Carol Giancana
September 25, 2008 notes
MAE 501
We began class by going over Homework #2 problems.
Problem 1g: We’re asked to give an example of a bijection function f: N → Z.
Bethany found the function f(x) = -½ (x + 1) if x is odd and f(x) = x/2 when x is even or 0.
We have to prove both surjectivity and injectivity.
First, prove surjectivity : (For all y, elements of Z, there is an x in N such that f(x) = y.)
Pick any y an element of Z. Find x with f(x) = y (for the following 3 cases).
Suppose y>0.
Then let x = 2y
So x is even
So f(x) = f (2y) = 2y/2 = y
Suppose y<0
Then x = -2y – 1
Then x is odd
So f(x) = -½(x+1)
= -½ (-2y – 1 + 1) = y
Suppose y = 0
Let x=y
So f(x) = f(y) = f(0) = 0 = y
Now, to prove injectivity (a function f: A→B for which every element of B appears at
most once as a second coordinate)
Our goal is to show that if f(x₁) = f(x₂), then x₁ = x₂.
Cases:
x₁, x₂ are both even
x₁, x₂ are both odd
x₁, x₂ where one is even and one is odd
Case 1:
Suppose
f(x₁) = f(x₂) and both x₁, x₂ are even
Then
½ x₁ = ½ x₂
So
x₁ = x₂
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f(x₁) = f(x₂) and both x₁, x₂ are odd
-½ (x₁ + 1) = -½ (x₂ + 1)
-½ x₁ - ½= -½ x₂ - ½
-½ x₁ = - ½ x₂
x₁ = x₂
Case 2:
Suppose
Then
So
So
So
Case 3:
Suppose
f(x₁) = f(x₂), where f(x1) is even and f(x2) is odd
Then
½ x₁ = -½ (x₂ + 1) (both x₁, x₂ are positive because of N)
This case could not exist.
1e. We looked at Christina’s solution to 1e. Find a function
f: Z → Z that is surjective and not injective.
Her function was f(j) = ½ j rounded to the nearest integer (.5→1 and -.5→ -1)
Some ordered pairs would be (0,0), (1,1), (-1,-1), (2,1), (-2,-1), (3,2), (-3, -2), (4,2),
(-4,-2), (5,3), (-5,-3).
Prove surjectivity:
For surjectivity, we want to show that for every k in the codomain, there exists a j in the
domain such that f(j) = k, where k = ½ j rounded to the nearest integer.
To start, let’s take k that is in the codomain and suppose there is no j such that f(j) = k.
k is an element of Z. Either 2|k or 2 doesn’t divide k.
Then, if 2|k we have k = 2q for some q, an element of Z.
So choose j = 4q
Then f(j) = 4q⁄2 = 2q = k which is an integer, so it remains as is (i.e. it need not be
rounded). So, for every even k in the codomain there is a j in the domain with f(j) = k.
Now, suppose 2 does not divide kf
So k = 2p + 1 for some p, an element of Z.
Let j = 4p + 2
Then f (j) = f(4p + 2) = ½(4p + 2) = 2p + 1 = k which is an odd integer. So for every odd
k in the codomain there exists a j in the domain with f(j) = k.
Therefore, the function is surjective.
Prove f(j) = ½ j rounded to the nearest integer in not injective by showing a
counterexample:
f(1) = ½(1) rounded = 1
f(2) = ½(2) rounded = 1
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The function is not injective since an element in the codomain appears more than once.
Problem 2a :
For which values m,b is f(x) = mx + b a group homomorphism from ( R, +) → (R, +)
f ( a₁ + a₂ ) = m(a₁ + a₂ ) + b
= ma₁ + ma₂ + b
f(a₁) = ma₁ + b
f(a₂) = ma₂ + b
f(a1) + f(a2) = ma1 + ma2 + 2b. When b=0,
f(a1) + f(a2) = f(a1 + a2). So, this is a homomorphism if b = 0 and no restriction on m.
h(x) = x²
h(a + b) = (a + b)²= a²+ 2ab + b²
h(a) + h(b) = a² + b²
It is correct to say h is not a homomorphism since h(a) + h(b) = h(a + b) only when a or b
is 0.
2b:
Some students wrote that h(x) is a homomorphism provided a or b = 0. This is not
correct. Here, b is not a parameter; it’s a variable over the domain. By definition of
homomorphism it must assume every value of the domain so we cannot limit it to b=0.
