Negative Real Number
We say that x ∈ R is negative when x < 0
Positive Real Number
We say that x ∈ R is positive when 0 < x
A Negative Number is Smaller than a Positive One
Suppose that a < 0 and that b > 0 then a < b
This follows because < is transitive.
Multiplying by -1 Inverts the Sign
Suppose x ∈ R is positive iff − x is negative
Observe that x ≥ 0 iff x + ( − x ) ≥ 0 + ( − x ) iff 0 ≥ − x as needed.
Negative Zero is Zero
For 0 ∈ R we have that − 0 = 0
Since 0 is an identity element with respect to + then x + 0 = x , specifically for x = 0 by definition this means that − 0 = 0 as needed.
A Positive Number is Greater than the Negative Version of Itself
Suppose x ∈ R where x ≥ 0 then x ≥ − x
If x = 0 then 0 = − 0 so trivially 0 ≥ − 0 . On the other hand if x > 0 then 0 ≥ − x therefore x ≥ − x
Multiplying by -1 Changes the Inequality Direction
For any x , y ∈ R x ≥ y ⟺ − x ≤ − y
Observe that x ≥ y is the same as x − y ≥ 0 is the same as 0 ≥ y − x is the same as − x ≤ − y as needed.
Absolute Value
the absolute value function | ⋅ | : R → R ≥ 0 is defined by
| x | = { x  if  x ≥ 0 − x  if  x < ; 0
Absolute Value Differs By Sign
∀ x ∈ R , x = | x | ∨ x = − | x |
Let x ∈ R if x > ; 0 , then x = | x | by definition, if x < ; 0 , then | x | = − x , or rather x = − | x |
Absolute Value Equals Max of Itself and it's Negation
Let x ∈ R then | x | = max ( x , − x )

Recall the definition of max ( x , − x ) which is that it's value is x if x ≥ − x and − x if x ≤ − x . Suppose that x ≥ 0 then | x | = x and since x ≥ − x then we know that max ( x , − x ) = x and so in this case we see that | x | = max ( x ) .

Now in the case that x < 0 then − x > 0 and therefore − x > x therefore max ( x , − x ) = − x and since x < 0 then | x | = − x .

Therefore it's been verified in any case so that | x | = max ( x , − x ) .

Multiplication in Absolute Value
| x ⋅ y ∣= | x | ⋅ ∣ y |
  • If both x , y are positive then | x ⋅ y | = x ⋅ y = | x | ⋅ | y |
  • if both x , y are negative, then their product is positive so it follows as above
  • if one of x , y is positive and the other negative, without loss of generality assume that x < ; 0 and y > ; 0 then their product is negative so | x ⋅ y | = − 1 ⋅ x ⋅ y = | x | ⋅ | y |
  • if any one of x , y is zero, then | x ⋅ y | = 0 = | x | ⋅ | y |
Value Between Positive and Negative Equivalence
Let a , x ∈ R where x ≥ 0 − x ≤ a ≤ x ⟺ | a | ≤ x

⟹ we want to prove that | a | ≤ x . Case one, if a ≥ 0 then | a | = a and since we already know that a ≤ x then we obtain that | a | ≤ x as needed. In the second case if a < 0 then | a | = − a so we want to show that − a ≤ x which is the same as a ≥ − x which we know by assumption as well.

⟸ Suppose that | a | ≤ x if a ≥ 0 then | a | = a and so we're really assuming that a ≤ x therefore − a ≥ − x since a ≥ − a then by chaining inequalities we get x ≥ a ≥ − a ≥ − x which implies x ≥ a ≥ − x as needed. Now for the case of a ≤ 0 then we've assumed that − a ≤ x now − a is positive and by symmetrically following our previous case with − a in place of a we see that x ≥ − a ≥ a ≥ − x and so x ≥ a ≥ − x again, as needed.

Maximum is Always Bigger than It's arguments
Let x , y ∈ R then x ≤ max ( x , y )  and  y ≤ max ( x , y )
Value Between Implies It's Absolute Value is Less than the Max of the Others
Suppose that a ≤ b ≤ c then | b | ≤ max ( | a | , | c | )
Let's recall that c ≤ | c | ≤ max ( | c | , | a | ) , and on the other hand that a ≥ − | a | ≥ − max ( | a | , | c | ) combining inequalities, that says − max ( | a | , | c | ) ≤ b ≤ max ( | a | , | c | ) therefore | b | ≤ | max ( | a | , | c | ) | = max ( | a | , | c | ) as needed.
Absolute Value Bounds Distance
Suppose that a , b , c ∈ R with c ≥ 0 then | a − b | ≤ c ⟺ b − c ≤ a ≤ b + c
We know that | a − b | ≤ c iff − c ≤ a − b ≤ c which is the same as b − c ≤ a ≤ b + c as needed.
absolute value greater inequality
Suppose that x ∈ R and that c ∈ R such that | c | ≥ 1 then | x | ≤ | c x |
Since 1 < | c | then 1 ⋅ | x | ≤ | c | ⋅ | x | = | c x |
absolute value equality
Suppose that a , b ∈ Z and that | a | = | b | , then a = b or a = − b
TODO
not equal range
Suppose that a , b ∈ Z then if − | b | < a < | b | then a ≠ b
Suppose that a = b for contradiction, then − | a | < a < | a | , but either a = | a | or a = − | a | , so this yields a contradiction.
changing sign doesn't change absolute value
∀ x ∈ R , | x | =∣ − x |
Let x ∈ R then | − x | = | − 1 ⋅ x | = | − 1 | ⋅ | x | = 1 ⋅ | x | = | x |
absolute value is greater than itself
∀ x ∈ R , x ≤ | x |
  • if x > ; 0 then | x | = x
  • if x < ; 0 then | x | = − x so | x | ≥ − x > ; x this is true since x < ; 0 and 0 < ; − x
absolute value is greater than the negative version of itself
∀ x ∈ R , − x ≤ | x |
using the fact y < ; | y | and subbing in − x we obtain the result
| a | − 1 = | 1 a |
| a | ⋅ | 1 a ∣= | a ⋅ 1 a | =∣ 1 | = 1
| x + y | ≤ | | x | + | y | |
TODO
Triangle Inequality
Suppose that x , y ∈ R , then we have | x + y | ≤ | x | + | y |
Generalized Triangle Inequality
Suppose that X ⊆ R is finite then | ∑ X | ≤ ∑ x ∈ X | x |
We prove it true by induction based on the size of X . Base case | X | = 2 is covered by the regular triangle inequality. Now suppose that it holds true for | Y | = k we'll show it holds true for | Z | = k + 1 , let z ∈ Z , then we have | ∑ ( Z ∖ { z } ) | ≤ ∑ a ∈ Z ∖ { z } | a | since | Z ∖ { z } | = k . Now by the regular triangle inequality we obtain that | ∑ Z | = | ∑ ( Z ∖ { z } ) + z | ≤ | ∑ ( Z ∖ { z } ) | + | z | ≤ ∑ a ∈ Z ∖ { z } | a | + | z | = ∑ z ∈ Z | z |
Isolate Absolute value Inequality
Let a , b , c ∈ R then | a − b | ≤ c implies that | a | ≤ c + | b |
We have the following: | a − b | ≤ c ⇕ | a − b | + | b | ≤ c + | b | ⇕ | a − b + b | ≤ | a − b | + | b | ≤ c + | b | ⇕ | a | ≤ c + | b | Where in the second last line we used the triangle inequality