Evaluation of Proofs See the instructionsfor Exercise (19) on page 100 from Section 3.1. (a) Proposition. If m is an odd integer, then .mC6/ is an odd integer. Proof. For m C 6 to be an odd integer, there must exist an integer n such that mC6 D 2nC1: By subtracting 6 from both sides of this equation, we obtain m D 2n6C1 D 2.n3/C1: By the closure properties of the integers, .n3/ is an integer, and hence, the last equation implies that m is an odd integer. This proves that if m is an odd integer, then mC6 is an odd integer

(A) Assume m is an odd integer.

Therefore m is of the m=2n-1 type where n is any number.

We have m + 6 = 2n-1 + 6 = 2n+5=2n+4 + 1 = 2(n+2) + 1, in which n + 2 is an integer, adding 6 to both ends.

Because m+6 is of the 2x + 1 type, where x =  n + 2; then m + 6 is an odd integer too.

(B) Provided that mn is an integer also. For some integer y and m, n are integers, this means mn is of the form mn = 2y.

And y = mn/2.

Therefore either m is divided by 2 or n is divided by 2 since y, m, n are all integers. To put it another way, either m is a multiple of 2 or n is a multiple of 2, which means that m is even or n is even.

answer: finding the extract measurements and knowing the points

step-by-step explanation:

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm