I have a sketch of proof of irrationality of $\sqrt2$ due to Marcin Mazur. It is asupposed that $\sqrt2=\frac ab$ where $a$ is the smallest positive numerator that $\sqrt 2$ can have as a fraction. Since $\sqrt2=\frac{-4\sqrt2+6}{3\sqrt2-4}$, we have $\sqrt 2=\frac{4a-6b}{4b-3a}$. In order to get a contradiction we use the fact that $1<\sqrt2=\frac ab<2$ or $b<a<2b$. Now, $a<2b$ implies that

$$3a<6b\implies4a<6b+a\implies 4a-6b<a$$

I've tried a lot to prove that $0<4a-6b$, but couldn't arrive to any result. Could you help me, please?

  • possible duplicate https://math.stackexchange.com/questions/2440864/proving-that-sqrt2-is-irrational-with-a-math-level-of-a-middle-school-stude/2440906#2440906 – Vasili Dec 27 '17 at 18:24
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    As written, the infinite descent is good starting with the fraction with smallest positive **de**nominator $b$. Then $3a-4b –  Dec 27 '17 at 18:35
  • @Vasya not really as the method here is different to the one you assumed. – For the love of maths Dec 27 '17 at 18:36
  • @user486600 I think Marcin Mazur´s proof doesn´t work this way and You found out why, $4a-6b<0$ but $a$ was assumed to be the samllest $\textbf{positive}$ numerator. Any way despaigne showed how this can work – Peter Melech Dec 27 '17 at 18:42
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    @Useless $\frac{a}{b}=\sqrt{2}$ and $\frac{a}{b}=\sqrt{2}>\frac{4}{3}$, since $2>\frac{16}{9}$, since $18>16$. –  Dec 27 '17 at 18:44

2 Answers2


I think you have a mistake in the argument. You need to start with the equation $$\sqrt{2}=\frac{6b-4a}{3a-4b}\tag{1}$$ Note that $4/3<\sqrt{2}=a/b<3/2$ and hence the integers $6b-4a$ and $3a-4b$ are positive. Further it should be noticed that $3a-4b<b$ because $a/b<5/3$. It follows that if we have positive integers $a, b$ with $a/b=\sqrt{2}$ then we have another positive integer $3a-4b$ less than $b$ and from equation $(1)$ we can see that the argument can be repeated indefinitely to get a sequence of decreasing positive integers. This is absurd and the proof is complete via this contradiction.

Paramanand Singh
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I believe this is how one could do it. Assume that $0 < 4a − 6b$. Then,

\begin{align*} 0 &< 4a - 6b \\ 6b &< 4a \\ \frac{3}{2} &< \frac{a}{b} = \sqrt{2} \\ \implies \frac{9}{4} &< 2 = \frac{8}{4} \end{align*}

which is clearly not true. Thus, $0 \nless 4a − 6b$.

Skeleton Bow
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