# Math Help - Proving a number is irrational by contradiction

1. ## Proving a number is irrational by contradiction

Hi

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

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

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

$\sqrt{2} = \frac{a}{b}$, b not equal to 0.
Squaring both sides: $2 =\frac{a^2}{b^2}$

Step 3: $2b^2 = a^2$
This means that $a^2$ must be even and thus $a$ must be even.
i.e. $a = 2k$

Step 4:
$2b^2 = (2k)^2$
$2b^2 = 4k^2$
$b^2 = 2k^2$
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 $\sqrt{2}$ 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 $\sqrt{3}$ example.

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

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

$\sqrt{3} = \frac{a}{b}$, b not equal to 0.
Squaring both sides: $3 =\frac{a^2}{b^2}$

Step 3: $3b^2 = a^2$
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:
$3b^2 = (2k + 1)^2$
......

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

David.

2. From Step 3, you have $\displaystyle 3b^2 = a^2$. So $\displaystyle a^2$ is a multiple of $\displaystyle 3$. This means $\displaystyle a$ is a multiple of $\displaystyle 3$. This can be proven using the contrapositive.

So we can let $\displaystyle a = 3k$ and therefore

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

$\displaystyle 3b^2 = 9k^2$

$\displaystyle b^2 = 3k^2$.

Since $\displaystyle b^2$ is a multiple of $\displaystyle 3$, so is $\displaystyle b$.

3. Originally Posted by Bucephalus
This means that $a^2$ must be even and thus $a$ must be even.
i.e. $a = 2k$
we can come to the same conclusion by $p|a^2\implies p|a$ where $p$ is a prime

this can be used to prove $\sqrt{3}$ is not a rational number since 3 is also a prime

4. ## Thankyou both for your help

Cheers.

5. Here is a proof that $a^2$ a multiple of 3 implies $a$ is a multiple of 3 (by contrapositive).

Suppose a is not a multiple of 3. Then $a=3k+1$ or $a=3k=2$ for some integer $k$. If $a=3k+1$, then $a^2=(3k+1)^2=9k^2+6k+1=3(3k^2+2k)+1$. $3k^2+2k$ is an integer because the integers are closed under addition and multiplication. So $a^2$ is not divisible by 3. The second case is similar.