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This question already has an answer here:

Show that, if $x$ satisfies $x^2 = 3$, then $c := 2x$ satisfies $c^2 = 12$. Using this fact, show that $c$ is an irrational number.

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marked as duplicate by MJD, kingW3, Shailesh, Leucippus, Siong Thye Goh Sep 16 '17 at 1:11

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  • $\begingroup$ Have you already proved that $x$ is irrational? $\endgroup$ – Lord Shark the Unknown Sep 15 '17 at 20:11
  • $\begingroup$ Have not proved that x is irrational yet. $\endgroup$ – Christopher Sep 15 '17 at 20:13
  • $\begingroup$ Sorry, I am fairly new to analysis. I will look at the question tagged above. $\endgroup$ – Christopher Sep 15 '17 at 20:14
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$$\begin{align} x^2 &= 3 \\ \left( \frac c2 \right) ^2 &= 3 \\ \frac{c^2}{4} &= 3 \\ c^2 &= 12 \\ c &= \pm2\sqrt 3 \\ \end{align}$$

Since $\sqrt 3$ is irrational, so is $\pm2\sqrt 3$ (this has already been proven on MSE).

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  • $\begingroup$ Maybe add the fact that any non-zero rational number times an irrational number is irrational as well, which isn't hard to prove either., $\endgroup$ – WaveX Sep 15 '17 at 20:37
  • $\begingroup$ @WaveX I stated that in there, but I figured the OP could easily Google the proof. $\endgroup$ – Chase Ryan Taylor Sep 15 '17 at 20:45
  • $\begingroup$ I reckon that $c=2\sqrt3$ (or $-2\sqrt3$). $\endgroup$ – Lord Shark the Unknown Sep 15 '17 at 20:47
  • $\begingroup$ @LordSharktheUnknown Ah duh thank you $\endgroup$ – Chase Ryan Taylor Sep 15 '17 at 20:47

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