Section 2.5 - Concordia University
... Exercise 2 Cont... Determine whether each of these sets is finite, countably infinite, or uncountable. For those that are countably infinite, exhibit a one-to-one correspondence between the set of positive integers and that set. d) the real numbers between 0 and 2 Answer: S = (0, 2), S is uncountab ...
... Exercise 2 Cont... Determine whether each of these sets is finite, countably infinite, or uncountable. For those that are countably infinite, exhibit a one-to-one correspondence between the set of positive integers and that set. d) the real numbers between 0 and 2 Answer: S = (0, 2), S is uncountab ...
Proof
... It is also important to discuss why proof by contradiction is useful and the conditions under which it can be most helpfully employed. For instance, when trying to prove a result involving infinity or irrationality it is natural to employ proof by contradiction as it is far easier to deal with finit ...
... It is also important to discuss why proof by contradiction is useful and the conditions under which it can be most helpfully employed. For instance, when trying to prove a result involving infinity or irrationality it is natural to employ proof by contradiction as it is far easier to deal with finit ...
Even Perfect Numbers and Sums of Odd Cubes Exposition by
... How you would derive this We are not going to do a standard proof by induction. Instead we discuss how you might derive this by hand with a minimum of calculation. ...
... How you would derive this We are not going to do a standard proof by induction. Instead we discuss how you might derive this by hand with a minimum of calculation. ...
The Unit Distance Graph and the Axiom of Choice.
... A proof of this theorem is very far beyond the scope of this class. A rough idea, however, is the following: given any set, we can approximate its measure by coming up with a collection of intervals that cover the set, and using this as an upper bound. One theorem you can prove is that if a set has ...
... A proof of this theorem is very far beyond the scope of this class. A rough idea, however, is the following: given any set, we can approximate its measure by coming up with a collection of intervals that cover the set, and using this as an upper bound. One theorem you can prove is that if a set has ...
Induction and Recursion - Bryn Mawr Computer Science
... Base case: [Proof of P (0) goes here.] Induction step: [Proof of ∀n ≥ a(P (n) → P (n + 1)] Definition 4 (Closed Form) If a sum with a variable number of terms is shown to be equal to a formula that does not contain either an ellipsis or a summation symbol, we say that it is written in closed form. E ...
... Base case: [Proof of P (0) goes here.] Induction step: [Proof of ∀n ≥ a(P (n) → P (n + 1)] Definition 4 (Closed Form) If a sum with a variable number of terms is shown to be equal to a formula that does not contain either an ellipsis or a summation symbol, we say that it is written in closed form. E ...
Bertrand`s Theorem - New Zealand Maths Olympiad Committee online
... This proves the induction step and, hence, the theorem. Proof of Bertrand’s Postulate. We will assume that there are no primes between n and 2n and obtain a contradiction. We will obtain that, under this assumption, the binomial coefficient 2n n is smaller than it should be. Indeed, in this case we ...
... This proves the induction step and, hence, the theorem. Proof of Bertrand’s Postulate. We will assume that there are no primes between n and 2n and obtain a contradiction. We will obtain that, under this assumption, the binomial coefficient 2n n is smaller than it should be. Indeed, in this case we ...
Comparing Contrapositive and Contradiction Proofs
... hypothesis seems to give more information to work with. Contradiction Assume P Λ Q', deduce Try this approach when a contradiction Q says something is not true. Proof by Cases Break the domain into Try this for proving two or more subsets properties of numbers and prove PQ for the where odd and eve ...
... hypothesis seems to give more information to work with. Contradiction Assume P Λ Q', deduce Try this approach when a contradiction Q says something is not true. Proof by Cases Break the domain into Try this for proving two or more subsets properties of numbers and prove PQ for the where odd and eve ...
Math Review
... Proofs: Contradiction • Assume the theorem is false. Show this implies something contradictory and/or stupid (like 1=2). • Theorem: There are an infinite number of primes. • Proof: Assume false. Then all the primes are P1, P2, P3, P4, …, Pk and Pk is the biggest . Consider N=1+(P1P2…Pk). Clearly N ...
... Proofs: Contradiction • Assume the theorem is false. Show this implies something contradictory and/or stupid (like 1=2). • Theorem: There are an infinite number of primes. • Proof: Assume false. Then all the primes are P1, P2, P3, P4, …, Pk and Pk is the biggest . Consider N=1+(P1P2…Pk). Clearly N ...
