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1. Prove using induction that for any natural number n we have: -1^2 + 2^2 - 3^2 + """ + (-1)^n n^2 = ((-1)^n n(n+1))/2 2. Let the Fibonacci sequence be defined by a1 = 1, a2 = 1, an+2 = an + an+1. Prove that for all natural numbers n, we have an < 2^n 3. Suppose that A is a set with n elements. Prove that A has 2^n subsets. 4. Point out the error in the proof of the following false statement. False Statement: All positive even numbers are a power of 2. faulty proof: An even number is divisible by two, so it is of the form 2n for some positive integer n. We prove the statement by induction on n. We proceed by induction on n. Base case: For n = 1, we have 2n = 2 = 2^1, which is indeed a power of 2. Induction step: Suppose the statement has been proved for all even numbers up to 2N. The next even number is B = 2(N + 1). But since B is even, we have B = 2C, and we have C ? 2N, so C is a power of two, say 2^e. But then B = 2C = 2^{e+1} is also a power of 2, showing that the statement also holds for all even numbers up to 2(N + 1). 5. We have seen two principles: The Principle of mathematical induction: For any X ? N satisfying the following two properties • 1 ? X, • for any n ? N, if {1, """, n} ? X then it follows that n + 1 ? X, it holds that X = N The Well-ordering principle: Any nonempty set Y ? N contains a least element. Prove that the principle of mathematical induction implies the well-ordering principle.

          1. Prove using induction that for any natural number n we have:
-1^2 + 2^2 - 3^2 + """ + (-1)^n n^2 = ((-1)^n n(n+1))/2
2. Let the Fibonacci sequence be defined by a1 = 1, a2 = 1, an+2 = an + an+1. Prove that for all natural numbers n, we have an < 2^n
3. Suppose that A is a set with n elements. Prove that A has 2^n subsets.
4. Point out the error in the proof of the following false statement.
False Statement: All positive even numbers are a power of 2.
faulty proof: An even number is divisible by two, so it is of the form 2n for some positive integer n. We prove the statement by induction on n.
We proceed by induction on n.
Base case: For n = 1, we have 2n = 2 = 2^1, which is indeed a power of 2.
Induction step: Suppose the statement has been proved for all even numbers up to 2N. The next even number is B = 2(N + 1). But since B is even, we have B = 2C, and we have C ? 2N, so C is a power of two, say 2^e. But then B = 2C = 2^{e+1} is also a power of 2, showing that the statement also holds for all even numbers up to 2(N + 1).
5. We have seen two principles:
The Principle of mathematical induction: For any X ? N satisfying the following two properties
• 1 ? X,
• for any n ? N, if {1, """, n} ? X then it follows that n + 1 ? X,
it holds that X = N
The Well-ordering principle: Any nonempty set Y ? N contains a least element.
Prove that the principle of mathematical induction implies the well-ordering principle.

1. Prove using induction that for any natural number n we have:
-1^2 + 2^2 - 3^2 + """ + (-1)^n n^2 = ((-1)^n n(n+1))/2
2. Let the Fibonacci sequence be defined by a1 = 1, a2 = 1, an+2 = an + an+1. Prove that for all natural numbers n, we have an < 2^n
3. Suppose that A is a set with n elements. Prove that A has 2^n subsets.
4. Point out the error in the proof of the following false statement.
False Statement: All positive even numbers are a power of 2.
faulty proof: An even number is divisible by two, so it is of the form 2n for some positive integer n. We prove the statement by induction on n.
We proceed by induction on n.
Base case: For n = 1, we have 2n = 2 = 2^1, which is indeed a power of 2.
Induction step: Suppose the statement has been proved for all even numbers up to 2N. The next even number is B = 2(N + 1). But since B is even, we have B = 2C, and we have C ? 2N, so C is a power of two, say 2^e. But then B = 2C = 2^e+1 is also a power of 2, showing that the statement also holds for all even numbers up to 2(N + 1).
5. We have seen two principles:
The Principle of mathematical induction: For any X ? N satisfying the following two properties
• 1 ? X,
• for any n ? N, if 1, """, n ? X then it follows that n + 1 ? X,
it holds that X = N
The Well-ordering principle: Any nonempty set Y ? N contains a least element.
Prove that the principle of mathematical induction implies the well-ordering principle.

Added by Josefina L.

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