What a bizarre question to ask, whether 3 is a rational or irrational number. We know 3 is rational. So, why are we even discussing the question?
In a first number theory course, there’s typically a proof that $\sqrt{2}$ is irrational. Or in more formal notation, we prove that $\sqrt{2}\not\in\mathbb{Q}$, meaning that $\sqrt{2}$ can’t be expressed as a fraction $a/b$ for any $a, b \in \mathbb{Z}$.
The traditional proof (shown below) is a clever one that uses the properties of divisibility to arrive at a contradiction. You can even take the exact same proof, swap in a 5 in place of the 2, and prove that $\sqrt{5}\not\in\mathbb{Q}$.
But, could you swap yet another number such as 9 into the proof, and “prove” that $\sqrt{9}\not\in\mathbb{Q}$? Doing so would “prove” that 3 is irrational, since $\sqrt{9} = \pm3$.
Today’s practice problem presents an incorrect “proof” that 3 is irrational. As a practice problem for your number theory or proofs course, your goal is to find the flaw in the “proof” which purports to show $\sqrt{9}\not\in\mathbb{Q}$.
First, for reference, here’s a correct but incomplete proof that $\sqrt{2}\not\in\mathbb{Q}$:
There aren’t any incorrect statements in this proof. But, I feel it’s missing details it would need to get full marks. Try to spot any unjustified steps in the proof.
Assume that $\sqrt{2}\in\mathbb{Q}$.
By the definition of the rational numbers $\mathbb{Q}$, we can express $\sqrt{2} = n/m$ where $n, m \in \mathbb{Z}$ and $m \neq 0$.
If the numerator $n$ and the denominator $m$ share any common factors, we can divide them off to get reduced numerator $a$ and reduced denominator $b$. That is,
\begin{equation*}
\sqrt{2} = \frac{n}{m} = \frac{a}{b},
\end{equation*}
where $a, b \in \mathbb{Z}$, $b \neq 0$, and $a$ and $b$ share no common factors. So,
\begin{align*}
\sqrt{2} &= \frac{a}{b} \\
\Rightarrow~~~~2 &= \frac{a^2}{b^2} \\
\Rightarrow~~2b^2 &= a^2.
\end{align*}
Because $a^2$ is equal to 2 times some integer, $a^2$ is divisible by 2. Thus, $a$ is divisible by 2.
Given that $a$ is divisible by 2, we can rewrite $a = 2k$ for some $k\in\mathbb{Z}$ and return to our equations:
\begin{align*}
2b^2 &= a^2 \\
\Rightarrow~~2b^2 &= (2k)^2 \\
\Rightarrow~~2b^2 &= 2\cdot 2k^2 \\
\Rightarrow~~~b^2 &= 2k^2.
\end{align*}
As before: because $b^2$ is equal to 2 times some integer, $b^2$ is divisible by 2. So, $b$ is divisible by 2.
This means that both $a$ and $b$ are divisible by 2. However, our construction of $\sqrt{2} = a/b$, above, guaranteed that $a$ and $b$ shared no common factors.
We’ve arrived at a contradiction, and the only assumption we made in order to reach that contradiction is that $\sqrt{2}\in\mathbb{Q}$. This means that assumption must be false, and $\sqrt{2}\not\in\mathbb{Q}$.
We can run the same proof on, as one example, $\sqrt{5}$. You may find it helpful, before moving on incorrect “proof” below, to run through the correct proof once more to show $\sqrt{5}\not\in\mathbb{Q}$.
At this point, you may be thinking: wait a second! If we can prove $\sqrt{5}\not\in\mathbb{Q}$ just by plugging a 5 into the proof in place of the 2, doesn’t that mean we could plug any other number we like into the proof? Let’s try it with 9, and see if we can’t “prove” that $\sqrt{9}\not\in\mathbb{Q}$ using the same proof:
Here is an incorrect “proof” that $\sqrt{9}\not\in\mathbb{Q}$:
It bears repeating: the “proof” below is incorrect (3 is most definitely not irrational). The exercise, for students in a number theory or proofs course, is to spot the error.
