$\sqrt{2}$
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
- 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]