Question

Prove that $3 + \sqrt 5 $ is an irrational number.

Answer 1

Hint: We have to prove $3 + \sqrt 5 $ is an irrational number firstly we assume to the contrary its rational number. Then we get the right hand side rational number and we will prove the right hand side number is irrational. We know that irrational numbers are not equal to the irrational number. Then our assumption will be wrong.

Complete step-by-step answer:
Step1: Let us assume, to the contrary, that $3 + \sqrt 5 $ is ration. So, we can find integers $r$ and $s( \ne 0)$ such that $3 + \sqrt 5 = \dfrac{r}{s}$
That i.e. \[\sqrt 5 = \dfrac{{r - 3s}}{s}\]
Rearranging this equation, we get \[\sqrt 5 = \dfrac{{r - 3s}}{s}{\text{ - - - - - - - - - (a)}}\]. Since $r$ and $s$ are integers, we get \[\dfrac{{r - 3s}}{s}\] is rational
So $\sqrt 5 $ is rational.
Step2: Lemma: Prove $\sqrt 5 $ is an irrational number. Let us assume, to the contrary, that $\sqrt 5 $ is rational. So, we can find integers $a{\text{ and }}b$ such that \[\sqrt 5 = \dfrac{a}{b}\] suppose $a{\text{ and }}b$ are co-prime.
So, \[b\sqrt 5 = a{\text{ - - - - - - - - - (B)}}\]
Squaring on both sides and rearranging, we get $5{b^2} = {a^2}$. Therefore, $\;{\text{5 }}divides\;{\text{ }}{a^2}$ . Now, $\;{\text{5 }}divides\;{\text{ }}a$. So, we can write $a = 5c$ for some integer$c$. Substitute for a in Equation (B) we get $5{b^2} = 25{c^2}.$
That is, ${b^2} = 5{c^2}.$
This means that $\;{\text{5 }}divides\;{\text{ }}{b^2}$ and so $\;{\text{5 }}divides\;{\text{ }}b$
Therefore, $a{\text{ and }}b$ have at least $5{\text{ as }}a$ common factor.
But this contradicts the fact $a{\text{ and }}b$ have no common factors other than $1$ .
This contradiction $ha$ arisen assumption is wrong. So, we conclude that \[\sqrt 5 \] is irrational.
Step3: From step2, $Js$ is irrational and from equation (a) $\dfrac{r}{3} - 3$ is rational.
From (1) step. This contradiction has arisen because of our incorrect assumption that $5 - \sqrt 3 $ is rational. So, we conclude that $5 - \sqrt 3 $ is irrational.

Note: Let $b$ be a prime number, if $b$ divides ${a^2}$ , the divides$a$, where $a$ is a positive integer - $A$ number $s$ is called irrational, if it cannot be written in the form $\dfrac{p}{q}$ , where $p$ and $q$are integers and $q \ne 0$ . Some examples of irrational numbers, with which we are already familiar, are…
\[\sqrt {2,} {\text{ }}\sqrt {3,} {\text{ }}....{\text{ }}\pi ,0.10110111011110.......{\text{etc}}\]