So I was thinking about how to prove the $\sqrt{2}$ is irrational while on holiday. (Bear in mind that at this point, I only knew that it was indeed irrational, not how to prove it.)

Anyway, I came up with my own proof with no knowledge of what "contradiction" was, this was just the easiest way.

My proof begins familiarly:

$\sqrt{2} = \frac{a}{b}$ ==> Suppose $\sqrt{2}$ is indeed rational, it can be expressed as the ratio of $a$ to $b$ in lowest form. $$ \begin{split} 2 &= a^2/b^2 \\ 2b^2 &= a^2 \end{split} $$ At this point, my proof deviates from the usual inductive process of finding $a$ and $b$, finding that they do indeed share a common factor greater than $1$ and so the original statement is false.

I reasoned that because $1^2 = 1$ and $2^2 = 4$, $\sqrt{2}$ must be between $1$ and $2$. Therefore, $a > b$ because $1 < \frac{a}{b} < 2$. As they are both positive integers, we can rewrite the equation: $$ 2(a-x)^2 = a^2 $$ (Where $x$ is the integer difference between $a$ and $b$. In addition, $x, a,$ and $b$ share no common factors.) It will also become important to recognize that $x < b$, and so $x$ is not a multiple of $b$. This can be proven by reasoning:

if $x \ge b$ then we could substitute in a value for $x$ into $\frac{a}{b} > 1$ but $\frac{a}{b} < 2$ (which we showed to be true earlier.)

Therefore, if $x \ge b$, $a$ could be rewritten as $(b + x)$ to give $(b + x)/b$. But if $x$ could be $b$ we can substitute $x$ as $b$ to give $2b/b = 2$. But $\sqrt{2} \ne 2$ because $2 = \left(\sqrt{2}\right)^2 \ne 2 = 4$.

Therefore, if $x \ge b$, then we get $\sqrt{2} \ge 2$, which is incorrect.

So we have shown that our equation holds based on our original assumption ($\sqrt{2} = a/b$).

If we expand our equation and simplify:

$$2(a-x)^2 = a^2$$

$$2(a^2 -2ax +x^2) = a^2$$

$$2a^2 -4ax + 2x^2 = a^2$$

$$a^2 -4ax + 2x^2 = 0$$

If we can prove this has solutions $a, x \in \Bbb Z^+$, then our original assumption ($\sqrt{2} = \frac{a}{b}$) must be invalid, as $a$ and $b$ must be integers.

I thought about how to prove this, and was stuck until I remembered completing the square. So that is what we shall do.

$$(a - 2x)^2 - 4x^2 + 2x^2 = 0$$

$$(a - 2x)^2 = 2x^2$$

$$a - 2x = \sqrt{2}(x)$$

$$a = \sqrt{2}(x) + 2x$$

We can assume that because $x$ is an integer $2x$ is also an integer.

But wait. We already know $x$ is not a factor or multiple of $b$, and $\sqrt{2}$ is not an integer. And since $a$ and $b$ are in simplest form, the smallest number we can times $\sqrt{2}$ by to produce an integer is $b$ to produce $a$. (Obviously assuming $\sqrt{2} is rational.)

But $x < b$ and not a multiple of $b$. Therefore $\sqrt{2}(x)$ does not produce integer values of $a$ or $x$.

Meaning $\sqrt{2}(x) + 2x \not \in \Bbb Z^+$. Therefore, $a \notin \Bbb Z^+$.

Meaning that there is no way of expressing $\sqrt{2}$ as a ratio of integers $\frac{a}{b}$.

Therefore $\sqrt{2} ≠ \frac{a}{b}$.

$\sqrt{2}$ is irrational.

That was quite long, but I hope everyone understands my proof. I was just wondering if anyone had seen this before and this is not original. So if you have seen it already please tell me. Also, more importantly, does it have any holes or mistakes? I understand that this may not be the simplest or best solution, but I prefer it to the usual one as I think it relies less on induction and more on algebra to produce a result.

Thank you.

