Marcin Mazur proof on irrationality of $\sqrt2$

Question

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
As written, the infinite descent is good starting with the fraction with smallest positive denominator $b$. Then $3a-4b<b$, since this is equivalent to $\frac{a}{b}=\sqrt{2}<\frac{5}{3}$. And $0<3a-4b$, since this is equivalent to $\frac{4}{3}<\sqrt{2}=\frac{a}{b}$. — , 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
@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

score 1 · Accepted Answer · answered Dec 28 '17 at 08:32

1

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.

answered Dec 28 '17 at 08:32

Paramanand Singh

87,309

Actually this is exactly @despaigne ´s comment – Peter Melech Dec 28 '17 at 11:08
@PeterMelech: Ok got your point! Saw the comment just now. Had I seen it earlier I would not have answered. But I think we can keep it so that it can be seen as a full exposition of the comment from despaigne. – Paramanand Singh Dec 28 '17 at 11:57
I agree. OP´s question is answered anyway. – Peter Melech Dec 28 '17 at 12:08

score 0 · Answer 2 · answered Dec 27 '17 at 18:26

0

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, $0 \nless 4a − 6b$.

answered Dec 27 '17 at 18:26

Skeleton Bow

2,101

but then where is the contradiction in Mazurs proof that the OP is asking for?@Skeleton Bow – Peter Melech Dec 27 '17 at 18:36
I'm not really sure to be honest, but I'll keep this answer up as someone else seems to have referenced it. E: the answer was deleted so I might delete this answer as well. – Skeleton Bow Dec 27 '17 at 18:38
better not to delete it.it shows why the proof doesn´t work this way. – Peter Melech Dec 27 '17 at 18:49

Marcin Mazur proof on irrationality of $\sqrt2$

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