Six more gems that every FP1 teacher should know
... The proof that there are an infinite number of prime numbers (mentioned as an investigation in the MEI C1/C2 textbook). The proof that the square root of 2 is irrational (appears in the MEI C3/C4 textbook, but a more powerful extension is discussed here). ...
... The proof that there are an infinite number of prime numbers (mentioned as an investigation in the MEI C1/C2 textbook). The proof that the square root of 2 is irrational (appears in the MEI C3/C4 textbook, but a more powerful extension is discussed here). ...
Document
... Theorem: Every n ∈ ℕ is the sum of distinct powers of two. Proof: By strong induction. Let P(n) be “n is the sum of distinct powers of two.” We prove that P(n) is true for all n ∈ ℕ. As our base case, we prove P(0), that 0 is the sum of distinct powers of 2. Since the empty sum of no powers of 2 is ...
... Theorem: Every n ∈ ℕ is the sum of distinct powers of two. Proof: By strong induction. Let P(n) be “n is the sum of distinct powers of two.” We prove that P(n) is true for all n ∈ ℕ. As our base case, we prove P(0), that 0 is the sum of distinct powers of 2. Since the empty sum of no powers of 2 is ...
Sample - University of Utah Math Department
... Note that the x-coordinates 0 ≤ m0 ≤ m0 + 1 ≤ m0 + 2 are all nonnegative and the ycoordinates 0 ≤ n0 − 1 ≤ n0 ≤ n0 + 2 are also nonnegative. In both cases, these moves stay in the board. Also, there were three additional moves, so the total is at most `k+1 = `k + 3 = 3k + 3 = 3(k + 1). The induction ...
... Note that the x-coordinates 0 ≤ m0 ≤ m0 + 1 ≤ m0 + 2 are all nonnegative and the ycoordinates 0 ≤ n0 − 1 ≤ n0 ≤ n0 + 2 are also nonnegative. In both cases, these moves stay in the board. Also, there were three additional moves, so the total is at most `k+1 = `k + 3 = 3k + 3 = 3(k + 1). The induction ...
Evaluating the exact infinitesimal values of area of Sierpinski`s
... introduced in [14] and built using Cantor’s ideas. It is important to emphasize that our point of view on axiomatic systems is also more applied than the traditional one. Due to Postulate 2, mathematical objects are not define by axiomatic systems that just determine formal rules for operating with ...
... introduced in [14] and built using Cantor’s ideas. It is important to emphasize that our point of view on axiomatic systems is also more applied than the traditional one. Due to Postulate 2, mathematical objects are not define by axiomatic systems that just determine formal rules for operating with ...
Inductive reasoning
... • Last year was a bad hurricane season and so was this year. Therefore next year will be a bad hurricane season. • In decimals, the number starts 3.1415. Therefore the next two digits in are 16. • The sun has come up every day for as long as anyone remembers. Therefore it will come up tomorrow. ...
... • Last year was a bad hurricane season and so was this year. Therefore next year will be a bad hurricane season. • In decimals, the number starts 3.1415. Therefore the next two digits in are 16. • The sun has come up every day for as long as anyone remembers. Therefore it will come up tomorrow. ...
Proofs by induction - Australian Mathematical Sciences Institute
... on the web, but I think it is a better exercise to do this by hand, as suggested above.) Rather surprisingly, the number n 2 − n + 41 is prime for n = 1, 2, . . . , 40. But Claim(41) is false, since 412 −41+41 = 412 = 1681, which is not prime, as it is obviously divisible by 41. ...
... on the web, but I think it is a better exercise to do this by hand, as suggested above.) Rather surprisingly, the number n 2 − n + 41 is prime for n = 1, 2, . . . , 40. But Claim(41) is false, since 412 −41+41 = 412 = 1681, which is not prime, as it is obviously divisible by 41. ...
Combinatorial formulas connected to diagonal
... The theory of symmetric functions arises in various areas of mathematics such as algebraic combinatorics, representation theory, Lie algebras, algebraic geometry, and special function theory. In 1988, Macdonald introduced a unique family of symmetric functions with two parameters characterized by ce ...
... The theory of symmetric functions arises in various areas of mathematics such as algebraic combinatorics, representation theory, Lie algebras, algebraic geometry, and special function theory. In 1988, Macdonald introduced a unique family of symmetric functions with two parameters characterized by ce ...
Solutions to Homework 6 Mathematics 503 Foundations of
... Proposition. For all real numbers x and y, |xy| = |x||y|. Proof. There are several cases to consider. Clearly, if either x or y is equal to 0, then |xy| = |x||y|. The equation is clearly true if x > 0 and y > 0, for |xy| = xy = |x||y| in this case. Suppose that x > 0 and y < 0. Then, |xy| = −xy. On ...
... Proposition. For all real numbers x and y, |xy| = |x||y|. Proof. There are several cases to consider. Clearly, if either x or y is equal to 0, then |xy| = |x||y|. The equation is clearly true if x > 0 and y > 0, for |xy| = xy = |x||y| in this case. Suppose that x > 0 and y < 0. Then, |xy| = −xy. On ...
Proof by Induction
... Let’s first rewrite the indirect proof slightly, to make the structure more apparent. First, we break the assumption that n is the smallest counterexample into three simpler assumptions: (1) n is an integer greater than 1; (2) n has no prime divisors; and (3) there are no smaller counterexamples. Se ...
... Let’s first rewrite the indirect proof slightly, to make the structure more apparent. First, we break the assumption that n is the smallest counterexample into three simpler assumptions: (1) n is an integer greater than 1; (2) n has no prime divisors; and (3) there are no smaller counterexamples. Se ...
