131A Week 2 Discussionazhou/teaching/20F/131a-week...131A Week 2 Discussion Alan Zhou Ordered elds Basic inequalities Theorems 3.1 and 3.2 establish most of the usual rules for manipulating

131A Week 2Discussion

Alan Zhou

Ordered fields

131A Week 2 DiscussionOrdered Field Axioms

Alan Zhou

October 15, 2020


Alan Zhou

Ordered fields

Example: Difference of squares

Let R = Z or R = ordered field. For all x , y ∈ R,

(x − y)(x + y) = x2 − y2.

Proof

(x − y)(x + y) = (x + (−y))(x + y)

= x(x + y) + (−y)(x + y) (DL)

= (x2 + xy) + ((−y)x + (−y)y) (DL)

= (x2 + (xy − yx)) + (−y)y (A1 and 3.1(iii))

= (x2 + (xy − xy))− y2 (M2 and 3.1(iii))

= (x2 + 0)− y2 (A4)

= x2 − y2. (A3)


Alan Zhou

Ordered fields



(x − y)(x + y) = x2 − y2.

Proof

(x − y)(x + y) = (x + (−y))(x + y)

= x(x + y) + (−y)(x + y) (DL)

= (x2 + xy) + ((−y)x + (−y)y) (DL)

= (x2 + (xy − yx)) + (−y)y (A1 and 3.1(iii))

= (x2 + (xy − xy))− y2 (M2 and 3.1(iii))

= (x2 + 0)− y2 (A4)

= x2 − y2. (A3)


Alan Zhou

Ordered fields



(x − y)(x + y) = x2 − y2.

Proof

(x − y)(x + y) = (x + (−y))(x + y)

= x(x + y) + (−y)(x + y) (DL)

= (x2 + xy) + ((−y)x + (−y)y) (DL)

= (x2 + (xy − yx)) + (−y)y (A1 and 3.1(iii))

= (x2 + (xy − xy))− y2 (M2 and 3.1(iii))

= (x2 + 0)− y2 (A4)

= x2 − y2. (A3)


Alan Zhou

Ordered fields

Basic inequalities

Theorems 3.1 and 3.2 establish most of the usual rules formanipulating equations and inequalities, starting from theaxioms. We strengthen a few of them here, so that we canuse them freely later.

TheoremLet K = (K , 0, 1,+, ·,≤) be an ordered field.

1. For all x ∈ K , we have x2 ≥ 0, with equality iff x = 0.

2. For all w , x , y , z ∈ K , if w ≤ x and y ≤ z , then

w + y ≤ x + z .

3. For all w , x , y , z ∈ K , if 0 ≤ w ≤ x and 0 ≤ y ≤ z , then

0 ≤ wy ≤ xz .


Alan Zhou

Ordered fields

Basic inequalities





w + y ≤ x + z .


0 ≤ wy ≤ xz .


Alan Zhou

Ordered fields

Basic inequalities





w + y ≤ x + z .


0 ≤ wy ≤ xz .


Alan Zhou

Ordered fields

Basic inequalities

Theorem (1)

For all x ∈ K , we have x2 ≥ 0, with equality iff x = 0.

Proof.The inequality is proved as 3.2(iv).For the equality case, if x = 0, then x2 = 0 · 0 = 0 (3.1(ii)).Conversely, if x2 = x · x = 0, then x = 0 (3.1(iv)).


Alan Zhou

Ordered fields

Basic inequalities

Theorem (1)

For all x ∈ K , we have x2 ≥ 0, with equality iff x = 0.

Proof.The inequality is proved as 3.2(iv).For the equality case, if x = 0, then x2 = 0 · 0 = 0 (3.1(ii)).Conversely, if x2 = x · x = 0, then x = 0 (3.1(iv)).


Alan Zhou

Ordered fields

Basic inequalitiesTheorem (2)For all w , x , y , z ∈ K , if w ≤ x and y ≤ z , then

w + y ≤ x + z .

Proof.We apply the additive order axiom (O4)

a ≤ b =⇒ a + c ≤ b + c

twice. First, since w ≤ x ,

w + y ≤ x + y .

Second, since y ≤ z ,x + y ≤ x + z .

Now by transitivity (O3), w + y ≤ x + z .


Alan Zhou

Ordered fields

Basic inequalitiesTheorem (2)For all w , x , y , z ∈ K , if w ≤ x and y ≤ z , then

w + y ≤ x + z .

Proof.We apply the additive order axiom (O4)

a ≤ b =⇒ a + c ≤ b + c

twice. First, since w ≤ x ,

w + y ≤ x + y .

Second, since y ≤ z ,x + y ≤ x + z .

Now by transitivity (O3), w + y ≤ x + z .


Alan Zhou

Ordered fields

Basic inequalitiesTheorem (3)For all w , x , y , z ∈ K , if 0 ≤ w ≤ x and 0 ≤ y ≤ z , then

0 ≤ wy ≤ xz .

Proof.To show that 0 ≤ wy ≤ xz , we must show that 0 ≤ wy and wy ≤ xz .The first statement is 3.2(iii). For the second, we apply themultiplicative order axiom (O5)

(a ≤ b and c ≥ 0) =⇒ ac ≤ bc

twice. First, since w ≤ x and y ≥ 0,

wy ≤ xy .

Second, since y ≤ z and x ≥ 0,

xy ≤ xz .

Now by transitivity, wy ≤ xz .


Alan Zhou

Ordered fields

Basic inequalitiesTheorem (3)For all w , x , y , z ∈ K , if 0 ≤ w ≤ x and 0 ≤ y ≤ z , then

0 ≤ wy ≤ xz .

