Abstract Algebra real

Abstract Algebra

Algebraic and Ordering Properties of Real Number

(A1) Properties of equality:

(a) For everya∈R, a= a (Reflexive property); (b) Ifa= b, then b= a (Symmetric property);

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

(A2) Addition is well defined: That is, if a,b,c,d∈R,where a= b and c= d, then a+ c= b+ d. (A3) Closure property of addition: For every a,b∈R, a+ b∈R.

(A4) Associative property of addition: For every a,b,c∈R, (a+ b)+ c= a+ (b+ c) (A5) Commutative property of addition: For everya,b∈R, a+ b= b+ a.

(A6) Existence of an additive identity: There exists an element 0∈R with the property that a+ 0= a for everya∈R.

(A7) Existence of additive inverses: For everya∈R, there exists someb∈Rsuch That a+ b= 0. Such an element bis called anadditive inverseofa, 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, ifa,b,c,d∈R,wherea = band c= d, thenac= bd.

(A9) Closure property of multiplication: For all a,b∈R,a·b∈R. The closure property of multiplication also holds forN,W,Z, andQ.

(A10) Associative property of multiplication: For everya,b,c∈R, (a·b)·c= a·(b·c)or(ab)c= a (bc).

(A11) Commutative property of multiplication: For everya,b∈R,ab= ba.

(A12) Existence of a multiplicative identity: There exists an element 1∈R with the property that a·1= a for everya∈R.

(A13) Existence of multiplicative inverses: For everya∈Rexcepta= 0, there exists some b∈R such that ab= 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 doesnothave 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)= (ab)+ (ac)= ab+ ac,where the multiplication is assumed to be done before addition in the absence of parentheses.

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

a+c+ (-c) = b+c(-c) (A7, A2)

a+0 = b+0 (A7)

a = b (A6)

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

Pick a∈R

a.0 = a.(0+0) (A7)

a.0 = a.0+a.0 (A14)

0+a.0 = a.0+a.0 (theorem 2.3.1)

0=a.0

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

Pick a∈R

Suppose b, c ∈ R inverse of a Then, a + b = 0 and a + c = 0

Show that b = c

a + b = a + c (A1, transitive)

b = c (Theorem 2.3.1)

Thus additive inverse is unique

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

Pick (–a)∈R, it means –(-a), such that (-a)+ (-(-a)) = 0 (-a) + (-(-a)) = (-a) + a

a + (-a)+ (-(-a)) = a + (-a) + a

0 + (-(-a)) = 0 +a

(-(-a)) = a

Theorem 2.3.5 – 0 =0 Proof:

Since – 0 is additive inverse of 0, then

0 + (- 0) = 0 (A7)

But, for every a∈R, 0 + a = a. Thus 0 + (-0) = - 0 (A6) We have 0 + (-0) = 0 and 0 + (-0) = - 0

Thus 0 = - 0 (A1, transitive)

Theorem 2.3.6: If a,b∈R, then:

(a) (−a)b = −(ab).

(b) (−a)(−b) = ab.

Corollary 2.3.7 � ∈ ℝ, ℎ −1 =−

Theorem 2.3.9 � = ≠ 0, ℎ =

Theorem 2.3.10 ℎ � � � � ≠0 � �

Theorem 2.3.11 � ≠0, −1 −1 =

Theorem 2.3.12 � , ∈ ℝ, −1 = −1 −1

Theorem 2.3.13 � ∈ ℝ, − −1 =− −1

2.3.2 Ordering of Real Numbers Theorem 2.3.14 � > , ℎ − <−

Corollary 2.3.15 ∈ ℝ. ℎ > 0 � � − < 0

2.3.3 Absolute Value

Definition 2.3.16 ∈ ℝ, � , ℎ

� = , � 0;

− � < 0

Theorem 2.3.17 � ∈ ℝ, − =

Theorem 2.3.18 0. ℎ = � � = ±

Theorem 2.3.19 0. ℎ < � � − < <

Corollary 2.3.20 0. ℎ � � −

Theorem 2.3.21 0. ℎ > � � � ℎ > <−

Theorem 2.3.22 (N3:Triangle Inequality) � , ∈ ℝ, + +

Theorem 2.3.24 � , ∈ ℝ, − −

Some theorems will prove in the following exercise.

