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I would like to know if the following proof of $\sqrt{3}$ being a irrational number is correct.

For the sake of contradiction, assume $\sqrt{3}$ is a rational number. Therefore there exists numbers $a$ and $b$, both rationals, for which $\sqrt{3} = \frac ab$

$\therefore a = \sqrt{3} \times b$

Since $a$ is a rational number, it can be expressed as the product of two irrational numbers

$\therefore \frac {b\sqrt{3}}{\sqrt{2}} \sqrt{2} = a$

We know that$\sqrt{2}$ is an irrational number. We must show that $\frac {b\sqrt{3}}{\sqrt{2}}$ is also an irrational number.

Now, let's asume for the sake of contradiction that $\frac {b\sqrt{3}}{\sqrt{2}}$ is a rational number.

$\Rightarrow \exists c \land \exists d \in \mathbb{Q} : \frac {b\sqrt{3}}{\sqrt{2}} = \frac cd$

Which is the same as $\sqrt{3} = \frac {c\sqrt{2}}{db}$

Now, $\frac {c}{db}$ is a rational number. But $\sqrt{2}$ is an irrational. We stated at the beggining that $\sqrt{3}$ is a rational number, but this contradicts $\sqrt{3} = \frac {c\sqrt{2}}{db}$ which states that $\sqrt{3}$ must be an irrational number

Therefore we have shown by contradiction that $\sqrt{3}$ is an irrational number.

I hope the formatting and wording is ok. If I am mistaken, in what part have I committed the mistake and how should it avoid in the future?

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    $\begingroup$ You've shown (or at least tried to show; I haven't looked through all the steps) that under the assumption that $\sqrt2$ is irrational, then $\frac{b\sqrt3}{\sqrt2}$ is also irrational. This does not allow you to conclude that $\sqrt3$ is irrational. What step in your proof would fail if you swapped $\sqrt3$ for $\sqrt4$? $\endgroup$ – Arthur Oct 2 '17 at 5:21
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    $\begingroup$ No. Not at all. Witht the same argument, that is using that $\frac{b\sqrt 9}{\sqrt 2}$ is irrational for every rational $b$, your method would prove that $\sqrt 9$ is irrational. $\endgroup$ – Hagen von Eitzen Oct 2 '17 at 5:21
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    $\begingroup$ As you already know from the comments, your method is wrong. But why you are not mimicing the proof that $\sqrt{2}$ is irrational? $\endgroup$ – Przemysław Scherwentke Oct 2 '17 at 5:23
  • $\begingroup$ Arthur and Hagen: Now I understood where I made the mistake. Przemysław: I tried at first and came up with problem of how to handle the number 3 regarding the even/odd aspect so I tried to find another way. Thanks for the help $\endgroup$ – Emannuel Weg Oct 2 '17 at 5:27
  • $\begingroup$ @EmannuelWeg One thing worth pointing out is that only integers and irrational numbers are solutions to polynomials with integer coefficients and a leading coefficient of 1. if you can somehow prove that fact or call upon it within the scope of whatever you are doing, then technically ruling that sqrt(3) is irrational is as easy as just stating that it isn't an integer. $\endgroup$ – The Great Duck Oct 2 '17 at 5:33
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Since $a$ is a rational number, it can be expressed as the product of two irrational numbers

Actually this is true. Every number except $0$ can be expressed as a product of two irrational number. But where do you use this statement? At least not in the statement that follows

$\therefore \frac {b\sqrt{3}}{\sqrt{2}} \sqrt{2} = a$

This follows simply by the fact that $\frac{\sqrt{2}}{\sqrt{2}}=1$.

If you state that a rational can be expressed as the product of two irrational numbers and you found two numbers that are not both irrational and their product is the given rational, then this is not a contradiction. You did not state that only the product of two irrationals can give this rational.

So no, that is no proof at all. Find a proof for the irrationality of $\sqrt{2}$ and try to construct an analogous proof for the irrationality of $\sqrt{3}$.


How can you avoid such mistakes?

In your proof replace $\sqrt{3}$ by $3$. Does your proof work for this number too? If so, then it is not a valid proof, because $3$ is not irrational. Use this to find out where your proof went wrong.

In the standard proof for the irrationality of $\sqrt{2}$ also replace $\sqrt{2}$ by $\sqrt{3}$, $3$, $\sqrt{9}$, $\sqrt[3]{3}$, $\pi$ and check if it works for these numbers.

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