November 3rd, 2010 (test solutions)
MAE-501 Course Notes
Notes by Kevin Maritato
1.)
a.) Explain what it means for two integers to be “equivalent modulo 5.”
b.) Take the set of integers modulo this equivalence relation. Describe the equivalence
classes. How many equivalence classes are there?
c.) Now consider Q[x], the ring of polynomials with rational coefficients. We say that
the polynomials f(x) and g(x) are equivalent modulo (x-5) if (x-5)|(f(x)-g(x)). Prove
that this is an equivalence relation.
d.) Describe the classes in Q[x] modulo the polynomial x-5. How many classes are there?
a.) If it is said that a and b, 𝑎, 𝑏 ∈ 𝑍 are equivalent modulo 5, it means that a=q * 5 + n,
and b = r * 5 + n, where 𝑞, 𝑟, 𝑛 ∈ 𝑍, n<5. In other words, when a and b are divided by
5, they both give the same remainder, n.
b.) Each equivalence class is for a different n, as described above. In simpler terms, each
equivalence class is defined by the remainder when 5 divides is members. The
classes are represented by 0,1,2,3, and 4, giving five equivalence classes total.
c.) Let f(x) ~ g(x) mean that f(x) and g(x) are equivalent modulo (x-5).
Reflexivity:
f(x) – f(x) = 0
Therefore, (x-5)|(f(x) – f(x))
Therefore, f(x) ~ f(x)
November 3rd, 2010 (test solutions)
MAE-501 Course Notes
Notes by Kevin Maritato
The relation is reflexive.
Symmetry:
Suppose that f(x) ~ g(X)
Then, (x-5)|(f(x) – g(x))
(x-5)|(-1)(f(x) – g(x))
(x-5)|(-f(x) + g(x))
(x-5)|(g(x) – f(x))
g(x) ~ f(x)
Therefore, the relation is symmetric.
Transitivity:
Suppose that f(x) ~ g(x) and g(x) ~ h(x).
Then, (x-5)|(f(x) – g(x)) and (x-5)|(g(x) – h(x))
Then, n(x)(x-5) = f(x) – g(x) and m(x)(x-5) = g(x) – h(x), for some 𝑛(𝑥), 𝑚(𝑥) ∈
𝑄[𝑥]
-(n(x)(x-5) – f(x)) = m(x)(x-5) + h(x) = g(x)
f(x) – n(x)(x-5) = m(x)(x-5) + h(x)
f(x) – h(x) = m(x)(x-5) + n(x)(x-5)
November 3rd, 2010 (test solutions)
MAE-501 Course Notes
Notes by Kevin Maritato
f(x) – h(x) = (m(x) + n(x))(x-5)
since 𝑛(𝑥), 𝑚(𝑥) ∈ 𝑄[𝑥], (𝑚(𝑥) + 𝑛(𝑥)) ∈ 𝑄[𝑥]
Therefore, (x-5)|(f(x) – h(x)).
Therefore, the relation is transitive.
Since the relation is reflexive, symmetric, and transitive, the relation is an
equivalence relation.
d.) The equivalence classes are the set of remainders when a polynomial is divided by
(x-5). Each rational number represents an equivalence class, since the function
𝑓(𝑥) = 𝑞 ∈ 𝑄[𝑥], for any 𝑞 ∈ 𝑄, and when a rational number is divided by a
polynomial of degree greater than 0 ((x-5) is of degree 1), the remainder is the initial
rational number. All of the equivalence classes will be represented by polynomials of
degree zero, since any polynomial of degree n can be divided (with a remainder) by
any polynomial of degree ≤ n, and (x-5) is of degree 1, meaning that any polynomial
of degree >0 can be divided (with a remainder) by (x-5). Irrational numbers cannot
represent equivalence classes, because the last step of the long division of a
polynomial by (x-5) will always leave the difference of the coefficient of x0 and a
multiple of 5 as a remainder, and because the coefficient and any multiple of five
must both be rational numbers, their difference must also be a rational number.
Therefore, the equivalence classes are represented by each of the rational numbers.
Because there are an infinite number of rational numbers, there are an infinite number
of equivalence classes.
November 3rd, 2010 (test solutions)
MAE-501 Course Notes
Notes by Kevin Maritato
2.) Consider the ring of integers modulo n. In class we conjectured that this ring forms a field if
and only if n is a prime. Prove this statement.
only if: If n is not prime, then there exists in the ring of integers from 0 to n-1 a pair of factors
of n, c and p, such that neither c nor p is a unit, by the definition of a prime number. The
product of c and p mod n is then 0.
