# Intro to Math 135

May 02, 2016

### Course Info

Instructor: Mukto Akash
Office/Extension: MC 6508 x36655
Email: mukto.akash@uwaterloo.ca
Office Hours:
Tuesdays noon - 2pm
Thursdays 2 - 3pm
Textbook: Reading, Discovering, and Writing Proofs Version 0.5 (0.42-0.5 work)
Can be Acquired: at SCH as courseware

The content below is heavily influenced on the courseware.

### Notes on Learn

The Course Notes are titled
– Reading, Discovering, and Writing Proofs

Statement:
If n is an integer, then $$n^2 ≥ n$$.

Proof:
Since n is an integer, either n > 1, n = 0 or n < 0.
When n > 1, multiply both sides by n to get $$n^2 > n$$.
When n = 0, we have $$n^2$$ =0,so $$n^2 = n$$.
Finally, when n < 0,we know that $$n^2 >0$$, so $$n^2 > n$$.
In all cases, n2 ≥ n is satisfied.

A statement is a sentence that has a definite state of being true or false.

ex. 2 + 2 = 4 (true)
3 + 2 < 5 (false)

not a statement: “Is 7 = 5?”

## Definitions

Proposition: A mathematical claim that is posed in the form of a statement that needs to be proven true or demonstrated false by a valid argument.

Theorem: A particularly significant proposition.

Lemma: A proposition that is used to help prove the theorem.

Corollary: A proposition that follows a theorem.

Proposition 1:
For every real number x, $$x^2$$ + 1 ≥ 2x

Proof: a series of convincing arguments that leaves no doubt that the stated proposition is true.

The Proof:
Suppose x is a real number.
Therefore, x - 1 must be a real number, and hence
$$(x-1)^2$$ ≥ 0
Expanding the terms on the left gives $$x^2$$ - 2x + 1 ≥ 0. Adding 2x to both sides yields $$x^2$$ + 1 ≥ 2x.

Axiom: A statement that is assumed to be true.

# Flawed Proofs and Thoughts

May 03, 2016

### Flawed Proofs (some examples)

Identify the flaw in the proof
Statement: 1 = 2.
Proof: Let a=b
then $$a^2$$ = ab
then $$a^2−b^2 = ab−b^2$$
then (a+b)(a−b) = b(a−b)
then a+b = b
then 2b = b
hence 2 = 1

In this case, (a-b) can be 0, and you cannot divide both sides by 0. The mistake happens on the third line.

Statement: 4 = 5
−20 = −20
16 − 36 = 25 − 35
$$4^2 −9*4=5^2 −9*5\\ 4^2−2*4*(\frac{9}{2})^2+ 2 = 5^2−2*5*\frac{9}{2}+\frac{9}{2}\\ (4−\frac{9}{2})^2 = (5-\frac{9}{2})^2\\ 4 - \frac{9}{2} = 5 - \frac{9}{2}\\ 4 = 5$$

In this case you cannot get rid of the exponents on both sides. The mistake happens on the fifth line.

# Truth Tables

May 04, 2016

### Negation

Truth Table

A B A $$\land$$ B A $$\lor$$ B
True True True True
True False False True
False True False True
False False False False

The $$\land$$ acts as “AND” and it is False when there is at least one False, and True otherwise.
The $$\lor$$ acts as “OR” and it is True when there is at least one True, and False otherwise.

$$\lnot$$ denotes a “NOT” sign

A B $$\lnot$$A
True True False
True False False
False True True
False False True

## De Morgan’s Laws:

### Law 1:

$$\lnot$$ ($$\lnot$$ A) = A

### Law 2:

($$\lnot$$A)∨($$\lnot$$B) ≡ $$\lnot$$(A ∧ B)
($$\lnot$$A)∧($$\lnot$$B) ≡ $$\lnot$$(A ∨ B)

### Law 3:

(A ∧ B) ∧ C ≡ A ∧ (B ∧ C)
(A ∨ B) ∨ C ≡ A ∨ (B ∨ C)
associativity law

Proving Law 2

Start out with: $$\lnot$$(($$\lnot$$A)∧($$\lnot$$B))
$$\lnot$$(($$\lnot$$A)∧($$\lnot$$B)) ≡ ($$\lnot$$($$\lnot$$A))∨($$\lnot$$($$\lnot$$B))
The ($$\lnot$$A)∧($$\lnot$$B) undergoes negation
≡ (A) ∨ (B)
This yields just (A) ∨ (B)

