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 Founded in 1884 by Gottlob Frege and Richard Dedekind  Math can be reduced to 9 simple logical axioms  2 axioms are not logical  Axiom of infinity.

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Presentation on theme: " Founded in 1884 by Gottlob Frege and Richard Dedekind  Math can be reduced to 9 simple logical axioms  2 axioms are not logical  Axiom of infinity."— Presentation transcript:

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2  Founded in 1884 by Gottlob Frege and Richard Dedekind  Math can be reduced to 9 simple logical axioms  2 axioms are not logical  Axiom of infinity  1st crisis in mathematics

3  There will always be another axiom required to describe some property of mathematics  Existing Axioms cannot prove all statements in math

4  Founded in 1908 by Luitzen Egbertus Jan Brouwer  Math is a mental activity  ABSTRACT  2 nd crisis in mathematics

5  Founded 1910, by German Mathematician David Hilbert  Basis for modern math  Opposite of Intuitionism  Put Theory ‘T’ into the formal language ‘L’

6  Formalize all of math, so as to prove all of math to be free of contradictions  Issues- we must use abstract language to define terms used in the formal language  FAILED  3 rd Crisis in mathematics

7  A paradox. Consider N = {1, 2, 3…}. Suppose we can only use 12 or fewer words to define natural numbers; since there are infinitely many natural numbers but only finitely many phrases, there exist some natural numbers that cannot be defined in English phrases. Let N be the smallest such natural number.

8  But there is something wrong with N. Say we use this phrase to define N: “the smallest natural number not definable in twelve or fewer words.” This has eleven words, so we have in fact defined N in twelve or fewer words. This is a contradiction.

9  Suppose there is a town with just one barber, who is male. In this town, every man keeps himself clean-shaven, and he does so by doing exactly one of two things:  shaving himself  being shaved by the barber.  The barber is a man in town who shaves all those, and only those, men in town who do not shave themselves.  Who shaves the barber?

10 ??

11  Let A be the set of all sets without itself in it. If A is not a member of itself, then it is a member of A


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