CHAPTER 8 Hilbert Proof Systems, Formal Proofs, Deduction
... to denote that a formula A has a formal proof in H2 (from the set of logical axioms A1, A2, A3). We write Γ `H2 A to denote that a formula A has a formal proof in H2 from a set of formulas Γ (and the set of logical axioms A1, A2, A3). Observe that system H2 was obtained by adding axiom A3 to the sy ...
... to denote that a formula A has a formal proof in H2 (from the set of logical axioms A1, A2, A3). We write Γ `H2 A to denote that a formula A has a formal proof in H2 from a set of formulas Γ (and the set of logical axioms A1, A2, A3). Observe that system H2 was obtained by adding axiom A3 to the sy ...
100.39 An olympiad mathematical problem, proof without words and
... May and October 1895 and totalled 60 pages. Volume 1, with the same sort of page size that we now have, was made up of 18 issues which appeared at intervals throughout 1896, 1897, 1898, 1899 and 1900. The 18 issues contained a very respectable 422 pages. It must have been quite a challenge for the t ...
... May and October 1895 and totalled 60 pages. Volume 1, with the same sort of page size that we now have, was made up of 18 issues which appeared at intervals throughout 1896, 1897, 1898, 1899 and 1900. The 18 issues contained a very respectable 422 pages. It must have been quite a challenge for the t ...
1 Proof by Contradiction - Stony Brook Mathematics
... This simple example shows the general form of a contradiction argument. • First we clue the reader in to the fact that the proof is operating by contradiction, “Assume for the sake of contradiction. . . ” is standard. • We assume the negation of the statement to be proved. • We do work to obtain a c ...
... This simple example shows the general form of a contradiction argument. • First we clue the reader in to the fact that the proof is operating by contradiction, “Assume for the sake of contradiction. . . ” is standard. • We assume the negation of the statement to be proved. • We do work to obtain a c ...
+ 1 - Stanford Mathematics
... The proof is again by induction; call the nth proposition Pn . The basis for induction P1 is the statement that 13 = 12 , which is obviously true. For the induction step, we assume that Pn is true. We’d like to show that Pn+1 is true, namely that: 13 + 23 + . . . + n3 + (n + 1)3 = (1 + 2 + . . . + n ...
... The proof is again by induction; call the nth proposition Pn . The basis for induction P1 is the statement that 13 = 12 , which is obviously true. For the induction step, we assume that Pn is true. We’d like to show that Pn+1 is true, namely that: 13 + 23 + . . . + n3 + (n + 1)3 = (1 + 2 + . . . + n ...
Mathematics in Context Sample Review Questions
... From this triangle construct two squares with sides of length a + b, also shown above. The two squares have the same lengths for their sides, so their areas must be equal. a. Calculate the area of the left-hand square by adding up the areas of the triangles and squares that compose it. (4) ...
... From this triangle construct two squares with sides of length a + b, also shown above. The two squares have the same lengths for their sides, so their areas must be equal. a. Calculate the area of the left-hand square by adding up the areas of the triangles and squares that compose it. (4) ...
Hilbert`s Program Then and Now
... logical grounding of our knowledge the axiomatic method deserves the first rank” [Hilbert, 1900b, 1093]. Hilbert thus was after a direct consistency proof of analysis, i.e., one not based on reduction to another theory. He proposed the problem of finding such a proof as the second of his 23 mathemat ...
... logical grounding of our knowledge the axiomatic method deserves the first rank” [Hilbert, 1900b, 1093]. Hilbert thus was after a direct consistency proof of analysis, i.e., one not based on reduction to another theory. He proposed the problem of finding such a proof as the second of his 23 mathemat ...
Axioms and Theorems
... uncovered. And since these can’t be together, they cannot be covered by one domino. Therefore it is impossible. ...
... uncovered. And since these can’t be together, they cannot be covered by one domino. Therefore it is impossible. ...
Sequences, Sums and Mathematical Induction Computer Science
... Why is this called Strong Induction? You might need one or every previous case to prove the k th case. This is like saying that each domino is bigger than the previous one, so the combined weight of all the previous dominoes is needed to push over the next one. A Proof by Strong Induction only diffe ...
... Why is this called Strong Induction? You might need one or every previous case to prove the k th case. This is like saying that each domino is bigger than the previous one, so the combined weight of all the previous dominoes is needed to push over the next one. A Proof by Strong Induction only diffe ...