Number Theory Midterm

nostrilswelderElectronics - Devices

Oct 10, 2013 (4 years and 4 days ago)

75 views

Number Theory Midterm

Study Guide

Review Session


Monday March 1 in IT 222

Some key definitions/Theorems to know:

a. Well Ordering Principle (We assumed this, we never proved it. It’s the assumption the
whole course comes from.)

b. First and Second Princi
ples of Finite Induction

c. Pascal’s Rule

d. Binomial Coefficient

e. Binomial Theorem

f. Triangular Number

g. Division Algorithm

h. Greatest Common Divisor

i. Relatively Prime

j. Euclid’s Algorithm


In addition, you should know theorems 2.2
-
2.5 Note:

2.4 t
ells us a and b are relatively prime iff there exist integers x and y with ax+by=1.

2.5 says that if a|bc with a and b relatively prime, then a|c

You do not need to study 2.6 for the midterm.


You should be able to apply the Euclidean Algorithm to do both:

a. Given integers a and b, find their gcd

b. Given integers a and b, express the gcd of a and b as a linear combination.


You should also recall from page 18 that the square of any integer is of the form 4k or 4k+1
for some k and that the square of any od
d integer is of the form 8j+1 for some j. (You
should also be able to prove these, but unless asked to prove, you can assume them. )


Some example problems: (Two of the test questions will come from these. I expect to add
one more problem on Monday to thi
s list.)


1. Show that 2.6.10… (4n
-
2)=(2n)!/n!

** Proof by induction

For the basis, let n=1. The left side is 2. The right side is (2x1)!/1!=2

For the induction step, assume the statement is true for n=k. So:

2.6….(4k
-
2)=(2k)!/k!

Then for n=k+1, the left s
ide:


by the induction hypothesis. The right side is:



2. More than any other method, we’ve used the well ordering principle to prove theorems.
Don’t complete the proof, but explain how we used the WOP to set
up the proof that given 2
integers, not both 0, there exists x and y such that gcd(a,b)=ax+by (Theorem 2.3)

** We begin by defining the set S as follows:


Then we prove S is not the empty set. The W.O.P.
tells us that there is a lea
st element in S. We then show that this element divides both a and
b using the fact that it is the least element of S .


3. Show that the cube of any integer has one of the forms 9k, 9k+1 or 9k
-
1 for some k.

**From the DA, every integer can be written as
3k or 3k+1 or 3k
-
1 (any 3k
-
1=3(k
-
1)+2)

The cubes of these are 27k, 27k
3
+3(27k
2
)+3(3k)+1=9m+1,

and 27k
3
-
3(27k
2
)+3(3k)
-
1=9n
-
1 for appropriate m or n.


4. Expand (a+b)
7

**


5. Find the gcd(1617,525)

**Use EA:

1617= 3x525+42

525=12x42
+21

42=2x21+0

gcd(1617,525)=21


6. Find x and y so the gcd(24,138)=24x+138y using the Euclidean Algorithm

**use EA to find gcd

138=5x24+18

24=1x18+6

18=3x6+0 so gcd=6

Then 6=24
-
1x18

=24
-
1x(138
-
5x24) = 6x24
-
1x138


7. For any integer x, show a|(ax+b) iff a|
b

**Part 1. Assume a|ax+b Then there exists m such that am=ax+b So am
-
ax=b

So a(m
-
x)=b So a|b

Part 2. Assume a|b Then there exists some integer m such that am = b Then

ax+b = ax+am = a(x+m) So a|ax+b



8. Show that if a|bc then a| gcd(a,b)gcd(a,c)

** Let

d=gcd(a,b) and e=gcd(a,c). Then by Theorem 2.3, there exist integers x,y,m, and n
such that d=ax+by and e=am+cn.

Then gcd(a,b)gcd(a,c)=d.e=(ax+by)(am+cn)=a
2
xm+axcn+abmy+bcny. Since a|bc, a divides
every term of this expression so a|

gcd(a,b)gcd(a,c)


9. (
Added Monday night)

The nth Catalan number for n


0, is defined to be

Prove

for n

1.


** Proof:

Calculate out what the left side is and what the right side is using the definition above.

Left side is

Right side is

Note the second last step involves multiplying both the numerator and denominator by n.