Discrete Mathematics
Name
Institution
Question 1.
We will proceed by induction on the numbers E of edges in the tree. If E (T) = 1, then we know exactly what T looks like, so we find that V (T) – E (T) = 2-1 = 1 as required.
Hence suppose we know the result for any tree with no more than n edges, and let T be a tree with nt, edges. Let T’ be a subtree of T with n edges. Then V (T) – E (T’) = 1. Now clearly we have E (T) = E (T’) + 1. On the other hand, the edge in T – T’ must share at least one vertex with T’ (because otherwise T would not be connected), but cannot share both with it (because otherwise there would be a loop formed by the new edge and any path in T’ joining the two vertices to which the ends of the new edge are attached (such a path exists because T’ is connected). Thus V (T) = V (T’) + 1 the result follows.
The existence of such a subtree requires some just fication in particulars. We cannot just remove any old edge, for this may disconnect T, so that T’ is not a tree, but rather a disjoint union of two trees. To show that it exists we need to show that there is at least one edge is T that has a free vertex; i.e. a vertex not contained in any other edge. To find it, pick any edge E, if it has a free vertex, you are done. If not, move to an adjacent edge E, repeat being role never to reverse direction. These are only n + 1.
Question 2.
(a).
= 1
k n Now for k = n +1
= + +
=
= * *
=
Hence, the result is thus by induction.
(b).
/
When k = 1, the result is thus, /
(Hence, we use the identity of Fibonacci number for n m II, n m 1, +, = )
Now suppose we assume that divides and we prove that divides
= – n+1 + – n -1
= +
= + q by (induction hypothesis)
= [ + q ]
Hence divides Therefore, is true for all by induction.
(c)
=
n = 0, then, + + +
= 1*1 + 1*2 + 2*3 + 3*5
= 1+2+6+15
= 0*1 + 1*1 + 1*2 + 2*3
=
n = 1, then, + + +
= 0+1+2+6 = 9 =
Hence because is valid we assume that;
= and show that
=
= + +
= + +
= +
=
=
Hence the result is true by induction.
Question 3.
(a)
p(n) = + )
Step-1
p(1) = + )
p(1) = +16)
p(1) = +16)
p(1) = ) is divided by 241
p(1) is true
Step-2
Let p(k) is true
p(k) = + ) is divided by 241
We shall now show that p(k+1) is true. It is shown that + -1
p(k+1) = +
= +
= .15+ – +
= 15 + + -15
= 15 + +
= 15 + + *241
Both terms are divisible by 241 so their sum is also divided by 241
Hence, p(k+1) is true
Therefore, p(n) is true for all n
(b)
1+nx + , hold for n 3
p(n) = 1+nx +
p(3) = 1+3x +
= + 3 + 3n 1+3n +3
p(3) is true for n = 3
Step-2
Put n = k
p(k) = 1+kx + is true
We shall now show that p(k+1) is true. It is shown that 1+ (k+1)n + is true.
p(k) is true,
1+ kn + multiply (1+n) on both sides,
(1+ kn + ) (1+n)
1+ (k+1)n + +
Therefore;
(1+ (k+1)n + )
Given is right for )
So, given declaration is correct for all
(c)
p(n) = + , given that (x + ) is an integer.
Step i
p(1) =( + = (x + )
p(1) is correct
Step ii
Let the given declaration is correct for h = k
p(k) =( + is an integer
We shall now show that p(k+1) is true
p(k+1) =( +
= + + +
= x + +
p(k+1) is correct
Given declaration is correct for all n
Question 4
(a)
In order to compute the value of f (5)
n = 5 (here)
So,
f (0) = -1
f (1) = 0
f (n) = n.f (n-1) +
f (2) = 2.f (1) +
= 2*0 – (-1)2
= 1
f (3) = 3.f (2) +
= 3
f (4) = 4.f (3) +
= 4*3 + (1)2
= 13
f (5) = 5.f (4) +
= 5*13 + (3)2
= 74
f (5) = 74
(b)
g (5, -1, 0, 0) =?
g (0, m, x, k) =m
g (n, m, x, k) =g (n-1, x, (k+2)x +m2, k+1)
g (5, -1, 0, 0) =g (4, 0, 1, 1)
We will compute the corresponding value of m for which n = 0
g (4, 0, 1, 1) =g (3, 1, 3, 2) for n = 4, x = 1, m = 0, k = 1
g (3, 1, 3, 2) =g (2, 3, 13, 3)
n m x k
g (2, 3, 13, 3) =g (1, 13, 74, 4)
n m x k
g (1, 13, 74, 4) =g (0, 74, 613, 5)
n m x k
Now on comparing it with base case
for n = 0
g (0, m, x, k) =m
g (0, 74, 613, 5) = 74
Therefore;
g (5, -1, 0, 0) = 74 all the computation give
(c)
To evaluate the expansion of
g (n, m, x, k) for some fixed natural numbers n, m, x, k we require (n-1) computations on steps and then finally use base step case for the computations.
As in (b) for n = 5, four computation steps were required.
(d)
First, we need to know what recursive function theory which is a functional approach to computation. Each computational theory start with computation is described in term of what is to accomplished instead of how to accomplish g may be written as g( o) = F( ),
g( , n+1) = H( ,n,g( ,n)) this is exactly what is formed hence, we say that g is obtained from F and H by the permutive recursive restating the above we get formal.
(e)
Definition; permutive recursion
For k, o a function
f NK+1 Nm
Is said to be constructed using permutive from the function
f( , o) = g( ,)
A function f(n) is permutive recursive function if and only (f) either is one of the initial function.
(f)
For h(n, m, r, k) = h(n-1, r, (k+2)r + m2,k+1)
n! 1n
Multiply both sides with (n+1) and we then know
(n+1) h(n, m, r, k) h(n-1, r, (k+2)r + m2,k+1)
Note that (n+1) 2, so for sure (n+1)2n we can therefore conclude that
(n+1)! 2n+1
Thus we have proved
h(n, m, r, k) = h(n-1, r, (k+2)r + m2,k+1)
(g)
Functions h and g are equal because they have same the same domain and codomain i.e. h(a) = g(a)
(h)
In this function dominion and codomain are the set means (the natural number)
(i)
Expression in form of g(n, -1, 0, 0) cannot be used in coding because it is in conical form.
