Gaining useful understanding of axiomatic systems is like

seeing the forest for the trees.

It has been said that God gave us 1 .then man created the rest.

Thus, we begin from 1.

1.1 Peanos PostulatesPeanos Postulates or Axioms were due to Giuseppe Peano. Using the Peanos Postulates, we derive a set of numbers that will serve as the building block of the real number system.

1.1 Peanos PostulatesP1. 1 X. P1 P2. For each n X, there is a unique P2 n* X called the successor of n. P3. For each n* X, n* 1. P3 P5. If S X, 1 S and p S implies p* S, P5 then S = X.

P4. For each pair n, m X with n m, n* m* P4

By P1, 1 X. P1 By P2, 1* X. Let 1* = 2. P2 By P3, 1 and 1* = 2 are different natural P3 numbers. Again by P2, 2* X. P2 Since 1 2, it follows from P4 that 1* 2*. By P3, 2* 1. Let 2* = 3. P3 Proceeding in this manner, we get X = {1,2,3, }. This set we denote by N.

1.2 Principle of Mathematical Induction (PMI)Let S be a subset of N satisfying the properties: 1. 1 S 2. For all n N, if n S n+1S. Then S = N.Note: PMI is appropriate when showing that statements/properties are true for all natural numbers n such as summations, generalizations, and divisibility, among others.

1.2 Principle of Mathematical Induction (PMI)Steps: 1. Show that the statement/formula holds for n=1. 2. Assume that the statement/formula holds for n=k. This is the Hypothesis of Induction or HOI. Then, show that if the formula holds for n=k it also holds for n=k+1. Hence, by the PMI, the formula is true for all n.

1.2 Principle of Mathematical Induction (PMI)Example: 1. Show that 1+2+3++n = n(n+1)/2 for all for all n N . Proof: b.(1)(1+1)/2=2/2=1 which is the sum of the first 1 natural number. b. Let k be given such that 1+2+3++k = k(k+1)/2.

1.2 Principle of Mathematical Induction (PMI)Then, 1+2+3++k +(k+1) = [k(k+1)/2] + (k+1) (by HOI) = [k(k+1)/2] + 2(k+1)/2 = [(k2+k) +(2k+2)]/2 = (k2+3k+2)/2 = (k+1)(k+2)/2 = (k+1)[(k+1)+1]/2 Which is the formula for n=k+1.

1.2 Principle of Mathematical Induction (PMI)Thus, by the PMI, the formula n(n+1)/2 gives the sum of the first n natural numbers for any n.

1.2 Principle of Mathematical Induction (PMI)Example: 2. 4 divides 5n -1 for all n N. Proof: 5(1) -1=5-1=4, which is divisible by 4. b. Suppose that the formula is true for n=k. Then, 4 divides 5k -1.

1.2 Principle of Mathematical Induction (PMI)5k+1 -1 = 5(5k)-1 (using law on exponents) = (4+1)(5k)-1 (since 5 =4+1) = 4(5k) + 5k 1 (distributive property) By HOI, since 5k -1is divisible by 4 and also 4(5k), then 5k+1 -1 is divisible by 4. Hence, the formula is true for n=k+1. Thus, by the PMI, 5n -1 is divisible by 4 for all n.

1.2 Principle of Mathematical Induction (PMI)Example: 3. Show that xn - yn is divisible by x-y for all n N

Proof: x(1) - y(1) =x-y is divisible by x-y. b. Suppose that xk - yk is divisible by x-y.

1.2 Principle of Mathematical Induction (PMI)So, xk+1 yk+1 = xxk yyk xyk + xyk (add and subtract xyk) = xxk xyk yyk + xyk (rearranging the terms) = x (xk yk ) - yk (x-y) (by grouping terms)

1.2 Principle of Mathematical Induction (PMI)By HOI, since xk yk is divisible by x-y then xk+1 yk+1 is divisible by x-y. Hence, the formula is true for n=k+1. Thus, by the PMI, xn yn is divisible by x-y for all n.

1.2 Principle of Mathematical Induction (PMI)Caveats in using the PMI/ Clarifications: 2. The plausibility that a statement/formula is true may be shown by trying out several cases first. However, keep in mind that no (necessarily finite) number of verifications of special cases is ever enough to prove that a statement is true for every natural number.

1.2 Principle of Mathematical Induction (PMI)Caveats in using the PMI/ Clarifications: 2. Assumption is made in the HOI but we want to show that the statement is true for all natural numbers not just for an instance. 3. Always verify the statement for n=1. Never omit or gloss over the step even if its trivial.

1.2 Principle of Mathematical Induction (PMI)Exercises: Test the truth of each of the following statements for at least 5 particular elements of N. Then, show that each statement is true for all natural numbers. 1. 2+4+6++2n = n(n+1) 2. 6 divides 7n -1 3. 2n-1 n!

1.3 Well-ordering Principle for N (WOP)Every non-empty subset of N has a least element. Proof: If WOP is false, then there is a nonempty subset A of N having no smallest element. Hence, if a natural number m has the property that m 0. We have to find a number n with the property that n > n an s < . Since s = sup { an } , there is an element an such that s < an s. Since ( an ) is increasing n > n s < an an s. Hence n > n an s < . This means that lim a = s. nn

Example 1.5.6 Show that the sequence 4 2n 4 2 = 2, 2, , , 15 , 3 3 n! converges using Thm 1.5. Solution: Is the sequence bounded? Give a lower bound and an upper bound.2n . 0 is a lower bound since 0 n!

Now, we will show that 2 is an upper bound.

n!= 1 2 3 ... n 1 2 2 ... 2 = 2 Thus, 2n 1

n 1

n!

2n 1 1 n! 2n 2 n!

Therefore, 2 is an upper bound of the sequence Hence, the sequence is bounded since it has a lower bound and an upper bound.

Now we show that the sequence 4 2n 4 2 = 2, 2, , , 15 , 3 3 n! is (monotonic) decreasing.2n+ 1 2n We wish to show that for all n, ( n + 1) ! n! 2 2 2n 2n 1 1 2 n+ 1 n+ 1 n + 1 n! n! n+ 1 n 2 2 ( n + 1) ! n! an an + 1 .

Since the sequence is bounded and monotonic, by Theorem, the sequence is convergent.

Limit of SumsTheorem

Assume that the limits lim xn = x and lim y n = y are finite. Then lim ( xn + y n ) = x + y .n n n

Proof

Let > 0 be given.We have to find a number n with the property n > n xn + y n x y < .

To that end observe that also

> 0. 2

Limit of SumsProof

Hence there are numbers n1 and n2 such that

n > n1 xn x < and n > n2 y n y < . 2 2 + = . 2 2

Let now n =max ( n1, n2 ) . We have n > n xn + y n x y xn x + y n y