Induction - Mathematical Institute
... Induction, or more exactly mathematical induction, is a particularly useful method of proof for dealing with families of statements which are indexed by the natural numbers, such as the last three statements above. We shall prove both statements B and C using induction (see below and Example 6). Sta ...
... Induction, or more exactly mathematical induction, is a particularly useful method of proof for dealing with families of statements which are indexed by the natural numbers, such as the last three statements above. We shall prove both statements B and C using induction (see below and Example 6). Sta ...
slides03 - Duke University
... • What do we know about y? Only that y is irrational: ¬ integers i,j: y = i/j. • But, it’s difficult to see how to use a direct proof in this case. We could try indirect proof also, but in this case, it is a little simpler to just use proof by contradiction (very similar to indirect). • So, what ar ...
... • What do we know about y? Only that y is irrational: ¬ integers i,j: y = i/j. • But, it’s difficult to see how to use a direct proof in this case. We could try indirect proof also, but in this case, it is a little simpler to just use proof by contradiction (very similar to indirect). • So, what ar ...
Full text
... ax = 130, ah = fc3; ax = 136, a 5 = Z?^. Thereafter, there are no more integer values of a± that yield ak = bk_1. Thus 100, 130, 136, 145, and 190 are the only values of a± (100 < ax < 191) for which {wn(l, a±; 1, -1)} n{w„(100, 191; 1, -1)} 4 0. Also, (y(4 + log 90)) C. ...
... ax = 130, ah = fc3; ax = 136, a 5 = Z?^. Thereafter, there are no more integer values of a± that yield ak = bk_1. Thus 100, 130, 136, 145, and 190 are the only values of a± (100 < ax < 191) for which {wn(l, a±; 1, -1)} n{w„(100, 191; 1, -1)} 4 0. Also, (y(4 + log 90)) C. ...
An Introduction to Discrete Mathematics: how to
... underlying ideas. Sometimes, when a theorem is applied to a problem, it is not the statement of the theorem that is applied, but rather, the main ideas that appear in the proof. So, to apply the theorem, it is not always enough just to know the statement. One must also understand why the theorem hol ...
... underlying ideas. Sometimes, when a theorem is applied to a problem, it is not the statement of the theorem that is applied, but rather, the main ideas that appear in the proof. So, to apply the theorem, it is not always enough just to know the statement. One must also understand why the theorem hol ...
eprint_4_1049_36.doc
... then 3 = 2 + 1, then 4 = 3 + 1, and so on. The principle makes precise the vague phrase “and so on.” Principle of Mathematical Induction: Let S be a set of positive integers with the following two properties: (i) 1 belongs to S. (ii) If k belongs to S, then k + 1 belongs to S. Then S is the set of a ...
... then 3 = 2 + 1, then 4 = 3 + 1, and so on. The principle makes precise the vague phrase “and so on.” Principle of Mathematical Induction: Let S be a set of positive integers with the following two properties: (i) 1 belongs to S. (ii) If k belongs to S, then k + 1 belongs to S. Then S is the set of a ...
Chapter 3 Proof
... Indeed there does. Unravel the proof of any mathematical theorem and eventually you will find yourself with certain unproved assumptions at its foundation. What good is a logical proof if it is based on things we can not prove? After all, if our theorem is proved on the basis of assumptions that are ...
... Indeed there does. Unravel the proof of any mathematical theorem and eventually you will find yourself with certain unproved assumptions at its foundation. What good is a logical proof if it is based on things we can not prove? After all, if our theorem is proved on the basis of assumptions that are ...
The Gödelian inferences - University of Notre Dame
... according to whom the various ways of stating the consistency of S are equivalent regardless of whether S proves such equivalences. Are the conditions sufficient for G2 also sufficient for S to be able to prove all such equivalences? If not, then it is consistent with G2 to assume that S proves its ...
... according to whom the various ways of stating the consistency of S are equivalent regardless of whether S proves such equivalences. Are the conditions sufficient for G2 also sufficient for S to be able to prove all such equivalences? If not, then it is consistent with G2 to assume that S proves its ...
N - 陳光琦
... After player 2 moves, there will then be either 11, 10, or 9 stones left. In any of these cases, player 1 can then reduce the number of stones to 8. Then, player 2 will reduce the number to 7, 6, or 5. Then, player 1 can reduce the number to 4. Then, player 2 must reduce them to 3, 2, or 1. Player 1 ...
... After player 2 moves, there will then be either 11, 10, or 9 stones left. In any of these cases, player 1 can then reduce the number of stones to 8. Then, player 2 will reduce the number to 7, 6, or 5. Then, player 1 can reduce the number to 4. Then, player 2 must reduce them to 3, 2, or 1. Player 1 ...
Proof Methods Proof methods Direct proofs
... • Proof: The only two perfect squares that differ by 1 are 0 and 1 – Thus, any other numbers that differ by 1 cannot both be perfect squares – Thus, a non-perfect square must exist in any set that contains two numbers that differ by 1 – Note that we didn’t specify which one it was! ...
