Real Numbers-Exercise 1.2
Maths - Exercise
1. Prove that \(\sqrt{5}\) is irrational.
Proof (by contradiction):Assume, to the contrary, that \(\sqrt{5}\) is a rational number. Then, we can write
\[\sqrt{5} = \frac{a}{b}, \quad b \neq 0, where~ a, ~b \in \mathbb{Z} ~and ~\\\\\gcd(a,b) = 1\text{ (i.e., a and b are coprime)}.\] Square both sides
\[\begin{align}\left(\sqrt{5}\right)^2 &= \left(\frac{a}{b}\right)^2\\\\ 5 &= \frac{a^2}{b^2}\\\\ a^2 &= 5b^2 \tag{1}\\\end{align}\] From (1),
\(a^2\) is divisible by 5.
Hence, \(a\) must also be divisible by 5.
we can find some integer \(c\) such that \(a = 5c\)
\[\begin{align}a^2 &= (5c)^2\\ &= 25c^2\end{align}\] Substituting into (1): \[\begin{align}25c^2 &= 5b^2\\ b^2 &= 5c^2 \tag{2}\end{align}\] From (2), \(b^2 \) is divisible by 5.
Thus, \(b\) must also be divisible by 5.
From (1) and (2), we find that both \(a\) and \(b\) are divisible by 5. This contradicts our assumption that \(\gcd(a,b)=1\\\) Conclusion:
Our assumption that \(\sqrt{5}\) is rational is false. \[\\\therefore \; \sqrt{5} \text{ is irrational.}\]
To prove: \(3 + 2\sqrt{5}\) is irrational.
Method 1:-
Using properties of rationals and irrationals
- The sum of a rational number and an irrational number is always irrational.
- The product of a non-zero rational number and an irrational number is also irrational.
- Here, 3 and 2 are rational numbers.
- Since \(\sqrt{5} \) is irrational, \(2\sqrt{5}\) is irrational.
- Therefore, \(3 + 2\sqrt{5}\) (sum of a rational number and an irrational number) is irrational.
Proof by contradiction
Assume, to the contrary, that \(3 + 2\sqrt{5} \) is rational.
Then, there exist integers \[a, b ~(with \gcd(a,b) = 1, b \neq 0)\] (gcd - Greatest Common Divisor) such that
\[3 + 2\sqrt{5} = \frac{a}{b}\] Rearranging, \[\begin{align}2\sqrt{5} &= \frac{a}{b} - 3 \\\\&= \frac{a - 3b}{b}\\\\ \sqrt{5} &= \frac{a - 3b}{2b}\end{align}\] Since a and b are integers, the right-hand side is rational.
This implies that \(\sqrt{5}\) is rational, which contradicts the known fact that \(\sqrt{5}\) is irrational.
Hence, our assumption is false.
\[\therefore \; 3 + 2\sqrt{5} \text{ is irrational.}\]
To prove: \(\dfrac{1}{\sqrt{2}}\) is irrational.
Suppose, to the contrary, that \(\dfrac{1}{\sqrt{2}}\) is a rational number.
Then there exist integers \(a\) and \(b\) (with \(\gcd(a,b)=1 \text{ and } b \neq 0) \) such that \[\frac{1}{\sqrt{2}} = \frac{a}{b}\] Squaring both sides, we get \[\begin{align}\left(\frac{1}{\sqrt{2}}\right)^2& = \left(\frac{a}{b}\right)^2\\\\ \frac{1}{2} &= \frac{a^2}{b^2}\\\\ b^2 &= 2a^2\end{align}\] This implies that \(b^2\) is divisible by 2
so, \(b\) must also be divisible by 2
Let \(b = 2c\), where \(c\) is an integer.
Substituting back,
\[\begin{aligned}b^2 &= (2c)^2 \\&= 4c^2,\\ 4c^2 &= 2a^2 \\\quad \Rightarrow a^2 &= 2c^2\end{aligned}\] Thus, \(a^2\) is divisible by 2
which means \(a\) is also divisible by 2.
Hence, both \(a\) and \(b\) are divisible by 2.
This contradicts our initial assumption that \(a\) and \(b\) are coprime (having no common factor other than 1).
Therefore, our assumption was false.
\[\therefore \; \frac{1}{\sqrt{2}} \text{ is irrational.}\]
Prove that \(7\sqrt{5}\) is an irrational number.
Let us assume, for the sake of contradiction, that \(7\sqrt{5}\) is a rational number.Then, there exist integers \[a \text{ and } b \text{ (with }\gcd(a,b)=1 \text{ and } b \neq 0\text{ )}\] such that \[7\sqrt{5} = \frac{a}{b}\] Dividing both sides by 7, we get \[\sqrt{5} = \frac{a}{7b}\] Here, \(a\) and \(7\)b are integers, so \(\dfrac{a}{7b}\) is a rational number.
This implies that \(\sqrt{5}\) is rational.
But it is well known that \(\sqrt{5}\) is irrational.
This contradiction arises from our assumption that \(7\sqrt{5}\) is rational. \[\therefore \; 7\sqrt{5} \text{ is irrational.}\]
To prove: \(6 + \sqrt{2}\) is irrational.
Proof: (by contradiction):Assume, to the contrary, that \(6 + \sqrt{2}\) is a rational number.
Then, there exist integers \[a, b (\gcd(a,b)=1, b \neq 0)\] such that \[6 + \sqrt{2} = \frac{a}{b}\] Rearranging, \[\begin{aligned}\sqrt{2} &= \frac{a}{b} - 6 \\&= \frac{a - 6b}{b}\end{aligned}\] Since \(a, b \in \mathbb{Z}\), the right-hand side is a rational number.
This implies that \(\sqrt{2}\) is rational, which contradicts the well-known fact that \(\sqrt{2}\) is irrational.
Hence, our assumption was wrong. \[\therefore \; 6 + \sqrt{2} \text{ is irrational.}\]
