THAI NGUYEN UNIVERSITY OF EDUCATION

MATHEMATICS

PROBLEM SEMINAR

Problem Set 1: Proof by contradiction.

Supervisors: PhD. TRAN NGUYEN AN

Author: Group 1

Unit: English for students of mathematics (NO2)

September, 2017

CONTENTS

MEMBER OF GROUP 1:

1. Bùi Thúy Hiền

2. Đoàn Thị Hoa

3. Nguyễn Thị Liên

4. Dương Lan Phương

5. Nguyễn Thị Ngọc Tú

2

1. Introduction

In mathematics, we are interested in objects (e.g. integers, real numbers, sets,

vector spaces, functions, . . . ) and by statements about these objects. To describe an

object, we need a definition, which has, in mathematics, to be unambiguous (for

example, the sentence “a number is called small if it is close to 0” is not a

definition, as it is not possible with this definition to determine if 0.5 is small).

A (mathematical) statement is a sentence or a sequence of mathematical symbols

which has a well defined and unambiguous meaning. It can be true or false. For

example:

•

•

•

•

•

•

“4 = 2 + 2” is a statement (which is true);

“5 = 2 + 7” is a statement (which is false);

“9 is not a prime number” is a statement (which is true);

“3 + 4” is not a statement;

“4 + 5 = 9 +” is not a statement;

“0.5 is small” is not a statement (except if “small” is well defined).

Given statements P and Q, we can form derived statements. The most common are

•

•

•

•

¬P: not P

P ∧ Q: P and Q

P ∨ Q: P or Q

P ⇒ Q: P implies Q

The statement we will be trying to prove will often be either simply “P is true” or

“P implies Q” i.e. we will be trying to prove either P or P ⇒ Q. For the rest of the

section, the statement we wish to prove will be “P implies Q” unless stated

otherwise.

The main aim of mathematicians is to prove theorems. A theorem is a statement

which has been proved to be true. A proof of a statement is a sequence of sentences

with logical connections which ensure logically the truth of the statement. If such a

proof exist, we say that the statement is proved and it becomes a theorem. Here is

an example of a theorem and its (correct) proof:

Theorem 1.1. For three integers a, b and c, if a divides b and b divides c then a

divides c.

Proof: By definition, a divides b means that there exists an integer k such that . In

the same way, b divides c means that there exists an integer such that . We deduce

that . As the product of two integers is an integer, �k is an integer.

Therefore, by definition, a divides c.

3

∎

- divides: chia hết cho

- deduce: kết luận

Note that sentences like “it is obvious” or “everybody knows that” are not proofs

of this theorem.

Usually, we distinguish in a theorem

(i) The hypotheses: these are the conditions which are supposed to be true (for

example, in Theorem 1.1, it is “a, b and c are integers” and “a divides b and b

divides c”);

(ii) the conclusion: in Theorem 1.1, “a divides c”.

Finally, note that the fact that, for some reason, we are not able to find a proof

does not necessarily imply that the statement is false. For example:

Theorem 1.2. If n is an integer greater or equal to 3, then the equation

admits no (strictly) positive integer solution.

(this theorem, known as Fermat’s Last Theorem has been proved by Andrew Wiles

in 1994).

4

2. Proof by contradiction

Proof by contradiction (also known as indirect proof or the method of reductio ad

absurdum) is a common proof technique that is based on a very simple principle:

something that leads to a contradiction can not be true, and if so, the opposite must

be true.

G.H. Hardy (1877-1947) called proof by contradiction "one of a mathematician's

finest weapons" saying, "It is a far finer gambit than any chess gambit: a chess

player may offer the sacrifice of a pawn or even a piece, but a mathematician offers

the game."

2.1. Principle

Proof by contradiction is based on the law of noncontradiction as first formalized

as a metaphysical principle by Aristotle. Noncontradiction is also a theorem

in propositional logic. This states that an assertion or mathematical statement cannot

be both true and false. That is, a proposition Q and its negation ¬

Q ("not-Q") cannot both be true. In a proof by contradiction it is shown that the

denial of the statement being proved results in such a contradiction. It has the form

of a reductio ad absurdum argument. If P is the proposition to be proved:

P is assumed to be false, that is

1.

