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How would I go about proving $1+2\sqrt 3$ is not a rational number assuming $\sqrt 3$ is not a rational?

Would direct proof be the easiest? Total beginner here, any insight would be appreciated.

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    $\begingroup$ indirectly, suppose it's equal to $\frac{p}{q}$. Solve for $\sqrt{3}$. $\endgroup$
    – MAM
    Mar 2, 2015 at 6:52
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    $\begingroup$ Actually, if x is not rational then 1 + 2x isn't rational either. $\endgroup$
    – gnasher729
    Mar 2, 2015 at 11:57

6 Answers 6

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Say $1 + 2 \sqrt{3} $ is rational by contradiction. So we have

$$ 1 + 2 \sqrt{3} = \frac{p}{q} $$

for $p,q \in \mathbb{Z}$. Notice we can move $1$ to the other side to obtain

$$ 2 \sqrt{3} = \frac{p}{q} - 1 = \frac{p-q}{q}$$

Now, divide by $2$ to obtain

$$ \sqrt{3} = \frac{p-q}{2q} $$

But this is a rational number. Hence $\sqrt{3} \in \mathbb{Q}$. And we have reached a contradiction since $\sqrt{3}$ is not rational.

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  • $\begingroup$ the trick here is to isolate the number that we "smell" irrational $\endgroup$
    – Ooker
    Jan 14, 2016 at 3:28
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From the question you made I assume you are beginning your studies in mathematics, so additionally to Willie Rosario's answer you will need to prove that $\sqrt{3}$ is irrational as well, and that's how you do:

Assume $\sqrt{3} \in \mathbb{Q}$. Then, $\sqrt{3}=p/q$, where $\gcd\{p,q\}=1$ (indeed, if this was not true, you could just simplify the fraction). This means that if $$ p=p_1p_2\ldots p_n, \quad q=q_1q_2\ldots q_m $$ are the decompositions of $p,q$ on prime factors, then $p_j\neq q_k$ for every $j,k$, since $\gcd\{p,q\}=1$. But now, $$ 3=\frac{p^2}{q^2} $$ means that $q^2$ divides $p^2$, which is impossible since $$ p^2=p_1^2p_2^2\ldots p_n^2, \quad q^2=q_1^2q_2^2\ldots q_m^2, $$ and we know that $p_j\neq q_k$ for all $j\neq k$ (the fact that $p^2$ divides $q^2$ implies that there exist $j,k$ such that $p_j=q_k$). So, we arrive at a contradiction, and we conclude that $\sqrt{3}\notin \mathbb{Q}$.

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  • $\begingroup$ You assume $m$ is a natural number, but $m=0$ if $\sqrt 3$ is an integer. You must prove $\sqrt 3 \notin \Bbb{Z}$ first in order for this proof to be valid. $\endgroup$ Apr 18, 2016 at 15:12
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$x = 1 + 2\sqrt{3} \Rightarrow (x-1)^2 = 12 \Rightarrow x^2-2x-11 = 0$. Using the rational root test, there is no rational root, hence the conclusion...

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If it was rational then $1+2 \sqrt{3} = { p \over q}$ and so $\sqrt{3} = {1 \over 2} ({ p \over q} -1)$, which would mean that $\sqrt{3}$ is rational.

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$\,r = 1+2\sqrt{3}\,\Rightarrow\, (r-1)^2 = 12.\,$ By the Rational Root Test, if it $\,r\,$ is rational then it is an integer. Therefore $\,n = r-1\,$ is an integer and $\,n^2 = 12,\,$ contra $\, 3^2 < 12 < 4^2,\,$ and $\,x^2\,$ is increasing (or contra $\,12\,$ is not a square mod $5)$

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PA $1+2\sqrt 3=x$ is rational. Therefre $\frac{x-1}{2}=\sqrt 3$ is rational because addition and multiplication is stable, but this is absurde. Therefore $1+2\sqrt 3$ isn't rational

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    $\begingroup$ I didn't down vote. What does " addition and multiplication is stable" means? $\endgroup$
    – jimjim
    Mar 2, 2015 at 7:40
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    $\begingroup$ That an operatiob between 2 items of some type leed to an item of thar type. In this situation that product and sum of rationals is rational $\endgroup$
    – Davide F.
    Mar 2, 2015 at 7:41
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    $\begingroup$ 25 years ago one used to say rationals are closed under addition and multiplication, is this new terminology now? $\endgroup$
    – jimjim
    Mar 2, 2015 at 7:45
  • $\begingroup$ Umh, well, it come from regular language theory where it's standard terminology. I'm quite used to it so I used it without thinking to much, but closed is used normaly still today $\endgroup$
    – Davide F.
    Mar 2, 2015 at 7:48
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    $\begingroup$ this answer is $\mathbf{not}$ helpful for a beginner who is learning the material for the first time. It is very advanced terminology used here. $\endgroup$
    – user203867
    Mar 2, 2015 at 10:00

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