Here's my proof for the irrationality of $\sqrt{6}$:


Let $\sqrt{6}$ be a rational number. Then

$\frac{a}{b}$ = $\sqrt{6}$ , where $a$ and $b$ are coprime.


$a^2 = 6 b^2$.

Here $a^2$ is divisible by both $2$ and $3$. Which means, $a$ too is divisible by both $2$ and $3$ (fundamental theorem of arithmetic).

Thus, $6b^2=(3\cdot2\cdot k)^2$ ; where $k$ is an integer.

$6b^2=36k^2$ or $b^2=6k^2$.

Here $b^2$ is divisible by both $2$ and $3$. Which means, $b$ too is divisible by both $2$ and $3$. Therefore, $a$ and $b$ have common factors, which is a contradiction. Therefore $\sqrt{6}$ is not rational.

Is the above proof correct? Is there an easier way to prove this?

  • $\begingroup$ What precisely is the fundamental principle of arithmetic? $\endgroup$ – Andrés E. Caicedo Mar 28 '17 at 16:55
  • $\begingroup$ Looks correct.. $\endgroup$ – user395952 Mar 28 '17 at 16:55
  • $\begingroup$ @AndrésE.Caicedo Sorry for my mistake. I meant "Fundamental theorem of arithmetic". I edited the correction in. $\endgroup$ – Rio1210 Mar 28 '17 at 16:57
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    $\begingroup$ You may be amused by the following snippet from Plato's dialogue Theaetetus: "Theodorus here was drawing some figures for us in illustration of roots, showing that squares containing three square feet and five square feet are not commensurable in length with the unit of the foot, and so, selecting each one in its turn up to the square containing seventeen square feet and at that he stopped." That is, Theodorus proved, one by one, the irrationality of $\sqrt3$, $\sqrt5$, $\sqrt6$, etc., on up to $\sqrt{17}$. (cont.) $\endgroup$ – Barry Cipra Mar 28 '17 at 17:31
  • $\begingroup$ It's anybody's guess (and in fact a topic of lively debate in certain circles) how Theodorus went about the proofs and why he stopped at $17$. $\endgroup$ – Barry Cipra Mar 28 '17 at 17:32

This is correct! This is pretty much the standard way to show that a number of this form in not rational and can actually be adopted to all non-perfect squares if you're a little careful (hint: all non-perfect squares are the product of a square-free number and a perfect square).

(these comments have been adopted into the OP) Two notational comments: The theorem that refer to is more commonly called the "fundamental theorem of arithmetic" or simply "the unique prime factorization theorem." Secondly, using a period to denote multiplication is something to be very much avoided. Use \cdot or\times to produce $\cdot$ and $\times$ respectively instead.

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    $\begingroup$ Thanks for the kind suggestions and hints. I have already edited in the \cdot as suggested. $\endgroup$ – Rio1210 Mar 28 '17 at 17:06
  • $\begingroup$ @Rio1210 No problem! I'm glad that I can be helpful. $\endgroup$ – Stella Biderman Mar 28 '17 at 17:08

That works fine, but you don't need to do both $2$ and $3$. If:


Then $a$ must be even, $a=2a_0$, and then: $4a_0^2=6b^2$ or $$2a_0^2=3b^2.$$

So $3b^2$ must be even, so by the fundamental theorem of arithmetic, $b$ must be even, and you are done.


Yes, your proof is valid. You have used a standard technique.


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