170- Midterm 3

November 21, 2010

The third midterm is here.

Solutions follow.

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170- Quiz 10

November 16, 2010

Quiz 10 is here.

Solutions follow.

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170- Quiz 9

November 11, 2010

Quiz 9 is here.

Solutions follow.

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170- Extra credit problem

November 9, 2010

This problem is due Tuesday November 30, and will replace your lowest quiz score, if you choose to turn it in. Additional extra credit is possible depending on the quality of your work.

Please work on your own. If somebody helps you, or you find ideas or the solution either in a book or online, please mention this in what you turn in, including the name of the person, the name of the book, or a link to the relevant website, together with any additional information that may be useful to identify the sources.

In lecture we explained how Newton’s method can be used to approximate \sqrt2 numerically: We start with a guess x_0, and then define

\displaystyle x_1=x_0-\frac{x_0^2-2}{2x_0},

\displaystyle x_2=x_1-\frac{x_1^2-2}{2x_1},

\displaystyle x_3=x_2-\frac{x_2^2-2}{2x_2},

etc.

The goal here is to see how good these approximations are.

  1. Show that if x_0>0, then x_1,x_2,\dots are all larger than 1.
  2. Show that if x_0\ge1, then {}|x_1^2-2|<|x_0^2-2|^2/4, {}|x_2^2-2|<|x_1^2-2|^2/4, etc.
  3. If we want to compute \sqrt2 with an accuracy of 12 significant digits, and we begin with x_0=1, how many iterations do we need to perform?

Woodin’s proof of the second incompleteness theorem for set theory

November 4, 2010

As part of the University of Florida Special Year in Logic, I attended a conference at Gainesville on March 5–9, 2007, on Singular Cardinal Combinatorics and Inner Model Theory. Over lunch, Hugh Woodin mentioned a nice argument that quickly gives a proof of the second incompleteness theorem for set theory, and somewhat more. I present this argument here.

The proof is similar to that in Thomas Jech, On Gödel’s second incompleteness theorem, Proceedings of the American Mathematical Society 121 (1) (1994), 311-313. However, it is semantic in nature: Consistency is expressed in terms of the existence of models. In particular, we do not need to present a proof system to make sense of the result. Of course, thanks to the completeness theorem, if consistency is first introduced syntactically, we can still make use of the semantic approach.

Woodin’s proof follows.

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