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131A Week 2Discussion
Alan Zhou
Ordered fields
131A Week 2 DiscussionOrdered Field Axioms
Alan Zhou
October 15, 2020
131A Week 2Discussion
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)
131A Week 2Discussion
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)
131A Week 2Discussion
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)
131A Week 2Discussion
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 .
131A Week 2Discussion
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 .
131A Week 2Discussion
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 .
131A Week 2Discussion
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)).
131A Week 2Discussion
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)).
131A Week 2Discussion
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 .
131A Week 2Discussion
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 .
131A Week 2Discussion
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 .
131A Week 2Discussion
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 .
131A Week 2Discussion
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.
131A Week 2Discussion
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.
131A Week 2Discussion
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.
131A Week 2Discussion
Alan Zhou
Ordered fields
Absolute value and inequalities
Proposition (Exercise 3.5(a))
For all x , y ∈ R,
|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.
131A Week 2Discussion
Alan Zhou
Ordered fields
Absolute value and inequalities
Proposition (Exercise 3.5(a))
For all x , y ∈ R,
|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.
131A Week 2Discussion
Alan Zhou
Ordered fields
Absolute value and inequalities
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.
131A Week 2Discussion
Alan Zhou
Ordered fields
Absolute value and inequalities
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.
131A Week 2Discussion
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.
131A Week 2Discussion
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.
131A Week 2Discussion
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.
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