Assume that $\sqrt{9}\in\mathbb{Q}$.
By the definition of the rational numbers $\mathbb{Q}$, we can express $\sqrt{9} = n/m$ where $n, m \in \mathbb{Z}$ and $m \neq 0$.
If the numerator $n$ and the denominator $m$ share any common factors, we can divide them off to get reduced numerator $a$ and reduced denominator $b$. That is,
\begin{equation*}
\sqrt{9} = \frac{n}{m} = \frac{a}{b},
\end{equation*}
where $a, b \in \mathbb{Z}$, $b \neq 0$, and $a$ and $b$ share no common factors. So,
\begin{align*}
\sqrt{9} &= \frac{a}{b} \\
\Rightarrow~~~~9 &= \frac{a^2}{b^2} \\
\Rightarrow~~9b^2 &= a^2.
\end{align*}
Because $a^2$ is equal to 9 times some integer, $a^2$ is divisible by 9. Thus, $a$ is divisible by 9.
Given that $a$ is divisible by 9, we can rewrite $a = 9k$ for some $k\in\mathbb{Z}$ and return to our equations:
\begin{align*}
9b^2 &= a^2 \\
\Rightarrow~~9b^2 &= (9k)^2 \\
\Rightarrow~~9b^2 &= 9\cdot 9k^2 \\
\Rightarrow~~~b^2 &= 9k^2.
\end{align*}
As before: because $b^2$ is equal to 9 times some integer, $b^2$ is divisible by 9. So, $b$ is divisible by 9.
This means that both $a$ and $b$ are divisible by 9. However, our construction of $\sqrt{9} = a/b$, above, guaranteed that $a$ and $b$ shared no common factors.
We’ve arrived at a contradiction, and the only assumption we made in order to reach that contradiction is that $\sqrt{9}\in\mathbb{Q}$. This means that assumption must be false, and $\sqrt{9}\not\in\mathbb{Q}$. And, since $\sqrt{9}=\pm3$, that means $3\not\in\mathbb{Q}$.
There’s clearly an error in any “proof” that claims to demonstrate $3\not\in\mathbb{Q}$, given that we can express 3 as the fraction $3/1$. So, as practice problems:
- Where is the error, or where are the errors, in the “proof” that $\sqrt{9}\not\in\mathbb{Q}$?
(So that you don’t go down the wrong path with this practice problem: it has nothing to do with the $\pm$.) - What’s different about $\sqrt{9}$, versus $\sqrt{2}$ or $\sqrt{5}$, that makes the “proof” fall apart for $\sqrt{9}$? Put another way: what type of number $n$ can we use this proof for, to correctly show that $\sqrt{n}\not\in\mathbb{Q}$?
- Return to the $\sqrt{2}$ proof and add the missing details. There are no incorrect statements in the proof as it’s written above. But, the fact the proof seems to work for both $\sqrt{2}$ and $\sqrt{9}$ demonstrates that the $\sqrt{2}$ proof has to be missing something. There needs to be at least one “…because…” statement in that proof, to differentiate the $\sqrt{2}$ proof from the $\sqrt{9}$ “proof”.
I’ve also written a (related) practice problem that uses proof by contradiction, to help solidify students’ understanding of the situation:
- Prove that $\sqrt{8}\not\in\mathbb{Q}$. We’ve already demonstrated that $\sqrt{2}\not\in\mathbb{Q}$, and you should make use of that fact in your proof.
Be careful: you won’t be able just to plug an 8 into our previous proof, in place of the 2 or 5. That’d work just as poorly as plugging a 9 into the proof did, despite $\sqrt{8}$ actually being irrational (unlike $\sqrt{9}$).
Working together on practice problems like this one can be part of any tutoring session. For more discussions, and to arrange for personalized tutoring for yourself or your study group, check out Vancouver Computer Science Tutoring.