  • $\begingroup$ " It will also become important to recognize that x<b, and so x is not a multiple of b. This can be proven by reasoning:" In can be proven by $1 < \frac ab < 2$ so $b < a < 2b$ and $0 < a - b = x < b$. $\endgroup$
    – fleablood
    Jul 17, 2018 at 5:40

3 Answers 3


Your calculation shows that, if $a^2=2b^2$ then, with $x=a-b$, you also have $(a-2x)^2=2x^2$, i.e. $(2b-a)^2=2(a-b)^2$. This last equation could be checked more simply by expanding both sides, cancelling a few terms, and remembering that $a^2=2b^2$. So, in particular, you've shown that, from any purported rational representation $a/b$ of $\sqrt2$, you can produce another one, $(2b-a)/(a-b)$, with a smaller denominator. So no such rational representation can be in lowest terms, and therefore no rational representation exists. (The last step could be replaced by an appeal to the least number principle: If there's a denominator that works, then there's a smallest one.)

This proof is correct, and I've seen it a few times before, but I don't remember where.


Your proof is correct, but there is a better way to complete it.

Once you arrive at the equation,


you must note that $2$ divides $a^2$ (in other words, $2$ is a factor of $a^2$). This means that $a^2$ is an even number. Now the question is,

What impact does this have on $a$ alone?

So $a$ is a number. What properties does a random (arbitrary) number have? We ask this question to see what properties of $a$ in particular might be affected by $a^2$ being even.

The main property that any number must have, is...

A parity; being odd or even.

Therefore, $a$ must be odd or even. Assume, that $a$ is odd. Then can $a^2$ be even? The answer is,


Because odd numbers are not divisible by $2$, so when you multiply them by themselves, the result will also not be divisible by $2$, and hence not even.

Thus the only option left is that $a$ is even. That means, there exists a number $c$ such that $2c=a$. Knowing that this is definitely true, substitute $a=2c$ into your given equation.

$$\begin{align}2b^2&=a^2\tag{given} \\ &\downarrow \\2b^2&=(2c)^2\tag{$a=2c$} \\ &= 2^2c^2 \\ &=4c^2 \\ &\downarrow \\ b^2&=2c^2\tag{dividing both sides by $2$}\end{align}$$

Now the question is,

What impact does this have on $b$ alone?

Can you take it from here?

  • 1
    $\begingroup$ Thanks, but that was what I was originally trying to avoid. I wanted to use less induction in my proof. $\endgroup$
    – Roskiller
    Jul 17, 2018 at 11:29

No. This is correct but a terrible proof.

Here is my favorite way to prove it:

We let $\mathbb{Q}(\sqrt{2})$ denote the extension field.

consider the automorphism $ \varphi : \mathbb{Q(\sqrt{2})} \to \mathbb{Q(\sqrt{2})} $.

Let $\varphi(a+b\sqrt{2}) = a - b\sqrt{2}$ be any such automorphism for $a,b \in\mathbb{Q}$.

Then it is trivial that $\forall q \in \mathbb{Q},~~ \varphi(q) = q$.

(if you can't see this immediately I leave the proof as an exercise for the reader)

However for the automorphism $\varphi$ we see $\varphi(\sqrt{2}) = -\sqrt{2}$.

Thus the automorphism does not send $\sqrt{2}$ to $\sqrt{2}$, and any automorphism from a field extension to itself must send rationals to rationals.

Thus $\sqrt{2}$ is irrational.

EDIT: I'm adding the proof which I initially left as exercise for the reader (lmao)

WTS: If $\varphi : \mathbb{F} \to \mathbb{F}$ is an automorphism on $\mathbb{Q}(\sqrt{2})$ Then $\forall h \in \mathbb{Q}$ $\varphi(h) = h$

Pick $p,q \in \mathbb{Z}$ arbitrarily and let $h = \frac{p}{q} \in \mathbb{Q}$.

Since $\varphi$ is an automorphism it must be a homomorphism then $\varphi(q) \varphi(\frac{p}{q}) = \varphi(q \cdot \frac{p}{q}) = \varphi(p) = p$ since $p \in \mathbb{Z}$.

Thus $\varphi(q) \cdot \varphi(\frac{p}{q}) = q \cdot \varphi(\frac{p}{q}) = p \implies \varphi(\frac{p}{q}) = \frac{p}{q}$

Thus for any $h \in \mathbb{Q}$ we know $\varphi(h) = h$ since all elements of $\mathbb{Q}$ can be expressed in the form $\frac{p}{q}$ where $p,q \in \mathbb{Z}$ and $q \neq 0$.


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