The Natural Number System: Induction and Counting
... need not be the case. Remember that in the inductive step, we would not be proving q(i) directly, but rather the conditional statement q(n−1) ⇒ q(n). Thus, a stronger property q(i) gives us a stronger inductive hypothesis, q(n − 1), to work with! Of course, q(n) is a stronger inductive conclusion, b ...
... need not be the case. Remember that in the inductive step, we would not be proving q(i) directly, but rather the conditional statement q(n−1) ⇒ q(n). Thus, a stronger property q(i) gives us a stronger inductive hypothesis, q(n − 1), to work with! Of course, q(n) is a stronger inductive conclusion, b ...
powerpoint
... P(n): Some bag of 2-cent and 5-cent coins has sum n. We prove that P(n) holds for all n>=4. Base case n= 4. A bag that contains 2 2-cent coins and 0 5-cent coins sums to 4. Inductive case. We prove P(k+1), for k>=4, assuming P(k). Since P(k) holds, there is a bag of 2-cent and 5cent coins that sums ...
... P(n): Some bag of 2-cent and 5-cent coins has sum n. We prove that P(n) holds for all n>=4. Base case n= 4. A bag that contains 2 2-cent coins and 0 5-cent coins sums to 4. Inductive case. We prove P(k+1), for k>=4, assuming P(k). Since P(k) holds, there is a bag of 2-cent and 5cent coins that sums ...
Proof Technique
... odd integers is n2. Proof: Let P(n) be the proposition that the sum of the first n odd integers is n2. Then to proof that P(n) is true for all n ≥ 1, we have to show that P(1) is true and P(k+1) is true if P(k) is true for k ≥ 1. Basic step: P(1) is true, because the sum of the first 1 odd integer ...
... odd integers is n2. Proof: Let P(n) be the proposition that the sum of the first n odd integers is n2. Then to proof that P(n) is true for all n ≥ 1, we have to show that P(1) is true and P(k+1) is true if P(k) is true for k ≥ 1. Basic step: P(1) is true, because the sum of the first 1 odd integer ...
(n!)+
... Now, just from this, what do we know about x and y? You should think back to the definition of rational: “… Since x is rational, we know (from the very definition of rational) that there must be some integers i and j such that x = i/j. So, let ix,jx be such integers …” We give them unique names so w ...
... Now, just from this, what do we know about x and y? You should think back to the definition of rational: “… Since x is rational, we know (from the very definition of rational) that there must be some integers i and j such that x = i/j. So, let ix,jx be such integers …” We give them unique names so w ...
Arithmetic Polygons
... We would conclude that eπi/2 ∈ Q[eπi/2 ], so that Q[eπi/2 ] = Q[eπi/2 ]. But this is well-known to be false (for instance, the dimension over Q of the first field is 2n−1 while that of the second is 2n−2 .) Thus the theorem holds for n + 1, and by induction for all n ∈ N. Lemma 1 gives us an immedia ...
... We would conclude that eπi/2 ∈ Q[eπi/2 ], so that Q[eπi/2 ] = Q[eπi/2 ]. But this is well-known to be false (for instance, the dimension over Q of the first field is 2n−1 while that of the second is 2n−2 .) Thus the theorem holds for n + 1, and by induction for all n ∈ N. Lemma 1 gives us an immedia ...
Section 9.3: Mathematical Induction
... THEN the sentence P (n) is true for all natural numbers n. The Principle of Mathematical Induction, or PMI for short, is exactly that - a principle.1 It is a property of the natural numbers we either choose to accept or reject. In English, it says that if we want to prove that a formula works for al ...
... THEN the sentence P (n) is true for all natural numbers n. The Principle of Mathematical Induction, or PMI for short, is exactly that - a principle.1 It is a property of the natural numbers we either choose to accept or reject. In English, it says that if we want to prove that a formula works for al ...
Theorem If p is a prime number which has remainder 1 when
... First we need to define even and odd. Definition: We define an integer n to be even if there exists an integer k such that n = 2k. We define an integer n to be odd if there’s an integer k such that n = 2k + 1. Proof of theorem: Write n2 + n = n(n + 1). It is well-known that every integer is either e ...
... First we need to define even and odd. Definition: We define an integer n to be even if there exists an integer k such that n = 2k. We define an integer n to be odd if there’s an integer k such that n = 2k + 1. Proof of theorem: Write n2 + n = n(n + 1). It is well-known that every integer is either e ...
Ramsey Theory
... we may assume k ≥ 1 and that the empty word is not in W at all. Take an arbitrary u ∈ W , as (i) fails, w 6v u. This allows the following factorization of w w = [...]1 `1 [...]2 `2 [...]3 · · · `i [...]i+1 , ...
... we may assume k ≥ 1 and that the empty word is not in W at all. Take an arbitrary u ∈ W , as (i) fails, w 6v u. This allows the following factorization of w w = [...]1 `1 [...]2 `2 [...]3 · · · `i [...]i+1 , ...
Provability as a Modal Operator with the models of PA as the Worlds
... If we consider the pointed submodels of M (which include some node A and all of its decendants) which such submodels are isomorphic to each other (in the sense of graph structure alone), how many isomorphism classes are there? We know that if B is sound and satisfies the Hilbert Bernays conditions, ...
... If we consider the pointed submodels of M (which include some node A and all of its decendants) which such submodels are isomorphic to each other (in the sense of graph structure alone), how many isomorphism classes are there? We know that if B is sound and satisfies the Hilbert Bernays conditions, ...