# How to prove $\sqrt3$ is irrational? [duplicate]

Possible Duplicate:
How can you prove that the square root of two is irrational?

How to prove $$\sqrt3$$ is irrational using Fermat's infinite descent method?

Like says in Carl Benjamim Boyer's book.

Isnt the same prove to $$\sqrt2$$, in Boyer's book says something like this.

$$\sqrt3=a1/b1$$

$$1/(\sqrt3-1)=(\sqrt3+1)/2$$

$$\sqrt3=(3b1-a1)/(a1-b1)$$

• Quite similarly to this answer: math.stackexchange.com/questions/5/… – Asaf Karagila Oct 9 '11 at 20:38
• Exactly the same way you prove it for $\sqrt{2}$: write $\sqrt{3}=\frac{a}{b}$, or $b\sqrt{3}=a$. Square both sides, conclude you can find $a'$, $b'$, $a'\lt a$, $b'\lt b$ with $\frac{a'}{b'} = \frac{a}{b}$. – Arturo Magidin Oct 9 '11 at 20:41

Because the greatest power of $3$ that divides $p^2$ must be even, whereas the greatest power of $3$ that divides $3q^2$ must be odd. So $(p/q)^2$ can't equal $3$.