Exercise Page 59

1. Prove Theorem 2.3.6: If a , b ∈ℝ, then: (a) (−a )b = −(a b).

(b) (−a )(−b) = a b. Answer :

(a) If a, b ∈ℝ, then (−a )b = −(a b) Proof :

 Pick (-a)b ∈ℝ , then

(-a)b + (-(-a )b) = 0…..(A7)

(-a)b + ab = 0…..(Theorem 2.3.4)  Pick –(ab) ∈ℝ , then

–(ab) + ab = 0…..(A7)

(-a)b + ab = - (ab) + ab…..(Theorem 2.3.1) (-a )b = -(ab)

(b) _{If a, b ∈ℝ, then} (−a )(−b) = a b Proof :

 For every ab ∈ℝ, there is –(ab) ∈ℝ is an inverse of ab such that (ab) + (–(ab))= 0

 (–a )(–b) + (–a )(b) = –a(–b+ b) …(A14)

= (–a ).0 ….(Theorem 2.3.2) = 0

(–a )(–b) + (–(ab)) = (ab) + (–(a b)) ….(Theorem 2.3.1) (–a )(–b) = (ab)

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

Proof :

Suppose ac = bc for a, b, c ∈R and c ≠ 0 c ∈ℝ, there is a −1, such that . −1 = 1

ac = bc

. −1 = . −1 . 1 = . 1…(�12)

=

3. Prove Theorem 2.3.10: The multiplicative inverse of a = 0 is unique. Proof:

Pick a, b, c ∈ℝ

Supposed that b, c are multiplicative inverse of a.

a . b = 1 a . c = 1 a . b = a . c b = c

4. Using reasoning similar to the argument for Theorem 2.3.4, prove Theorem 2.3.11: For all a = 0, (a−1 )−1 = a.

Proof:

Pick a ∈ℝ, a ≠ 0. a . a– 1= 1

a– 1. (a– 1)– 1= 1

a . a– 1= a– 1. (a– 1)– 1…(A11)

a– 1 . a = a – 1. (a– 1)– 1… (Theorem 2.3.1) a = (a – 1) – 1 …(A11)

(a – 1) – 1= a

5. Prove Theorem 2.3.12: For all nonzero a , b ∈ℝ, (a b)−1 = a−1 b−1 . Proof :

Pick a,b ∈ℝ, a ≠ 0, b ≠ 0.

−1 = 1 −1 −1 ₌ −1_{. 1}

−1_. −1 ₌ −1 _…(�12) −1 ₌ −1

−1 −1 ₌ −1 −1

−1 = −1 −1 …. (A11)

−1 = −1 −1

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

a b = 0

a b = a . 0 ….(Theorem 2.3.2) b = 0 ….(Theorem 2.3.9) or

a b = 0

a b = b . 0 ….(Theorem 2.3.2) a = 0 ….(Theorem 2.3.9)

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

(a + b)(c + d ) = ((a + b)c) + ((a+b)d) …(A14) = (c(a+b)) + (d(a+b)) ….(A11) = ca + cb + da + db …(A14) = ac + bc + ad +bd … (A5) = ac + ad + bc + bd

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 ∈ ℝ, ≠0, suppose 1 = 0

a. 1 = a (A12)

a. 0 = a (1 = 0)

0 = a (Theorem 2.3.2)

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 ₊ c.

b. If a < 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.

c. If a > 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 a b < 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 a b > 0.

Proof :

If a < 0 and b < 0  0 - a > 0 and 0 - b > 0  -a (– b) > 0

 ab > 0 g. If a < b and c > 0, then a c < bc.

Proof :

If a < b and c > 0 b - a > 0 and c > 0  ( b - a) c > 0  bc – ac > 0

 bc > ac or ac < bc h. If a < b and c < 0, then a c > bc.

Proof:

If a < b and c < 0 b - 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 < b  0 < a and a < b and also b > 0 a > 0 and a < b  a2 < ab ... (1) b > 0 and a < b  ab < b2 ...…(2) From equation (1) and equation (2) we have

a2 < ab < b2, so a2 < b2

j. If a < b < 0, then a2 > b2 . Proof:

If a < b < 0 0 > a , 0 > b and a < b

-a > 0 and a < b  -a (b-a) > 0  a2> ab…… (1) -b > 0 and a < b  - b (b –a) > 0  ab > b2 ... (2) From equation (1) and equation (2) we have

a2 > ab > b2, so a2 > b2

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 a 0a.1a.0 or a 0 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.

Suppose x2 = 1 has a solution, suppose a is the solution, so we get

If a<0, a2=a.a=-1 (contradict 9f)

If a=0, a.a=-1 (contradict thoe 2.3.2)

Thus, x2=-1 has no solution.

