Exercise 1.2.1

Question

\begin{enumerate}[label=(\alph*)] 
  \item Prove that $\sqrt 3$ is irrational. Does a similar similar argument work to show $\sqrt 6$ is irrational?
  \item Where does the proof break down if we try to prove $\sqrt 4$ is irrational?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item
    Suppose for contradiction that $p/q$ is a fraction in lowest terms, and that $(p/q)^2 = 3$.
    Then $p^2 = 3q^2$ implying $p$ is a multiple of $3$ since $3$ is not a perfect square. Therefore we can write $p$ as $3r$ for some $r$, substituting we get $(3r)^2 = 3q^2$ and $3r^2 = q^2$ implying $q$ is also a multiple of $3$ contradicting the assumption that $p/q$ is in lowest terms. \
    For $\sqrt 6$ the same argument applies, since $6$ is not a perfect square.

\item $4$ is a perfect square, meaning $p^2 = 4q^2$ does not imply that $p$ is a multiple of four as $p$ could be $2$.
   \end{enumerate}

Richard Ganaye · Answer

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\begin{proof} 
\begin{enumerate}
\item[(a)]Suppose for contradiction that $\sqrt{3} = p/q$, where $p,q$ are positive integers and $p/q$ is reduced in its lowest terms. Then $p^2 = 3q^2$, where $p,q$ are relatively prime.

Since $3$ is prime, $3 \mid p^2 \Rightarrow 3 \mid p$, so $p = 3r$ for some integer $r$. Then $9p^2 = 3q^2$, so $3p^2 = q^2$. The same reasoning shows that $3 \mid q$. This is a contradiction, since $p,q$ are relatively prime.

\bigskip For $p^2 = 6q^2$ (where $\mathrm{gcd}(p,q) = 1$), we must change slightly the proof, since $6$ is not prime. From $2 \mid 6q^2 = p^2$, we obtain as above that $2 \mid p$, so $p = 2r$ for some integer $r$. Then
$4r^2 = 6q^2$, $2r^2 = 3q^2$. From $2 \mid 3 q^2$, where $\mathrm{gcd}(2,3) = 1$, we obtain $2 \mid q^2$, where $2$ is prime, therefore $2 \mid q$. Here $p,q$ are even, in contradiction with $\mathrm{gcd}(p,q) = 1$.

(Alternatively, we can reason with the factor $3$, or note that, using $\mathrm{gcd}(2,3)=1$,
 $$6 \mid p^2 \Rightarrow (2 \mid p^2, 3 \mid p^2) \Rightarrow (2 \mid p, 3 \mid p) \Rightarrow 6 \mid p.)$$ 
\item[(b)]  From $p^2 = 4q^2$ (where $\mathrm{gcd}(p,q) = 1$, we deduce $4 \mid p^2$. But $4 \mid p^2$ doesn't imply $4 \mid p$, because $4$ is not a prime. If we use $2 \mid p$, we obtain $p = 2r$ and $r^2 = q^2$, which gives no contradiction.
\end{enumerate}
\end{proof}

Note: Platon, in Theaetetus, tells that Socrates meats the young Theaetetus, which relates that in the morning, they prove with their teacher the irrationality of $\sqrt{n}$ for $n = 2,3,5,\ldots$ up to $17$, using each time a different argument.

If we want to prove the general theorem ``$\sqrt{n}$ is irrational if $n$ is not a perfect square'', we must use the notion of decomposition of an integer in prime factors, which was not a greek notion.

Exercise 1.2.1

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Exercise 1.2.1

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