we have 10–3 > 5–3 (or 7 > 2). If x + 6 < 9 and 6 = 6, then x + 6–6 < 9–6 or x < 3.
AXIOM 7: If unequals are subtracted from equals, the differences are unequal in the opposite order.
Since 10 = 10 and 5 > 3, we have 10–5 < 10–3 (or 5 < 7). If x + y = 12 and y > 5, then x + y–y < 12–5 or x < 7.
AXIOM 8: If unequals are multiplied by the same positive number, the products are unequal in the same order.
Thus if 
x < 5, then 4(x) < 4(5) or x < 20.
AXIOM 9: If unequals are multiplied by the same negative number, the results are unequal in the opposite order.
Thus if 
x < 5, then (–2)(x) > (–2)(5) or–x >–10 or x < 10.
AXIOM 10: If unequals are divided by the same positive number, the results are unequal in the same order.
Thus if 4x > 20, then 
or x > 5.
AXIOM 11: If unequals are divided by the same negative number, the results are unequal in the opposite order.
Thus if–7x < 42, then 
or x >–6.
POSTULATE 1: The length of a line segment is the shortest distance between two points.
PRINCIPLE 1: The sum of the lengths of two sides of a triangle is greater than the length of the third side. (Corollary: The length of the longest side of a triangle is less than the sum of the lengths of the other two sides and greater than their difference.)
Thus in Fig. 13-2, BC + CA > AB and AB > BC–AC.
Fig. 13-2
PRINCIPLE 2: In a triangle, the measure of an exterior angle is larger than the measure of either nonad-jacent interior angle.
Thus in Fig. 13-2, m∠BCD > m∠BAC and m∠BCD > m∠ABC.
PRINCIPLE 3: If the lengths of two sides of a triangle are unequal, the measures of the angles opposite these sides are unequal, the larger angle being opposite the longer side. (Corollary: The largest angle of a triangle is opposite the longest side.)
Thus in Fig. 13-2, if BC > AC, then m ∠A > m ∠B.
PRINCIPLE 4: If the measures of two angles of a triangle are unequal, the lengths of the sides opposite these angles are unequal, the longer side being opposite the larger angle. (Corollary: The longest side of a triangle is opposite the largest angle)
Thus in Fig. 13-2, if m∠A > m∠B, then BC > AC.
PRINCIPLE 5: The perpendicular from a point to a line is the shortest segment from the point to the line.
Thus in Fig. 13-3, if 
⊥ 
and 
is any other line from P to , then PC < PD.
Fig. 13-3
Fig. 13-4
PRINCIPLE 6: If two sides of a triangle are congruent to two sides of another triangle, the triangle having the greater included angle has the greater third side.
Thus in Fig. 13-4, if BC = B'C', AC = A'C', and m∠C > m∠C, then AB > A'B'.
PRINCIPLE 7: If two sides of a triangle are congruent to two sides of another triangle, the triangle having the greater third side has the greater angle opposite this side.
Thus in Fig. 13-4, if BC = B'C', AC = A'C', and AB > A'B', then m∠C > m∠C'.
PRINCIPLE 8: In the same or equal circles, the greater central angle has the greater arc.
Thus in Fig. 13-5, if m∠AOB > m∠COD, then m
> m
.
PRINCIPLE 9: In the same or equal circles, the greater arc has the greater central angle. (This is the converse of Principle 8.)
Thus in Fig. 13-5, if m >, m then m∠AOB > m∠COD.
Fig. 13-5
PRINCIPLE 10: In the same or equal circles, the greater chord has the greater minor arc.
Thus in Fig. 13-6, if AB > CD, then m > m.
Fig. 13-6
PRINCIPLE 11: In the same or equal circles, the greater minor arc has the greater chord. (This is the converse of Principle 10.)
Thus in Fig. 13-6, if m > m then AB > CD.
PRINCIPLE 12: In the same or equal circles, the greater chord is at a smaller distance from the center.
Thus in Fig. 13-7, if AB > CD, then OE < OF.