10.

a. 0−1does not exist in ℝ ans.

b. if > 1 then −1 > 0 − → ≡ ∧

Pick > 0, we suppose that −1 0 meaning that −1 < 0 or −1 = 0, we will prove that −1 > 0

 −1_{. = 1 (Property of A13)}

0. = 1

0 = 1 This contradicts the Property of A15

 −1_{. = 1 We know that 1 > 0 meaning that 1 is positive} if > 0 and −1 < 0 then . −1 < 0

−1_{. = 1 , it contradicts the proof of 9.e} c. if < 0 , then −1 < 0

We can prove this by using its contraposition

if −1 > 0, then > 0(we don’t use −1 = 0since we’ve already proven 10.a)

Pick −1 > 0, we suppose that 0 meaning that < 0 or = 0, we will prove that > 0

 −1_{. = 1 (Property of A13)}

−1_{. 0 = 1}

0 = 1 This contradicts the Property of A15

 −1_{. = 1 We know that 1 > 0 meaning that 1 is positive} if −1 > 0 and < 0 then . −1 < 0

−1_{. = 1 , this contradicts the proof of 9.e} Since the contraposition is correct, then

firstly > 1→ −1 > 0 −1 < 1

 > 1→ −1 > 0 has been already proven by 10.b  > 1→ −1 < 1

> 1

. −1 > 1. −1

1 > −1

Secondly, 0 < −1 < 1→ > 1 −1 _{> 0} −1 _{< 1 → > 1}

−1 _{> 0 then we directly know (has been proven) that > 1}

−1 _{< 1}

. −1 < . 1 Property of A8

1 <

e. if > 0 and > 1 , then < > 1

. > . 1 (A18)

. . −1 > . 1. −1 (A18)

. 1 > . −1

>

f. if , ∈ ℤ and = 1 , then = = ±1

there are five cases : = −1, = 0, = 1, > 1, and < −1

case 1

> 1

. = 1 > 0

. = 1 = . −1 > 0 , this

11. If , ∈ ℝ\{0} and < , does it follow that 1 < 1 ? Use results from Exercises 9 and 10 to state and prove the relationship between 1 and 1 depending on the signs of and

 < where < 0 and < 0

1. < 1. 1

. . < 1. . 1

( ) <1( ) 1

< 1 (9.f)

 < where < 0 and > 0

1. < 1. 1

. . < 1. . 1

> 1( ) (9.h)

 < where > 0 and > 0

1. < 1. 1

. . < 1. . 1

( ) <1( ) 1

< 1 (9.f)

Since the values of 1 and 1 vary upon the cases, it is obvious that the relationship between depends on the signs of and

12. Prove that if < are real numbers , then < +

2 < . how do you know that 2 > 0

1 > 0 or 0 < 1

0 + 1 < 1 + 1

1 < 1 + 1 = 2

0 < 1 < 2

0 < 2 or 2 > 0 (9.c) then 2−1 =1

2 > 0 (10.b)

<

.1

2< . 1

2 (9.g)

/2 < /2

< +

2 ……(i) 2+2 < 2+2 (9.a)

+

2 < ……(ii)

(i) and (ii) using 9.b we get

< +

2 <

13. For all ∈ ℝ, − =

let 0 then – < 0

− = − − if – < 0 (def. 2.3.16) − = , 0 then =

let < 0, =− (def. 2.3.16)

− 0 so − = − then from def. above − =− =

− =

14. For all ∈ ℝ, −

For 0, = (def.) 0,then − 0, 0 so that −

For 0 =− then − = 0 , 0 0, 0 so that

− and

We can say that −

15. Suppose x, y ∈ , ∈ ℝ. Proof the following. a. =

b. � ≠0, ℎ −1 = −1

c. � ≠0, ℎ / = /

a. =

 0, 0, = , =

0 . = . = .  0, < 0, = , = −

. < 0 . = − . = (− ) = .  < 0, 0, = − , =

< 0 . =− =− . = .  < 0, < 0, = − , =−

. 0 . = . =− .− = .

b. � ≠0, ℎ −1 = −1

Suppose 0, = −1 = −1 −1 ₀ −1 _{= 1}

−1_{= .} −1

−1. . −1 = −1. . −1

−1 = −1

Suppose < 0, = − −1 = − −1 −1 = 1

− −1 _{= .} −1 − −1 = − .− −1

− −1 _. − _. −1 ₌ − −1_. − _. −1

−1 = −1

c. � ≠0, ℎ / = /

. −1 = −1

For −1 0

. −1 = −1

. −1 = . −1. −1

. −1 = . −1. . −1

. −1 = . −1. −1

. −1 = . −1

For −1 < 0

. −1 = −1

. −1 = . − −1 . 1

. −1 = . − −1 . −1

. −1 = . − −1 . − . −1

. −1 = . −1. −1

. −1 = . −1

16. Prove the ⇐ direction of Theorem 2.3.19: Suppose 0. Then < if – < <

Proof:

Let 0

< 0 >−

If = 0 means that –0 < < 0 or 0 < < 0 , so = 0 is impossible If 0, = , because < , so <

