G30 MATH SEMINAR 1 - PROOFS BY CONTRADICTION 1
... The example we saw in previous section is called a direct proof. We will see a useful tool when direct proofs do not work. The idea of a proof by contradiction is the following one. If we want to prove that a statement is true, we suppose that its hypotheses are true and its conclusion is false. The ...
... The example we saw in previous section is called a direct proof. We will see a useful tool when direct proofs do not work. The idea of a proof by contradiction is the following one. If we want to prove that a statement is true, we suppose that its hypotheses are true and its conclusion is false. The ...
Generating Functions for the Digital Sum and Other Digit Counting
... ways in which we might extend the idea to k-ary numbers, either by summing digits, or by counting non-zeros. We use the notation sk (n) (again following [1]) for the sum-of-digits function and ck (n) for the counting non-zeros function. ...
... ways in which we might extend the idea to k-ary numbers, either by summing digits, or by counting non-zeros. We use the notation sk (n) (again following [1]) for the sum-of-digits function and ck (n) for the counting non-zeros function. ...
QED - Rose
... Note that part of what makes this proof work is the careful way in which we defined the set S. The only element in S which is allowed to be 1 is the first element. Every other element must be 2 or greater. If we allowed other 1’s into the set then case III could possibly fail. If p(k+n)=1 for all n> ...
... Note that part of what makes this proof work is the careful way in which we defined the set S. The only element in S which is allowed to be 1 is the first element. Every other element must be 2 or greater. If we allowed other 1’s into the set then case III could possibly fail. If p(k+n)=1 for all n> ...
Domain Restrictions
... the function increases. Consider the graph of q ( x ) above. The function q ( x ) increases as the x-values increase beginning with zero and extending forward. Thus, q ( x ) increases along the interval (0, ∞). Likewise, a function decreases along an interval if the function’s values decrease as the ...
... the function increases. Consider the graph of q ( x ) above. The function q ( x ) increases as the x-values increase beginning with zero and extending forward. Thus, q ( x ) increases along the interval (0, ∞). Likewise, a function decreases along an interval if the function’s values decrease as the ...
A 1 ∪A 2 ∪…∪A n |=|A 1 |+|A 2 |+…+|A n
... part if only if they correspond to the same circular r-permutations . Thus the number of circular r-permutations equals the number of parts. Since each part contains r linear rpermutations, the number of parts is the number p(n,r) of linear r-permutations divided by r. ...
... part if only if they correspond to the same circular r-permutations . Thus the number of circular r-permutations equals the number of parts. Since each part contains r linear rpermutations, the number of parts is the number p(n,r) of linear r-permutations divided by r. ...
March - The Euler Archive - Mathematical Association of America
... The factors of the infinite product can be rewritten as ...
... The factors of the infinite product can be rewritten as ...
Solutions to Test 2 Mathematics 503 Foundations of Mathematics 1
... 1. (a) Some examples of McNugget numbers are 6, 9, 12 = 2·6, 15 = 6+9, and 18 = 3·6 = 2·9. Some examples of non-McNugget numbers are 1, 2, 3, 4, and 5. (b) There exists a McNugget number m such that for all natural numbers n ≥ m, n is also a McNugget number. Namely, m = 44 is such a number. Proof. W ...
... 1. (a) Some examples of McNugget numbers are 6, 9, 12 = 2·6, 15 = 6+9, and 18 = 3·6 = 2·9. Some examples of non-McNugget numbers are 1, 2, 3, 4, and 5. (b) There exists a McNugget number m such that for all natural numbers n ≥ m, n is also a McNugget number. Namely, m = 44 is such a number. Proof. W ...
OF DIOPHANTINE APPROXIMATIONS
... is true if either p = 1 or v = 1. Hence we may assume 0 = /i< 1, 0 = v < 1. Pick two (not necessarily distinct) primes P = PieS1, Q = QseS2 from the sets (1.8). We now restrict our attention to p' = Pp, q' = Q". It suffices to show ...
... is true if either p = 1 or v = 1. Hence we may assume 0 = /i< 1, 0 = v < 1. Pick two (not necessarily distinct) primes P = PieS1, Q = QseS2 from the sets (1.8). We now restrict our attention to p' = Pp, q' = Q". It suffices to show ...
