16 ms·
Yes, he sure is controversial. He does not acknowledge the existence of irrational numbers for example. In his lectures he avoids the use of any transcendental
by blux 10y ago
Yes, he sure is controversial. He does not acknowledge the existence of irrational numbers for example. In his lectures he avoids the use of any transcendental functions for the same reason, so no square roots, sine, or cosine.
I'm no mathematician, so I'm not qualified to comment on this either. Would love to hear other opinions on this too.
My personal opinion on this is that his rational approach does produce beautiful math, in the sense that it is really simple and intuitive.
- nabla9 10y ago>the existence of irrational numbers for example. I think most of cases of "non acknowledging" are deeply personal aesthetic considerations. One likes to play with one set of objects and only with them. In the widest sense existence in mathematics means that mathematical object is well-defined and the system used is consistent (you can't derive contradictions). Of course mathematicians are free do limit themselves into any subset of axioms or concepts they feel is "natural" or "real" and work only with them. But if they make philosophical arguments against other using irrational numbers, axiom of choice etc. I think they should argue that they are not consistent or well defined.
- jcranmer 10y ago> In his lectures he avoids the use of any transcendental functions for the same reason, so no square roots, sine, or cosine. Square roots are algebraic, not transcendental.