Prove that \( \sqrt{5} \) is irrational.
Let \( \sqrt{5} \) be rational. Then it can be written as \( \sqrt{5} = \dfrac{a}{b} \), where \(a\) and \(b\) are integers having no common factor other than 1 (i.e., \(\gcd(a,b)=1\)) and \(b \neq 0\).
Squaring both sides, we get
\[ 5 = \frac{a^2}{b^2} \]
which gives
\[ a^2 = 5b^2. \]
This means that \(a^2\) is divisible by 5. Hence \(a\) must also be divisible by 5 (because if a prime divides the square of a number, it divides the number itself). So we can write \(a = 5k\) for some integer \(k\).
Substituting \(a = 5k\) in \(a^2 = 5b^2\), we get
\[ (5k)^2 = 5b^2 \Rightarrow 25k^2 = 5b^2 \Rightarrow b^2 = 5k^2. \]
Thus, \(b^2\) is also divisible by 5, and hence \(b\) is divisible by 5.
Therefore, both \(a\) and \(b\) are divisible by 5, which means they have a common factor 5. This contradicts our assumption that \(a\) and \(b\) have no common factor other than 1.
Hence, our initial assumption that \( \sqrt{5} \) is rational is false. Therefore, \( \sqrt{5} \) is irrational.
Prove that \( 3 + 2\sqrt{5} \) is irrational.
Assume that \( 3 + 2\sqrt{5} \) is rational. Then there exists a rational number \(r\) such that
\[ 3 + 2\sqrt{5} = r. \]
Rewriting, we get
\[ 2\sqrt{5} = r - 3. \]
Since \(r\) and 3 are rational numbers, their difference \(r - 3\) is also rational. Thus, the right-hand side \(r - 3\) is rational. Dividing both sides by 2, we obtain
\[ \sqrt{5} = \frac{r - 3}{2}. \]
The right-hand side is a quotient of rational numbers, so it is rational. Hence \( \sqrt{5} \) would be rational.
But from the previous result, \( \sqrt{5} \) is irrational. This is a contradiction.
Therefore, our assumption that \( 3 + 2\sqrt{5} \) is rational is false. Hence \( 3 + 2\sqrt{5} \) is irrational.
Prove that the following are irrationals:
(i) \( \dfrac{1}{\sqrt{2}} \)
(ii) \( 7\sqrt{5} \)
(iii) \( 6 + \sqrt{2} \).
(i) Irrationality of \( \dfrac{1}{\sqrt{2}} \)
Assume that \( \dfrac{1}{\sqrt{2}} \) is rational. Then there exists a rational number \(k\) such that
\[ \frac{1}{\sqrt{2}} = k. \]
Taking reciprocals on both sides (which is allowed because \(k \neq 0\)), we get
\[ \sqrt{2} = \frac{1}{k}. \]
The right-hand side is the reciprocal of a rational number and hence is rational. Thus \( \sqrt{2} \) would be rational, which contradicts the known fact that \( \sqrt{2} \) is irrational. Therefore, \( \dfrac{1}{\sqrt{2}} \) must be irrational.
(ii) Irrationality of \( 7\sqrt{5} \)
Assume that \( 7\sqrt{5} \) is rational. Then there exists a rational number \(m\) such that
\[ 7\sqrt{5} = m. \]
Dividing both sides by 7 (a non-zero rational number), we get
\[ \sqrt{5} = \frac{m}{7}. \]
The right-hand side is a quotient of two rational numbers, so it is rational. Hence \( \sqrt{5} \) would be rational, which contradicts the fact that \( \sqrt{5} \) is irrational. Therefore, \( 7\sqrt{5} \) is irrational.
(iii) Irrationality of \( 6 + \sqrt{2} \)
Assume that \( 6 + \sqrt{2} \) is rational. Then there exists a rational number \(n\) such that
\[ 6 + \sqrt{2} = n. \]
Rearranging, we get
\[ \sqrt{2} = n - 6. \]
Since \(n\) and 6 are rational, their difference \(n - 6\) is rational. Thus, \( \sqrt{2} \) would be rational, which contradicts the fact that \( \sqrt{2} \) is irrational.
Hence the assumption is false, and \( 6 + \sqrt{2} \) is irrational.
Therefore, each of \( \dfrac{1}{\sqrt{2}} \), \( 7\sqrt{5} \) and \( 6 + \sqrt{2} \) is irrational.