Can this proof that $\sqrt{2}$ is irrational be rewritten using only integers?

Question

Can this proof that $\sqrt{2}$ is irrational be rewritten using only integers?

Most proofs that $\sqrt{2}$ is irrational start with assuming that $2=\dfrac{a^2}{b^2}$ and derive a contradiction.

For a nice collection, see my question here: What is the most unusual proof you know that $\sqrt{2}$ is irrational?

These proofs generally only use properties of the integers.

However the following proof uses the actual $\sqrt{2}$ and thus requires the actual square roots between integers.

So I wondered if this could be modified so it did not require non-integers.

I will write the proof in this more general form:

If $n$ and $k$ are positive integers such that $k^2 < n < (k+1)^2$ then $\sqrt{n}$ is irrational.

Proof:

Suppose $\sqrt{n}=\dfrac{a}{b}$, so $a=b\sqrt{n}$ and $a\sqrt{n}=nb$.

Then

$\begin{array}\\ \sqrt{n} &=\dfrac{a}{b}\\ &=\dfrac{a}{b}\dfrac{\sqrt{n}-k}{\sqrt{n}-k}\\ &=\dfrac{a\sqrt{n}-ka}{b\sqrt{n}-kb}\\ &=\dfrac{nb-ka}{a-kb}\\ \end{array} $

Since $k^2 < n < (k+1)^2$, $k \lt \sqrt{n}=\dfrac{a}{b} \lt k+1 $ so $kb < a < kb+b$ or $0 < a-kb < b $.

This new fraction for $\sqrt{n}$ has a smaller denominator than $\dfrac{a}{b}$, and this can be arbitrarily continued.

By infinite descent, this is impossible, so there is a contradiction.

---

To not use the actual $\sqrt{n}$, it seems to me that we have to start with $n=\dfrac{a^2}{b^2}$ and then derive a contradiction.

I will now try to mimic the proof above and show where it fails.

Again, assume $k^2 < n < (k+1)^2$.

Suppose $n=\dfrac{a^2}{b^2}$, so $a^2=nb^2$ and $na^2=n^2b^2$.

Then

$\begin{array}\\ n &=\dfrac{a^2}{b^2}\\ &=\dfrac{a^2}{b^2}\dfrac{n-k^2}{n-k^2}\\ &=\dfrac{a^2n-k^2a^2}{b^2n-k^2b^2}\\ &=\dfrac{n^2b^2-k^2a^2}{a^2-k^2b^2}\\ \end{array} $

What we want at this point, to apply infinite descent, is that $a^2-k^2b^2 \lt b^2$.

However, all I can get is this:

Since $k^2 < n=\dfrac{a^2}{b^2} < (k+1)^2$, $k^2b^2 < a^2 < (k+1)^2b^2$ so $0 < a^2-k^2b^2 \lt (2k+1)b^2 $ and this is not good enough.

So, can this be done?

Thank you.

Answer 1

Both $k^2$ and $n$ are integers, with $k^2 < n$. Thus $n - k^2 \geq 1$. Therefore, $a^2 - k^2b^2 \geq b^2$.

Final edit: I am certain that $a^2 - k^2b^2 \geq b^2$, and I can now see that your identity comes from: $$a = \sqrt{n}b \\ b = \frac{a}{\sqrt{n}}$$ To be clear, your proof cannot work in the second case because: $$a^2 - k^2b^2 = b^2(n - k^2) \geq b^2$$ But this trick functions in the first proof as: $$a-kb = b(\sqrt{n}-k) < b$$ You don't actually need to use the inequality manipulation that you have at the end, once you can show that $a-kb$ is an integer, and it is less than $b$ (as you multiplied $b$ by a number smaller than $1$), your conclusion follows immediately.