2.3 Algebraic and Ordering Properties of R · 2.3 Algebraic and Ordering Properties of R Assumptions A1-A18 will be used to derive basic and familiar algebraic and ordering properties

2.3 Algebraic and Ordering Properties of R

Assumptions A1-A18 will be used to derive basic and familiar algebraic and ordering properties

of the real numbers. The theorems and examples that follow are designed to do two things. First,

they will give a feel for how to write proofs of this sort. Second, they will serve as assumptions

for the exercises and theorems from later sections.

These are the (A1 – A18):

(A1) Properties of equality;

(a) For every aaRa , (Reflexive property);

(b) If a = b, then b = a (Symmetric property);

(c) If a = b and b = c, then a = c (Transitive property).

(A2) Addition is well defined;

That is, if dbcadcbaRdcba then , and where,,,, .

(A3) Closure property of addition;

For every RbaRba ,, .

(A4) Associative property of addition;

For every cbacbaRcba ,,, .

(A5) Commutative property of addition;

For every abbaRba ,, .

(A6) Existence of an additive identity;

There exists an element 0 ∈ R with the property that a + 0 = a for every a ∈ R.

(A7) Existence of additive inverses;

For every a ∈ R, there exists some b ∈ R such that a + b = 0. Such an element b is called

an additive inverse of a, and is typically denoted −a to show its relationship to a. We do

not assume that only one such b exists.

(A8) Multiplication is well defined;

That is, if a, b, c, d ∈ R, where a = b and c = d, then ac = bd.

(A9) Closure property of multiplication;

For all a, b ∈ R , a∙b ∈ R. The closure property of multiplication also holds for N, W, Z,

and Q.

(A10) Associative property of multiplication;

For every a, b, c ∈ R, (a·b)·c = a·(b·c) or (a∙b)∙c = a∙(b∙c).

(A11) Commutative property of multiplication;

For every a, b ∈ R, a·b = b·a.

(A12) Existence of a multiplicative identity;

There exists an element 1 ∈ R with the property that a·1 = a for every a ∈ R.

(A13) Existence of multiplicative inverses;

For every a ∈ R except a = 0, there exists some b ∈ R such that a·b = 1. Such an element

b is called a multiplicative inverse of a and is typically denoted a-1

to show its

relationship to a. As with additive inverses, we do not assume that only one such b exists.

Furthermore, the assumption that a-1

exists for all a ≠ 0 does not assume that zero does

not have a multiplicative inverse. It says nothing about zero at all.

(A14) Distributive property of multiplication over addition;

For every a, b, c ∈ R, a (b + c) = (a∙b) + (a∙c) = a∙b + a∙c, where the multiplication is

assumed to be done before addition in the absence of parentheses.

(A15) 1 ≠ 0

(A16) Trichotomy law;

For any a ∈ R, exactly one of the following is true:

(a) a > 0, in which case we say a is positive;

(b) a = 0;

(c) 0 > a, in which case we say a is negative.

(A17) If a > 0 and b > 0, then a + b > 0.

That is, the set of positive real numbers is closed under addition.

(A18) If a > 0 and b > 0, then a∙b > 0.

That is, the set of positive real numbers is closed under multiplication

2.3.1 Basic algebraic properties of real numbers

Theorem 2.3.1. (Cancellation of addition) For all a, b, c ∈ R, if a + c = b + c, then a = b.

Proof:

Suppose a + c = b + c. By property A7, there exists −c ∈ R such that c + (−c) = 0. Since addition

is well defined (A2), we have a + c + (−c) = b + c + (−c), which yields a + 0 = b + 0 or a = b.

Theorem 2.3.2. For every a ∈ R, a ·0 = 0

Proof:

Pick a ∈ R. By properties A6 and A14, we have that 0 + a ·0 = a ·0 = a · (0 + 0) = a ·0 + a · 0.

By theorem 2.3.1, we may cancel a · 0 from both sides of the equation to have a · 0 = 0.

Theorem 2.3.3. The additive inverse of a real number is unique.

Proof:

Pick a ∈ R. Suppose b, c ∈ R are both additive inverses of a. Then a + b = 0 and a + c = 0. We

must show that b = c. Now since a + b = 0 and a + c = 0, then by assumption A1, the transitive

property of equality, a + b = a + c. But by Theorem 2.3.1, b = c. Thus the additive inverse of a is

unique.

Theorem 2.3.4. For every a ∈ R, −(−a) = a

Proof:

Pick a ∈ R. By assumption A7, the existence of additive inverses, a + (−a) = 0. Thus, −(−a) is an

inverse of –a. Hence, we can write −a + − (−a) = 0 and by A5 it may be written as −(−a) + (−a)

= 0. Then by assumption A1, the transitive property of equality, a + (−a) = −(−a) + (−a). But by

Theorem 2.3.1, −(−a) = a.

