MthT 430 Notes Chapter 1a More on Basic Properties of Numbers
MthT 430 Notes Chapter 1a More on Basic Properties of Numbers
Properties (P1) - (P9) give the following Theorems:



Theorem. Assuming P1 - P9, for all a, a ·0 = 0 .
The proof is given on page 7 of Spivak's book.
Proof.
a ·0
= a ·(0 + 0)
= a ·0 + a·0.        
(P9)
Now subtract a ·0 from both sides of the equation
a ·0
= a ·0 + a·0.



Theorem. If
a ·b = 0,
then either
a = 0 or b = 0.
Proof. If a ≠ 0 and a ·b = 0, then
a−1 ·(a ·b)
= a−1 ·0,
(a−1·a)·b
= 0,
1 ·b
= 0,
b
= 0.
Similarly, if b ≠ 0, then a = 0.



Theorem. 
− (a ·b)
= (−a)·b
= a ·(− b).
Proof. We will show that
(a ·b) + (−a)·b
= 0.

(a ·b) + (−a)·b
= ( a + (−a))·b    
(byP9)
= 0 ·b    
(byP3)
= 0    
(byTheorem).
The characterization of − (a ·b) as a ·(−b) is shown by repeating the proof or interchanging the roles of a and b and using (P8).


Inequalities


Inequalities are defined in terms of of the positive numbers P. We say that a > 0 or 0 < a if a is in P. We that a < b [b > a] if 0 < b − a or b − a is in P.
Note that
•
If a < b , then for all c, a + c < b + c.
•
If a < b , and 0 < c, then a c < b c.
•
If a < b , and c < 0, then a c > b c.
We shall also say that a is nonnegative [a ≥ 0] if and only if a = 0 or a > 0.


•
For all a, a2 is nonnegative.
•
If a and b are nonnegative, then a ≤ b iff a2 ≤ b2.


Absolute Value


Definition. The absolute value of a number a is defined as
|a| = ⎧
⎪
⎨
⎪
⎩
a,
a ≥ 0
−a,
a ≤ 0.
Properties of absolute value often involve a proof by cases .

Theorem. For all a,
− |a| ≤ a ≤ |a|.
Also
− |a| ≤ − a ≤ |a|.
Proof. If a ≥ 0, then |a| = a, and
− |a| ≤ 0 ≤ a = |a|.
If a ≤ 0, then − |a| = a, and
− |a| = a ≤ 0 ≤ |a|.
The second statement is shown in a similar manner, or by multiplying each expression in the first statement by −1.



Theorem. (The triangle inequality) For all numbers a, b,
|a + b| ≤ |a|+ |b|.

Proof. We have
−|a|
≤ a ≤ |a|,
−|b|
≤ b ≤ |b|,
so that
−(|a| + |b|) ≤ a + b ≤ |a| + |b|.
Now break into the two cases a + b ≥ 0 and a + b ≤ 0.
Another Proof. Since |a + b| and |a| + |b| are both nonnegative, compare |a + b|2 and (|a| + |b|)2.
|a + b|2
= (a + b)2
= a2 + 2 a b + b2,
(|a| + |b|)2
= a2 + 2 |a| |b| + b2.
We know(?) that |a b| = |a| |b| so that
2 a b
≤ 2 |a| |b|,
|a + b|2
≤ (|a| + |b|)2.
Fractions
For this section assume (P1) - (P9).


If b ≠ 0, we define
1
b
≡ b−1,
a
b
≡ a ·b−1
=a · 1
b
 .
Of course we hope that for b ≠ 0, d ≠ 0,
a
b
 + c
d
 = a d + b c
b d
 .

Theorem. For b ≠ 0, d ≠ 0,
a
b
 + c
d
 = a d + b c
b d
 .
 Proof. We show that
(a d + b c)·(b d)−1
= a b−1 + c d−1.
(a d + b c)·(b d)−1
= (a d + b c)· d−1b−1
why?
= a ·d ·d−1 ·b−1 + b ·c ·d−1 ·b−1
by P9
= a ·1 ·b−1 + 1 ·c ·d−1
by P8 and P7
= a b−1 + c d−1.
by P9


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On 22 Aug 2014, 10:44.