DIscrete math, Please do both parts! Thank you :) Let A and B be two natural num

Question

DIscrete math, Please do both parts! Thank you :)

Let A and B be two natural numbers. Follow the proof given below and identify which steps) is (are) not valid Step # Justification We start with this assumption Multiply by A on each side uation (commutativity); substract B on both side (A-B)(A+B)=(A-B)B A+B=B B+B=B 2B = B |Identity: A--B-=(A-B)(A+B); factor B on right side Simplify: divide by A-B on each side From step 1,A -B, therefore A+B=B+B By definition, B+B-2B Simplify by B Hints -An integer number N is odd if it can be written in the form N 2q +1, where q is - An integer number N is even if it can be written in the form N 2q, where q is an - An integer number N is a multiple of an integer number k if there exists an integer an integer number integer number number q such that N kq Prove the following statements a) The sum of any three consecutive even numbers is always a multiple of 6 b) The sum of any four consecutive odd numbers is always a multiple of8 c) Prove that if you add the squares of two consecutive integer numbers and then add one, you always get an even number

Explanation / Answer

1st part

step 5 is invalid as there we are divinding the expression by A-B

In the 1st step we have assumed that A= B and so A-B is 0 . We cannot divide the expression by 0 .

Hence step 5 is invalid.

part 2

a)
let the 3 consecutive even number be as follows:

2q -2 , 2q , 2q+2
now adding the 3 numbers as follows:

2q -2 + 2q + 2q+2 = 6q

Since 6q will always be divisible by 6 we can say that it will always be multiple of 6

b)

let the 4 consecutive odd number be as follows:

2q -3 , 2q-1 , 2q+1 , 2q+3
now adding the 4 numbers as follows:

2q -3 + 2q-1 + 2q+1 + 2q+3 = 8q

Since 8q will always be divisible by 8 we can say that it will always be multiple of 8

c)
let 2 consecutive integers be as follows:
q , q+1

The sum of their square is as follows:

q^2 + q^2 +2q + 1 = 2q^2 +2q +1

adding 1 to it
=2q^2 +2q +1+1 = 2q^2 +2q +2

we know that sqaure of even number is always even and since we have taken 2 consecutive integers 1 is even. hence we take q is even
so q^2 is even
2q^2 is even
2q is even
2 is even

hence on adding even to even the result is also an even number.
Hence we can say adding 1 to sqaure of q , q+1 is even.

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