My proposed question:
How can we define a ring homomorphism?
Problem 3. The problem stated that an ordered field is a set F closed under two
operations, + and ∙ , satisfying the field axioms stated in class, with a relation, denoted
<, (“<” is an arbitrary symbol), that satisfies the following properties:
1.)
2.)
3.)
4.)
For every x and y in F, exactly one of the following is true: x<y, y<x, or x=y.
The relation is transitive: x<y and y<z imply that x<z.
x<y implies that x + z < y + z.
x<y and 0< z implies xz <yz.
Part 3c. We are supposed to show that if 0<x<y then 0<⅟y<⅟x.
Many students incorrectly wrote something like the following:
Since 0<⅟x and 0<⅟y, using the rule of multiplication, we can write
0<x<y → 0∙⅟x∙⅟y<x∙⅟x∙⅟y<y∙⅟x∙⅟y → 0<⅟y<⅟x.
What is wrong with this?
We need to show the first step, i.e., that 0<⅟x and 0<⅟y.
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(If we know 0<x how can we show 0<⅟x?)
⅟x ∙ x = 1 (field axiom states that all non-zero elements have multiplicative
inverses)
So,
⅟x ≠0, or else the above statement is false
Suppose
⅟x <0; let’s arrive at a contradiction by multiplying through by x
x ∙ ⅟x < x ∙ 0 or 1< 0
We proved that 0<1 (in 3b) so 1<0 is not true.
Assume
In 3d. we have to prove that complex numbers of a field with the four properties for “<”
listed above is not an ordered field (ie we can’t have the symbol “<”).
One common mistake was to assume i< 0 or 0<i. (We must show this is true.)
Instead, we can say, i≠ 0 so either i<0 or 0<i.
Suppose i < 0
Note: We can not say 1/i ∙ i < 0 → -i ∙ i = 1 because we do not know if 0 < -i.
(According to our property, before we multiply both sides, we have to know that 0< z.)
Then i + -i < -i + 0 (by property (3))
Then 0 < -i
Note: At this point why is it wrong to multiply by -1? Because by property (4),
multiplication over “<” is only defined if 0<z.
Since 0 < -i, let’s use property (4):
Multiply through by -i to get –i ∙ 0 < -i ∙ -i
0< -1
1+ 0< 1 + -1 (property (3))
1<0 shows a contradiction
Now, suppose 0 < i
So
i ∙ 0<i ∙ i
Then
0< -1
0 + 1 < -1 + 1 or 1<0, which we know is wrong because we already proved
that 0<1.
Since we do not have i < 0 and we do not have 0 < i, this is not an ordered field . The
notion of ordering the elements of the field does not make sense.
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Next, we reviewed the Euclidean Algorithm of finding the GCD in a ring. In an integral
domain (ring with no zero divisors), R , d is the GCD of a and b if d|a and d|b and if c|a
and c |b then c|d.
For example, find the GCD of (24, 18).
24 = 18(1) + 6
18 = 6(3) + 0
GCD = 6, however c may be 2 where 2|24, 2|18, and 2|6.
We claim that by repeating the algorithm we will find the GCD.
a,b
b≠0
a = bq₁ + r₁
b = r₁q₂ + r₂
r₁ = r₂q₃ + r₃
.
.
.
r(n-3) = r(n-2)q(n-1) + r(n-1), (where r(n-1) is the GCD)
r(n-2) = r(n-1)qn + 0
Hints for next week’s homework: Prove r(n-1) |a and r(n-1)|b and if anything else divides a
and b it must divide r(n-1).
Assume r(n-1) |r(n-2) because r(n-2) = r(n-1)qn + 0
Then r(n-2) = r(n-1)qn
r(n-3) = r(n-1) qnq(n-1) + r(n-1) = r(n-1)(qnq(n-1) + 1)
Therefore, r(n-1) | r(n-3)
Working backwards to the first line we can show r(n-1) | b and a.
Now if c|a and c|b, then c|r(n-1)
To prove this, try letting a =cm and b=cn and we get cm = cnq₁ + r₁
Then we must prove c|r(n-1). . . .
References
Dubisch, Roy and Weiss, Marie J., Higher Algebra for the Undergraduate, New York,
John Wiley & Sons, Inc., 1962.
Eggen, Maurice, St. Andre, Richard, and Smith, Douglas, A Transition to Advanced
Mathematics, California, Thompson Brooks/Cole, 2006.
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