... • Proof: The only two perfect squares that differ by 1 are 0 and 1 – Thus, any other numbers that differ by 1 cannot both be perfect squares – Thus, a non-perfect square must exist in any set that contains two numbers that differ by 1 – Note that we didn’t specify which one it was! ...
An Example of Induction: Fibonacci Numbers
... Theorem 1. The Fibonacci number F5n is a multiple of 5, for all positive integers n. Proof. Proof by induction on n. Since this is a proof by induction, we start with the base case of n = 1. That means, in this case, we need to compute F5×1 = F5 . But it is easy to compute (from the definition) that ...
... Theorem 1. The Fibonacci number F5n is a multiple of 5, for all positive integers n. Proof. Proof by induction on n. Since this is a proof by induction, we start with the base case of n = 1. That means, in this case, we need to compute F5×1 = F5 . But it is easy to compute (from the definition) that ...
Introduction to Logic for Computer Science
... trying to symbolise the whole of mathematics could be disastrous as then it would become quite impossible to even read and understand mathematics, since what is presented usually as a one page proof could run into several pages. But at least in principle it can be done. Since the latter half of the ...
... trying to symbolise the whole of mathematics could be disastrous as then it would become quite impossible to even read and understand mathematics, since what is presented usually as a one page proof could run into several pages. But at least in principle it can be done. Since the latter half of the ...
Section.1.1
... Example 1. There is a prime number between 200 and 220. Proof: Check exhaustively and find that 211 is prime. QED. Example 2. Each of the numbers 288, 198, and 387 is divisible by 9. Proof: Check that 9 divides each of the numbers. QED. Conditional Proof Most statements we prove are conditionals. We ...
... Example 1. There is a prime number between 200 and 220. Proof: Check exhaustively and find that 211 is prime. QED. Example 2. Each of the numbers 288, 198, and 387 is divisible by 9. Proof: Check that 9 divides each of the numbers. QED. Conditional Proof Most statements we prove are conditionals. We ...
Lacunary recurrences for Eisenstein series
... University of Cologne postdoc grant DFG Grant D-72133-G-403-151001011, funded under the Institutional Strategy of the University of Cologne within the German Excellence Initiative. 1Note that for k = 2, we have to fix a certain order of summation to ensure convergence of the defining double series a ...
... University of Cologne postdoc grant DFG Grant D-72133-G-403-151001011, funded under the Institutional Strategy of the University of Cologne within the German Excellence Initiative. 1Note that for k = 2, we have to fix a certain order of summation to ensure convergence of the defining double series a ...
Prove if n 3 is even then n is even. Proof
... • Basic idea is to assume that the opposite of what you are trying to prove is true and show that it results in a violation of one of your initial assumptions. • In the previous proof we showed that assuming that the sum of a rational number and an irrational number is rational and showed that it re ...
... • Basic idea is to assume that the opposite of what you are trying to prove is true and show that it results in a violation of one of your initial assumptions. • In the previous proof we showed that assuming that the sum of a rational number and an irrational number is rational and showed that it re ...
Direct proof and disproof
... in the variable declarations and the hypothesis of the if/then statement) and moved gradually towards the information that needed to be proved (the conclusion of the if/then statement). This is the standard “logical” order for a direct proof. It’s the easiest order for a reader to understand. When w ...
... in the variable declarations and the hypothesis of the if/then statement) and moved gradually towards the information that needed to be proved (the conclusion of the if/then statement). This is the standard “logical” order for a direct proof. It’s the easiest order for a reader to understand. When w ...
i(k-1)
... integers (or a subset like integers larger than 3) that have appropriate self-referential structure— including both equalities and inequalities—using either weak or strong induction as needed. – Critique formal inductive proofs to determine whether they are valid and where the error(s) lie if they a ...
... integers (or a subset like integers larger than 3) that have appropriate self-referential structure— including both equalities and inequalities—using either weak or strong induction as needed. – Critique formal inductive proofs to determine whether they are valid and where the error(s) lie if they a ...
EppDm4_05_04
... by transitivity of divisibility, k + 1 is divisible by the prime number p. Therefore, regardless of whether k + 1 is prime or not, it is divisible by a prime number [as was to be shown]. [Since we have proved both the basis and the inductive step of the strong mathematical induction, we conclude tha ...
... by transitivity of divisibility, k + 1 is divisible by the prime number p. Therefore, regardless of whether k + 1 is prime or not, it is divisible by a prime number [as was to be shown]. [Since we have proved both the basis and the inductive step of the strong mathematical induction, we conclude tha ...
n is even
... With examples from Number Theory (Rosen 1.5, 3.1, sections on methods of proving theorems and fallacies) ...
... With examples from Number Theory (Rosen 1.5, 3.1, sections on methods of proving theorems and fallacies) ...