If < 0, = − → = −

− <

− −1 < − −1

> Theorem number 19h

17. Proof Theorem 2.3.21: Suppose 0. Then > if and only if either > 0 <

− . Proof:

(→) Let >

If 0, = , = > , >

If < 0, = − , >

− = , − < − < −

(←) > <− → >

For > , =

So, >

18. Proof Theorem 2.3.22: � , ∈ ℝ, + +

Proof:

−

− (+)

− − + + − + + + We use theorem no.16/2.3.19

So, we get

19. Proof Theorem 2.3.23: � , ∈ ℝ, − −

Proof:

Let = = −

− −

+ − + − + − − −

20. Proof Theorem 2.3.24: � , ∈ ℝ, − −

Proof:

− −

Suppose =− = −

− − − − − − − − + − − − − − − − − − −

− − − ……….. (1)

In Theorem in number 19, we have − − ………(2)

From (1) and (2), we get

− − − By Theorem 2.3.19

− −

21. There are two theorems and one exercise in this section besides Corollary 2.3.8 that can

this form.

Ans:

a) Corollary 2.3.8. A negative of addition is equal to an addition of negative b) Theorem 2.3.13. An inverse of a negative = is equal to negative of an inverse. c) Exercise 15.b. An absolute of an inverse is equal to An inverse of an absolute.

< +

2 ……(i) 2+2 < 2+2 (9.a)

+

2 < ……(ii)

(i) and (ii) using 9.b we get

< + 2 <

13. For all ∈ ℝ, − =

let 0 then – < 0

− = − − if – < 0 (def. 2.3.16)

− = , 0 then =

let < 0, =− (def. 2.3.16)

− 0 so − = − then from def. above − =− = − =

14. For all ∈ ℝ, −

For 0, = (def.) 0,then − 0, 0 so that −

For 0 =− then − = 0 , 0 0, 0 so that

− and

We can say that −

15. Suppose x, y ∈ , ∈ ℝ. Proof the following. a. =

b. � ≠0, ℎ −1 = −1

c. � ≠0, ℎ / = /

a. =

 0, 0, = , =

0 . = . = .

 0, < 0, = , = −

. < 0 . = − . = (− ) = .

 < 0, 0, = − , =

< 0 . =− =− . = .

 < 0, < 0, = − , =−

. 0 . = . =− .− = . b. � ≠0, ℎ −1 = −1

Suppose 0, = −1 = −1 −1 ₀ −1 _{= 1}

−1_{= .} −1 −1. . −1 = −1. . −1

−1 = −1

Suppose < 0, = − −1 = − −1 −1 = 1

− −1 _{= .} −1

− −1 = − .− −1

− −1 _{. − .} −1 _{= −} −1_{. − .} −1 −1 = −1

c. � ≠0, ℎ / = /

. −1 = −1

For −1 0

. −1 = −1

. −1 = . −1. −1

. −1 = . −1. . −1

. −1 = . −1. −1

. −1 = . −1

For −1 < 0

. −1 = −1

. −1 = . − −1 . 1

. −1 = . − −1 . −1

. −1 = . − −1 . − . −1

. −1 = . −1. −1

. −1 = . −1

16. Prove the ⇐ direction of Theorem 2.3.19: Suppose 0. Then < if – < <

Proof: Let 0

< 0 >−

If = 0 means that –0 < < 0 or 0 < < 0 , so = 0 is impossible If 0, = , because < , so <

If < 0, = − → = −

− < − < −

− −1 < − −1

> Theorem number 19h

17. Proof Theorem 2.3.21: Suppose 0. Then > if and only if either > 0 < − .

Proof:

(→) Let >

If 0, = , = > , >

If < 0, = − , >

− = , − < − < −

(←) > <− → >

For > , =

So, >

18. Proof Theorem 2.3.22: � , ∈ ℝ, + +

Proof:

−

− (+)

− − + +

− + + + We use theorem no.16/2.3.19

So, we get

19. Proof Theorem 2.3.23: � , ∈ ℝ, − −

Proof:

Let = = − − −

+ − + − + − − −

20. Proof Theorem 2.3.24: � , ∈ ℝ, − −

Proof:

− −

Suppose =− = − − − − − − −

− − + − −

− − − −

− − − −

− − − ……….. (1)

In Theorem in number 19, we have − − ………(2) From (1) and (2), we get

− − − By Theorem 2.3.19

− −

21. There are two theorems and one exercise in this section besides Corollary 2.3.8 that can

this form. Ans:

a) Corollary 2.3.8. A negative of addition is equal to an addition of negative b) Theorem 2.3.13. An inverse of a negative = is equal to negative of an inverse. c) Exercise 15.b. An absolute of an inverse is equal to An inverse of an absolute.