¬P is true.

It is shown that

2.

¬P implies two mutually contradictory assertions, Q and ¬

Q.

3.

Since Q and

¬Q cannot both be true, the assumption that P is false must be wrong, and P must

be true.

An alternate form derives a contradiction with the statement to be proved itself:

1.

P is assumed to be false.

2.

It is shown that

¬P implies P.

3.

Since P and ¬P cannot both be true, the assumption must be wrong

and P must be true.

5

2.2. Relationship with other proof techniques:

What is the relationship between a "proof by contradiction" and "proof by

contrapositive"? When we compare an exercise, one person proves by contradiction,

and the other proves the contrapositive, the proofs look almost exactly the same.

For example, say I want to prove:

- When I want to prove by contradiction, I would say assume this is not true.

Assume Q is not true, and P is true. Then, this implies P is not true, which is a

contradiction.

- When I want to prove the contrapositive, I say. Assume Q is not true. Then, this

implies P is not true.

Proof by contradiction is closely related to proof by contrapositive, and the two are

sometimes confused, though they are distinct methods. The main distinction is that

a proof by contrapositive applies only to statements of the form P→Q (i.e.,

implications), whereas the technique of proof by contradiction applies to statements

¬Q of any form

In the case where the statement to be proven is an implication , let us look at the

differences between direct proof, proof by contrapositive, and proof by

contradiction:

• Direct proof: assume P and show Q.

• Proof by contrapositive: assume and show

This corresponds to the equivalence

•

Proof by contradiction: assume P and and derive a contradiction.

This corresponds to the equivalences

P → Q ≡ ¬¬ ( P → Q) ≡ ¬ ( P → Q) →⊥≡ ( P ∧ ¬Q) →⊥

2.3. Examples:

2.3.1. Arithmetic (Số học):

Proof by contradiction is common in Arithmetic because many proofs require

some kind of binary choice between possibilities.

Example 1: If a and b are consecutive integers, then the sum a + b is odd.

Proof: Assume that a and b are consecutive integers. Assume also that the sum a +

b is not odd.

Because the sum a + b is not odd, there exists no number k such that

6

However, the integers a and b are consecutive, so we may write the sum a + b as

2a +1.

Thus, we have derived that for any integer k and also that

This is a contradiction. If we hold that a and b are consecutive then we know that

the sum a + b must be odd.

∎

- consecutive integers : số nguyên kề nhau

- exist /ig'zist/: tồn tại

- derive: suy ra

Example 2: There do not exist integers m and n such that 14m + 21n = 100.

Proof: Suppose 14m + 21n = 100.

Since 14m +21n = 7(2m + 3n), we have that 7 divides 100.

This is not true, so the hypothesis 14m + 21n = 100 is not true.

∎

- hypothesis: giả thiết

Example 3: There is no rational number (fraction) a such that . That is: is not a

fraction.

Proof: Since a is a fraction, we can find integers n and m such that , and n and m

are not both even. We have .

Then , so is even. Thus n is even, or n = 2p for some integer p.

Thus or . Therefore is even, so m is even.

Hence m and n are both even, which contradicts the assumption that n and m are

not both even.

Class prove: is not a fraction.

∎

Example 4: Suppose n is a positive integer. If n is odd then n + 1 is even.

Proof: Suppose n and are both odd. Since 0 is even and , there is an even integer

less than n.

Thus there is a largest even integer 2p such that . Also, since , there is an even

integer larger than . Thus there is a smallest even integer such that . Thus

7

⇒

⇔

⇔

⇔

⇔

Since is the largest even integer smaller than n, . Since is the smallest even

integer greater than . Thus , which is impossible.

∎

Example 5: Prove that there are infinitely many prime numbers.

