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This question already has an answer here:

I spend a lot of time on math.se and even though I don't understand many of the questions posed, I try to understand what is being said or atleast wiki something to get some gist of the question and the answers. However, I never understood what "rigour" meant or what people mean when they say "is this a rigorous solution?". So can someone please explain to me what rigour in mathematics is?.

Thanks!

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marked as duplicate by rschwieb, Daniel Fischer, Jeel Shah, Omnomnomnom, Amzoti Aug 22 '13 at 18:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Ahh I didn't see that. I have voted to close as duplicate also. $\endgroup$ – Jeel Shah Aug 22 '13 at 18:17
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"Rigor" is a subjective term, but basically it means that the intended reader shouldn't have to spend too much time at any step of your argument figuring out "why is this step justified?"

Here's an example: Suppose we are given a function on the natural numbers which is defined by $$ f(0)=0,\quad f(n)=f(n-1)+n, \text{ for } n>0 $$ Find a closed form expression for $f(n)$.

Here's a potentially non-rigorous (but correct) solution: $$ \begin{align} f(n)&=f(n-1)+n\\ &=(f(n-2)+(n-1))+(n) = f(n-2)+(n-1)+(n)\\ &=f(n-3)+(n-2)+(n-1)+(n) \end{align} $$ Continuing, we see that $$ f(n)=f(0)+1+2+\cdots+(n-1)+(n)=\frac{n(n+1)}{2} $$

Here's what I meant by "intended reader" in my first line: if the audience were professional mathematicians, this would be an acceptable proof, since the implicit justification for each step would be perfectly clear to them. However, if this was an exercise in my introductory Discrete Mathematics course, it would fail to meet my standards of rigor in at least two ways, since in this case the intended audience would be their peers:

  1. The part below "Continuing, we see that" is a guess, based on an apparent pattern developing in the lines above it. The pattern is indeed correct, but it's still a guess and will remain so until it has been proven (by induction, say).

  2. Depending on what we had covered up to the time, the last equality, while again correct, might not meet the standard of rigor if the students hadn't seen the expression for the sum of an arithmetic series. If they hadn't, then a citation (or a proof) would be necessary.

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  • $\begingroup$ Very well explained! Thank you. This really clears it up. $\endgroup$ – Jeel Shah Aug 22 '13 at 19:19
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    $\begingroup$ @gekkostate Glad it helped. Be sure to look at the link to the duplicate for another perspective. $\endgroup$ – Rick Decker Aug 22 '13 at 19:22

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