Rosen 1pt5 p75. 21. Theorem: “If n is an integer and n + 5 is odd
... 25. Theorem: “The sum of an irrational number and a rational number is irrational.” Proof by contradiction: Let x be irrational, and let r = p/q be rational, and let x + r = s. Suppose the sum s is not irrational, then the difference s – r which equals x must also be rational. This is a contradictio ...
... 25. Theorem: “The sum of an irrational number and a rational number is irrational.” Proof by contradiction: Let x be irrational, and let r = p/q be rational, and let x + r = s. Suppose the sum s is not irrational, then the difference s – r which equals x must also be rational. This is a contradictio ...
3. Recurrence 3.1. Recursive Definitions. To construct a
... (1) Here and in the previous exercise we see the slight variation in the basis step from the ones encountered in Module 3.3 Induction; there may be more than one initial condition to verify before proceeding to the induction step. (2) Notice that in this example as well as in some of the other examp ...
... (1) Here and in the previous exercise we see the slight variation in the basis step from the ones encountered in Module 3.3 Induction; there may be more than one initial condition to verify before proceeding to the induction step. (2) Notice that in this example as well as in some of the other examp ...
HILBERT`S TENTH PROBLEM: What can we do with Diophantine
... where D is a polynomial with integer coecients. These equations were named after greek mathematician Diophantus who lived in the 3rd century A.C. So Hilbert's tenth problem can be also viewed as a collection of problems but there are essential di erences with the other problems: First, these indi ...
... where D is a polynomial with integer coecients. These equations were named after greek mathematician Diophantus who lived in the 3rd century A.C. So Hilbert's tenth problem can be also viewed as a collection of problems but there are essential di erences with the other problems: First, these indi ...
Prime Numbers are Infinitely Many: Four Proofs from
... crucial point is that they were proposed in very different historical and socio-cultural contexts. So the main question is the following: is it correct to consider the quoted proofs as four different proofs of the same theorem? In our opinion the answer is: no. The celebrated proposition according t ...
... crucial point is that they were proposed in very different historical and socio-cultural contexts. So the main question is the following: is it correct to consider the quoted proofs as four different proofs of the same theorem? In our opinion the answer is: no. The celebrated proposition according t ...
lecture notes on mathematical induction
... the methods of infinite descent and minimal counterexample, which we shall discuss later – and the technique seems to have become commonplace by the end of the 18th century. Not having an formal understanding of the relationship between mathematical induction and the structure of the natural numbers ...
... the methods of infinite descent and minimal counterexample, which we shall discuss later – and the technique seems to have become commonplace by the end of the 18th century. Not having an formal understanding of the relationship between mathematical induction and the structure of the natural numbers ...
continued fractions - University of Hawaii Mathematics
... However, we require that an > 1, where an is the last element of a finite continued fraction. Then the answer is ”yes”. Exercise 1. Prove that under the assumption an > 1 the continued fraction representation given in Proposition 1 is unique. In other words, the correspondence between • finite conti ...
... However, we require that an > 1, where an is the last element of a finite continued fraction. Then the answer is ”yes”. Exercise 1. Prove that under the assumption an > 1 the continued fraction representation given in Proposition 1 is unique. In other words, the correspondence between • finite conti ...
Formal methods: lecture notes no
... The first and quite familiar way to look at functions is to examine their graphical representations. With every ordered pair (x, y) in the relation f we will associate a point in the Cartesian plane. This view is believed to be first suggested by Derichlet. However, as stated earlier, today’s lectur ...
... The first and quite familiar way to look at functions is to examine their graphical representations. With every ordered pair (x, y) in the relation f we will associate a point in the Cartesian plane. This view is believed to be first suggested by Derichlet. However, as stated earlier, today’s lectur ...
