Proof by contradiction

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

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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.
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∎
- 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).

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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.

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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
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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

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⇒
⇔
⇔
⇔
⇔
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
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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

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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.

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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

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Now these congruences can be observed:
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•
•

, 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.

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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.

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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.

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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.

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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 ..."
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