Proof.To show that 0 ≤ wy ≤ xz , we must show that 0 ≤ wy and wy ≤ xz .The first statement is 3.2(iii). For the second, we apply themultiplicative order axiom (O5)

(a ≤ b and c ≥ 0) =⇒ ac ≤ bc

twice. First, since w ≤ x and y ≥ 0,

wy ≤ xy .

Second, since y ≤ z and x ≥ 0,

xy ≤ xz .

Now by transitivity, wy ≤ xz .


Alan Zhou

Ordered fields

Absolute value and inequalities

Proposition (Exercise 3.5(a))

For all x , y ∈ R,

|x | ≤ y ⇐⇒ −y ≤ x ≤ y .

Proof ( =⇒ ).

First we prove the ( =⇒ ) direction. Suppose |x | ≤ y . Since|x | ≥ 0 (3.5(i)), we have y ≥ 0. To show that −y ≤ x ≤ y ,we must show that −y ≤ x and x ≤ y . We consider twocases and show that in each case, both inequalities hold.If x ≥ 0, then |x | = x , so we know x ≤ y . Since y ≥ 0, wehave −y ≤ 0, so −y ≤ x .Otherwise, x < 0, so |x | = −x and −x ≤ y . This rearrangesto −y ≤ x . Since y ≥ 0, we have x ≤ y as well.Thus we have shown the required conclusion in all cases.


Alan Zhou

Ordered fields




|x | ≤ y ⇐⇒ −y ≤ x ≤ y .

Proof ( =⇒ ).



Alan Zhou

Ordered fields




|x | ≤ y ⇐⇒ −y ≤ x ≤ y .

Proof ( =⇒ ).



Alan Zhou

Ordered fields




|x | ≤ y ⇐⇒ −y ≤ x ≤ y .

Proof (⇐= ).

Now we prove the (⇐= ) direction. Suppose −y ≤ x ≤ y .If x ≥ 0, then |x | = x ≤ y . Otherwise, x < 0, so |x | = −x .Since −y ≤ x , we have −x ≤ y , so |x | ≤ y .Thus we have shown the required conclusion in all cases.


Alan Zhou

Ordered fields




|x | ≤ y ⇐⇒ −y ≤ x ≤ y .

Proof (⇐= ).

Now we prove the (⇐= ) direction. Suppose −y ≤ x ≤ y .If x ≥ 0, then |x | = x ≤ y . Otherwise, x < 0, so |x | = −x .Since −y ≤ x , we have −x ≤ y , so |x | ≤ y .Thus we have shown the required conclusion in all cases.


Alan Zhou

Ordered fields


Corollary (Exercise 3.7(c))

For all x , `, ε ∈ R,

|x − `| ≤ ε ⇐⇒ `− ε ≤ x ≤ `+ ε.

Proof.By the proposition,

|x − `| ≤ ε ⇐⇒ −ε ≤ x − ` ≤ ε.

For the first half of the inequality, −ε ≤ x − ` if and only if`− ε ≤ x (why?). For the second half of the inequality,x − ` ≤ ε if and only if x ≤ `+ ε. Thus

−ε ≤ x − ` ≤ ε ⇐⇒ `− ε ≤ x ≤ `+ ε,

and the proof is complete.


Alan Zhou

Ordered fields


Corollary (Exercise 3.7(c))

For all x , `, ε ∈ R,

|x − `| ≤ ε ⇐⇒ `− ε ≤ x ≤ `+ ε.

Proof.By the proposition,

|x − `| ≤ ε ⇐⇒ −ε ≤ x − ` ≤ ε.

For the first half of the inequality, −ε ≤ x − ` if and only if`− ε ≤ x (why?). For the second half of the inequality,x − ` ≤ ε if and only if x ≤ `+ ε. Thus

−ε ≤ x − ` ≤ ε ⇐⇒ `− ε ≤ x ≤ `+ ε,

and the proof is complete.


Alan Zhou

Ordered fields

Reverse triangle inequality

Proposition (Exercise 3.5(b))

If x , y ∈ R, then ||x | − |y || ≤ |x − y |.

Proof.By the corollary, the conclusion holds if and only if

|y | − |x − y | ≤ |x | ≤ |y |+ |x − y |.

To prove the first inequality, the triangle inequality gives

|y | = |(y − x) + x | ≤ |y − x |+ |x | = |x − y |+ |x |,

so |y | − |x − y | ≤ |x |. For the second inequality, the triangleinequality gives

|x | = |(x − y) + y | ≤ |x − y |+ |y |.

Thus the double inequality holds, as required.


Alan Zhou

Ordered fields



If x , y ∈ R, then ||x | − |y || ≤ |x − y |.


|y | − |x − y | ≤ |x | ≤ |y |+ |x − y |.


|y | = |(y − x) + x | ≤ |y − x |+ |x | = |x − y |+ |x |,


|x | = |(x − y) + y | ≤ |x − y |+ |y |.



Alan Zhou

Ordered fields



If x , y ∈ R, then ||x | − |y || ≤ |x − y |.


|y | − |x − y | ≤ |x | ≤ |y |+ |x − y |.


|y | = |(y − x) + x | ≤ |y − x |+ |x | = |x − y |+ |x |,


|x | = |(x − y) + y | ≤ |x − y |+ |y |.


Documents

131A Week 2 Discussionazhou/teaching/20F/131a-week...131A Week 2 Discussion Alan Zhou Ordered elds Basic inequalities Theorems 3.1 and 3.2 establish most of the usual rules for manipulating