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Proving "for all" statements
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Prove the result for some cases and deduce it for the rest
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Make use of special cases and transitivity
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[QUICK DESCRIPTION] To establish a result for all objects X, it sometimes suffices to establish it for some relatively simple special cases from which the full result can more straightforwardly be deduced. A typical situation might be that one wants to show that two collections of objects X and Y are 'equivalent'; the easiest way to do so might be to show that they are both 'equivalent' to a third, simple collection of objects Z. [note article contributions wanted]More examples needed. [/note] [PREREQUISITES] Complex numbers for an example, and the idea of an [[w:Equivalence_relation|equivalence relation]] to set things in context. [EXAMPLE mobius|M\"obius transformations and triples of points] A ''M\"obius map'', or ''[[w:Möbius_transformation|M\"obius transformation]]'', is a function from $\C$ to $\C$ given by a formula of the form $z\mapsto\frac{az+b}{cz+d}$, where $a$, $b$, $c$ and $d$ are complex numbers with $ad-bc\ne 0$. Such maps form a group under composition, as discussed in an example in the article [[Using generators and closure properties]]. We will consider M\"obius transformations acting on complex numbers, as well as on the special symbol $\infty$; indeed, they act on the [[w:Riemann_sphere|Riemann sphere]] $\C \cup \{ \infty \}.$ Consider the collection of triples $(x,y,z)$ of distinct points in $\C \cup \{ \infty \}.$ It turns out that any such triple of points can be mapped to any other triple $(x',y',z')$ of distinct points by a M\"obius transformation. If one attempts to show this directly by finding suitable values for $a,b,c,d$ by substitution, however, the algebra quickly gets quite messy. This is where the technique of this article comes in: if the triple $(x,y,z)$ is supposed to be mappable to any other triple $(x',y',z')$, then one must certainly be able to map it to the simple triple $(0,1,\infty)$. But if one can do that for any triple $(x,y,z)$, then to show that $(x,y,z)$ can be mapped to $(x',y',z')$ in general one can simply take the composition of a map that takes $(x,y,z)$ to $(0,1,\infty)$ with the inverse of one that takes $(x',y',z')$ to $(0,1,\infty).$ Thus once one has established that all triples are 'equivalent' to the simple case $(0,1,\infty)$, one is done. Furthermore, establishing this last fact is relatively simple, by virtue of having chosen a simple representative. [GENERAL DISCUSSION] In the above example, there was a natural choice for the 'simple special case' for which it sufficed to establish the result. A choice of such an object may not always be so obvious, but can often be found after a little thought. The technique can also be applied in situations where the notion of 'equivalence' is slightly more rough, not actually being an equivalence relation but just a [add]transitive relation{{, meaning that if X is related to Y and Y is related to Z, then X is related to Z}}[/add].<comment thread="15" />
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