- i)
- .
- ii)
- for all , .

If is an inductive subset of , then

etc. Hence every inductive set contains

If is an inductive subset of
, then

so the only inductive subset of is itself. The set

is an inductive subset in .

- a)
- Find an inductive subset of , such that and .
- b)
- Find an inductive subset of , such that and .

Proof: Since is in every inductive set, . Let be an inductive subset of . Then for all ,

Hence

Hence is inductive.

(whatever ``" might mean).

Proof: Let be a proposition form on
satisfying (3.11) and
(3.12). Let

I want to show that is inductive. Well, , by (3.11). Let be any element in .

**Case 1.**-

**Case 2.**-
If , then is false, so is true.

This shows that is inductive. Since every inductive set contains , ; i.e., for all , is true.

Proof: Let be the proposition form on
defined by

Then says `` " which is true. For all ,

By the induction theorem, is true for all ; i.e.,

Now define a proposition form on by

Then says `` " which is true. For all ,

By the induction theorem, is true for all ; i.e.,

Proof: Define a proposition form on
by

Clearly is true. let . To show that , I'll show that and that . Well

and

Hence , and by induction is true for all .

Proof: Define a proposition form on
by

and

Hence , and by induction, is true for all .

Proof: For each
define a proposition form on
by

I'll show that for each , is true for all . Now says `` or " which is true, since . Now let . To show that , I'll show that

and that

By the previous lemma

Also

This completes the proof that , so by induction is true for all .

Proof: Suppose

Since , the previous theorem says . This contradicts corollary 3.15, so (3.22) is false.

Proof: I will show that if is a subset of having no least element, then .

Let be a subset of
having no least element. For each
define
a proposition by

Now , since if were in it would be a least element for . Hence all elements in are greater than , and is true. Now let be a generic element of . Then

since if were in , it would be a least element for . Thus

and by induction, is true for all . It follows that , since if contained an element , then would say that .

(3.26) |

Hence for all . Now note:

Since is false, I cannot apply the induction theorem. Notice that when I prove I do

. Suppose

Proof: Let be the proposition form on
defined by

(observe that so is defined). Then , so is true by (3.28). For all ,

so

By the induction theorem, is true for all ; i.e., is true for all . To complete the proof, I need to show that

It is clear that

To show the opposite inclusion, observe that if and , then , and by theorem 3.19, .

In example 3.25, we showed that for all and that is true. Hence, by our generalized induction theorem we can conclude that is true for all with .

- a)
- What are the even numbers in ?
- b)
- What are the odd numbers in ?

- a)
- Let be a field. Prove that every element in is either even or odd. HINT: Let is even or is odd".
- b)
- Let be an ordered field. Prove that no element of is both even and odd. Why doesn't this contradict the result of exercise 3.31?

for ``one" signifies a measure of some plurality, and `` a number" signifies a measured plurality or a plurality of measures. Therefore, it is also with good reason that unity is not a number; for neither is a measure measures, but a measure is a principle, and so is unity . [5, page 237, N, 1, 1088a5]

Zero was first recognized to be a number around the ninth century. According to [31, page 185] Mahavira (ninth century) noted that any number multiplied by zero produces zero, and any number divided by zero remains unchanged! Bhaskara (1114-1185) said that a number divided by is called an infinite quantity.

Although arguments that are essentially arguments by induction appear in Euclid, the first clear statement of the induction principle is usually credited to Blaise Pascal. (1623-1662) who used induction to prove properties of Pascal's Triangle.[36, page 73]

I believe that the idea of defining the natural numbers to be things that are in every inductive set should be credited to Giuseppe Peano [37, page 94, Axiom 9]. In. 1889, Peano gave a set of axioms for natural numbers (starting with ), one of which can be paraphrased as: If is any set, such that and for all , , then .