Fig. 13-7
PRINCIPLE 13: In the same or equal circles, the chord at the smaller distance from the center is the greater chord. (This is the converse of Principle 12.)
Thus in Fig. 13-7, if OE < OF, then AB > CD.
13.1 Selecting inequality symbols
Determine which inequality symbol, > or <, makes each of the following true:
(a) 5 
3
(b) 6 9
(c)–5 3
(d)–5 –3
(e) If x = 3, then x2 x.
(f) If x > 10, then 10 x.
Solutions
(a) >
(b) <
(c) <
(d) <
(e) >
(f) <
13.2 Applying inequality axioms
Complete each of the following statements:
(a) If a > b and b > 8, then a 8.
(b) If x > y and y = 15, then x 15.
(c) If c < 20 and d < 5, then c + d 25.
(d) If x > y and y > 6, then x y 6.
(e) If x > y, then x y.
(f) If e < f, then 4e f.
(g) If–y < z then y – z.
(h) If–4x > p, then x –p.
(i) If Paul and Jack have equal amounts of money and Paul spends more than Jack, then Paul will have than Jack.
(j) If Anne is now older than Helen, then 10 years ago, Anne was than Helen.
Solutions
(a) >
(b) >
(c) <
(d) >, >
(e) >
(f) <
(g) >
(h) <
(i) less
(j) older
13.3 Applying triangle inequality theorems (Fig. 13-8)
(a) Determine the integer values that the length of side a of the triangle can have if the other two sides have lengths 3 and 7.
(b) Determine which is the longest side of the triangle if two angles have measures 59° and 60°.
(c) Determine which is the longest side of parallelogram ABCD if E is the midpoint of the diagonals and m∠AEB > m∠AED.
Fig. 13-8
Solutions
(a) Since a must be less than 3 + 7 = 10 and greater than 7–3 = 4, a can have the integer values of 5, 6, 7, 8, 9.
(b) Since m∠B = 59° and m∠C = 60°, m∠A = 180°–(59° + 60°) = 61°. Then the longest side is opposite the largest angle, ∠A, so the longest side is BC.
(c) In 
ABCD, AE = CE and DE = EB. Since m∠AEB > m∠AED, AB > AD or AB (= DC) is the longest side (Principle 6).
13.4 Applying circle inequality theorems
In Fig. 13-9, compare
(a) OD and OF if ∠C is the largest angle ofΔ ABC
(b) AC and BC if m
m
(c) m and m if OF > OE
(d) m∠AOB and m∠BOC if m m
Fig. 13-9
Solutions
(a) Since ∠C is the largest angle of the triangle, is the longest side, or AB > BC; hence, OD < OF by Principle 12.
(b) Since m > m, AC > BC, the greater arc having the greater chord .
(c) Since OF > OE, BC < AC by Principle 13; hence, m < m by Principle 10.
(d) Since m m, m∠AOB > m∠BOC, the greater arc having the greater central angle.
13.5 Proving an inequality problem
PROOF:
We often arrive at a correct conclusion by indirect reasoning. In this form of reasoning, the correct conclusion is reached by eliminating all possible conclusions except one. The remaining possibility must be the correct one. Suppose we are given the years 1492, 1809, and 1960 and are assured that one of these years is the year in which a president of the United States was born. By eliminating 1492 and 1960 as impossibilities, we know by indirect reasoning that 1809 is the correct answer. (Had we known that 1809 was the year in which Lincoln was born, the reasoning would have been direct.)
In proving a theorem by indirect reasoning, a possible conclusion may be eliminated if we assume it is true and that assumption results in a contradiction of some given or known fact.
13.6 Applying indirect reasoning in life situations
Explain how indirect reasoning is used in each of the following situations:
(a) A detective determines the murderer of a slain person.
(b) A librarian determines which volume of a set of books is in use.
Solutions
(a) The detective, using a list of all those who could have been a murderer in the case, eliminates all except one. He or she concludes that the remaining one is the murderer.
(b) The librarian finds all the books of the set except one by looking on the shelf and checking the records. He or she concludes that the missing one is the one in use.