Proof: Assume there are finitely many prime numbers. Then, we can say that there

are n prime numbers, and we can write them down, in order:

Let be a set of all the prime numbers. Let N be the product of all these primes

plus 1, i.e.

Since , and is the largest prime number, N is not prime.

However, according to the Fundamental Theorem of Arithmetic, N must be

divisible by some prime,This means one of the primes in our list must divide N. In

other words, there exists an integer i with such that divides N.

Since divides both N and the product of all the primes, it must also divide

Since , it is impossible that divides one, so we have a contradiction.

Hence, our assumption that there are finitely many prime numbers must have been

false.

∎

- prime number: số nguyên tố

2.3.2. Algebra (Đại số):

Example 1: Prove that the function cannot have more than one root

In this proof, I’ll use Rolle’s Theorem, which says: If f is continuous on the

interval , diﬀerentiable on the interval (a,b), and then for some

Proof: Suppose on the contrary that has more than one root. Then has at least two

roots. Suppose that a and b are (diﬀerent) roots of with

8

Since f is a polynomial, it is continuous and diﬀerentiable for all x. Since a and b

are roots, I have

By Rolle’s Theorem, for some c such that . However,

Since is a sum of even powers and a positive number (17), it follows that for all

x. This contradicts .

Therefore, does not have more than one root.

∎

- polynomial: đa thức

- contrary: ngược lại

- root: nghiệm

- diﬀerentiable: khả vi

- continuous: liên tục

Example 2: If a, b and c are all odd integers, prove that can't have a rational

solution.

Proof: Assume that a rational number is the solution to with and co-prime. This

means that both of and can't be even. At least one of them has to be odd.

If is indeed a solution to the quadratic, then

Now the right hand side is even, so the left hand side has to be even too. But since

are all odd, that can happen only if both and are even, and it was made very clear

that they are not.

Therefore, a rational number can't be a solution to if and are all odd.

∎

- a rational solution : nghiệm hữu tỉ

- quadratic: bậc hai, toàn phương, phương trình bậc hai

- indeed: quả thực, thực lại là

Example 3: Prove that the harmonic series diverges.

The harmonic series is

9

Proof: Suppose that the harmonic series converges, and consider the following

series:

where the grouping symbols denote the ceiling function.

Each term in this sequence is positive and less than or equal to the corresponding

term in the harmonic series:

Then if the harmonic series converges, this series converges as well.

However, this series does not converge. Grouping the like terms gives a repeated

sum of .

The fact that this series diverges is a contradiction. Therefore, the harmonic series

diverges.

∎

- harmonic: điều hòa

- diverge: phân kì

- converge: hội tụ

2.3.3. Geometry (Hình học):

When proof by contradiction are used for geometry, it often leads to figures that

look absurd. This is to be expected, because a proof by contradiction always begins

with a premise that goes against what is believed to be true.

Example 1: Prove that a line tangent to a circle is perpendicular to the radius of

the circle that contains the point of tangency.

10

Proof: We are given circle O tangent line m and point of tangency A

Suppose that the tangent line is not perpendicular to the radius containing the point

of tangency. Then there exists some other point, B on line m such that

OB ⊥ m

Since A is the point of tangency, B must be outside of the circle. Therefore,

However, is a right triangle, and is the hypotenuse of this triangle. Therefore, .

This contradicts the previous assertion that

Therefore, a line tangent to a circle is always perpendicular to the radius of the

circle that contains the point of tangency.

∎

- tangent: tiếp xúc

- point of tangency: tiếp điểm

- hypotenuse: cạnh huyền

11

Example 2: Given the Pythagorean theorem, prove the converse of the

Pythagorean theorem. That is, if a triangle contains side lengths a, b and c such

that then the triangle is a right triangle.

Proof: Suppose that there exists a triangle with that is not a right triangle. The

triangle can either be acute or obtuse.

Assume that the triangle is acute.

Let point D be the point such that and

is a right triangle with side lengths and . It is given that so by the Pythagorean

theorem, .

12

Note that and are isosceles triangles. Thus, and

It is also apparent that and However, this cannot be true given the previous

assertion about congruent angles.

