User:Prime.mover

MATHEMATICIANS ARE FUCKWITS

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 * Mathematics is the best...

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THE MOST COMMON ERRORS IN UNDERGRADUATE MATHEMATICS by

In particular I need to look at:

"Some more advanced students (e.g., college seniors) use the implication symbol ($\implies$) as a symbol for the phrase "the next step is." A string of statements of the form


 * $A \implies B \implies C \implies D$

should mean that A by itself implies B, and B by itself implies C, and C by itself implies D; that is the coventional interpretation given by mathematicians. But some students use such a string to mean merely that if we start from A, then the next step in our reasoning is B (using not only A but other information as well) and then the next step is C (perhaps using both A and B), etc.

Actually, there is a symbol for "the next step is." It looks like this: $\leadsto$ It is also called "leads to," and in the LaTeX formatting language it is given by the code \leadsto. However, I haven't seen it used very often."
 * Such a distinction seems unnecessary, especially considering theorems such as Extended Rule of Implication. Surely at each step of the way we have an implied $\land$ going on everything before it? --GFauxPas 04:23, 12 February 2012 (EST)
 * Just a comment...I've seen $\leadsto$ to also mean "converges to" for sequences. Andrew Salmon 17:56, 12 February 2012 (EST)
 * Every notation has been used for more than one thing. The point is we need to choose one and stick to it. --prime mover 18:22, 12 February 2012 (EST)
 * What about $\therefore$, i.e. 'therefore' (given by \therefore)? It is relatively unambiguous, but little known/used. I have come to prefer words when other people have to read my work as well; in all other cases, $\implies$ suffices instead. --Lord_Farin 19:03, 12 February 2012 (EST)
 * I've seen cogent arguments elsewhere as to why not to use $\therefore$ but, shrug, do I care nowadays ...--prime mover 01:08, 13 February 2012 (EST)

Okay, see Definition:Distinction between Logical Implication and Conditional where I have laid down the law.

the generalization of the tarski-vaught test is also a standard statement of the theorem, and while slightly more general, it does not really require much more of a proof, hence the modication.--Yaddie 17:33, 6 March 2012 (EST)


 * ... but as I say, please make the statement as a separate section of this page. As soon as I've finished what I'm currently doing I'll get onto it. --prime mover 17:36, 6 March 2012 (EST)

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