### Corollary

Let a and b be integers. If a | b and b | a, then a = ± b.

Can we use Bounds by Divisibility (BBD) if we did not say a $$\ne$$ 0 and b $$\ne$$ 0

The answer to that is we must consider a,b = 0 as a special case, and use BBD for other cases.

Proof:

Case 1: Since 0 | 0 and 0 $$\nmid$$ n for any non-zero integer n,
then when a = 0, b must be zero and vice versa

Case 2:
When a $$\ne$$ 0, and b $$\ne$$ 0
From a | b and b $$\ne$$ 0, we can use BBD to get |a| $$\le$$ |b|
From b | a and a $$\ne$$ 0, we can use BBD to get |b| $$\le$$ |a|
Hence |a| = |b|
Therefore a = ± b

## Negating an Implication

A $$\Rightarrow$$ B and $$\lnot$$A v B are the same

A B A $$\Rightarrow$$ B $$\lnot$$A not A v B
T T T F T
T F F F F
F T T T T
F F T T T

$$\lnot$$(A $$\Rightarrow$$ B) $$\equiv$$ $$\lnot$$(($$\lnot$$A) v B)
$$\equiv$$ (($$\lnot$$($$\lnot$$A)) $$\land$$ ($$\lnot$$ B))
$$\equiv$$ A $$\land$$ ($$\lnot$$ B)

In A then B Implication
A and not B Negated Implication

If a | b and b|c then a|c.
Negated Implication:
a | b and b | c and a $$\nmid$$ c

If a | bc then a|b or a|c.
Negated Implication:
a | bc and a $$\nmid$$ b and a $$\nmid$$ c

If a | b and b $$\ne$$ 0, then|a|≤|b|.
Negated Implication:
a | b and b $$\ne$$ 0 and |a| > |b|

## Divisibility of Integer Combinations.

“If m is an integer and 14 | m,
prove that 7 | 135m + 693”.
Note that 7 $$\nmid$$ 135, so this is not a straight-forward result.

Recall previously: Proposition (Divisibility of Integer Combinations (DIC))
Let a, b and c be integers.
If a | b and a | c, then for any integers x and y, a | (bx + cy).

Proof: Assume 14 | m
Since 7 | 14, by Transivity of Divisibility 7 | m
m = 7k, for some integer k

Now 135m + 693 = 135(7k) + 7(99) = 7(135k + 99)
As 135k + 99 is an integer, 7 | (135m + 693)

## Converse

Given an implication:

If A then B.

the converse of “If A then B” (A $$\Rightarrow$$ B) is “If B then A” (B $$\Rightarrow$$ A)

An implication and its converse are not logically related.

##Sets

Sets are collection of objects, called elements or members

∅ = { }. The empty-set has nothing in it.
{∅} has one element, namely the empty set, so
{∅} ≠ ∅.

$$\mathbb{R}$$ is the set of real numbers. This is essentially the set that contains every possible number we have seen (unless you have seen complex numbers).

$$\mathbb{Q}$$ is the set of rational numbers. These are numbers of the type 2, 1, 5 , 2, etc. The members are fractions, with numerator and denominator both being integers, and denominator cannot be zero.

$$\mathbb{Z}$$ = {…,−3,−2,−1,0,1,2,3,…}, the set of integers (whole numbers)

$$\mathbb{N}$$ = {1,2,3,…}, the set of natural numbers (positive integers). Note that in MATH 135, N starts at 1 and 0 ∉ N. This may be different that what you learn in CS.