Knowing Law 1
($$\lnot$$A)∧($$\lnot$$B) ≡ $$\lnot$$($$\lnot$$($$\lnot$$A)∧($$\lnot$$B))) ≡ $$\lnot$$(A ∨ B)

By negating (A) ∨ (B), we can get: ($$\lnot$$A)∧($$\lnot$$B) ≡ $$\lnot$$(A ∨ B)

## Implication

Hypothesis: If … then

Conclusion: then…

A B A $$\Rightarrow$$ B
True True True
True False False
False True True
False False True

If A, then B

Think of this like “If you give me $100, I will give you$200 tomorrow.” The implication is similar to if I keep my promise in this situation.

If the hypothesis is correct (I gave you $100 to begin with), then the implication is true if I give you back$200 tomorrow, or false I accept the money and do not give you back the money.

The hypothesis is false if you do not give me $100. I may give you$200 regardless tomorrow, making the implication true.

The hypothesis can also be false if you do not give me money, and I give you no money. In this case, the implication is also true.

Here are four implications. Only one is false. Identify the false implication.

1. If x is a real number and $$x^2$$ < 0, then $$x^4$$ ≥ 0.
2. If x is a real number and $$x^2$$ < 0 then $$x^4$$ < 0.
3. If n is a positive integer, then $$n^2$$ +13 is not a perfect square.
4. If n is an integer and $$2^{2n}$$ is an odd integer, then $$2^{−2n}$$ is an odd integer.

Only the implication “If n is a positive integer, then $$n^2$$ + 13 is not a perfect square.” is false.

# Divisibility

May 06, 2016

## Divisibility of integers

An integer m divides an integer n, and we write m | n, when there exists an integer k so n = km.

If m | n, then we say that m is a divisor or a factor of n, and that n is a multiple of m or than n is divisible by m.

3 | 6 since 6 = 3 * 2. That is, there exists an integer k such that 6 = 3k.

For all integers a, a | 0 since 0 = 0 * a. This is true for a = 0 as well.

## Transivity of Divisibility

Let a, b, and c be integers. If a | b and b | c, then a | c.

Proof:
Assume a | b and b | c.
Since a | b there exists an integer r so that ra = b.
Since b | c there exists an integer s so that sb = c.
Substituting ra for b in the previous equation, we get (sr)a = c.
Since sr is an integer, a |c.

##Divisibility of Integer Combinations (DIC)

Let a, b, and c be integers. If a | b and a | c, then for any integers x and y, a | (bx + cy)

Since we are proving an implication, not using it we assume the hypothesis is true to demonstrate the conclusion is true. This is called a Direct Proof.

Proof:
Assume a | b and a | c.
Since a | b there exists an integer m so that ma = b
Since a | c there exists an integer j so that ja = c
Let x and y be integers.
bx + cy = (ma)x + (ja)y = a (mx +jy)
Since mx + jy is an integer, it follows the definition of divisibility that a | (bx + cy)

##Proof of Bounds by Divisibility

###Bounds by Divisibility (BBD)

Let a and b be integers, If a | b and b $$\ne$$ 0, then |a| $$\le$$ |b|.

Proof:
Since a | b, there exists an integer k so that b = ka.
Since b $$\ne$$ 0, k $$\ne$$ 0 But if k $$\ne$$ 0, |k| $$\ge$$ 1
Hence |b| = |ka| = |k||a| $$\ge$$ |a|
(You can multiply |k| $$\ge$$ 1 by |a| to get |k||a| $$\ge$$ |a|)

# Negation, Start of Sets

May 09, 2016

### Corollary

Let a and b be integers. If a | b and b | a, then a = ± b.

Can we use Bounds by Divisibility (BBD) if we did not say a $$\ne$$ 0 and b $$\ne$$ 0

The answer to that is we must consider a,b = 0 as a special case, and use BBD for other cases.

