Contradiction Method: Proving Statements Through Contradiction and Supposition

[Devil Moo]

Proof by contradiction starts by supposing a statement, and then shows the contradiction.

1. Homework Statement

Now, there is a statement $A$.
Suppose $A$ is false.
It leads to contradiction.
So $A$ is true.

My question:
There are two statements $A$ and $B$.
Suppose $A$ is true.
Further suppose $B$ is false.
It leads to contradiction.
Can I conclude that "if $A$ is true, then $B$ is false."

Moreover, how do I distinguish between "Proof by Contradiction" and "Proof by Exhaustion"?

 

[andrewkirk]

Can I conclude that "if A is true, then B is false."

No. But you can conclude that if A is true then B is true.

Moreover, how do I distinguish between "Proof by Contradiction" and "Proof by Exhaustion"?

Proof by exhaustion does not necessarily involve any contradictions. Here's the definition from wiki:

Proof by exhaustion, also known as proof by cases, proof by case analysis, perfect induction, or the brute force method, is a method of mathematical proof in which the statement to be proved is split into a finite number of cases or sets of equivalent cases and each type of case is checked to see if the proposition in question holds.

 

[Mark44]

Moreover, how do I distinguish between "Proof by Contradiction" and "Proof by Exhaustion"?

Here's an example of a proof by exhaustion.
Assume that n is a positive integer. Then n(n + 1)(n + 2) is divisible by 3.

Case 1: n = 3k, for some integer k
Then n(n + 1)(n + 2) = 3k(3k + 1)(3k + 2), which clearly has a factor of 3, and so is divisible by 3

Case 2: n = 3k + 1, for some integer k
Then n(n + 1)(n + 2) = (3k + 1)(3k + 2)(3k + 3) = (3k + 1)(3k + 2)3(k + 1), which has a factor of 3, so is divisible by 3.

Case 3: n = 3k + 2, for some integer 8
Then n(n + 1)(n + 2) = (3k + 2)(3k + 3)(3k + 4) = (3k + 2)3(k + 1)(3k + 4), also has a factor of 3, so is divisible by 3.

There are no other possible cases. Any integer n falls into one of three equivalence classes: $n \equiv 0 (\mod 3)$, or $n \equiv 1 (\mod 3)$, or $n \equiv 2 (\mod 3)$. IOW, when an integer n is divided by 3, the remainder will be 0, 1, or 2. The three cases above are based on this fact.

 

[Mark44]

Reopened...