A similar contradiction arises if one assumes that the triangle is obtuse.

Thus, if a triangle has side lengths a, b and c such that then the triangle is a right

triangle.

∎

- acute: nhọn

- obtuse: tù

- congruent angles: các góc bằng nhau

Example 3: Prove that if two angles of a triangle are congruent, then the sides

opposite them are congruent.

Proof: Suppose that there exists a triangle with two congruent angles, but the sides

opposite those angles are not congruent. If the sides are not congruent, then one of

them must be longer.

In the triangle above, and .

Since is longer than let D be placed on such that

13

Now these congruences can be observed:

•

•

•

, which is given by how point D was defined.

, also given.

, by the reflexive property of congruence.

These congruences suggest that by SAS triangle congruence. However, this

cannot be true, because is clearly larger.

Hence, if two angles of a triangle are congruent, then the sides opposite them are

congruent.

∎

- the reflexive property of congruence: tính phản xạ của sự đồng dạng.

2.3.4. Combinatorics (Tổ hợp):

Example: Pigeonhole Principle: 5 points are placed within a unit equilateral

triangle. Prove that two of those points must be a maximum distance of from each

other.

14

Proof: Suppose that for any pair of points among the 5, the distance between the

points is greater than . Now consider the partition of the triangle into 4 smaller

equilateral triangles.

Note that the maximum distance between two points within one of these smaller

triangles is .

A point could be placed within each of the triangles (the 4 blue points) such that

each point is further than from the other points. However, the 5 th point (the red one)

would have to be placed within the same triangle as another point.

15

Thus, if 5 points are placed within a unit equilateral triangle, two of those points

must be at most away from each other.

∎

Example 2: Ramsey's Theorem:

Prove that out of a party of 6 people, there exists a group of 3 mutual friends or a

group of 3 mutual non-friends.

Proof: Suppose that given any group of 3 people among the 6, there are at most 2

friendships or 2 non-friendships. Let the 6 people be labeled with . The possible

relationships are shown with dashed lines below.

16

Begin with person A. Suppose that A is friends with at least 3 people, and suppose

3 of those people are B, C and D.

Let a friendship be denoted with a blue line, and let a non-friendship be denoted

with a red line. If any pair of B, C or D are friends, then they form a group of 3

mutual friends with A.

If B, C and D are all non-friends, then they form a group of 3 mutual non-friends.

Without loss of generality, this same contradiction arises with any combination of

4 people including A.

17

When you consider what happens when A has at least 3 non-friends among the

group, a similar contradiction arises. Note that A must have at least 3 friends or have

at least 3 non-friends.

Therefore, given any party of 6 people, there exists a group of 3 mutual friends or

a group of 3 mutual non-friends.

∎

18

3. Conclusion

Proof by contradiction isn't very useful for proving formulas or equations. Proof

by contradiction needs a specific alternative to whatever you are trying to prove.

For example, you wouldn't use proof by contradiction to prove the quadratic

formula. There isn't any specific alternative equation to the quadratic equation, so

proof by contradiction doesn't help to prove it.

However, proof by contradiction can sometimes be used to prove the converse of a

formula or equation. The proof of the converse of the Pythagorean theorem is an

example of this.

References

[1] G. H. Hardy, A Mathematician's Apology; Cambridge University Press, 1992.

ISBN 9780521427067. PDF p.19.

[2] S. M. Cohen, "Introduction to Logic", Chapter 5 "proof by contradiction ...

Also called indirect proof or reductio ad absurdum ..."

[3] http://www.math.nagoya-u.ac.jp/~demonet/semi/1contradiction.pdf

[4] https://brilliant.org/wiki/contradiction/

19

MATHEMATICS

PROBLEM SEMINAR

Problem Set 1: Proof by contradiction.

