8. (8 marks) prove that for all integers m and n, m + n and m-n are either both even or both odd

Answer with explanation:

Let m and n are integers

To prove that m+n and m-n are either both even or both odd.

1. Let m and n are both even

We know that sum of even number is even and difference of even number is even.

Suppose m=4 and n=2

m+n=4+2=6 =Even number

m-n=4-2=2=Even number

Hence, we can say m+n and m-n are both even .

2. Let m and n are odd numbers .

We know that sum of odd numbers is even and difference of odd numbers is even.

Suppose m=7 and n=5

m+n=7+5=12=Even number

m-n=7-5=2=Even number

Hence, m+n and m-n are both even .

3. Let m is odd and n is even.

We know that sum of an odd number and an even number is odd and difference of an odd and an even number is an odd number.

Suppose m=7 , n=4

m+n=7+4=11=Odd number

m-n=7-4=3=Odd number

Hence, m+n and m-n are both odd numbers.

4.Let m is even number and n is odd number .

Suppose m=6, n=3

m+n=6+3=9=Odd number

m-n=6-3=3=Odd number

Hence, m+n and m-n are both odd numbers.

Therefore, we can say for all inetegers m and n , m+n and m-n are either both even or both odd.Hence proved.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet