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There are two general methods for showing that an object is unique.
Suppose that *P(x)* is some predicate and we wish to establish
that ∃! *x, P(x)*. Both methods start by establishing
that *P(x)* holds for some particular *x*. In the first
method, uniqueness is proven by showing that all *y*, *y
≠ x, P(y)* is false. In the second method, we show that for
all *y* such that *P(y)* is true, *y* is actually
equal to *x*. Which method works better depends on the situation
at hand. Let us consider an example of each method.

Claim:Letaandbbe two odd integers such thata ≠ b. Then, there exists a unique integercsuch that|a − c| = |b − c|.Proof:First we need to show that such acexists. Without loss of generality assume thata < b. Now, letc=(a + b)/2, that is,cis the average ofaandb. Sinceaandbare both odd,a + bmust be even, which makes(a + b)/2an integer.The arithmetic below shows that |

a − c| = |b − c|.Sincec − a=(a + b)/2 − a=(a + b − 2a)/2=(b − a)/2.(b−a)/2 > 0, |a − c| =c − a=(b − a)/2.Similarly,

Againb − c=b − (a + b)/2=(2b − a − b)/2=(b − a)/2.(b−a)/2 > 0, so |b − c| =b − c=(b − a)/2. Thus, |a − c| =(b − a)/2= |b − c|.Now, we want to show that

cis unique. So, suppose that for some other integerc'that |a − c'| = |b − c'|. We will show thatc'is, in fact, equal toc.

Case 1:Suppose thatc' < a < b.Then |

a − c'| =a − c'and |b − c'| =b − c'. So, we have|But, this contradicts the hypothesis in the claim thata − c'| = |b − c'|

⇒a − c'=b − c'

⇒a=ba ≠ b. So, Case 1 cannot occur.

Case 2:Suppose thata < b < c'.Then |

a − c'| =c' − aand |b − c'| =c' − b. So, we have|Again, this contradicts the hypothesis in the claim thata − c'| = |b − c'|

⇒c' − a=c' − b

⇒a=ba ≠ b. So, Case 2 also cannot occur.

Case 3:Suppose thata ≤ c' ≤ b.Then |

a − c'| =c' − aand |b − c'| =b − c'. So, we have|But, then botha − c'| = |b − c'|

⇒c' − a=b − c'

⇒2 c'=b + a

⇒c'=(b + a)/2candc'are equal to(b + a)/2. Thus,c=c'.QED

Claim:LetGbe a connected undirected graph without any cycles. Then, for every pair of verticesuandvinG, there is an unique path fromutov.

Proof:First, since the graphGis connected, we know there is a path fromutov. Call this pathπ_{1}. Suppose that there is a different pathπ_{2}fromutov:Consider π_{1}andπ_{2}as they go from vertexuto vertexv. Sinceπ_{1}andπ_{2}aredifferentpaths, at some point the two paths must diverge. That is, from some common vertexx,π_{1}proceeds to vertexy, butπ_{2}goes to a different vertexz. (If such verticesyandzdo not exist, thenπ_{1}andπ_{2}must be identical paths.) The vertexxmight be actually beu, but we won't need to assume thatx≠uso this does not matter. Afteryandz, the two paths will eventually converge, since they both eventually reach the vertexv. Letwbe the first vertex onπ_{1}afterythat also appears onπ_{2}afterz. (It might be the case thatw=v, but this does not matter.)So, there exists a path from

xtoythen towalong pathπ_{1}. The path can continue fromwtozin the reverse direction ofπ_{2}and finally fromzback tox. This path forms a cycle fromxtowand then back tox. This contradicts the hypothesis that the graphGhas no cycle. Therefore, the assumption that there exists a second pathπ_{2}fromutovmust be false andπ_{1}is the unique path fromutov.One final note: it is possible that

w=x. This does not invalidate our proof. The cycle traced out will simply go back toxtwice.QED

⇐ Proof by Cases | Equivalence Proofs ⇒ |

Last Modified: 30 Sep 2010 15:33:40 EDT by Richard Chang to Fall 2010 CMSC 203 Homepage