Supervisors: PhD. TRAN NGUYEN AN

Author: Group 1

Unit: English for students of mathematics (NO2)

September, 2017

CONTENTS

MEMBER OF GROUP 1:

1. Bùi Thúy Hiền

2. Đoàn Thị Hoa

3. Nguyễn Thị Liên

4. Dương Lan Phương

5. Nguyễn Thị Ngọc Tú

2

1. Introduction

In mathematics, we are interested in objects (e.g. integers, real numbers, sets,

vector spaces, functions, . . . ) and by statements about these objects. To describe an

object, we need a definition, which has, in mathematics, to be unambiguous (for

example, the sentence “a number is called small if it is close to 0” is not a

definition, as it is not possible with this definition to determine if 0.5 is small).

A (mathematical) statement is a sentence or a sequence of mathematical symbols

which has a well defined and unambiguous meaning. It can be true or false. For

example:

•

•

•

•

•

•

“4 = 2 + 2” is a statement (which is true);

“5 = 2 + 7” is a statement (which is false);

“9 is not a prime number” is a statement (which is true);

“3 + 4” is not a statement;

“4 + 5 = 9 +” is not a statement;

“0.5 is small” is not a statement (except if “small” is well defined).

Given statements P and Q, we can form derived statements. The most common are

•

•

•

•

¬P: not P

P ∧ Q: P and Q

P ∨ Q: P or Q

P ⇒ Q: P implies Q

The statement we will be trying to prove will often be either simply “P is true” or

“P implies Q” i.e. we will be trying to prove either P or P ⇒ Q. For the rest of the

section, the statement we wish to prove will be “P implies Q” unless stated

otherwise.

The main aim of mathematicians is to prove theorems. A theorem is a statement

which has been proved to be true. A proof of a statement is a sequence of sentences

with logical connections which ensure logically the truth of the statement. If such a

proof exist, we say that the statement is proved and it becomes a theorem. Here is

an example of a theorem and its (correct) proof:

Theorem 1.1. For three integers a, b and c, if a divides b and b divides c then a

divides c.

Proof: By definition, a divides b means that there exists an integer k such that . In

the same way, b divides c means that there exists an integer such that . We deduce

that . As the product of two integers is an integer, �k is an integer.

Therefore, by definition, a divides c.

3

∎

- divides: chia hết cho

- deduce: kết luận

Note that sentences like “it is obvious” or “everybody knows that” are not proofs

of this theorem.

Usually, we distinguish in a theorem

(i) The hypotheses: these are the conditions which are supposed to be true (for

example, in Theorem 1.1, it is “a, b and c are integers” and “a divides b and b

divides c”);

(ii) the conclusion: in Theorem 1.1, “a divides c”.

Finally, note that the fact that, for some reason, we are not able to find a proof

does not necessarily imply that the statement is false. For example:

Theorem 1.2. If n is an integer greater or equal to 3, then the equation

admits no (strictly) positive integer solution.

(this theorem, known as Fermat’s Last Theorem has been proved by Andrew Wiles

in 1994).

4

2. Proof by contradiction

Proof by contradiction (also known as indirect proof or the method of reductio ad

absurdum) is a common proof technique that is based on a very simple principle:

something that leads to a contradiction can not be true, and if so, the opposite must

be true.

G.H. Hardy (1877-1947) called proof by contradiction "one of a mathematician's

finest weapons" saying, "It is a far finer gambit than any chess gambit: a chess

player may offer the sacrifice of a pawn or even a piece, but a mathematician offers

the game."

2.1. Principle

Proof by contradiction is based on the law of noncontradiction as first formalized

as a metaphysical principle by Aristotle. Noncontradiction is also a theorem

in propositional logic. This states that an assertion or mathematical statement cannot

be both true and false. That is, a proposition Q and its negation ¬

Q ("not-Q") cannot both be true. In a proof by contradiction it is shown that the

denial of the statement being proved results in such a contradiction. It has the form

of a reductio ad absurdum argument. If P is the proposition to be proved:

P is assumed to be false, that is

1.

¬P is true.

It is shown that

2.

¬P implies two mutually contradictory assertions, Q and ¬

Q.

3.

