Exercise 1.2.1.
- Prove that \(\sqrt{3}\) is irrational. Does a similar argument work to show
\(\sqrt{6}\) is
irrational?
- Where does the proof of Theorem 1.1.1. break down if we try to use it to prove
\(\sqrt{4}\) is
irrational?
Solution (a):
Assume, for contradiction, that there exist integers \(p\) and \(q\) such that \(\gcd(p,q) =1 \) and
\[ \Bigg( \, \frac{p}{q} \, \Bigg)^2 = 3 \tag{1}\]
Next, equation (1) implies
\[p^2 = 3q^2 \tag{2} \]
This shows that \(p^2\) is a multiple of 3. Now, the only way for a perfect square to be
a multiple of 3 is if its factors also include 3. Therefore, \(p\) must also be a multiple of 3. Let \(p=3k\),
where \(k\) is an integer. Substituting this into (2) gives:
\[ (3k)^2 = 3q^2 \]
\[ 9k^2 = 3q^2 \]
\[ 3k^2 = q^2 \tag{3} \]
This shows that \(q^2\) is a multiple of 3, and hence, \(q\) is a multiple of 3. This contradicts
our hypothesis that \(\gcd(p,q) = 1\). From this reasoning, we can only assume that equation (1)
cannot hold for any integers \(p\) and \(q\). Thus, \(\sqrt{3}\) is irrational.
Q.E.D.
Yes, a similar argument works to show that \(\sqrt(6)\) is irrational. We can replace equation (2) with
\[ p^2 = 6q^2 \tag{4} \]
which shows that \(p\) is a multiple of 6. Let \(p = 6k \) and substitute into (4).
\[ (6k)^2 = 6q^2 \]
\[ 36k^2 = 6q^2 \]
\[ 6k^2 = q^2 \]
This tells us that \(q\) is a multiple of 6. By the same reasoning as above, \(\sqrt{6}\) is
irrational.
Solution (b):
From \(p^2 = 4q^2\), we can only conclude that \(p\) is even.