This type of exercise tends to be easier, as it tells you and advance what method you should use.
Remember the beauty of proving a theorem, is that you have an arsenal of methods to use and most of the time, your skill identifying what method to use is what will take you to a successful proof.
a) Using a proof by contraposition
In a proof by contraposition, we use the fact that p->q is logically equivalent to its contrapositive ¬q -> ¬p.
In this case, the contrapositive is: if n is an integer and it is not even, then n3 +5 is not odd.
In other words: if n is an integer and n is odd, then n3 +5 is even.
If n is odd, then n=2k+1 for a certain integer k.
Substituting n=2k+1 in n3 +5 we get the following:
n3 +5 = (2k+1)3 + 5 = (8k3 + 12k2 + 6k + 1) + 5 = 8k3 + 12k2 + 6k + 6 = 2(4k3 + 6k2 + 3k + 3)
Therefore, n3 +5 is even because it is two times an integer number ∎
b) Using a proof by contradiction
To use proof by contradiction to prove a conditional statement(p->q), we do the following:
- Assume the negation of q (the conclusion of the conditional statement).
- Use p and ¬q to arrive at a contradiction.
This type of proofs is valid because p->q ≡ (p^q)-> F.
p= “n3 +5 is odd”
¬q = “n is odd”
Because n is odd, n=2k+1
Substituting n=2k+1 in n3 +5 we get the following:
n3 +5 = (2k+1)3 + 5 = (8k3 + 12k2 + 6k + 1) + 5 = 8k3 + 12k2 + 6k + 6 = 2(4k3 + 6k2 + 3k + 3)
Therefore, n3 +5 is even because it is two times an integer number.
In other words, p is false.
Because p is true (premise) and ¬p is also true, we have a contradiction.
This completes the proof by contradiction ∎
Related posts:
- Use a direct proof to show that the sum of two even integers is even
- Use a direct proof to show that the sum of two odd integers is even
- Use a proof by contradiction to prove that the sum of an irrational number and a rational number is irrational
- Prove or disprove that the product of two irrational numbers is irrational
- Prove that if m and n are integers and mn is even, then m is even or n is even
