Let a be a finite, non-empty subset of r. prove that a has a maximum and a minimum. (recall that a maximum of a set a is an upper bound for the set that belongs to a, and a minimum of a set is a lower bound for the set that belongs to the set.) hint: this result may seem so obvious that it isn't clear h one way is to use induction on the number of elements of the set.

Let us use mathematical induction to prove the statement. So, we are going to start checking the statement for the first natural numbers.

: Our set is . So, obviously, is the maximum and minimum of our set. Then, the statement is true for .

: Our set is . Necessarily, or . In both cases, there is a minimum and a maximum.

Once we have our statement checked for the initial cases, we state our induction hypothesis:

For every finite set A of elements there exists a maximum and a minimum.

Now, let us prove the that the above assertion is true for sets with elements.

Our set is and we want to find

.

Notice that this problem is equivalent to solve

,

i.e, to find the maximum among numbers, we can find first the miximum among and then compare with the other one.

Now, using our induction hypothesis we know that there is a maximum in the set , because it has elements. Let us write

.

So, in order to find the maximum of A, we have to find the maximum of . As we have checked at the beginning, there is a maximum in , and it is the maximum of A.

Hence, we have completed the prove for the existence of the maximum of a set with elements. The prove for the existence of the minimum is analogue, we just need to change ‘‘maximum’’ for ‘‘minimum’’.

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

i think is g(1)=-1.

hope that i can