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This question is somewhat of a continuation of the very interesting question and its responses:

Can every proof by contradiction also be shown without contradiction?

I did a rough count of proofs in my primary source for study of real analysis, notes from lectures by Vaughan Jones (Fields Medal).

https://sites.google.com/site/math104sp2011/lecture-notes

He did not explicitly follow a text. In the beginning of the course which dealt primarily with metric spaces, 21 of the first 55 proofs (rough count) were by contradiction. When the material transitioned to calculus, virtually all the proofs were direct.

In the algebra lecture series by Benedict Gross,

http://www.extension.harvard.edu/open-learning-initiative/abstract-algebra

unless I missed something, the first proof by contradiction came in lecture 19 (out of 36). Showing a group of order $p^{2}$, $p$ a prime, is abelian.

I was wondering if there is something inherent in various topics - axioms, or something - that make of type of proof more natural, accessible, appropriate, whatever the correct word might be.

Thanks

share|improve this question
    
one natural class is the proofs where you pick a maximal element with a certain property and then contradict optimality, like the claim that integers are finite -- pick the maximal element $x$ then $x+1$ is also an integer... –  gt6989b Nov 26 '12 at 21:10
    
Good question and research. Do you know the book by Paul Taylor Practical Foundations of Mathematics where he writes (paraphrasing a bit) that excluded middle "so infests" foundations that it's necessary to start over. –  alancalvitti Nov 27 '12 at 3:48
    
@alancalvitti Thanks I'll check it out. –  Andrew Nov 27 '12 at 23:14

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