Theorem 2.3.5. −0 = 0

Proof:

Since −0 is the additive inverse of 0, by definition it satisfies 0 + (−0) = 0. But for any a ∈ R,

0 + a = a. Thus 0 + (−0) = −0. Thus we have 0 + (−0) = 0 and 0 + (−0) = −0. By the transitive

property of equality, −0 = 0.

Theorem 2.3.6. If a, b ∈ R then

1. (−a)b = −(ab)

2. (−a)(−b) = ab

Proof:

1. Since ab + −(ab) = 0 (A7, the existence of additive inverses), we have to prove that ab +

(−a)b = 0, so that by A1 and theorem 2.3.1, (−a)b = −(ab).

ab + (−a)b = (a + (−a))b (A14)

= 0 ∙ b (A5)

= b ∙ 0 (Theorem 2.3.2)

= 0

Thus, (−a)b = −(ab).

2. Since ab + −(ab) = 0 (A7, the existence of additive inverses), we have to prove that

(−a)(−b) + −(ab)= 0, so that by A1 and theorem 2.3.1, (−a)(−b) + −(ab).

(−a)(−b) + −(ab) = (−a)(−b) + (−a)b (The previous theorem)

= (−a) + (b + (−b)) (A14)

= (−a) ∙ 0 (Theorem 2.3.2)

= 0

Corollary 2.3.7. If b ∈ R then (−1)b = −b

Proof:

Let a = 1, using the first part of theorem 2.3.6., we will find

1∙b + (−1)b = (1+ (−1))b (A14)

= 0 ∙ b (A5)

= b ∙ 0 (Theorem 2.3.2)

= 0

Thus, (−1)b = −b.

Corollary 2.3.8. If a, b ∈ R, then −(a + b) = (−a) + (−b).

Proof:

By corollary 2.3.7 and the distributive property.

−(a + b) = (−1)(a + b) = (−1)(a) + (−1)(b) = (−a) + (−b).

Theorem 2.3.9. If ac = bc and c ≠ 0, then a = b

Proof:

Suppose ac = bc. By property A13, there exists c-1 ∈ R such that c∙ c-1

= 1. Since multiplication is

well defined (A8), we have that a c∙ c-1 = bc∙c-1

which yields a∙1 = b∙1, or a = b.

Theorem 2.3.10. The multiplicative inverse of a ≠ 0 is unique.

Proof:

Pick a ∈ R and a ≠ 0. Suppose b, c ∈ R are both multiplicative inverses of a. Then a∙b = 1 and

a∙c = 1. We must show that b = c. Now since a∙b = 1 and a∙c = 1, then by assumption A1, the

transitive property of equality, a∙b = a∙c. But by Theorem 2.3.9, b = c. Thus the multiplicative

inverse of a is unique.

Theorem 2.3.11. For all a ≠ 0, (a−1)−1 = a.

Proof:

Pick a ∈ R and a ≠ 0. By assumption A13, the existence of multiplicative inverses, a∙ a-1 = 1.

Thus, (a−1)−1 is an inverse of a−1. Hence, we can write a−1∙ (a−1)−1 = 1 and by A11 it may

be written as (a−1)−1∙ a−1 = 1. Then by assumption A1, the transitive property of equality, a∙ a-

1 = (a−1)−1∙ a−1. Thus by Theorem 2.3.9, (a−1)−1 = a.

Theorem 2.3.12. For all nonzero a, b ∈ R, (ab)−1 = a−1b−1

Proof:

Since ab∙(ab)-1

= 1 (A13, the existence of multiplicative inverses), we have to prove that

ab∙ a−1b−1 = 1, so that by A1 and theorem 2.3.9, (ab)−1 = a−1b−1.

ab∙ a−1b−1 = ba ∙ a−1b−1 (A11)

= b∙1∙ b−1 (A13)

= b∙ b−1 (A12)

= 1 (A13)

Thus, (ab)−1 = a−1b−1.

Theorem 2.3.12. For all nonzero a ∈ R, (−a)−1 = − (a)−1

Proof:

Suppose a = 0. Then there exists a−1 ∈ R, and by part 2 of Theorem 2.3.6,

1 = a∙a−1 = (−a)[−(a−1)]. Since a ≠ 0, then neither is −a, else a = −(−a) = 0 by Theorem 2.3.5.

Thus there exists (−a)−1∈R, and we may multiply both sides by (−a)−1 to have (−a)−1= −(a−1).