Proof:

Case 1: Since 0 | 0 and 0 $$\nmid$$ n for any non-zero integer n,
then when a = 0, b must be zero and vice versa

Case 2:
When a $$\ne$$ 0, and b $$\ne$$ 0
From a | b and b $$\ne$$ 0, we can use BBD to get |a| $$\le$$ |b|
From b | a and a $$\ne$$ 0, we can use BBD to get |b| $$\le$$ |a|
Hence |a| = |b|
Therefore a = ± b

## Negating an Implication

A $$\Rightarrow$$ B and $$\lnot$$A v B are the same

A B A $$\Rightarrow$$ B $$\lnot$$A not A v B
T T T F T
T F F F F
F T T T T
F F T T T

$$\lnot$$(A $$\Rightarrow$$ B) $$\equiv$$ $$\lnot$$(($$\lnot$$A) v B)
$$\equiv$$ (($$\lnot$$($$\lnot$$A)) $$\land$$ ($$\lnot$$ B))
$$\equiv$$ A $$\land$$ ($$\lnot$$ B)

In A then B Implication
A and not B Negated Implication

If a | b and b|c then a|c.
Negated Implication:
a | b and b | c and a $$\nmid$$ c

If a | bc then a|b or a|c.
Negated Implication:
a | bc and a $$\nmid$$ b and a $$\nmid$$ c

If a | b and b $$\ne$$ 0, then|a|≤|b|.
Negated Implication:
a | b and b $$\ne$$ 0 and |a| > |b|

## Divisibility of Integer Combinations.

“If m is an integer and 14 | m,
prove that 7 | 135m + 693”.
Note that 7 $$\nmid$$ 135, so this is not a straight-forward result.

Recall previously: Proposition (Divisibility of Integer Combinations (DIC))
Let a, b and c be integers.
If a | b and a | c, then for any integers x and y, a | (bx + cy).

Proof: Assume 14 | m
Since 7 | 14, by Transivity of Divisibility 7 | m
m = 7k, for some integer k

Now 135m + 693 = 135(7k) + 7(99) = 7(135k + 99)
As 135k + 99 is an integer, 7 | (135m + 693)

## Converse

Given an implication:

If A then B.

the converse of “If A then B” (A $$\Rightarrow$$ B) is “If B then A” (B $$\Rightarrow$$ A)

An implication and its converse are not logically related.

##Sets

Sets are collection of objects, called elements or members

∅ = { }. The empty-set has nothing in it.
{∅} has one element, namely the empty set, so
{∅} ≠ ∅.

$$\mathbb{R}$$ is the set of real numbers. This is essentially the set that contains every possible number we have seen (unless you have seen complex numbers).

$$\mathbb{Q}$$ is the set of rational numbers. These are numbers of the type 2, 1, 5 , 2, etc. The members are fractions, with numerator and denominator both being integers, and denominator cannot be zero.

$$\mathbb{Z}$$ = {…,−3,−2,−1,0,1,2,3,…}, the set of integers (whole numbers)

$$\mathbb{N}$$ = {1,2,3,…}, the set of natural numbers (positive integers). Note that in MATH 135, N starts at 1 and 0 ∉ N. This may be different that what you learn in CS.

# Sets, Set Builder Notation

May 10, 2016

### Set Builder Notation

There are two different styles in writing sets in set builder notation.

## Style 1

S = {x ∈ T : P(x)} or alternatively S = {x ∈ T | P(x)}.
Read as: x is in T such that P(x)is true.

P(x) is called the membership criteria.
P(x) is an example of an open sentence:
a mathematical sentence whose truth value depends on the value of x, e.g., $$x^2$$ −1 = 0(only true when x = ± 1).

For an object n to be in S, both n ∈ T and P(n) must be true.
If n ∈ T or P(n) is false, then n ∉ S.

Example: {n ∈ $$\mathbb{N}$$: n < 1000 ∧ 7|n}
The members are: 7, 14, 21, . . . , 994,
i.e., all positive integers less than 1000 that are divisible by 7.

## Style 2

S = {f(x):x ∈ T, P(x)} or alternatively S = {f (x) | x ∈ T, P(x)}.
Read as: f (x) is in T such that x satisfies P(x).

T has been already defined, and P(x) is still called the membership criteria.

f(x) is an expression in x, e.g., $$x^2$$ − 1.