Since Q and

¬Q cannot both be true, the assumption that P is false must be wrong, and P must

be true.

An alternate form derives a contradiction with the statement to be proved itself:

1.

P is assumed to be false.

2.

It is shown that

¬P implies P.

3.

Since P and ¬P cannot both be true, the assumption must be wrong

and P must be true.

5

2.2. Relationship with other proof techniques:

What is the relationship between a "proof by contradiction" and "proof by

contrapositive"? When we compare an exercise, one person proves by contradiction,

and the other proves the contrapositive, the proofs look almost exactly the same.

For example, say I want to prove:

- When I want to prove by contradiction, I would say assume this is not true.

Assume Q is not true, and P is true. Then, this implies P is not true, which is a

contradiction.

- When I want to prove the contrapositive, I say. Assume Q is not true. Then, this

implies P is not true.

Proof by contradiction is closely related to proof by contrapositive, and the two are

sometimes confused, though they are distinct methods. The main distinction is that

a proof by contrapositive applies only to statements of the form P→Q (i.e.,

implications), whereas the technique of proof by contradiction applies to statements

¬Q of any form

In the case where the statement to be proven is an implication , let us look at the

differences between direct proof, proof by contrapositive, and proof by

contradiction:

• Direct proof: assume P and show Q.

• Proof by contrapositive: assume and show

This corresponds to the equivalence

•

Proof by contradiction: assume P and and derive a contradiction.

This corresponds to the equivalences

P → Q ≡ ¬¬ ( P → Q) ≡ ¬ ( P → Q) →⊥≡ ( P ∧ ¬Q) →⊥

2.3. Examples:

2.3.1. Arithmetic (Số học):

Proof by contradiction is common in Arithmetic because many proofs require

some kind of binary choice between possibilities.

Example 1: If a and b are consecutive integers, then the sum a + b is odd.

Proof: Assume that a and b are consecutive integers. Assume also that the sum a +

b is not odd.

Because the sum a + b is not odd, there exists no number k such that

6

However, the integers a and b are consecutive, so we may write the sum a + b as

2a +1.

Thus, we have derived that for any integer k and also that

This is a contradiction. If we hold that a and b are consecutive then we know that

the sum a + b must be odd.

∎

- consecutive integers : số nguyên kề nhau

- exist /ig'zist/: tồn tại

- derive: suy ra

Example 2: There do not exist integers m and n such that 14m + 21n = 100.

Proof: Suppose 14m + 21n = 100.

Since 14m +21n = 7(2m + 3n), we have that 7 divides 100.

This is not true, so the hypothesis 14m + 21n = 100 is not true.

∎

- hypothesis: giả thiết

Example 3: There is no rational number (fraction) a such that . That is: is not a

fraction.

Proof: Since a is a fraction, we can find integers n and m such that , and n and m

are not both even. We have .

Then , so is even. Thus n is even, or n = 2p for some integer p.

Thus or . Therefore is even, so m is even.

Hence m and n are both even, which contradicts the assumption that n and m are

not both even.

Class prove: is not a fraction.

∎

Example 4: Suppose n is a positive integer. If n is odd then n + 1 is even.

Proof: Suppose n and are both odd. Since 0 is even and , there is an even integer

less than n.

Thus there is a largest even integer 2p such that . Also, since , there is an even

integer larger than . Thus there is a smallest even integer such that . Thus

7

⇒

⇔

⇔

⇔

⇔

Since is the largest even integer smaller than n, . Since is the smallest even

integer greater than . Thus , which is impossible.

∎

Example 5: Prove that there are infinitely many prime numbers.

Proof: Assume there are finitely many prime numbers. Then, we can say that there

are n prime numbers, and we can write them down, in order:

Let be a set of all the prime numbers. Let N be the product of all these primes

plus 1, i.e.

Since , and is the largest prime number, N is not prime.

However, according to the Fundamental Theorem of Arithmetic, N must be

divisible by some prime,This means one of the primes in our list must divide N. In

other words, there exists an integer i with such that divides N.