2.3.2 Ordering of the real numbers

If a > 0, we call a positive, and if 0 > a (or a < 0), we call a negative. Notationally, we write

the positive and negative real numbers as R+ and R−, respectively. Assumptions A17 and A18

describe some assumed behaviors of addition and multiplication in R+, namely, that it is closed

under addition and multiplication. By splitting R into these three pieces, that is, R+, {0}, and

R−, we can then assign meaning to the statement a > b by declaring a > b ⇔ a − b > 0.

Thus, anytime we see a statement a > b, it means precisely that a − b > 0. Similarly, a > b

means a − b > 0. This definition, with Theorem 2.3.4, allows us to prove the following.

Theorem 2.3.14. If a > b, then −a < −b.

Proof:

Suppose a > b. Then a − b > 0. But a = −(−a), so that −(−a) − b > 0, which we may write

as −(−a) + (−b) > 0. By commutativity, (−b) + [−(−a)] > 0, or −b − (−a) > 0, so that −b >

−a. Hence, −a < −b.

Corollary 2.3.15. Suppose c ∈ R. Then c > 0 if and only if −c < 0. Proof:

If c > 0, then let a = c and b = 0 in Theorem 2.3.14 to have −c < −0 = 0. Conversely, if −c <

0, then let b = −c and a = 0 in Theorem 2.3.14 to have −0 < −(−c), or c > 0. 2.3.2 Absolute Value

Definition 2.3.16. For x ∈ R, we define ,the absolute value of x by

x , x > 0

│x│=

-x , x < 0 Some problems in exercises which are not included in the resume:

6. Prove the principle of zero product: If ab = 0, then either a = 0 or b = 0.

7. Prove (a + b)(c + d) = ac + ad + bc + bd for all a, b, c, d ∈ R.

8. Suppose we replace assumption A15 with the assumption that 1 = 0. Show that, with

this assumption, there are no nonzero real numbers.

9. Prove the following.

a. If a < b, then a + c b and b > c, then a > c.

d. If a < 0 and b < 0, then a + b < 0.

e. If a > 0 and b < 0, then ab < 0.

f. If a < 0 and b < 0, then ab > 0.

g. If a 0, then ac < bc.

h. If a bc.

i. If 0 < a < b, then a2 < b2 .

j. If a b2 .

k. 1 > 0.

l. For a ∈ R, write a2 = a · a. Show that for every a ∈ R, a2 ≥ 0.

m. Explain why the equation x 2 = −1 has no solution x ∈ R.

10. Prove the following:

a. does not exist in R

b. If then

c. If then

d. if and only if

e. If and , then

f. If and , then

11. If and , does it follow ? Use results from exercise 9 and 10 to

state and prove the relationship between and depending on the signs of a and b.

12. Prove that if f are real numbers, then . (how do you know that

what exactly is 2 anyway?)

13. Prove the theorem 2.3.17: For all x ,

14. Prove that for all ,

15. Suppose . Prove the following.

a.

b. If , then

c. If , then

16. Prove the direction of Theorem 2.3.19: Suppose . Then if

17. Prove Theorem 2.3.21: Suppose Then if and only if either or

18. Prove Theorem 2.3.22: If , then

19. Prove Theorem 2.3.23: For all then

20. Prove Theorem 2.3.24: For all then

Solutions:

6. Prove the principle of zero product: If ab = 0, then either a = 0 or b = 0.

If a ≠ 0, ab = 0 = a ∙ 0, by theorem 2.3.9, thus b = 0

If b ≠ 0, ab = 0 = 0 ∙ b, by theorem 2.3.9, thus a = 0

7. Prove (a + b)(c + d) = ac + ad + bc + bd for all a, b, c, d ∈ R.

Proof:

ac + ad + bc + bd = a(c + d) + b(c + d) (A14)

= (a + b)(c + d) (A14)

8. Suppose we replace assumption A15 with the assumption that 1 = 0. Show that, with this

assumption, there are no nonzero real numbers.

Proof:

Pick , suppose 1 = 0

1 = a (A12)

0 = a (1 = 0)

0 = a (Theorem 2.3.2)

a = 0 contradict with

9. Prove the following.

a. If a < b, then a + c < b + c

Proof:

Suppose a < b. Then a − b < 0. So that (a + c) – (b + c) = a − b < 0. Hence, a + c b and b > c, then a > c.

Proof:

Suppose a > b and b > c . Then a − b > 0 and b − c > 0 . So that, by A16 we will find

(a − b) – (b − c) = a − c > 0. Hence, a > c.

d. If a < 0 and b < 0, then a + b < 0.

Proof :

If a < 0 and b < 0 0 – a > 0 and 0 - b > 0

( 0 - a) + ( 0 - b) > 0

0 – ( a + b) > 0

0 > (a + b) or a + b < 0

e. If a > 0 and b < 0, then ab < 0.

