Answer:
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 <em>induction hypothesis</em>:
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’’.