Problem 0.1. Prove that the Axiom of Induction and the Well Ordering Principle are equivalent (hint: the sets S and T play complementary roles). Here are some statements that we have wanted to prove and now can, give two proofs for

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Problem 0.1.

We need to come up with a way to state what we mean by the natural numbers. Informally we know that they are the numbers that you get by starting at 1 and counting up repeatedly by 1, we just need to figure out how to phrase this precisely to use in a proof. Counting up by 1 just means replacing \( n \) by \( n + 1 \) so that is clear, what we want to make precise is the claim that you can count up to any natural number eventually. The reason this is difficult is that the process goes on forever, and infinite things are always tricky. There are two common ways this is done:

- **(Axiom of Induction)** If \( S \) is a subset of \( \mathbb{N} \) such that \( 1 \in S \) and, for all \( n \in S \) we also have \( n + 1 \in S \) then \( S = \mathbb{N} \)
- **(Well Ordering Principle)** If \( T \) is a non-empty subset of \( \mathbb{N} \) then there is an element \( t_0 \in T \) so that \( t_0 \leq t \) for all \( t \in T \)

Although these look very different, they are equivalent. In fact, they are "contrapositives" of each other. If you have a statement "If \( P \) is true then \( Q \) is true" the contrapositive of "If \( Q \) is false then \( P \) is false". They are the same because they are both ways of saying that it is impossible for \( P \) to be true while \( Q \) is false. Think about some real world examples of \( P \) and \( Q \) and see how the two ways of phrasing things sound the same or different.

**Problem 0.1.** Prove that the Axiom of Induction and the Well Ordering Principle are equivalent (hint: the sets \( S \) and \( T \) play complementary roles).

Here are some statements that we have wanted to prove and now can, give two proofs for each, one each with the Axiom of Induction and the Well Ordering Principle.

(1) 1 is the smallest natural number

(2) For any natural number \( n \) there are
Transcribed Image Text:We need to come up with a way to state what we mean by the natural numbers. Informally we know that they are the numbers that you get by starting at 1 and counting up repeatedly by 1, we just need to figure out how to phrase this precisely to use in a proof. Counting up by 1 just means replacing \( n \) by \( n + 1 \) so that is clear, what we want to make precise is the claim that you can count up to any natural number eventually. The reason this is difficult is that the process goes on forever, and infinite things are always tricky. There are two common ways this is done: - **(Axiom of Induction)** If \( S \) is a subset of \( \mathbb{N} \) such that \( 1 \in S \) and, for all \( n \in S \) we also have \( n + 1 \in S \) then \( S = \mathbb{N} \) - **(Well Ordering Principle)** If \( T \) is a non-empty subset of \( \mathbb{N} \) then there is an element \( t_0 \in T \) so that \( t_0 \leq t \) for all \( t \in T \) Although these look very different, they are equivalent. In fact, they are "contrapositives" of each other. If you have a statement "If \( P \) is true then \( Q \) is true" the contrapositive of "If \( Q \) is false then \( P \) is false". They are the same because they are both ways of saying that it is impossible for \( P \) to be true while \( Q \) is false. Think about some real world examples of \( P \) and \( Q \) and see how the two ways of phrasing things sound the same or different. **Problem 0.1.** Prove that the Axiom of Induction and the Well Ordering Principle are equivalent (hint: the sets \( S \) and \( T \) play complementary roles). Here are some statements that we have wanted to prove and now can, give two proofs for each, one each with the Axiom of Induction and the Well Ordering Principle. (1) 1 is the smallest natural number (2) For any natural number \( n \) there are
Expert Solution
Step 1

The well-ordering principle is a property of the positive integers which is equivalent to the statement of the principle of mathematical induction. Every nonempty set S of non-negative integers contains a least element; there is some integer a in S such that a≤b for all b’s belonging. 

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