Square Root of Three is Irrational

$\sqrt{2}$ is irrational

Proving that $\sqrt{2}$ is irrational is a standard example of a proof by contradiction.

the proof

Let us assume the opposite (negation) that $\sqrt{2}$ is actually rational and we can write

$${\sqrt{2} = \frac{m}{n}}$$

where $n,m \in \mathbb{Z}^{+}$ and $n, m $ are not coprime. (This means m,n have no common prime factors).

Now, we begin some manipulation.

$${\sqrt{2} = \frac{m}{n}}$$

$${\sqrt{2} \cdot n = m}$$

$${2 \cdot n^2 = m^2}$$

If $m^2$ is a multiple of 2, then $m^2$ is even, which implies that m has to be even1 and can be written as

$$m = 2k$$

Substituting, $m = 2k$.

$${2n^2 = m^2 = (2k)^2 = 2(2k^2)}$$

$${n^2 = 2k^2}$$

This would mean that n is also even. But this cannot be the case as we assumed that m and n had no common factors.

Therefore, we have a contradiction. Hence the our assumption is wrong and $\sqrt{2}$ is irrational.

qed

the proof of $\sqrt 3$ is irrational

In class, we tried this proof of contradiction in the same manner as with $\sqrt{2}$ but realised that we rely on other axioms we may have not proven yet or did not know.

We started by assuming

$${\sqrt{3} = \frac{m}{n}}$$

where $n,m \in \mathbb{Z}^{+}$ and $n, m $ are not coprime.

This led to

$$m^2 = 3n^2$$

Here, the solutions provided suggested that

$$m^2 = 3n^2 \Rightarrow m = 3k , k \in \mathbb{Z}^{+}$$ This does not seem obvious at first.

So, we also have to prove this conjecture. Here, it makes sense to use the method of contrapositive.

$${P \Rightarrow Q}$$ is proven by showing $${\neg Q \Rightarrow \neg P}$$

Where, ${\neg Q}$ is the contrapositive of $Q$ and ${ \neg P}$ is the contrapositive of $P$.

Let, $m = 3k \pm 1$ (not a multiple of 3).

$$\Rightarrow m^2 = (3k \pm 1)^2$$ $$\Rightarrow m^2 = 9k^2 \pm 6k + 1$$ $$\Rightarrow m^2 = 3(3k^2 \pm 2k) + 1$$

Hence, $$m^2 = 3n^2 \Rightarrow m = 3k , k \in \mathbb{Z}^{+}$$

Back to the original proof ($\sqrt 3$ is irrational)

Let $m = 3k$

$$\Rightarrow m^2 = (3k)^2 = 3n^2$$ $$\Rightarrow 3(3k^2) = 3n^2$$ $$\Rightarrow 3k^2 = n^2$$

this means that n is also a multiple of 3 and so both m and n are multiples of 3 which is a contradiction to our original assumtion that m and n have no common factors.

qed


  1. only the square of an odd number is an odd number. (2k+1)(2k+1) = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1. Hence, odd. [return]

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