Assignment on Prove directly that If a is any odd integer and b is any even integer, then 2a + 3b is even.

Title: Prove directly that If a is any odd integer and b is any even integer, then 2a + 3b is even.

Math

APA

As we know, “a” is odd integer and therefore “a” for some integer becomes,

The even integer is “b” therefore for some integer “s” it becomes

Substituting the value of a and b in the equation 2a + 3b as,

Simplifying the equation gives

As 2r + 3s + 1 is an integer and it is twice than 2a + 3b therefore it is even. An alternative method is to simplify the equation as,  

This shows that the a + 3s is an integer that is twice than 2a + 3b therefore it is even.

Question 2

Prove directly that If r is a rational number then is rational.

Given that “r” is a rational number and all integers are rational numbers therefore it gives

Substituting the rational number, it gives

We know that

The equation 1 gives the form of p/q, therefore, the equation becomes rational for .

Question 3

Prove by contrapositive the following: ∀ x ∈ Z, if  is odd then x is odd.

According to given statement, ∀ x ∈ Z, if  is odd therefore x is also odd by the contrapositive method, suppose that x is an even integer therefore  is even. Since x is considered as even integer and it becomes x = 2r for some . now substituting the value in the equation gives,

Inserting the value in the equation 1 gives that x is even then  is even. Also, if  is odd then x is odd.

Question 4

Find an explicit formula for each of the following sequences

a.   

Question 5

Compute the following

a.       

Question 6

Prove the following by mathematical induction

For all integers 

Step 1

Checking if the equation is true for n = 1, n =  2

Step 2

Using the induction hypothesis and inserting  the equation becomes

Step 3

Considering the case of n =  k + 1 and the equation becomes P (k + 1) and extracting k + 1 term from the summation, extracting  from the summation

Applying the induction hypothesis by replacing  with

Simplifying gives

Therefore, 

for n > 0 has been proven true.

Question 7

Consider the sequence defined by

a.       Write out the first five terms.

For first 5 terms 

b.      Now find an explicit formula for this sequence.

Examining the following sequence with the values as follows,

The sequence becomes

Question 8

Consider the sequence defined by the explicit formula:

for all integers n ≥ 0.  Show that this sequence satisfies the recurrence relation:

                               for all integers k ≥ 1

Consider n = 1 for the initial condition and it should be according to the closed formula as,

The next process is to check if the proposed solution satisfies the recurrence relation as, 

The recurrence relation shows the solution is clear under the given relation.

Question 9

Then “s” is an integer and it becomes

The equation b satisfies the condition for being in B and every element in A is in B and it implies that  

Equating both sides of the equation gives

Inserting it back in the equation,

As all the elements satisfies the condition therefore all the element in B are in A that implies,

Question 10

Let the universal set be the set U = {a, b, c, d, e, f, g, h, i, j} and let A = {a, b, c, e, f} and B = {b, f, g, h}. Compute the following:

(a)