My just exposure come proofs was in a sdrta.net logic class I soaked up University. Ns was wondering if my attempt at proving that $\sqrt12$ is irrational is OK.

You are watching: Is the square root of 12 a rational number

$$\Big(\fracmn\Big)^2 = 12$$$$\Big(\fracm2n\Big)^2 = 3$$$$m^2=3*(2n)^2$$

This means $m$ is even and so $n$ have to be odd.

The problem can be lessened to:

$$\Big(\fracpn\Big)^2 = 3$$

Because $n$ is odd, $p^2$ is odd, so $p$ is odd.

This implies:$$4a+1 = 3(4b+1)$$$$4a - 12b = 2$$$$2a - 6b = 1$$

I"m kind of stuck at this point. I understand that this can"t be true but I don"t know exactly how to state it. Any kind of critiques or suggestions? Thanks!

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edited Nov 2 "15 at 6:27

young name Sleziak
request Feb 16 "13 in ~ 15:18

user21154user21154
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$\begingroup$ "... And so $n$ must be odd." doesn't follow. However, you presumably intended to begin the proof with "assume $\sqrt12$ is rational, and write it together $m/n$ is shortest terms". If so, climate it walk follow that $n$ have to be weird (because otherwise $m$ and $n$ would have actually a typical factor). What you say after this allude seems to come out of slim air. Where did $p$ come from? How can the trouble be reduced? then where execute $a$ and also $b$ and also those equations come from? $\endgroup$
–user14972
Feb 16 "13 in ~ 15:28

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If you’re willing to use the basic Theorem the Arithmetic, which states that the decomposition of any nonzero integer as a product of primes is unique, then this proof, and also all others for irrationality of $r$-th roots, drops best out.

Write $m^2=12n^2$. This contradicts FTA because there space evenly many $3$’s ~ above the left however oddly numerous on the right.

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reply Feb 16 "13 at 16:58

LubinLubin
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You made the too complicated in my opinion.

First we show that a reasonable number (different from $0$) time an irrational is irrational.

Proof through contradiction:Let $x\in \sdrta.netbbR\setminus\sdrta.netbbQ$, $a,c\in\sdrta.netbbZ\setminus\0\$, $b,d\in \sdrta.netbbN, d \neq 0$.$$x\cdot \fracab=\fraccd \iff x=\fracbcad,$$ for this reason $x$ would be rational.

Use$$\sqrt12=\sqrt4\cdot 3 = \sqrt4 \cdot \sqrt3 = 2 \cdot \sqrt3$$So $\sqrt12$ irrational $\iff \sqrt3$ is irrational.

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Now, we show $3|p^2 \implies 3|p$: We understand 3 is prime so through Euclid"s lemma we have$$3|p^2 \implies 3|p \lor 3|p \implies 3|p$$To prove that $\sqrt3$ is irrational, you have a contradiction:$$\sqrt3=\fracpq\iff 3q^2=p^2 \implies \exists k \in \sdrta.netbbN: 3k =p$$$$q^2=3k^2$$So $q$ and $p$ both have the divisor $3$. Currently there room two different ways to usage this information. The very first proceeds through contradiction, assuming the $p$ and also $q$ don"t have actually a usual divisor, yet as you can display they always have the divisor $3$, you can"t compose $\sqrt3$ as a fraction of number without common divisors. This is the much more elegant way in my opinion.The other means is to present that both $p$ and also $q$ can"t be finite, becauseby repeating this discussion we watch $3^n|p$ for every $n \in \sdrta.netbbN$ and likewise for $q$.But due to the fact that of $p,q\in \sdrta.netbbN$, $3^n> 1+2n$ and also the Archimedean principle we get a contradiction.