Proof :

If a > 0 and b < 0 a > 0 and 0 - b > 0

a (0 – b) > 0

a(0) – ab > 0

0 – ab > 0

– ab > 0

ab < 0

f. If a < 0 and b < 0, then ab > 0.

Proof :

If a < 0 and b < 0 0 - a > 0 and 0 - b > 0

-a (– b) > 0

ab > 0

g. If a 0, then ac < bc.

Proof :

If a 0 b - a > 0 and c > 0

( b - a) c > 0

bc – ac > 0

bc > ac or ac < bc

h. If a bc.

Proof:

If a 0 and 0 - c > 0

( b - a) (0 – c) > 0

(b – a) (-c) > 0

-bc + ac > 0

ac > bc

i. If 0 < a < b, then a2 < b2 .

Proof:

If 0 < a 0

a > 0 and a 0 and a b2 .

Proof:

If a a , 0 > b and a 0 and a 0 a2 > ab…… (1)

-b > 0 and a 0 ab > b2 ..... (2)

From equation (1) and equation (2) we have

a2 > ab > b

2, so a

2 > b

2

k. 1 > 0.

Proof:

Suppose 1 ≤ 0

There are two possibilities 1 = 0 or 1 < 0

1 = 0 is false because 1 is identity of multiplication and unique

1 < 0 if 00.1.0 aoraaa is false because a > 0

Therefore 1 > 0

l. For a ∈ R, write a2 = a · a. Show that for every a ∈ R, a2 ≥ 0.

Proof:

a R a2 ≥ 0

if a = 0 a. a = 0.0 = 0

if a < 0 -a(-a) = a2 > 0

if a > 0 a. a > 0

Therefore for every a R, a2 ≥ 0

m. Explain why the equation x 2 = −1 has no solution x ∈ R.

Answer : Because every a R, a2 ≥ 0

10. The solution:

a. From the theorem for all . Suppose 0 has an inverse 0-1

. It means for an

follows . Thus, . This is impossible to happen in real number, which means

that there does not exist inverse of 0.

b. By contradiction, suppose , which means and .

For , (a contradiction)

For , and (given), it gives ( a contradiction). Since

those two proofs show contradiction, it means that if then

c. Like the proof above, by contradiction, suppose , which means and

. For , (a

contradiction). For , and (given), it gives ( a

contradiction). Since those two proofs show contradiction, it means that if then

d.

(since )

…..2)

From 1) and 2), it gives

follows , which cause

(proven)

e. Since

, karena

atau

f. By contradiction, suppose . For and , we have

(contradict to the given that ). For and , we have

(contradict to the given that ). Thus, it gives that and

, then

11. There are three possibilities of elements and .

a) For

(given)

or

b) For

(given)

c) For

(given)

12. Since

Since

Why ? since which means , while , then

13. By definition of absolute value , we obtain

For all x ,

The result above gives a value as same as the result of

14. If , then (definition). If , then because │x│≥ 0. Thus, x ≤

│x│… (1). Since│x│≥ 0, then -│x│ 0. It shows that if or , then

… (2). From (1) and (2), we obtain ,

then

then

then

15. Proof:

a) If both x and y is 0, then . If and , then

.Thus, . If and , then

. If and , then .

Lastly, If and , then .

b) For all and , there exist such that . Using this result, we

have

(based on the theorem 15a)

c) If x is 0 and y ≠ 0, then . If and ,

then .Thus, . If and , then

. If and , then

. Lastly, If and , then

.

16. Suppose meaning that or . For we have

. Thus, there does not exist satisfying the latest condition 1). For with

, we have and .

For , …2)

For …3)

From 1), 2) and 3) we can conclude that for , if then

17. direction.

For , , since , then

For , since , then or . Thus we conclude that For

, if then or

direction.

Suppose , which means or . For we have or

Thus, there does not exist satisfying the latest condition. 1)

For , …2)

For …3)

From 1), 2) and 3) we can conclude that for , if or , then

18. From no 14, we have ..1). Likewise, we can also write ..2).

By adding 1) and 2), we get . Based on the previous

theorem, we obtain

19. From no 14, we have ..1). Likewise, we can also write ..2).

By subtracting 2) and 1), we get . Based on the previous

theorem, we obtain

20. We can write and use the theorem no 18 to show that

Since , we can subtract inequality above become ……1)

Then, write and use the theorem no 18 to show that

Since , we can subtract inequality above become

From 1) and 2), we have , which is equivalent with

2.3 Algebraic and Ordering Properties of R · 2.3 Algebraic and Ordering Properties of R Assumptions A1-A18 will be used to derive basic and familiar algebraic and ordering properties

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