### Example

all points on a circle of radius 8 centred at the origin

Solution:
S = {(x, y) ∈ $$\mathbb{R}$$ * $$\mathbb{R}$$ : $$x^2$$ + $$y^2$$ = 64} or S = {(8cos$$\Theta$$, 8sin$$\Theta$$) : $$\Theta$$ ∈ $$\mathbb{R}$$, 0 $$\le$$ $$\Theta$$ $$\le$$ 2$$\pi$$ }

# Sets Comparison

May 11, 2016

## Union

S ∪ T = {x | (x ∈ S ) ∨ (x ∈ T )}

## Intersection

S ∩ T = {x | (x ∈ S) ∧ (x ∈ T)}

## Set Difference

Set of all elements in S but not T.

S - T = {x | (x ∈ S) and ∉ T} = {x | (x ∈ S) ∧ (x ∉ T)}

_
S = {x | x ∉ S}

## CartesianProduct

S × T ={ (x,y) |x ∈ S,y ∈ T}

What common sets do the following sets refer to?
A {m ∈ $$\mathbb{Z}$$:2|m} ∪ {2k + 1:k ∈ Z}
B {m ∈ $$\mathbb{Z}$$:2|m} ∩ {2k + 1:k ∈ Z}

A This is the set of all integers

B This is an empty set

A set S is called a subset of a set T, written S ⊆ T, when the implication “if x ∈ S then x ∈ T” is true for every x in the universe of discourse.
To prove S ⊆ T , we prove the implication (x ∈ S) => (x ∈ T).

Prove:

{n ∈ $$\mathbb{N}$$ : 5 | n + 2} ⊆ {2k + 1 : k ∈ $$\mathbb{N}$$}

when n = 8

8 ∈ {n ∈ $$\mathbb{N}$$ : 5 | n + 2}

but 8 ∉ {2k + 1 : k ∈ $$\mathbb{N}$$}

## Converse of DIC (Divisibility of Integer Combinations)

If for any integers x and y, a | (bx + cy)
Then a | b and a | c

Proof: Assume for any integer x,y a | (bx + cy)
then when x = 1, y = 0
a | (b(1) + c(0)), i.e. a | b
Use specific scenarios when it is true
then when x = 0, y = 1
a | (b(0) + c(1)), i.e. a | c

## Converse in Sets

To prove S ⊆ T we prove the implication
x ∈ S and x ∈ T
Two sets S and T are equal i.e. S = T,
S and T have the same elements

[(x ∈ S) ∧ (x ∈ T) AND (x ∈ T) ∧ (x ∈ S)]

i.e. we prove
(1) S ⊆ T
(2) T ⊆ S

# Set Equality and Quantifiers

May 13, 2016

## If and Only If

A B A $$\leftrightarrow$$ B
T T T
T F F
F T F
F F T

When A = B, A $$\leftrightarrow$$ B is True.

## Proof

If S = T, then S ∪ T and S ∩ T

(“$$\rightarrow$$”)

Assume S = T
Let x ∈ S ∩ T, then (x ∈ S) ∧ (x ∈ T)
Hence (x ∈ S) ∧ (x ∈ T)
Therefore (x ∈ S) v (x ∈ T)

Therefore x ∈ S ∪ T, so S ∩ T ⊆ S ∪ T
On the other hand, let y ∈ S ∪ T
Then (y ∈ S) v (y ∈ T)

At least one of y ∈ S or y ∈ T is a true statement. Without loss of generality (WLOG), assume y ∈ S
From assumption S = T, so when y ∈ S, then y ∈ T
Thus S ∪ T ⊆ S ∩ T

(“$$\leftarrow$$”)

S ∪ T = S ∩ T then S = T
Assume S ∪ T = S ∩ T
let (x ∈ S) then (x ∈ S) v (x ∈ T)
Thus x ∈ S ∪ T.
Since S ∪ T = S ∩ T, x ∈ S ∩ T
Hence (x ∈ S) ∧ (x ∈ T)
Thus x ∈ T. Therefore S ⊆ T
Similarly, T ⊆ S
Therefore S = T

## Quantifiers

quantifier, variable, domain, open sentence

For all x ∈ $$\mathbb{R}$$, $$x^2$$ + 1 $$\ge$$ 2x

Universal Quantifer: “For all”
Variable: x
Domain: “$$\mathbb{R}$$”
Open Sentence: $$x^2$$ + 1 $$\ge$$ 2x