Since divides both N and the product of all the primes, it must also divide

Since , it is impossible that divides one, so we have a contradiction.

Hence, our assumption that there are finitely many prime numbers must have been

false.

∎

- prime number: số nguyên tố

2.3.2. Algebra (Đại số):

Example 1: Prove that the function cannot have more than one root

In this proof, I’ll use Rolle’s Theorem, which says: If f is continuous on the

interval , diﬀerentiable on the interval (a,b), and then for some

Proof: Suppose on the contrary that has more than one root. Then has at least two

roots. Suppose that a and b are (diﬀerent) roots of with

8

Since f is a polynomial, it is continuous and diﬀerentiable for all x. Since a and b

are roots, I have

By Rolle’s Theorem, for some c such that . However,

Since is a sum of even powers and a positive number (17), it follows that for all

x. This contradicts .

Therefore, does not have more than one root.

∎

- polynomial: đa thức

- contrary: ngược lại

- root: nghiệm

- diﬀerentiable: khả vi

- continuous: liên tục

Example 2: If a, b and c are all odd integers, prove that can't have a rational

solution.

Proof: Assume that a rational number is the solution to with and co-prime. This

means that both of and can't be even. At least one of them has to be odd.

If is indeed a solution to the quadratic, then

Now the right hand side is even, so the left hand side has to be even too. But since

are all odd, that can happen only if both and are even, and it was made very clear

that they are not.

Therefore, a rational number can't be a solution to if and are all odd.

∎

- a rational solution : nghiệm hữu tỉ

- quadratic: bậc hai, toàn phương, phương trình bậc hai

- indeed: quả thực, thực lại là

Example 3: Prove that the harmonic series diverges.

The harmonic series is

9

Proof: Suppose that the harmonic series converges, and consider the following

series:

where the grouping symbols denote the ceiling function.

Each term in this sequence is positive and less than or equal to the corresponding

term in the harmonic series:

Then if the harmonic series converges, this series converges as well.

However, this series does not converge. Grouping the like terms gives a repeated

sum of .

The fact that this series diverges is a contradiction. Therefore, the harmonic series

diverges.

∎

- harmonic: điều hòa

- diverge: phân kì

- converge: hội tụ

2.3.3. Geometry (Hình học):

When proof by contradiction are used for geometry, it often leads to figures that

look absurd. This is to be expected, because a proof by contradiction always begins

with a premise that goes against what is believed to be true.

Example 1: Prove that a line tangent to a circle is perpendicular to the radius of

the circle that contains the point of tangency.

10

Proof: We are given circle O tangent line m and point of tangency A

Suppose that the tangent line is not perpendicular to the radius containing the point

of tangency. Then there exists some other point, B on line m such that

OB ⊥ m

Since A is the point of tangency, B must be outside of the circle. Therefore,

However, is a right triangle, and is the hypotenuse of this triangle. Therefore, .

This contradicts the previous assertion that

Therefore, a line tangent to a circle is always perpendicular to the radius of the

circle that contains the point of tangency.

∎

- tangent: tiếp xúc

- point of tangency: tiếp điểm

- hypotenuse: cạnh huyền

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Example 2: Given the Pythagorean theorem, prove the converse of the

Pythagorean theorem. That is, if a triangle contains side lengths a, b and c such

that then the triangle is a right triangle.

Proof: Suppose that there exists a triangle with that is not a right triangle. The

triangle can either be acute or obtuse.

Assume that the triangle is acute.

Let point D be the point such that and

is a right triangle with side lengths and . It is given that so by the Pythagorean

theorem, .

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Note that and are isosceles triangles. Thus, and

It is also apparent that and However, this cannot be true given the previous

assertion about congruent angles.

A similar contradiction arises if one assumes that the triangle is obtuse.

Thus, if a triangle has side lengths a, b and c such that then the triangle is a right

triangle.

∎

- acute: nhọn

- obtuse: tù

- congruent angles: các góc bằng nhau

Example 3: Prove that if two angles of a triangle are congruent, then the sides

opposite them are congruent.

