This exercise came from my discrete mathematics book.
Determine which statements are true and which are false. Prove those that are true and disprove those that are false.
problem 6: $\frac{\sqrt 2}{6}$ is rational.
I assumed that $\frac{\sqrt 2}{6}$ being rational was true.
My first attempt was to use prove by contradiction. So, i assumed that $√2/6$ is irrational, which led to a dead-end since you can't proof irrationality.
My second attempt was a direct proof:
Proof: Suppose $\frac{\sqrt 2}{6}$ is rational. By definition of rational, there exists integers $p$ and $q$ such $\frac{\sqrt 2}{6}=\frac{q}{p}$ with $q≠0$.
Then, $\frac{\sqrt 2}{6}=\frac{q}{p}$$⟹$ $\sqrt{2}=\frac{q}{p}*6$ $⟹$ $\sqrt{2}=\frac{6q}{p}$
Note that $6p$ and $q$ are integers since $p$ and $q$ are integers. Also, $q≠0$ by the zero-product property. Hence, by definition of rational, there exists integers $6p$ and $q$ such that $\sqrt{2}=\frac{6q}{p}$; thus, $\sqrt{2}$ is rational. However, there is a contradiction since we know that $\sqrt{2}$ is irrational.
From here I discovered that the statement I am trying to prove is false.
- So, I figured that I should disprove the statement.
At the time the only method of disprove I know was counterexample. So, I tried to come up with a counterexample, but the problem is not knowing how to provide a counterexample while dealing with square root.
I searched online for other ways to disprove statements that didn't require a counterexample.
I came across this pdf:
"How to disprove statements"
In this they introduce the concept of disproof:
It turns out that there is a very simple and utterly convincing procedure that proves a statement is false. The process of carrying out this procedure is called disproof.
Suppose you want to disprove a statement $P$. In other words you want to prove that $P$ is false. The way to do this is to prove that $∼ P$ is true, for if $∼ P$ is true, it follows immediately that $P$ has to be false
They then separate them into universal disproof, which is just counterexample.
To disprove a universally quantified statement such as $∀x ∈ S,P(x)$
we must prove its negation. Its negation is $∼ (∀x ∈ S,P(x)) = ∃ x ∈ S,∼P(x)$. The negation is an existence statement. To prove the negation is true, we just need to produce an example of an x ∈ S that makes ∼ P(x) true.
Now Disproof by Contradiction is the part I am interested in:
Contradiction can be a very useful way to disprove a statement. To see how this works, suppose we wish to disprove a statement P. We know that to disprove $P$, we must prove $∼ P$. To prove $∼ P$ with contradiction, we assume $∼∼ P$ is true and deduce a contradiction. But since $∼∼ P = P$, this boils down to assuming P is true and deducing a contradiction.
So, my question is whether Disproof by Contradiction is valid method, or if there are better ways of disproving mathematical statements?