Running Time of Euclidean Algorithm
... Proof. Base cases (both needed as can’t apply induction step on f1 since f-1 is undefined) n = 0: f0 = 0 < 1 = 20 n = 1: f1 = 1 < 2 = 21 Induction n > 0: fn = fn-1+fn-2 < 2n-1 + 2n-2 applying both P (n-1) and P (n-2) which can be assumed by strong induction hypothesis. Doing more algebra: 2n-1+2 ...
... Proof. Base cases (both needed as can’t apply induction step on f1 since f-1 is undefined) n = 0: f0 = 0 < 1 = 20 n = 1: f1 = 1 < 2 = 21 Induction n > 0: fn = fn-1+fn-2 < 2n-1 + 2n-2 applying both P (n-1) and P (n-2) which can be assumed by strong induction hypothesis. Doing more algebra: 2n-1+2 ...
Induction
... L-shapes subject to conditions 1′ and 2′ . A 1 × 1 grid consists of only one square. This square is s, and we can leave it untiled. Since there are no other squares in the grid to tile, this gives us a valid tiling of a 1 × 1 grid. Now we prove the inductive step using a direct proof. Let n ∈ N and ...
... L-shapes subject to conditions 1′ and 2′ . A 1 × 1 grid consists of only one square. This square is s, and we can leave it untiled. Since there are no other squares in the grid to tile, this gives us a valid tiling of a 1 × 1 grid. Now we prove the inductive step using a direct proof. Let n ∈ N and ...
Arbitrarily Large Gaps Between Primes - PSU Math Home
... Using the V.I.F. that if d | a and d | b , then d | (a+b) , we know that the first number in that list is divisible by 2, the second number is divisible by 3, the third number is divisible by 4, … and the last number is divisible by n+1. So all of these numbers are non‐primes! And there ...
... Using the V.I.F. that if d | a and d | b , then d | (a+b) , we know that the first number in that list is divisible by 2, the second number is divisible by 3, the third number is divisible by 4, … and the last number is divisible by n+1. So all of these numbers are non‐primes! And there ...
Mathematical Induction - Penn Math
... For more examples of the constructive uses of induction, including the FFT, Quicksorting, finding maximum independent sets in graphs, etc., you can download Chapter 2 of Algorithms and Complexity from ...
... For more examples of the constructive uses of induction, including the FFT, Quicksorting, finding maximum independent sets in graphs, etc., you can download Chapter 2 of Algorithms and Complexity from ...
Outline for Chapter 10
... This is an outline of the material that I have covered (or will cover) from Chapter 10. It is almost all in the book, but I am skipping around a bit so I thought it would help to collect the relevant material here. I added a few things that I didn’t have time to discuss in the lecture but which are ...
... This is an outline of the material that I have covered (or will cover) from Chapter 10. It is almost all in the book, but I am skipping around a bit so I thought it would help to collect the relevant material here. I added a few things that I didn’t have time to discuss in the lecture but which are ...
Chapter 1-sec1.1
... The contrapositive of this statement is “if x is even, then x2 is even,” which is true by Example 5. QED. Example 7 If x2 is even then x is even. Proof: This is the contrapositive of Example 4, which has been shown to be true. QED. If And Only If (Iff) Proofs A statement of the form “A if and only i ...
... The contrapositive of this statement is “if x is even, then x2 is even,” which is true by Example 5. QED. Example 7 If x2 is even then x is even. Proof: This is the contrapositive of Example 4, which has been shown to be true. QED. If And Only If (Iff) Proofs A statement of the form “A if and only i ...
Lecture 4 - CSE@IIT Delhi
... Question: If a and b are irrational, can ab be rational?? We know that √2 is irrational, what about √2√2 ? Case 1: √2√2 is rational Then we are done, a=√2, b=√2. Case 2: √2√2 is irrational Then (√2√2)√2 = √22 = 2, a rational number So a=√2√2, b= √2 will do. So in either case there are a,b irrational ...