Suppose that c has an inverse mod n. Then, let 𝑏 ∈ {0,1, … , 𝑛 − 1}, be this inverse.
b*c= q*n + 1, 𝑞 ∈ 𝑍
b*c = q*p*c + 1
(b - q*p)c = 1
Since 𝑏, 𝑞, 𝑝 ∈ 𝑍, (𝑏 − 𝑞 ∗ 𝑝) ∈ 𝑍.
Then, since c is not a unit, (b - q*p)c ≠ 1.
This contradiction shows that c has no inverse mod n, so the integers mod n are not a
field.
if: If n is prime, then there are no two numbers, a,b in {1,2,…,n-1}, such that a*b = 0 mod n, by
the definition of a prime number (since n cannot divide a or b, but does divide itself).
Therefore, all products of two numbers a and b in the ring must be >0 mod n, as long as
a,b ≠ 0.
November 3rd, 2010 (test solutions)
MAE-501 Course Notes
Notes by Kevin Maritato
Let 𝑥, 𝑦, 𝑧 ∈ {1,2, … , 𝑛 − 1}. Suppose that (x*y) = (x*z) mod n.
Then, x*y – x*z = 0 mod n
x(y – z) = 0 mod n
Then, as shown above, x = 0 mod n, or (y – z) = 0 mod n.
Since 𝑥 ∈ {1,2, … , 𝑛 − 1}, x ≠ 0 mod n.
Therefore, y – z = 0 mod n
y = z mod n
Since 𝑦, 𝑧 ∈ {1,2, … , 𝑛 − 1}, and 1,2,…,n-1 define equivalence classes of the relation mod n,
y = z. Therefore, the product of 𝑥 ∈ {1,2, … , 𝑛 − 1} and any element 𝑦 ∈ {1,2, … , 𝑛 − 1} is
unique mod n. Since there are n-1 products (since there are n-1 possible values of y) and
there are n-1 possible values of the products mod n (also {1,2,…,n-1}, since none of the
products can be 0), one of the products must be 1, by the pigeonhole property. Therefore,
for any 𝑥 ∈ {1,2, … , 𝑛 − 1}, x has an inverse mod n. Since we already know that
multiplication is commutative, and 1 is a multiplicative identity for the ring of integers mod
n, since 1*x = x = x mod n, for any 𝑥 ∈ {0,1, … , 𝑛 − 1}, this means that the ring of integers
mod n is a field, if n is prime.
3.)
a) Give a clear, concise definition of a function.
b) Give a clear, concise definition of an injective function.
c) Give a clear, concise definition of a surjective function.
November 3rd, 2010 (test solutions)
MAE-501 Course Notes
Notes by Kevin Maritato
d) Give an example of a function studied in high school mathematics that is injective, but
that does not subject onto the real numbers.
e) Give an example of a function studied in high school mathematics that surgects onto the
real numbers but is not injective.
a.) A function, f(x): A → B, where A and B are sets, maps each 𝑎 ∈ 𝐴 to a single 𝑏 ∈ 𝐵
b.) A function, f(x): A → B, is called injective if f(a) = f(c) = b, 𝑎, 𝑐 ∈ 𝐴, 𝑏 ∈ 𝐵 iff a = c.
c.) A function, f(x): A → B, is called surjective if, for any 𝑏 ∈ 𝐵, there exists an 𝑎 ∈ 𝐴, such
that f(a) = b.
d.)
f(x): R→ R
𝑥 − 2,
𝑓(𝑥) = {
𝑥 + 2,
injectivity:
let f(a) = f(b)
if a,b ≤ 0
a–2=b–2
a=b
if a,b > 0
a+2=b+2
a=b
𝑥≤0
𝑥>0
November 3rd, 2010 (test solutions)
MAE-501 Course Notes
Notes by Kevin Maritato
if a ≤ 0, and b > 0
a–2=b+2
a=b+4
but, a < b, so this case cannot occur
if a > 0, and b ≤ 0
a+2=b–2
a=b–4
but, a > b, so this case cannot occur.
Since f(a) = f(b) iff a = b, the function is injective.
surjectivity:
Suppose that there exists 𝑐 ∈ 𝑅 , such that f(c) = 0
Then, if c > 0, f(c) = c + 2,
so c + 2 = 0
c = -2
but, c > 0, so this cases cannot occur.
if c ≤ 0, f(c) = c – 2,
so c – 2 = 0
November 3rd, 2010 (test solutions)
MAE-501 Course Notes
Notes by Kevin Maritato
c=2
but, c ≤ 0, so this cannot occur.
