Proposition 2.1 \(\sqrt{2}\) is not rational.
We'll use the proof of Proposition 2.1 to explore a number of notions about proofs.
First, some nomenclature:
Observe the following:
Proof: (by contradiction)
Let \(y = \sqrt{2}\) and assume \(y\) is rational. Since \(y\)
is rational, \(y\) can be expressed as a ratio of integers
in simplest form: \(y = \frac{p}{q}\). Which means that
\(\frac{p}{q} = \sqrt{2}\) or \(p^2 = 2q^2\) and thus
\(p^2\) is even. But even squares are squares of even numbers,
and thus \(p\) is even, which means we can write \(p\)
as \(p=2r\) for some number \(r\). From this,
\(p^2 = 4r^2\). Combining this with \(p^2=2q^2\), we see
that \(4r^2 = 2q^2\) or \(q^2 = 2r^2\), and thus \(q\) is even.
Since both \(p\) and \(q\) are even, the ratio \(\frac{p}{q}\)
is not in the simplest form, which leads to a contradiction.
\(\Box\)
Observe the following:
Proof: (by contradiction)
Let \(y = \sqrt{2}\)
$$
\begin{array}{lll}
\Rightarrow
\;\;\;\;\; & \;\;\;\;\;
y = \frac{p}{q}
\;\;\;\;\; & \;\;\;\;\;
\mbox{in simplied form for some integers } p,q \\
\Rightarrow
\;\;\;\;\; & \;\;\;\;\;
\frac{p}{q} = \sqrt{2}
\;\;\;\;\; & \;\;\;\;\;
\\
\Rightarrow
\;\;\;\;\; & \;\;\;\;\;
p^2 = 2q^2
\;\;\;\;\; & \;\;\;\;\;
\\
\Rightarrow
\;\;\;\;\; & \;\;\;\;\;
p^2 \mbox{ is even}
\;\;\;\;\; & \;\;\;\;\;
\\
\Rightarrow
\;\;\;\;\; & \;\;\;\;\;
p \mbox{ is even}
\;\;\;\;\; & \;\;\;\;\;
\mbox{because square roots of even squares are even}
\\
\Rightarrow
\;\;\;\;\; & \;\;\;\;\;
p = 2r
\;\;\;\;\; & \;\;\;\;\;
\mbox{for some integer } r
\\
\Rightarrow
\;\;\;\;\; & \;\;\;\;\;
p^2 = 4r^2
\;\;\;\;\; & \;\;\;\;\;
\\
\Rightarrow
\;\;\;\;\; & \;\;\;\;\;
2q^2 = 4r^2
\;\;\;\;\; & \;\;\;\;\;
\mbox{because \(p^2 = 2q^2\) from earlier}
\\
\Rightarrow
\;\;\;\;\; & \;\;\;\;\;
q^2 = 2r^2
\;\;\;\;\; & \;\;\;\;\;
\\
\Rightarrow
\;\;\;\;\; & \;\;\;\;\;
\mbox{\(q\) is even}
\;\;\;\;\; & \;\;\;\;\;
\\
\Rightarrow
\;\;\;\;\; & \;\;\;\;\;
\mbox{both \(p\) and \(q\) are even}
\;\;\;\;\; & \;\;\;\;\;
\\
\Rightarrow
\;\;\;\;\; & \;\;\;\;\;
\frac{p}{q} \mbox{ is not in simplest form}
\;\;\;\;\; & \;\;\;\;\;
\\
\end{array}
$$
This contradicts the earlier assertion that \(\frac{p}{q}\)
is in simplest form, that arose from the
assumption that \(y\) is rational.
\(\Box\)
Note:
Proof sketch: (by contradiction)
Let \(y = \sqrt{2}\) and assume \(y\) is rational. Since \(y\)
is rational, \(y\) can be expressed as a ratio of integers
in simplest form: \(y = \frac{p}{q}\). Squaring to
get \(p^2 = 2q^2\) shows that \(p^2\) and therefore
\(p\) is even. A similar argument shows that \(q\) is even,
which contradicts the simplest form assumption for \(\frac{p}{q}\).