... Question: If a and b are irrational, can ab be rational?? We know that √2 is irrational, what about √2√2 ? Case 1: √2√2 is rational Then we are done, a=√2, b=√2. Case 2: √2√2 is irrational Then (√2√2)√2 = √22 = 2, a rational number So a=√2√2, b= √2 will do. So in either case there are a,b irrational ...
What is a proof? - Computer Science
... Kemp gave a proof that was deemed false 11 years after it was published! His proof, however, contains the essential ideas that were used in subsequent proofs. In our case, we will not learn much from a false proof now, but it will give some insight about the nature of what a proof really is. Conside ...
... Kemp gave a proof that was deemed false 11 years after it was published! His proof, however, contains the essential ideas that were used in subsequent proofs. In our case, we will not learn much from a false proof now, but it will give some insight about the nature of what a proof really is. Conside ...
Erd˝os`s proof of Bertrand`s postulate
... a very strong conjecture of Cramér [2] would imply we could shorten it to (n, n+(1+) ln2 n]. Here is a very lovely open question much in the spirit of Bertrand’s postulate. Question 3.1 Is it true that for all n ≥ 2, there is always a prime p with n2 < p < (n + 1)2 ? As mentioned in the introducti ...
... a very strong conjecture of Cramér [2] would imply we could shorten it to (n, n+(1+) ln2 n]. Here is a very lovely open question much in the spirit of Bertrand’s postulate. Question 3.1 Is it true that for all n ≥ 2, there is always a prime p with n2 < p < (n + 1)2 ? As mentioned in the introducti ...
Diophantine Representation of the Fibonacci Numbers
... This completes the proof of the theorem. (Putnam's method [10] would produce a polynomial of degree 9.) The polynomial (1), which represents the set of Fibonacci numbers, assumes in addition certain negative values such as - 2 8 (x = 2, y = 21 The appearance of non-Fibonacci numbers cannot be preven ...
... This completes the proof of the theorem. (Putnam's method [10] would produce a polynomial of degree 9.) The polynomial (1), which represents the set of Fibonacci numbers, assumes in addition certain negative values such as - 2 8 (x = 2, y = 21 The appearance of non-Fibonacci numbers cannot be preven ...
Chapter1p3
... Example: Show that if x and y are integers and both x∙y and x+y are even, then both x and y are even. Proof: Use a proof by contraposition. Suppose x and y are not both even. Then, one or both are odd. Without loss of generality, assume that x is odd. Then x = 2m + 1 for some integer k. Case 1: y is ...
... Example: Show that if x and y are integers and both x∙y and x+y are even, then both x and y are even. Proof: Use a proof by contraposition. Suppose x and y are not both even. Then, one or both are odd. Without loss of generality, assume that x is odd. Then x = 2m + 1 for some integer k. Case 1: y is ...
Notes for week 11.
... false for each x in A. That is, the domain of P(x) is a set A, and the range is {true, false}. NOTE: Sometimes propositional function are called predicates. Observe that the set A can be partitioned into two subsets: • Elements with an image that is true. • Elements with an image that is false. In p ...
... false for each x in A. That is, the domain of P(x) is a set A, and the range is {true, false}. NOTE: Sometimes propositional function are called predicates. Observe that the set A can be partitioned into two subsets: • Elements with an image that is true. • Elements with an image that is false. In p ...
Introduction to Discrete Mathematics
... Question: If a and b are irrational, can ab be rational?? We (only) know that √2 is irrational, what about √2√2 ? Case 1: √2√2 is rational Then we are done, a=√2, b=√2. Case 2: √2√2 is irrational Then (√2√2)√2 = √22 = 2, a rational number So a=√2√2, b= √2 will do. So in either case there are a,b irr ...
... Question: If a and b are irrational, can ab be rational?? We (only) know that √2 is irrational, what about √2√2 ? Case 1: √2√2 is rational Then we are done, a=√2, b=√2. Case 2: √2√2 is irrational Then (√2√2)√2 = √22 = 2, a rational number So a=√2√2, b= √2 will do. So in either case there are a,b irr ...