MATH 201 Quiz I

1. State the well-ordering property of the set of natural numbers, N.

Every non-empty subset of N has a smallest element.

2. State the Archimedean property of the set of natural numbers, N.

Given any two positive integers a and b, there exists a positive integer n such that an > b.

3. State the first principle of mathematical induction.

Let S be a subset of N that satisfies the following two properties:

(A) 1 is a member of S

(B) if k is a member of S then k+1 is a member of S.

Then S = N.

4. Albertine, a student of Number Theory, is trying to prove that the well-ordering property of N implies the Archimedean property of N. Alas, her proof has several gaps because she is rushed for time. Please help her by filling in the missing steps indicated by the underscores:

Suppose that the Archimedean property is not true for N.

So there exist positive integers a and b for which it is not the case that na > b for some positive integer n. Thus na < b for all natural numbers n.

Let S be the set {b-na | n Î N}

Then S is a subset of N because b-na is an integer since the set of integers is closed under multiplication and subtraction, and b-na>0 by assumption. Thus b-na must be a positive integer.

Furthermore, S is non-empty because, for example, b-5a Î S.

Then, by the well-ordering property of N, we know that S possesses a smallest element. By definition of S, this smallest element may be written in the form b – n*a, where n* Î N.

But this leads to a contradiction because: b-(n* + 1)a is also a member of S (since it is of the form b-na where n is a natural number) and yet b-(n* + 1)a is actually smaller than the presumed smallest element of S because b-(n* + 1)a = b – n*a – a < b – n*a. Thus we have a contradiction.

Thus we conclude that our assumption that the Archimedean property is not true for N is false. Thus for any positive integers a and b there exists a positive integer n such that na ³ b.

Extra Credit:

Using mathematical induction, prove that

(1)(2) + (2)(3) + …+ n(n+1) = n(n+1)(n+2)/3 for all natural numbers n.

Let S = {nÎN | (1)(2) + (2)(3) + …+ n(n+1) = n(n+1)(n+2)/3 }.

We must show that 1ÎS and that kÎS implies k+1ÎS.

To show 1ÎS, we must verify that (1)(2) = 1(1+1)(1+2)/3. Note that the left-hand side of this equation simplifies to 2 and that the right-hand side of this equation also simplifies to 2.

Next, suppose that kÎS. Then

(*) (1)(2) + (2)(3) + …+ k(k+1) = k(k+1)(k+2)/3

We claim that k+1ÎS. In other words, we must show that

(**) (1)(2) + (2)(3) + …+ (k+1)(k+2) = (k+1)((k+1)+1)((k+1)+2)/3

Now the LHS of (**) =

(1)(2) + (2)(3) + …+ k(k+1) + (k+1)(k+2) = (by associativity of addition)

[(1)(2) + (2)(3) + …+ k(k+1)] + (k+1)(k+2) = (using (*))

k(k+1)(k+2)/3 + (k+1)(k+2) = (using algebra)

(k+1)(k+2)(k+3)/3

Now, the RHS of (**) simplifies to (k+1)(k+2)(k+3)/3, and thus we find that

The LHS of (**) = RHS of (**). So k+1 Î S.

Thus, by mathematical induction, we have shown that S = N.