Vachmi

1
Prove that $\sqrt{5}$ is irrational.


Solution

Let us assume that $\sqrt{5}$ is a rational number.
Therefore, we can find two integers a and b (b ≠ 0) such that $\sqrt{5}$ = $\dfrac{a}{b}$

Suppose a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime.
So a = $\sqrt{5}$b
Squaring both sides,
$a^{2} = 5b^{2}$
Therefore, $a^{2}$ is divisible by 5 and it can be said that a is divisible by 5.
Let a = 5c. where c is an integer
$(5c)^{2} = a^{2}$
$⇒ (5c)^{2} = 5b^{2}$
$⇒ b^{2} = 5c^{2}$
This means that $b^{2}$ is divisible by 5 and hence, b is divisible by 5.
This implies that a and b have 5 as a common factor. But this contradicts the fact that a and b are co-prime.
So we conculde that $\sqrt{5}$ is irrational.




2
Prove that $3 + 2\sqrt{5}$ is irrational.


Solution

Let us assume that $3 + 2\sqrt{5}$ is a rational number.
This means that we can find co prome numbers a and b (b ≠ 0) such that $3 + 2\sqrt{5}$ = $\dfrac{a}{b}$

$∴ \dfrac{a}{b} - 3$ = $2\sqrt{5}$
$⇒ \dfrac{a - 3b}{2b}$ = $\sqrt{5}$
Since a and b are integers, $\dfrac{a - 3b}{2b}$ as rational.
∴ $\sqrt{5}$ is also rational.
This contradiction has arisen because of our incorrect assumption that $3 + 2\sqrt{5}$ is rational.
So we conclude that $3 + 2\sqrt{5}$ is irrational.




3
Prove that the following are irrationals:
(i) $\dfrac{1}{\sqrt{2}}$ (ii) $7\sqrt{5}$ (iii) $6+\sqrt{2}$


Solution

(i) Let us assume that $\dfrac{1}{\sqrt{2}}$ is a rational number.
Therefore, we can find two integers a and b (b ≠ 0) such that $\dfrac{1}{\sqrt{2}}$ = $\dfrac{a}{b}$
So $\sqrt{2}$ = $\dfrac{b}{a}$
$\dfrac{b}{a}$ is rational as a and b are integers. Therefore $\sqrt{2}$ is rational which contradicts the fact that $\sqrt{2}$ is irrational.
Hence our assumption is false and $\dfrac{1}{\sqrt{2}}$ is irrational.


(ii) Let us assume that $7\sqrt{5}$ is a rational number.
Therefore, we can find two integers a and b (b ≠ 0) such that $7\sqrt{5}$ = $\dfrac{a}{b}$
So $\sqrt{5}$ = $\dfrac{a}{7b}$
$\dfrac{a}{7b}$ is rational as a and b are integers. Therefore $\sqrt{5}$ is rational which contradicts the fact that $\sqrt{5}$ is irrational.
Hence our assumption is false and $7\sqrt{5}$ is irrational.


(ii) Let us assume that $6+\sqrt{2}$ is a rational number.
Therefore, we can find two integers a and b (b ≠ 0) such that $6+\sqrt{2}$ = $\dfrac{a}{b}$
So $\sqrt{2}$ = $\dfrac{a}{b} - 6$
$\dfrac{a}{b} - 6$ is rational as a and b are integers. Therefore $\sqrt{2}$ is rational which contradicts the fact that $\sqrt{2}$ is irrational.
Hence our assumption is false and $6+\sqrt{2}$ is irrational.