As an undergraduate student, who wants to solidify his mathematical skills, I want to understand what exactly determines if a proof is rigorous.
I have heard several times that some mathematician has given another and more rigorous for an established theorem, but I don't know what does it really mean and what differences makes it to be more 'rigorous'. My understanding of a proof is that a proof is some explanation to convincing others that a statement is true. If some proof is not rigorous, does it mean that while the proof is acceptable for many but it is not for everybody? If so, the proof is not correct since it going to be rejected by some people and it is not a proof at all. Or, does rigorous mean it uses more advanced tools? If so, why not any alternative (but simpler) proof is also acceptable as rigorous (which is more 'valuable' as well since it's simpler!)?   Saying those limited insight of mine, my questions are:   $1-$ What does rigorous proof mean, in general definition?  $2-$ In order to understand your answers, as an example I am re-writing three of the proofs for the compactness of the torus. Which of the following proofs are most rigorous, 'normal' and least rigorous; and why so?    Theorem: The torus $T^2$ is compact.  First proof: The torus is homeomorphic to the product space $S^1 \times S^1$. The circle $S^1$ is compact. Therefore the torus is the product of two compact spaces, and thus it is compact.   Second proof: The torus $T$ is the subspace of $\mathbb{R}^3$ obtained by rotating a circle about the $z$-axis. The torus is closed since given any point in its complement, there is an open ball of sufficiently small radius centered at that point and disjoint from the torus. Also, the torus is bounded since it is contained in the ball of radius $4$ centered at the origin. Since the torus is closed and bounded in $\mathbb{R}^3$, it is compact.   Third proof: The torus is homeomorphic to a space obtained as a quotient  space of the square by identifying opposite edges. The square is a closed and  bounded subset of $\mathbb{R}^3$ and therefore is compact. Furthermore, a quotient map on the square is a continuous function, and the image of a compact space under a continuous function is compact. Therefore the torus is compact.   Thanks a lot for your guidance.
How can I prove the following?  $\gamma (t)$ is unit speed, $\dot \gamma(t) \not= 0 \Rightarrow \ddot \gamma(t)$ is perpendicular to $\gamma(t)$   I don't really see where a problem would arise when $\dot \gamma(t)=0$ would cause such a problem How many edges does an undirected tree with $n$ nodes have? I followed @Micha's instruction and successfully used the modules in GRASS. I use r.watershed to get the basin raster and the drainage line：  and I use r.water.outlet to get the watershed of one outlet：  I got the whole basin area of the green outlet point, but all I want is the little part of the original basin area, like this：part B is what I want  This schematic photo shows the idea of what I want：  Can I achieve this goal using r.water.outlet or other modules in GRASS？ Or all I can do is using the intersect tool in Geoprocessing tools in QGIS？  I want to find the watershed area, but using QGIS. 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How can I prevent SQL injection in PHP? I would like to show that   $$ \int_{0}^{1} \frac{x-1}{\ln(x)} \mathrm dx=\ln2 $$  What annoys me is that $ x-1 $ is the numerator so the geometric power series is useless.  Any idea? Proof of properties of injective and surjective functions. I'd like to see if these proofs are correct/have them critiqued.     Let $g: A \to B$ and $f: B \to C$ be functions.  Then:   (a) If $g$ and $f$ are one-to-one, then $f \circ g$ is one-to-one.   (b) If $g$ and $f$ are onto, then $f \circ g$ is onto.   (c) If $f \circ g$ is one-to-one, then $f$ is one-to-one?   (d) If $f \circ g$ is one-to-one, then $g$ is one-to-one?   (e) If $f \circ g$ is onto, then $f$ is onto?   (f) If $f \circ g$ is onto, then $g$ is onto?     (a)  Let $a,b \in A$.  If $(f \circ g)(a)=(f \circ g)(b)$, then $f(g(a))=f(g(b))$.  Since $f$ is one-to-one, we know that $g(a)=g(b)$.  And, since $g$ is one-to-one is must be that $a=b$.  Hence $f \circ g$ is one-to-one.    (b) Since $f$ is surjective we know that for all $c \in C$ there is a $b\in B$ such that $f(b)=c$.  Since $g$ is surjective, there is a $a \in A$ such that $g(a)=b$.  Hence, for all $c \in C$ there is an $a \in A$ such that $(f \circ g)(a)=f(g(a))=f(b)=c$.    (c)  This is false.  Let $A=\{1\}$, $B=\{1,2\}$, and $C=\{1\}$. Define $g(1)=1$ and $f(1)=f(2)=1$. Then $f \circ g$ is injective, but $f$ is not injective.    (d)  This is true.  Let $a,b \in A$ where $a \neq b$.  If $(f\circ g)(a)\neq(f\circ g)(b)$, then $f(g(a))\neq f(g(b))$.  Hence $g(a) \neq g(b)$.  (e)  Since $f \circ g$ is surjective, then for every $ c \in C$ there is an $a \in A$ such that $f(g(a))=c$.   Let $b = g(a)$.  Then $b \in B$ and $f(b)=g(f(a))=c$.  Thus $f(b)=c$.  Hence $f$ is surjective.  (f)  Using the same set up as part (c), we have that $g$ is not surjective since there is nothing in $A$ such that $g$ will go to 1. 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