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Different ways of writing a proof


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:


First version

 
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:


Second version

 
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.    \(\Box\)
 

Observe the following:


Third version

 
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 \(\frac{p}{q}\).
 

Note:

  • A proof sketch is not a complete proof. Many pieces will be missing or statements will be glossed over.

  • The purpose of a proof sketch is to outline a proof idea.


© 2015, Rahul Simha