Question

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

Answer 1

Hint:
To prove that a given number is an irrational number by supposing it a rational number. By considering it as a rational number we will prove that our assumption is wrong.

Complete step by step solution:
Let us assume the opposite, i.e., $\sqrt 5 $ is a rational number.
Hence, $\sqrt 5 $ can be written as in the form $\dfrac{a}{b}$ where a and $b\;(b \ne 0)$ are co-prime (no common factor other than 1 ).
Hence $\sqrt 5 = \dfrac{a}{b} \Rightarrow \sqrt 5 b = a$
Squaring both sides: ${(\sqrt 5 b)^2} = {(a)^2} \Rightarrow 5{b^2} = {a^2}$
$ \Rightarrow \dfrac{{{a^2}}}{5} = {b^2}$
Hence, 5 divide $a$
By theorem: if $p$ is a prime number and $p$ divides ${a^2}$, then $p$ divides $a$, where a in a positive number.
So, 5 shall divide $a$ also --$(1)$
Now, we know that $5{b^2} = {a^2}$ put $a = 5c \Rightarrow 5{b^2} = {(5c)^2} = 25{c^2}$
$ \Rightarrow {b^2} = \dfrac{1}{5} \times 25{c^2} = 5{c^2}$
$ \Rightarrow \;\;\;{\kern 1pt} {c^2} = \dfrac{{{b^2}}}{5}$
Hence, $5$ divides ${b^2}$
So, 5 divide $b$ also, --${\kern 1pt} (2)$
By $(1)$ and $(2)$ $5$ divides both a and b.
This means that 5 is a common factor of a and b.
Hence, our assumption is wrong.

$\therefore \sqrt 5 $ is irrational number.

Additional Information:
Numbers are defined into two main categories:
1) Rational Number
2) Irrational Number
1) Rational Number: A number is a rational number when the number is written in the form of $\dfrac{p}{q}$ such that $q \ne 0$. In simple words we rational number is the number when the number can be written in fraction such that when we divide the numerator with denominator the number after decimal ends not continuously occurring.
2) Irrational Number: A number which cannot be written in the form of $\dfrac{p}{q}$. In simple words irrational numbers have values after decimals are not ending.
These Rational Numbers and Irrational Numbers together are called Real Numbers.

Note:
While proving the irrational number you have to remember the theorem which says that: if $p$ is a prime number and $p$ divides ${a^2}$, then $p$ divides $a$, where a in a positive number. Without remembering this you can’t prove the number irrational.