Math 6b HW 1 Solutions
... (Problem 11, Chapter 4) Suppose n > srp then prove that any sequence of n real numbers must contain a strictly increasing subsequence of length s + 1, or strictly decreasing subsequence of length r + 1 or a constant subsequence of length p + 1. Solution. Assume that there does not exist a constant s ...
... (Problem 11, Chapter 4) Suppose n > srp then prove that any sequence of n real numbers must contain a strictly increasing subsequence of length s + 1, or strictly decreasing subsequence of length r + 1 or a constant subsequence of length p + 1. Solution. Assume that there does not exist a constant s ...
2.1 inductive reasoning and conjecture ink.notebook
... • To show that a conjecture is true, you must show that it is true for all cases. • To show that a conjecture is false, you must find one counterexample. • A counterexample is a specific case for which the conjecture is false. ...
... • To show that a conjecture is true, you must show that it is true for all cases. • To show that a conjecture is false, you must find one counterexample. • A counterexample is a specific case for which the conjecture is false. ...
Full text
... Combinatorial arguments are used to establish these results; hence, it would be helpful to recall that S(n, k) counts the number of ways to partition a set of n elements into k nonempty subsets. The first main result is Theorem 1: Let r = (r1?..., rm) be an /w-tuple of positive integers, and let N b ...
... Combinatorial arguments are used to establish these results; hence, it would be helpful to recall that S(n, k) counts the number of ways to partition a set of n elements into k nonempty subsets. The first main result is Theorem 1: Let r = (r1?..., rm) be an /w-tuple of positive integers, and let N b ...
Countable and Uncountable Sets
... f(m) = m/2 and f(n) = n/2, it follows that f(m)=f(n) implies m=n. Let m and n be two odd natural numbers, then f(m) = ‐(m‐1)/2 and f(n) = ‐(n‐1)/2, it follows that f(m)=f(n) implies m=n. Therefore, f is injective. We now show that f is surjective by case analysis on the sign of some integer ...
... f(m) = m/2 and f(n) = n/2, it follows that f(m)=f(n) implies m=n. Let m and n be two odd natural numbers, then f(m) = ‐(m‐1)/2 and f(n) = ‐(n‐1)/2, it follows that f(m)=f(n) implies m=n. Therefore, f is injective. We now show that f is surjective by case analysis on the sign of some integer ...
Collatz conjecture
The Collatz conjecture is a conjecture in mathematics named after Lothar Collatz, who first proposed it in 1937. The conjecture is also known as the 3n + 1 conjecture, the Ulam conjecture (after Stanisław Ulam), Kakutani's problem (after Shizuo Kakutani), the Thwaites conjecture (after Sir Bryan Thwaites), Hasse's algorithm (after Helmut Hasse), or the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers (because the values are usually subject to multiple descents and ascents like hailstones in a cloud), or as wondrous numbers.Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process (which has been called ""Half Or Triple Plus One"", or HOTPO) indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. The property has also been called oneness.Paul Erdős said about the Collatz conjecture: ""Mathematics may not be ready for such problems."" He also offered $500 for its solution.