Prove that if n ∈ ℕ where n > 1, then n + 1 is odd.

  • Thread starter Dustinsfl
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In summary, "n ∈ ℕ" means that n is a natural number, or a positive integer. This statement also specifies that n > 1 to ensure it only applies to natural numbers greater than 1. The proof for n + 1 being odd uses the definition of an odd number and an example is provided to illustrate this statement. This statement is always true for all natural numbers greater than 1, as it is a fundamental property of odd and even numbers.
  • #1
Dustinsfl
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Prove that if n ∈ ℕ where n > 1, then n! + 1 is odd.
n = k + 1, k = 1,...,∞
Now substitute n!+1 = (k+1)! + 1 = ?

I am not sure what to do next or if that is on the right path.
 
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  • #2
If n > 1 then n! is an even number since it has 2 as one of its factors, so n! is even for n > 1. Conclude.
 

1. What does "n ∈ ℕ" mean?

It means that n is a natural number, or a positive integer.

2. Why is n > 1 included in the statement?

This is included to specify that the statement only applies to natural numbers greater than 1, as the statement would not hold true for n = 1.

3. How do you prove that n + 1 is odd?

We can prove this by using the definition of an odd number, which states that an odd number is one that cannot be divided evenly by 2. Since n is a natural number, n + 1 will always be one more than n, and therefore, when divided by 2, n + 1 will always have a remainder of 1, making it odd.

4. Can you provide an example to illustrate this statement?

Sure, for example, if n = 5, then n + 1 = 6, which is an even number. However, if n = 9, then n + 1 = 10, which is also an even number. This pattern continues for all natural numbers, proving that n + 1 is always even.

5. Is this statement always true for all natural numbers?

Yes, this statement is always true for all natural numbers greater than 1. It is a fundamental property of odd and even numbers.

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