Show the following propositions using a direct proof.

(a)

If a is an odd integer show a² + 3a + 5 is odd

An odd integer is of the form 2k+1 for some integer k.

a = 2k + 1

a² = (2k + 1)² = 4k² + 4k + 1

3a = 6k+3

a² + 3a + 5 = 4k² + 4k + 1 + 6k+3 + 1 =  4k² + 10k + 4 + 1 = 2(2k² + 5k + 2) + 1

Since k is an integer, so is l=2k² + 5k + 2, so a² + 3a + 5  is in the form 2l+1 so is odd.  

QED

(b)

a | b

is read a divides b, aka b  is a multiple of a.  So there's some integer k such that

b = ak

Squaring,

b² = a²k²

That shows b² is an integer multiple of a² so we conclude

a² | b²    

QED

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