Proof: Suppose that there exists a triangle with two congruent angles, but the sides

opposite those angles are not congruent. If the sides are not congruent, then one of

them must be longer.

In the triangle above, and .

Since is longer than let D be placed on such that

13

Now these congruences can be observed:

•

•

•

, which is given by how point D was defined.

, also given.

, by the reflexive property of congruence.

These congruences suggest that by SAS triangle congruence. However, this

cannot be true, because is clearly larger.

Hence, if two angles of a triangle are congruent, then the sides opposite them are

congruent.

∎

- the reflexive property of congruence: tính phản xạ của sự đồng dạng.

2.3.4. Combinatorics (Tổ hợp):

Example: Pigeonhole Principle: 5 points are placed within a unit equilateral

triangle. Prove that two of those points must be a maximum distance of from each

other.

14

Proof: Suppose that for any pair of points among the 5, the distance between the

points is greater than . Now consider the partition of the triangle into 4 smaller

equilateral triangles.

Note that the maximum distance between two points within one of these smaller

triangles is .

A point could be placed within each of the triangles (the 4 blue points) such that

each point is further than from the other points. However, the 5 th point (the red one)

would have to be placed within the same triangle as another point.

15

Thus, if 5 points are placed within a unit equilateral triangle, two of those points

must be at most away from each other.

∎

Example 2: Ramsey's Theorem:

Prove that out of a party of 6 people, there exists a group of 3 mutual friends or a

group of 3 mutual non-friends.

Proof: Suppose that given any group of 3 people among the 6, there are at most 2

friendships or 2 non-friendships. Let the 6 people be labeled with . The possible

relationships are shown with dashed lines below.

16

Begin with person A. Suppose that A is friends with at least 3 people, and suppose

3 of those people are B, C and D.

Let a friendship be denoted with a blue line, and let a non-friendship be denoted

with a red line. If any pair of B, C or D are friends, then they form a group of 3

mutual friends with A.

If B, C and D are all non-friends, then they form a group of 3 mutual non-friends.

Without loss of generality, this same contradiction arises with any combination of

4 people including A.

17

When you consider what happens when A has at least 3 non-friends among the

group, a similar contradiction arises. Note that A must have at least 3 friends or have

at least 3 non-friends.

Therefore, given any party of 6 people, there exists a group of 3 mutual friends or

a group of 3 mutual non-friends.

∎

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3. Conclusion

Proof by contradiction isn't very useful for proving formulas or equations. Proof

by contradiction needs a specific alternative to whatever you are trying to prove.

For example, you wouldn't use proof by contradiction to prove the quadratic

formula. There isn't any specific alternative equation to the quadratic equation, so

proof by contradiction doesn't help to prove it.

However, proof by contradiction can sometimes be used to prove the converse of a

formula or equation. The proof of the converse of the Pythagorean theorem is an

example of this.

References

[1] G. H. Hardy, A Mathematician's Apology; Cambridge University Press, 1992.

ISBN 9780521427067. PDF p.19.

[2] S. M. Cohen, "Introduction to Logic", Chapter 5 "proof by contradiction ...

Also called indirect proof or reductio ad absurdum ..."

[3] http://www.math.nagoya-u.ac.jp/~demonet/semi/1contradiction.pdf

[4] https://brilliant.org/wiki/contradiction/

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## Ngữ pháp tiếng Anh by TK (phần 4)

## Ngữ pháp tiếng Anh by TK (phần 6)

## Ngữ pháp tiếng Anh by TK (phần 7)

## Ngữ pháp tiếng Anh by TK (phần 8)

## Ngữ pháp tiếng Anh by TK (phần 9)

## Ngữ pháp tiếng Anh by TK (phần 10)

## ngữ pháp tiếng Anh19 by TK

## Ex- Tuvung- Unit 10+ KEY by N.T.Nhan

## Ex-U1- E1CB- By N.T.Nhan

## Ex-U2- E10CB by N.T.Nhan

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