#import("@local/dtu-template:0.5.1"):*

#show: dtu-note.with(
  course: "01017",
  course-name: "Discrete Mathematics",
  title: "Polynomials and the Extended Euclidean Algorithm - Exercises",
  date: datetime(year: 2025, month: 12, day: 4),
  author: "Rasmus Rosendahl-Kaa",
  semester: "Fall 2025"
)

= Exercise 6.16
Let $p(x) = sum^n_(k=0) c_k x^k$ be a polynomial if the coefficients $c_0, dots , c_n$ are all integers where $c_0 eq.not 0$ as well as $c_n eq.not 0$.
Let $QQ$ denominate the set of rational number, meaning fractions with integers in the numerator and the denominator. Then the following theorem is true:

If $a/b in QQ$ with $"gcd"(a,b) = 1$, and if $p(a/b)=0$, then it is true that $a | c_0$ and $b | c_n$.

/ a): Show by the help of the above that the polynomial $p(x)=x^2-2$ does not have any rational roots.

#solution()[
For $a | c_0$ and $b | c_n$ to be true, $a = 1,2$ and $b = 1$.

If $p(x)$ has integer roots, they would divide $c_0 "and" c_n$

Here, we already have the only numbers that can divide $c_0 "and" c_n$. But neither $P(2/1) "nor" P(1/1)$ equals 0, so that must mean $p(x)$ does not have any rational roots. Only $a=b=1$ would have $"gcd"(a,b)=1$, but $p(1/1)$ still doesn't equal 0, so that must mean that $p(x)$ doesn't have rational roots.
]

/ b): Conclude that $sqrt(2) in.not QQ$

#solution()[
$p(x)$ has roots: $-sqrt(2)$ and $sqrt(2)$. Because we have what values $a,b$ ($a=b=1$) could be in $p(x)$, we can see that $1/1 in QQ$, $"gcd"(1,1)=1$, and that $a | c_0$ and $b | c_n$.

But since we know that $sqrt(2)$ and $-sqrt(2)$ are roots, this means $p(sqrt(2))=p(-sqrt(2))=0$. But since we have that $1/1 eq.not sqrt(2)$, we can conclude that $sqrt(2) in.not QQ$.
]

/ c): Conclude in a similar fashion that $sqrt(5) in.not QQ$

#solution()[
We can observe a similar equation: $p(x)=x^2-5$. We can also observe that for the given theorem, then $a=b=1$ are the only value that they could have. But for similar reasons as before, $1/1 eq.not sqrt(5)$, so $sqrt(5) in.not QQ$.
]

/ d): Is it possible that $sqrt(5) − sqrt(2) ∈ QQ$? We actually do not know that yet. Show that $sqrt(5) - sqrt(2)$ is a root of the polynomial $q(x) = x^4 − 14x^2 + 9$. Show that$sqrt(5) - sqrt(2) in.not QQ$.

#solution()[
$
p(sqrt(5)-sqrt(2)) = (sqrt(5)-sqrt(2))^4 - 14 dot (sqrt(5)-sqrt(2))^2 + 9\
$

$
(sqrt(5)-sqrt(2)) dot (sqrt(5)-sqrt(2)) &= sqrt(5)^2+sqrt(2)^2-2 dot sqrt(5) dot sqrt(2)\
&=5+2-2 dot sqrt(10) = 7 - 2 sqrt(10)
$

$
p(sqrt(5)-sqrt(2)) &= (7 - 2 sqrt(10)) dot (7 - 2 sqrt(10)) - 14 dot (7 - 2 sqrt(10)) + 9\
&= 49 + 4 sqrt(10)^2 - 28 sqrt(10) - 14 dot 7 + 28 sqrt(10) + 9\
&= 49+40-14 dot 7+9\
&=98 - 98 = 0
$
$sqrt(5)-sqrt(2)$ is a root.

The only $a,b$ we can have are $a=b=1$. 1 is not a root:
$
q(1)=1^4-14+9 = -4
$
Therefore $sqrt(5)-sqrt(2) in.not QQ$
]

/ e): (Extra, not in the curriculum) Prove the theorem in the beginning of the exercise. (Tip: Consider $p(a/b) = 0$ and multiply by the common denominator, such that all terms are integers. Thereafter use modulus arithmetic.)

= Exercise 6.17
We will examine the execution time of Euclid’s algorithm

/ a): Prove as function of $deg(f(x))$ and $deg(g(x))$, how many iteration Euclid's algorithm uses at most, when it is executed on $f(x)$ and $g(x)$.

#solution()[
Because we in the Euclidean Algorithm have that $deg(R_i) < deg(R_(i-1))$ aka the degree must always go down, and also have that $deg(R_1) < deg(M)$. Then the Euclidean Algorithm must use at most $deg(g(x))$ iterations, where $deg(g(x))<deg(f(x))$
]

/ b): Let $D(n)$ be an upper limit for the number of arithmetic operations it takes to execute a division of $f(x)$ by $g(x)$ with a remainder if $deg(f(x)), deg(g(x)) <= n$. By arithmetic operations we mean $+,−,· "or" "/"$ of elements from the field, thus $RR$ or $CC$. Argue that $D(n) <= 2n^2$.

#solution()[
Imagine for the upper bound that $deg(f(x))=deg(g(x))=n$.


]
/ c):
/ d):

= Exercise 6.18
As usual with fractions, they can be reduced such that given $p(x)/q(x)$ if you by chance know that there exists a $t(x)$ such that $p(x)=t(x)p_1 (x)$ and $q(x)=t(x)q_1 (x)$, then we have:
$
p(x)/q(x)=(t(x)p_1 (x))/(t(x)q_1 (x)) = (p_1(x))/(q_1(x))
$
If you are just given a rational function $p(x)/q(x)$, describe a course of action to calculate the completely reduced fraction.

#solution()[
Calculate the roots of each of the functions, then divide the two. You would be able to remove all their common roots, and be left with a completely reduced fraction
]

= Exercise 6.19
Let $p(x)$ and $d(x)$ be polynomials both different from zero. Assume that $d(x) = d_1(x)d_2(x)$, where $"gcd"(d_1(x), d_2(x)) = 1$, and assume that $deg(p(x)) < deg(d(x))$. Show that there exists polynomials $p_1(x)$ and $p_2(x)$ such that
$
deg(p_1(x)) < deg(d_1(x)) "and" deg(p_2(x)) < deg(d_2(x))
$
and
$
p(x)/d(x)=(p_1(x))/(d_1(x)) + (p_2(x))/(d_2(x))
$
Tip: First multiply the wanted equation by $d(x)$

#solution()[
$
p(x)=p_1(x)d_2(x) + p_2(x)d_1(x)
$
]