For every natural number n, 2$$n^2$$ + 11n + 15 is composite
Proof:
Let n ∈ $$\mathbb{N}$$
2$$n^2$$ + 11n + 15 = (2n + 5)(n + 3)
Since n $$\ge$$ 1, 2n + 5 $$\ge$$ 2 and (n + 3) $$\ge$$ 2
Also: n ∤ 3 $$\ne$$ (2$$n^2$$ + 11n + 15) as 2n + 5 $$\ne$$ 1
as 2n + 5 ∈ $$\mathbb{Z}$$, (n + 3) | (2$$n^2$$ + 11n + 15)
So 2$$n^2$$ + 11n + 15 is composite

$$\forall$$ x ∈ $$\mathbb{Z}$$, a | (bx + c)
Proof:

Pick x = 1
a | (b(1) + c)
a | b + c

In this case we use the substitution method

## The Select Method

Used to Prove $$\forall$$ x ∈ S, P(x):

Select a representative mathematical object x ∈ S. This cannot be a specific object. It has to be a placeholder, (a variable) that the argument would work for any specific member of S.

Then show that the open sentence P would work for any representative x.

For each x ∈ $$\mathbb{R}$$, x < x + 1
Proof:
Let x be a real number.
Therefore (x + 1) - (x) = 1
Since (x + 1) - (x) > 0
x < x + 1 is true
Thus, by select method, x ∈ $$\mathbb{R}$$, x < x + 1 is a true statement

For every natural number n, n $$\le$$ $$n^2$$
Proof:
Let P(x): x $$\le$$ $$n^2$$. We know that the set of natural numbers is given by

$$\mathbb{N}$$ = {1, 2, 3, 4, 5, …}

Let n ∈ $$\mathbb{N}$$.
Since n is a natural number, n $$\ge$$ 1 Multiply both sides of the inequality by n (note: n is positive) to generality

n * n $$\ge$$ n * 1
$$n^2$$ $$\ge$$ n

Thus, P(n) is true. Therefore, by the select method, $$\forall$$ n ∈ $$\mathbb{N}$$, P(n) is a true statement.

## Existential Quantifier

$$\exists$$ x ∈ S, P(x)

Also means:

1. There is an x ∈ S such that P(x) is true
2. P(x) is true for some x ∈ S
3. At least one x ∈ S satisfies P(x)
4. S has an element x such that P(x) is satisfied

## The Construct Method

Used to prove $$\exists$$ x ∈ S, P(x)

$$\exists$$ x ∈ {1,2,3,4}, such that x is even

Proof:
Consider x = 2. Since 2 is an element of the given domain, and 2 is even, therefore the above statement is true.

Quantified statement When True? When False
$$\forall$$ x ∈ S, P(x) P(x) is true for every element in S P(x) is false for at least one element in S
$$\exists$$ x ∈ S, P(x) P(x) is true for at least one element in S P(x) is false for every element in S

# Assuming a Quantified Statement is True

## The Substitution Method

Used to prove $$\forall$$ x ∈ S, P(x)

Substitute an appropriate value of x from S into the open sentence P, and use the fact that P(x) must be true to arrive at the desired conclusion.

If $$\forall$$ x ∈ $$\mathbb{N}$$, n | x then n = 1

The trick here is to use an appropriate value of x from $$\mathbb{N}$$ that helps give the conclusion of n = 1.
Proof:
Assume $$\forall$$ x ∈ $$\mathbb{N}$$, n | x.
Since x ∈ $$\mathbb{N}$$, consider x = 1. By our own assumption n | 1.
We know that the only positive divisor of 1 is 1 itself. Therefore n = 1.

## The Object Method

Used to prove $$\exists$$ x ∈ S, P(x), it true

Use a symbol such as s to denote an element of S for which P(x) is true. Since we do not know exactly which element of S satisfies P, we cannot assign a specific value to x. Instead we use x as a variable.

