CS1231 - Lecture 09
... We have proven that f is a bijection. Let’s pause for a moment to consider this propositoin that we have just proven. Z is infinite. Z+ is infinite. Which is more infinite? Most would answer intuitively that Z contains more elements than Z+. But your intuition must be in subjection to logic. And log ...
... We have proven that f is a bijection. Let’s pause for a moment to consider this propositoin that we have just proven. Z is infinite. Z+ is infinite. Which is more infinite? Most would answer intuitively that Z contains more elements than Z+. But your intuition must be in subjection to logic. And log ...
Fibonacci modk
... study. Below I present the last three cases tabulated above: k = 10, 11 and 12––these illustrate the three possibilities for the number of zeros per period that we've seen so far. I celebrate the "cyclical" nature of a period by putting the terms in a circle, the primary zero at the top, with the se ...
... study. Below I present the last three cases tabulated above: k = 10, 11 and 12––these illustrate the three possibilities for the number of zeros per period that we've seen so far. I celebrate the "cyclical" nature of a period by putting the terms in a circle, the primary zero at the top, with the se ...
ARITHMETIC TRANSLATIONS OF AXIOM SYSTEMS
... Co, Ci, • • • , C„_i, and such that ( 3x)F{x) was not formerly picked in defining a Cr, and take C„ to be F{um). If n is odd, say n = 2m + X, take C„ to be the first formula which is not already a Cr, and which involves only uu u2, ■ ■ ■ , um together with symbols of So, and such that no contradicti ...
... Co, Ci, • • • , C„_i, and such that ( 3x)F{x) was not formerly picked in defining a Cr, and take C„ to be F{um). If n is odd, say n = 2m + X, take C„ to be the first formula which is not already a Cr, and which involves only uu u2, ■ ■ ■ , um together with symbols of So, and such that no contradicti ...
Full text
... i is adjacent to j and ϕ(i), ϕ(j) ∈ {c1 , . . . , ck }, we have ϕ(i) 6= ϕ(j). Otherwise, we say that the map ϕ is improper. In somewhat looser terminology, one can think of {ck+1 , . . . , ck+` } as coloring “wildcards”. Let χG (x, y) be a function such that χG (k, `) is the number of proper (k, `)- ...
... i is adjacent to j and ϕ(i), ϕ(j) ∈ {c1 , . . . , ck }, we have ϕ(i) 6= ϕ(j). Otherwise, we say that the map ϕ is improper. In somewhat looser terminology, one can think of {ck+1 , . . . , ck+` } as coloring “wildcards”. Let χG (x, y) be a function such that χG (k, `) is the number of proper (k, `)- ...
