Assignment Set 1 (60 Marks)
Ques1. carry that the relationis a ? b(mod m ) equivalence relation. Ans.
Ques2. Ans.
Ques3. Prove by the mode of contradiction
that is not a rational heel.
Ans. A rational number is of the form p/q where , and p, q are not having any everyday factors. Assume that is a rational number. So it can be create verbally as
If p is even, then it can be written as p = 2k. and so 4 = 2 . Therefore q is even. This is a contradiction to our assumption that p and q have no common factors. Therefore rational number. Ques.4. Prove by numeric instalment
is not a
Ans. (i) Base Step: Let n = 0. and so the sum on the left is zero, since there is nothing to add. The mirror image on the right is also zero. If n=1, left berth = Right side = =1 = = 1.
Hence the allow is legitimate for n = 1. (ii) Induction Hypothesis: Assume that the result to be true for n = m because Adding the ?
th
term i.e.
2
to some(prenominal) sides of the above equation, we flummox,
=
=
= =
Therefore the result is true for n= m+1. Hence by mathematical induction the given result is true for all positive integers n.
Ques5.
Prove that The sum of the degrees of the vertices of a graph G is twice the number of edges Ans. (The proof is by induction on the number of edges e). Case-(i): Suppose e = 1. Suppose f is the edge in G with f = uv. Then d(v) = 1, d(u) = 1. Therefore
= 2 x (number of edges)
Hence by induction we get that the sum of the degrees of the vertices of the graph G is twice the poem of edges. Ques6. Prove that T is a tree there is wizard and only one path amid every straddle of vertices Ans. Part 1: Suppose T is a tree. Then T is a connected graph and contains no circuits.
Since T is connected, there exists at least one path between every pair of vertices in T. Suppose that between devil vertices a and b of T, there are two...If you want to get a full essay, order it on our website: Ordercustompaper.com
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