Factors of 0
- Every number is a factor of 0:
LaTeX:
d \mid 0
- Proof:
- If dk = n then there must always be a k such that dk = 0
- We can set k = 0
0 as a factor
- If 0 is a factor of n then n = 0:
LaTeX:
0 \mid n \implies n = 0
- Proof:
- If dk = n and d = 0 then 0k = n
- n must be 0
1 as a factor
- 1 is a factor of every number:
LaTeX:
1 \mid n
- Proof:
- If dk = n and d = 1 then 1k = n
- We can set k = n
n as a factor
- n is a factor of n:
LaTeX:
n \mid n
- Proof:
- If dk = n and d = n then nk = n
- We can set k = 1
factors of 1
- if d is a factor of 1 then d = 1 or d = -1:
LaTeX:
d \mid 1 \implies d = 1 \text{ or } d = -1
- Proof:
- If ab = 1 then either a = 1 and b = 1 or a = -1 and b = -1
- If dk = 1 then either d = 1 or d = -1
Transitivity
- if d is a factor of n and n is a factor of m then d is a factor of m:
LaTeX:
d \mid n \text{ and } n \mid m \implies d \mid m
- Proof:
- Set n = dk_{1} and m = nk_{2}
- In the second equality we can substitute n for dk_{1} to get m = dk_{1}k_{2}
- And set k_{3} = k_{1}k_{2} to get m = dk_{3}
- d is a factor of m
Multiplication
- if d is a factor of n then ad is a factor of an
LaTeX:
d \mid n \implies ad \mid an
- Proof:
Cancellation
- if ad is a factor of an where a is not equal to 0 then d is a factor of n
LaTeX:
ad \mid an, a \neq 0 \implies d \mid n
- Proof:
- We set adk = an
- And divide both sides by a to get dk = n
Linearity
- if d is a factor of n and m then d is a factor of an + bm for all a and b
LaTeX:
d \mid n \text{ and } d \mid m \implies d \mid an + bm \text{ for all a and b}
- Proof:
- Lets set n = dk_{1} and m = dk_{2}
- Then we can rewrite an + bm as adk_{1} + bdk_{2}
- And factorize: d(ak_{1} + bk_{2})
Comparison
- if d is a factor of n and both d and n < 0, then d ≤ n
LaTeX:
d \mid n \text{ and } d \mid m \implies d \mid an + bm \text{ for all a and b}
- Proof:
If dk = n, lets consider different cases for k:- k < 0
If k is less than 0 then dk is less than 0, this breaks the rule that n > 0 and can be ignored - k = 0
If k is 0 then dk is 0, this breaks the rule that n > 0 and can be ignored - k = 1
if k is 1 then d equals n, this fits our definition that d ≤ n - k > 1
if k is greater than 1 then d < n as dk = n
- k < 0