Extended Euclidean Algorithm
After finding the GCD with the Euclidean Algorithm, start
by solving for the remainder of the step next to the last
step.
From the previous example, it would be the 3rd step:
4=3(1)+1
Solve for 1.
1=4-3(1)
Next, you want to go to the step before the one that was
worked on and solve for that remainder.
For the example, solve for 3 from the second step.
2nd step: 7=4(1)+3
3=7-4(1)
Substitute the result into the 3 in the previous equation,
1=4-3(1)
1=4-(7-4)
1=4-3(1)
Keep working your way up until you get an equation with
your GCD equaling to your arguments
Modular Definition 1:
If a and b are integers with a≠0, we say that a divides b if there is
an intger c such that b =ac. When a divides b we say that a is a
factor of b and that b is a multiple of a. The notation a|b denotes
that a divides b. We write a b when a does not divide b.∤b when a does not divide b.
Ex1:
a|b a|bc
a|b: b=a(t) b is equal to a times some num t
b(c)=a(t)(c)
Therefore, a|bc
Ex2:
Premise:
a|b, b|a, a and b are some integers
Therefore, a=b or a=-b
For a|b, a=b(t)
For b|a, b=a(s)
a=a(s)(t), then divide both sides by a (a≠0)
1=st, with s=t=1, or s=t=-1
Then, go to the previous formulas.
a mod m a=q1m + r
Euclidean Algorithm
GCD(someNum1, someNum2)
Whichever number is greater, put it on the left hand side
of the equation. For the other number, have it multiply as
many times as it can into the greater number and add the
remainder.
Ex: GCD(7, 25)
1st step: 25=7(3) + 4
Afterwards, you want to shift everything to the left. Take
the 7 and put it on the left hand side and multiply 4 into 7
as many times as it can without going over and add the
remainder.
2nd step: 7=4(1) + 3
Just rinse and repeat from here until you get a remainder
of zero. Once you get to that point, then the number
added to zero is the GCD.
3rd step: 4=3(1) + 1
4th step: 3=1(3) + 0
GCD=1
Modular Theorems
1) Let a, b, and c be a set of all integers
a) a|b + a|c a|(b+c)
b) a|b a|bc for all integers c.
c) a|b + b|c a|c
2) The Division Algorithm:
Let a be an integer. Then there are unique integers q and
r, with 0≤r<d, such that a=dq+r
3) Let a and b be integers, and let m be a positive integer.
Then a≡b(mod m) if and only if a mod m = b mod m.
4) Let m be a positive integer. The integers a and b are
congruent to modulo m if and only if there is an integer k
such that a=b+km.
5) Let m be a positive integer. If a≡b(mod m) and c≡d(mod
m), then a+c≡b+d(mod m) and ac≡bd(mod m)
a≡b mod m, m|a-b
Summations
∑
k=0
n
a rk(r ≠ 0)
a r
n+1
−a
r−1,r ≠ 1
∑
k=1
n
k
n
(
n+1
)
2
∑
k=1
n
k2
n
(
n+1
) (
2n+1
)
6
∑
k=1
n
k3
n2
(
n+1
)
2
4
Prove by induction that
5|3^2n + 4^n+1 for n≥0
N=0, 3^2*0 + 4^0+1 = 5 5|5
Induction hypothesis
Assume that for n=k, kϵZ+ is followsZ+ is follows
5|[3^2k + 4^(k+1)]
Induction step
Based on the IH we have to show that 5|[3^2(k+1) + 4^(k+2)]
5|[3^(k+2)+4^(k+2)=3^2k * 3^2 + 4^(k+1) * 4]
5|[3^(k+2) + 4^(k+2) = 9*3^(2k) + 4^(k+1)*4]
(5+4)3^(2k) + 4^(k+1)*4
5(3^2k) + 3^(2k)*4 + 4^(k+1)*4
5[3^(2k) + 4(3^(2k) + 4^(k+1))]
5*3^(2k) + 4*5*someConstant
5(3^(2k) + 4*someConstant)
POOF
GCD(A, B) *LCM(A,B) = A*B
Set proof: Odd-positive integers
1 ε S
x ε S x + 2 ε S
∀xP(x)
Therefore, P(c) Universal Instantiation
P(c) for arbitrary c
Therefore, ∀xP(x) Universal generalization
∃xP(x)
Therefore, P(c) for some element C Existential
instantiation
P(c) for some element C
Therefore, ∃xP(x) Existential generalization
