Suppose f:[0,1]->R is continuous, f(0)>0, f(1)=0.
Prove that there is a X0 in (0,1] such that f(Xo)=0 & f(X) >0 for 0<=X<Xo (there is a smallest point in the interval [0,1] which f attains 0)
Since f is continuous, then there exist a sequence Xn converges to X0, and f(Xn) converges to f(Xo)...
Tell me if I'm not on the right track.
Use induction on k.
Pick a path P of maximum length, and suppose vertex vi is a vertex on this path, which has degree at least k, with a set of adjacent vertices {w1,w2,…,wj}, the adjacent vertex set must be on the path.
The minimum path length of a...
This is a graph theory related question.
Let G be a simple graph with min. degree k, where k>=2. Prove that G contains a cycle of length at least k+1.
Am I suppose to use induction to prove G has a path length at least k first, then try to prove that G has a cycle of length at least k+1...
Problem: "Let G be a simple graph on n vertices such that deg(v)>= n/2 for every vertex v in G. Prove that deleting any vertex of G results in a connected graph."
Well, I tried to find the min. case.
Let k be the min. deg. of vertex in a simple graph,
n is number of vertices in G
so k =...
Okay, so by looking at my original assumption P(n-1)=[(n-1)^2/4]=[(n^2-2n+1)/4]=[n^2/4+(1-2n)/4]
So now I need to prove that by adding additional one vertex in result of adding additional (2n-1)/4 edges for all n>5.
So P(n)=P(n-1)+(2n-1)/4= [n^2/4+(1-2n)/4+(2n-1)/4]=[n^2/4]
But how to...
How to prove that the number of edges in a simple bipartite graph with n vertices is at most n^2/4?
Definition of bipartite graph: a graph whose vertex-set can be partitioned into two subsets such that every edge has one endpoint in one part and one endpoint in the other part.
I try to...
Convergence of Sequence "Does {An^2} converges => {An} converges? How to prove it?"
Does sequence {An^2} converges implies to sequence {An} converges? True or False. How to prove it?
I kinda think it is false, but couldn’t think of any counterexample to directly proof it. So I try to use...
Yeah, my bet! a, b are real numbers, typo...
I got this one.
Let C be the set of all complex numbers
C={a+bi: a, b are real no.}
For any real number r can be mapped to a complex no. by r=r+0i, where r=a and is real no., b=0 is also real no.
Let R be set of all real numbers
R={r+0i: r...
Yeah, my bet! a, b are real numbers
I've constructed a linear function f: (0,1)->(0,2) defined by f(x)=2x
such that f(1/2)=1, when x=1/2 (mid point of domain), y=1 (mid point of range)
This linear function is certainly bijection, therefore |(0,1)|=|(0,2)|
But how to prove...
How to prove the open intervals (0,1) and (0,2) have the same cardinalities? |(0, 1)| = |(0, 2)|
Let a, b be real numbers, where a<b. Prove that |(0, 1)| = |(a, b)|
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|(0,1)| = |R| = c by Theorem
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I know that we...
How to prove the set of complex numbers is uncountable?
Let C be the set of all complex numbers,
So C={a+bi: a,b belongs to N; i=sqrt(-1)}
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set of all real numbers is uncountable
open intervals are uncountable...
(a) A nonempty set S is countable if and only if there exists surjective function f:N->S
(b) A nonempty set S is countable if and only if there exists a injective function g:S->N
There are two way proves for both (a) and (b)
(a-1) prove if a nonempty set S is countable, then there exists...
Okay, here is what I got so far.
There should be two steps that I need to prove to show |S|<|N|
step 1) to construct a injective function f:S->N
step 2) to prove the function f:S->N is NOT bijection (mainly NOT surjective function)
Step 1) I started with trying to contrust a injection f:S->N...