Therefore, there is no 𝑐 ∈ 𝑅, such that f(c) = 0, but 0 ∈ 𝑅, so the function is not surjective.
d.)
f(x): R→ R
𝑥 + 1,
𝑥 < −1
𝑓(𝑥) = {0, − 1 ≤ 𝑥 ≤ 1
𝑥 − 1,
𝑥>1
injectivity: f(-1) = f(0) = 0, but -1 ≠ 0, so the function is not injective.
surjectivity:
let 𝑦 ∈ 𝑅
then, y < 0, or y = 0, or y > 0
if y < 0,
y – 1 < 0, and 𝑦 − 1 ∈ 𝑅
let x = y – 1
f(x) = (y – 1) + 1
f(x) = y
if y = 0
0∈𝑅
November 3rd, 2010 (test solutions)
MAE-501 Course Notes
Notes by Kevin Maritato
f(0) = 0
f(0) = y
if y > 0
y + 1 > 0, and 𝑦 + 1 ∈ 𝑅
let x = y + 1
f(x) = (y + 1) – 1
f(x) = y
Therefore, for any 𝑦 ∈ 𝑅, there exists an 𝑥 ∈ 𝑅, such that f(x) = y.
Therefore, the function is surjective.
4.) Suppose a student asserts the following equalities on a homework assignment:
cos−1 (cos(𝑥)) = 𝑥
cos(cos−1(𝑥)) = 𝑥
Write a few paragraphs explaining how you would assess and respond to this student’s work.
The second statement is correct, since the inverse cosine of x will map to the angle
between 0 and π whose cosine is x, and the cosine of the angle between 0 and π whose cosine is
x, is x by definition.
November 3rd, 2010 (test solutions)
MAE-501 Course Notes
Notes by Kevin Maritato
The first statement is not correct for all x, as the student assumes it will be. For example,
cos−1 (cos(0)) = cos −1 (1) = 0
but, cos−1 (cos(2𝜋)) = cos−1 (1) = 0
and, 2π ≠ 0, despite the fact that 2π and 0 have the same cosine.
The cosine function is not injective, that is, there are multiple angles with the same cosine
(particularly, for two angles, x and y, cos(x) = cos(y) iff x = n*2π + y, where 𝑛 ∈ 𝑍). The range
of the inverse cosine function, however, only includes the values between 0 and π, and maps any
value, x, in its domain (between -1 and 1) to the angle between 0 and π that has x as its cosine.
This allows for the situation above, where two different x’s have the same inverse cosine, and
therefore give the same value when plugged into cos(cos−1(𝑥)), which will be equal to x if, and
only if 0 ≤ x ≤ π.
5.)
a) Write our definition of rational numbers.
b) Write our analogous definition of rational functions.
c) Give an example of two rational functions which are, according to our definition,
algebraically equivalent, but which are not equal as functions. Explain!
November 3rd, 2010 (test solutions)
MAE-501 Course Notes
Notes by Kevin Maritato
d) Give an example of a rational function that has a horizontal asymptote at y = b,
for some 𝑏 ∈ 𝑅, and such that, for some a in the domain of f, f(a) = b. Include a
clear graph of the function, appropriately scaled, on labeled coordinate axes.
𝑎
𝑎
a.) Q is the set of numbers of the form 𝑏, 𝑎, 𝑏 ∈ 𝑍, b ≠ 0, where 𝑏 ~
𝑎′
if ab’ = a’b
𝑏′
𝑓(𝑥)
b.) Q(x) is the set of functions of the form 𝑔(𝑥), 𝑓(𝑥), 𝑔(𝑥) ∈ 𝑄[𝑥], g(x) ≠ 0, where
𝑓(𝑥)
𝑔(𝑥)
~
𝑖(𝑥)
𝑗(𝑥)
if f(x)*j(x) = i(x)*g(x).
c.) Let f(x) = x2 – 1, g(x) = x + 1, i(x) = x – 1, and j(x) = 1.
f(x)*j(x) = (x2 – 1)(1), i(x)*g(x) = (x – 1)(x + 1)
f(x)*j(x) = (x – 1)(x + 1)(1), i(x)*g(x) = (x – 1)(x + 1)
f(x)*j(x) = (x – 1)(x + 1) = i(x)*g(x) so,
𝑓(𝑥)
𝑖(𝑥)
~
𝑔(𝑥) 𝑗(𝑥)
𝑓(𝑥)
𝑖(−1)
However, 𝑔(𝑥) is undefined at x = -1, since g(-1) = 0, and 𝑗(−1) =
𝑓(𝑥)
−2, so 𝑔(𝑥) ≠
d.) 𝑓(𝑥) =
𝑥−2
𝑥2− 3
−1 − 1
1
1
= 0.
−2
1
=
𝑖(𝑥)
𝑗(𝑥)
has a horizontal asymptote at y = 0, since lim
𝑥−2
𝑥→∞ 𝑥 2 − 3
= 0 (because x2
grows significantly faster than x, as x approaches infinity), but 𝑓(2) =
0
=
2−2
4−3
=
MAE-501 Course Notes
Notes by Kevin Maritato
November 3rd, 2010 (test solutions)