Let n be an integer. Prove the following statement about n:
If n is of the form 4l + 1 for some positive integer l, then 8 | ($$n^2$$ -1)

$$\exists$$ l ∈ $$\mathbb{Z}$$, such that n = 4l + 1

Our goal is to prove 8 | ($$n^2$$ - 1),
which is equivalent to proving that k ∈ $$\mathbb{Z}$$, 8k = ($$n^2$$ - 1)

Proof: Assume, by the object method, that we can write n = 4l + 1 for some integer l.
let k = $$2l^2$$ + l. Since l is an integer, therefore k must be also an integer.
In that case, $$n^2$$ - 1 = $$(4l + 1)^2$$ - 1 = $$16l^2$$ + 8l + 1 - 1
$$16l^2$$ + 8l = 8($$2l^2$$ + l) = 8k

Thus according to the construct method, we have shown that k ∈ $$\mathbb{Z}$$, 8k = ($$n^2$$ - 1)
Therefore 8 | ($$n^2$$ -1)

May 19, 2016

## Contrapositive

$$\not$$B $$\rightarrow$$ $$\not$$A $$\equiv$$ A $$\rightarrow$$ B

### When to use Contrapositive:

• when the statement Not A or Not B gives useful information
• if A or B have conditions that are only one of two possibilities (ex. odd or even)

A ∧ $$\not$$A must be false. “A ∧ $$\not$$A must be true” is a Contradiction

• when it is difficult to use direct method
• when NOT B gives more useful information than B

May 25, 2016

# Surjection/Onto, One-to-One

June 08, 2016

One-to-One: x has only one y value

Surjection/Onto: y has only one x value

A function f : S → T that is both one-to-one and onto is said to be a bijection

# List of Propositions

June 12, 2016

List Starting with GCD Propositions:

## GCD

The greatest common divisor of integers a and b,
where a $$\ne$$ 0 or b $$\ne$$ 0, is an integer d $$\gt$$ 0 such that

•  d a, d b, and
•  for all c ∈ $$\mathbb{Z}$$, if c a and c | b, then c $$\le$$ d

## GCD with Remainders (GCD WR)

If a,b,q,r are integers such that
a = qb+r, then gcd(a, b) = gcd(b, r)

## GCD Characterization Theorem (GCD CT)

Let a, b be integers, not both zero, and d ∈ $$\mathbb{N}$$

•  If d a and d b,and
• There exist x, y ∈ $$\mathbb{Z}$$ such that d = ax+by
Then d = gcd(a, b).

## EEA (EEA Theorem (a.k.a Bezout’s Lemma))

Let a, b ∈ Z. Then d = gcd(a, b) can be computed, and there exist x , y ∈ $$\mathbb{Z}$$
such that d = ax + by.

## EEA shows converse of GCD CT is true

Let a, b ∈ Z, not both zero, and d ∈ N.
If d = gcd(a, b)
then

•  d a and d b
• There exist x, y ∈ $$\mathbb{Z}$$ such that d = ax+by.

## GCD of One (GCD OO)

Let a and b be integers.
Then gcd(a, b) = 1
if and only if there exist integers x and y with ax + by = 1.

## GCD DB (DB GCD))

Let a and b be integers. If gcd(a, b) = d $$\ne$$ 0, then gcd( $$\frac{a}{d} , \frac{b}{d}$$ ) = 1.

## Coprime

We say integers a and b are coprime if and only if gcd(a, b) = 1.

Let a, b, c ∈ $$\mathbb{Z}$$. If c | ab and gcd( a , c ) = 1, then c | b.

## Primes

An integer p > 1 is said to be a prime if and only if the only positive divisors of p are 1 and p itself.

## Primes and Divisibilty

Let p be a prime and a, b ∈ $$\mathbb{Z}$$.
If p | (ab) then p | a or p | b.

## (Infinitely Many Primes (INF P))

There are infinitely many primes.

## (Prime Factorization (PF))

For every integer n > 1,
n can be written as a product of primes.

## Linear Diophantine Equation 1 (LDE1)

Let a, b ∈ $$\mathbb{Z}$$ and gcd(a, b) = d.
The linear Diophantine equation ax + by = c has a solution iff d | c.

## Linear Diophantine Equation 2 (LDE2)

Let gcd(a,b) = d where a $$\ne$$ 0 and b $$\ne$$ 0 are integers.
If x = $$x_0$$ and y = $$y_0$$ is one particular integer solution to the linear diophantine equation ax + by = c,

## Proposition (DAP)

If f (x), g(x) ∈ F[x] and g(x) is not the zero polynomial, then there exist unique polynomials q(x), r(x) ∈ F[x]
such that f (x) = q(x)g(x) + r(x) with r(x) being the zero polynomial or deg(r(x)) < deg(g(x))

## Remainder Theorem (RT)

Suppose f(x) ∈ F[x] and c ∈ F.
The remainder when the
polynomial f (x) is divided by x − c is f (c).