Proving a number is irrational by contradiction

Hi

I have this worked example to prove $\displaystyle \sqrt{2}$ is an irrational number via contradiction. I will put it down here now. But my question is how do I do this with $\displaystyle \sqrt{3}$ as I get stuck half way through this when we come to talking about even numbers.

Step 1: Suppose that $\displaystyle \sqrt{2}$ is rational.

Step 2: This means that its possible to find $\displaystyle a, b \in \mathbb{Z} $ where a and b have only a common factor of 1 such that:

$\displaystyle \sqrt{2} = \frac{a}{b}$, b not equal to 0.

Squaring both sides: $\displaystyle 2 =\frac{a^2}{b^2}$

Step 3: $\displaystyle 2b^2 = a^2$

This means that $\displaystyle a^2$ must be even and thus $\displaystyle a$ must be even.

i.e. $\displaystyle a = 2k$

Step 4:

$\displaystyle 2b^2 = (2k)^2$

$\displaystyle 2b^2 = 4k^2$

$\displaystyle b^2 = 2k^2$

This means that b^2 and b must be even.

Step 5: So both a and b are even. This means that a and b have a common factor or 2. This contradicts the original hypothesis and so the hypothesis must be false. This means that $\displaystyle \sqrt{2}$ is not able to be expressed in the form a/b and is therefore an irrational number.

The end.

So that is the example proof. Here is what I have done so far with the $\displaystyle \sqrt{3}$ example.

Step 1: Suppose that $\displaystyle \sqrt{3}$ is rational.

Step 2: This means that its possible to find $\displaystyle a, b \in \mathbb{Z} $ where a and b have only a common factor of 1 such that:

$\displaystyle \sqrt{3} = \frac{a}{b}$, b not equal to 0.

Squaring both sides: $\displaystyle 3 =\frac{a^2}{b^2}$

Step 3: $\displaystyle 3b^2 = a^2$

This condition is only met when a is odd and b is odd. I just did this with a little trial and error. So a is odd. Thus a = 2k + 1

Step 4:

$\displaystyle 3b^2 = (2k + 1)^2$

......

Ok this is where I come unstuck. Does anyone have any suggestions?

David.

Thankyou both for your help