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Limitations
Limitations

... Prove: forall a:aexp. eval(a) is finite Base 1: eval(var) is finite Base 2: eval(const) is finite Ind1: suppose eval(b) is finite and eval(c) is finite prove eval(b+c) is finite Ind2: suppose eval(b) is finite and eval(c) is finite prove eval(b-c) is finite ...
A Transition to Advanced Mathematics
A Transition to Advanced Mathematics

Untitled
Untitled

... actual homework exercises submitted by students, offers the reader the opportunity to solidify her facility for writing proofs by critiquing these submissions as if she were the instructor for the course. (7) The notation for images and inverse images of sets under a function, defined in Section 4.2 ...
Version 1.0 of the Math 135 course notes - CEMC
Version 1.0 of the Math 135 course notes - CEMC

... 39.3 Infinite Sets Are Even Weirder Than You Thought 39.4 Not All Infinite Sets Have the Same Cardinality . . ...
Book of Proof - people.vcu.edu
Book of Proof - people.vcu.edu

recursion
recursion

... Suppose the number of primes is finite, k. The primes are P1, P2….. Pk The largest prime is Pk Consider the number N = 1 + P1, P2….. Pk N is larger than Pk Thus N is not prime. So N must be product of some primes. ...
Book of Proof
Book of Proof

- ScholarWorks@GVSU
- ScholarWorks@GVSU

... The closure properties of the number systems discussed in Section 1.1 and the properties of the number systems in Table 1.2 on page 18 are being used as axioms in this text. A definition is simply an agreement as to the meaning of a particular term. For example, in this text, we have defined the ter ...
- ScholarWorks@GVSU
- ScholarWorks@GVSU

On Cantor`s First Uncountability Proof, Pick`s Theorem
On Cantor`s First Uncountability Proof, Pick`s Theorem

... coming from deeper and deeper in the sequence (an ). By construction of the b and c sequences, for every i ≤ max{k, r}, we have ai ≤ cm or ai ≥ bm . But from above, cm < L < bm . We have thus arrived at a contradiction and hence the conclusion of a rather ingenious proof. Let’s now run through Canto ...
N - University of Alberta
N - University of Alberta

... © Vadim Bulitko : CMPUT 272, Fall 2003, UofA ...
Those Incredible Greeks! - The Saga of Mathematics: A Brief History
Those Incredible Greeks! - The Saga of Mathematics: A Brief History

Recursive call to factorial(1)
Recursive call to factorial(1)

... The loop computes each Fibonacci number by starting at 2 and working its way upward Clearly, the number of iterations is bounded above by n The amount of space required is constant ...
the fundamentals of abstract mathematics
the fundamentals of abstract mathematics

... The first of these deductions is very famous, but the second one is lame. It may seem odd to even call it a deduction, because the two hypotheses have nothing at all to do with the conclusion, but, given our definition, it does count as a deduction. However, it is is a very poor one, so it cannot be ...
Full text
Full text

... ON DEDEKIND SUMS AND LINEAR RECURRENCES OF ORDER TWO Neville Robbins Mathematics Department, San Francisco State University, San Francisco, CA 94132 (Submitted September 2001-Final Revision December 2001) ...
MAD2104 Course Notes - FSU Math
MAD2104 Course Notes - FSU Math

... 1. INTRODUCTION TO SETS ...
An Introduction to Higher Mathematics
An Introduction to Higher Mathematics

... logical expressions similar to the algebra for numerical expressions. This subject is called Boolean Algebra and has many uses, particularly in computer science. If two formulas always take on the same truth value no matter what elements from the universe of discourse we substitute for the various v ...
Ordinal Arithmetic
Ordinal Arithmetic

... ordinals exist. Consider the set of all ordinals rigorously defined thus far — the empty set. Since the empty set has no elements, every element of the empty set is a subset of the empty set. Therefore, it is a transitive set of ordinals! Now that we have an ordinal, we can consider a new set of ord ...
New Generalized Cyclotomy and Its Applications
New Generalized Cyclotomy and Its Applications

... M0, 1, 2, 2 , n!1N with integer addition modulo n and integer multiplication modulo n as the ring operations. Here and hereafter a mod n denotes the least nonnegative integer that is congruent to a modulo n. As usual, we use Z* to n denote all the invertible elements of Z . n Let S be a subset of Z ...
History of Mathematics
History of Mathematics

Booklet of lecture notes, exercises and solutions.
Booklet of lecture notes, exercises and solutions.

... Is a building either right or wrong? No, of course not. If it falls down, we could probably say it’s “wrong”, but is it otherwise “right”? Likewise mathematics can be completely wrong, but it can also be all sorts of types of dodgy without being completely wrong. One of the most common ways that stu ...
Introduction to Programming Languages and Compilers
Introduction to Programming Languages and Compilers

... problems” at least if they require knowing if an expression is zero, and it could be from this class R. • It doesn’t enter into our programs, since the difficulty of simplifying sub-classes of this, or “other” classes is computationally hard and/or ill-defined, regardless of this result. Richard Fat ...
Strings with maximally many distinct subsequences and
Strings with maximally many distinct subsequences and

... Proof. Only the second and fourth points require proof, and we take them together. Recall that a de Bruijn graph Gk has a vertex for each (k − 1)-long string over [d], and for each k-long string, has a directed edge from the string’s (k −1)-prefix vertex to its (k −1)-suffix vertex. Gk is Eulerian, ...
Lecture Notes - School of Mathematics
Lecture Notes - School of Mathematics

... I have received this delightful email from David Rudling: I have been working through your lecture notes at home now that I am retired and trying to catch up on not going to university when younger. I have noticed that when introducing : as the symbol for “such that” in set theory you have not added ...
A. Pythagoras` Theorem
A. Pythagoras` Theorem

... Solution: Step 1: Consider the number 11 as the sum of perfect squares, i.e., 32  12  12  11. Step 2: Construct a right-angled triangle with legs 3 units and 1 unit. Then mark 10 ( 32  12 ) on the number line. Step 3: Continue to construct a right-angled triangle with legs 10 units and 1 unit. ...
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Brouwer–Hilbert controversy

In a foundational controversy in twentieth-century mathematics, L. E. J. Brouwer, a supporter of intuitionism, opposed David Hilbert, the founder of formalism.
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