# Direct Irrationality Proof for $\sqrt{3}$ and $\sqrt{6}$ [duplicate]

I am having trouble with proving this directly. I am currently learning about greatest common divisors and know that this has a role in the proof. However, I can only prove the two through contradiction and not directly.

• How would you prove it by contradiction? Often such proofs can be unwound into straightforward ones – Frederick Feb 28 '14 at 8:38

Suppose $\sqrt{3}$ is rational. We know any number $x$ is rational if it has the form $\frac{p}{q}$ for $p,q \in \mathbb{Z}$. Hence, by assumption, we have with $\gcd(p,q) = 1$

$$\sqrt{3} = \frac{p}{q} \implies (\sqrt{3})^2 = (\frac{p}{q})^2 \implies 3 = \frac{p^2}{q^2} \implies p^2 = 3 q^2$$

Hence, $3$ divides $p^2$. As an easy exercise, you should check that then we must have that $3$ divides $p$. Since $3 | p$, then we take an integer $r$ such that $p = 3r$. Substituting this into original equation, we have

$$9r^2 = 3q^2 \implies q^2 = 3 r^2$$.

As above, we also must have that $q$ divides $3$. Since both $p$ and $q$ divides $3$, then it is impossible to have $\gcd(p,q) = 1$. This is a contradiction. Hence, $\sqrt{3}$ must be irrational.

• I don't understand the part where 3|p. – mrQWERTY Feb 28 '14 at 9:29

We can use The Fundamental Theorem of Arithmetic (FTA) to prove these easily.

IF $\sqrt{3}$ is rational THEN $\sqrt{3} = \frac{a}{b}$. For integers a , b

Square both sides , $$3 = \frac{a^2}{b^2}$$ $$3b^2 = a^2$$

Now look at this carefully , using FTA , the RHS has every prime to an EVEN power but the LHS has the prime 3 to an ODD power therefore the equation cannot be true.

We can do the same analysis for $\sqrt{6}$ $$6n^2 = m^2$$ $$2 \cdot 3 \cdot n^2 = m^2$$

Now use FTA , what can you conclude?