THAI NGUYEN UNIVERSITY OF EDUCATION MATHEMATICS PROBLEM SEMINAR Problem Set 1: Proof by contradiction Supervisors: PhD TRAN NGUYEN AN Author: Group Unit: English for students of mathematics (NO2) September, 2017 CONTENTS MEMBER OF GROUP 1: Bùi Thúy Hiền Đoàn Thị Hoa Nguyễn Thị Liên Dương Lan Phương Nguyễn Thị Ngọc Tú 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” is a statement (which is true); “5 = + 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 + = +” 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 ơP: 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 ∎ - 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) 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 ¬P is true It is shown that ¬P implies two mutually contradictory assertions, Q and ¬ Q 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: P is assumed to be false It is shown that ¬P implies P Since P and ¬P cannot both be true, the assumption must be wrong and P must be true 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 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 - exist /ig'zist/: tồn - derive: suy Example 2: There 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 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 + is even Proof: Suppose n and are both odd Since 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 ⇒ ⇔ ⇔ ⇔ ⇔ 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 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, tồn phương, phương trình bậc hai - indeed: thực, thực lại Example 3: Prove that the harmonic series diverges The harmonic series is 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: góc 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 đồng dạng 2.3.4 Combinatorics (Tổ hợp): Example: Pigeonhole Principle: 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 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 blue points) such that each point is further than from the other points However, the th point (the red one) would have to be placed within the same triangle as another point 15 Thus, if 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 people, there exists a group of mutual friends or a group of mutual non-friends Proof: Suppose that given any group of people among the 6, there are at most friendships or non-friendships Let the 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 people, and suppose 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 mutual friends with A If B, C and D are all non-friends, then they form a group of mutual non-friends Without loss of generality, this same contradiction arises with any combination of people including A 17 When you consider what happens when A has at least non-friends among the group, a similar contradiction arises Note that A must have at least friends or have at least non-friends Therefore, given any party of people, there exists a group of mutual friends or a group of mutual non-friends ∎ 18 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 "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 ... 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: ... Theorem has been proved by Andrew Wiles in 1994) Proof by contradiction Proof by contradiction (also known as indirect proof or the method of reductio ad absurdum) is a common proof technique that... 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,