13.7 Proving an inequality theorem by the indirect method
Prove that in the same or equal circles, unequal chords are unequally distant from the center.
PROOF:
13.1. Determine which inequality symbol, > or <, makes each of the following true:
(13.1)
(a) If y > 15, then 15 y.
(b) If x = 2, then 3x–1 4.
(c) If x = 2 and y = 3, then xy 5.
(d) If a = 4 and b = , then a/b 15.
(e) If a = 5, then a2 4a.
(f) If b = , then b2 b.
13.2. Complete each of the following statements:
(13.2)
(a) If y > x and x = z, then y z.
(b) If a + b > c and b = d, then a + d c.
(c) If a < b and b < 15, then a 15.
(d) If z > y, y > x, and x = 10, then z 10.
13.3. Complete each of the following statements about Fig. 13-10:
(13.2)
(a) BC BD
(b) m∠BAD m∠BAC
(c) ΔADC ΔABC
(d) If m∠A = m∠C, then AB BD.
Fig. 13-10
13.4. Complete each of the following statements:
(13.2)
(a) If Mary and Ann earn the same weekly wage and Mary is to receive a larger increase than Ann, then Mary will earn Ann earns.
(b) If Bernice, who is the same weight as Helen, loses more weight than Helen, then Bernice will weigh Helen weighs.
13.5. Complete each of the following statements:
(13.2)
(a) If a > 3, then 4a 12.
(b) If x–3 > 15, then x 18.
(c) If 3x < 18, then x 6.
(d) If f > 8, then f + 7 15.
(e) If x = y, then x + 5 y + 6.
(f) If g = h, then g– 10 h–9.
13.6. Which of the following sets of numbers can be the lengths of the sides of a triangle?
(13.3)
(a) 3, 4, 8
(b) 5, 7, 12
(c) 3, 4, 6
(d) 2, 7, 8
(e) 50, 50, 5
13.7. What integer values can the length of the third side of a triangle have if the two sides have lengths
(13.3)
(a) 2 and 6;
(b) 3 and 8;
(c) 4 and 7;
(d) 4 and 6;
(e) 4 and 5;
(f) 7 and 7?
13.8. In Fig. 13-11, arrange, in descending order of size, (a) the angles of ΔABC; (b) the sides of Δ DEF; (c) the angles 1, 2, and 3.
(13.3)
Fig. 13-11
13.9. (a) In quadrilateral ABCD of Fig. 13-12, compare m∠BAC and m∠ACD if AB = CD and BC > AD.
(b) In ΔABC of Fig. 13-13, compare AB and BC if 
is the median to 
and m∠AMB > m∠BMC.
(13.3)
Fig. 13-12
Fig. 13-13
13.10. Arrange, in descending order of magnitude,
(13.4)
(a) The sides of ΔABC in Fig. 13-14
(b) The central angles AOB, BOC, and AOC in Fig. 13-14
(c) The sides of trapezoid ABCD in Fig. 13-15
(d) The distances of the sides of ΔDEF from the center in Fig. 13-16
Fig. 13-14
Fig. 13-15
Fig. 13-16
13.11. Provide the proofs requested in Fig. 13-17.
Fig. 13-17
13.12. Explain how indirect reasoning is used in each of the following situations:
(13.6)
(a) A person determines which of his ties has been borrowed by his roommate.
(b) A girl determines that the electric motor in her train set is not defective even though her toy trains do not run.
(c) A teacher finds which of his students did not do their assigned homework.
(d) A mechanic finds the reason why the battery in a car does not work.
(e) A person accused of a crime proves her innocence by means of an alibi.
13.13. Prove each of the following:
(13.7)
(a) The base angles of an isosceles triangle cannot be right angles.
(b) A scalene triangle cannot have two congruent angles.
(c) The median to the base of a scalene triangle cannot be perpendicular to the base.
(d) If the diagonals of a parallelogram are not congruent, then it is not a rectangle.
(e) If a diagonal of a parallelogram does not bisect a vertex angle, then the parallelogram is not a rhombus.