Question

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

Answer 1

Hint: First we have to prove $\sqrt{5}$ is irrational. Then we can prove the difference between 2 rational numbers is always rational. Then it would be easier to prove $3-\sqrt{5}$ an irrational number. $3-\sqrt{5}$ has to be an irrational number because of the above 2 proofs.

Complete step-by-step solution:
We can prove $3-\sqrt{5}$ is an irrational number by proving $\sqrt{5}$ is an irrational number.
We can prove $\sqrt{5}$ is an irrational number by the contradictory method.
Let’s assume $\sqrt{5}$ is a rational number that means we can write $\sqrt{5}$ in the form $\dfrac{a}{b}$ where b is not equal to 0 and a and b have no common factor other than 1(co-prime).
So, we have $\sqrt{5}$=$\dfrac{a}{b}$
$\Rightarrow 5=\dfrac{{{a}^{2}}}{{{b}^{2}}}$
$\Rightarrow $${{a}^{2}}=5{{b}^{2}}$
${{a}^{2}}$ is divisible by 5.
Given p is a prime number if ${{a}^{2}}$ is divisible by p then $a$ will also be divisible by p.
From the above statement, we can tell $a$ is divisible by 5.
If we take $a=5c$
Then $25{{c}^{2}}=5{{b}^{2}}$
$\Rightarrow $ $5{{c}^{2}}={{b}^{2}}$
${{b}^{2}}$ is divisible by 5 that implies $b$ is divisible by 5.
So $a$ and $b$ have a common factor 5. But we assume a and b have no common factor thus $\sqrt{5}$ is an irrational number.
Now we have to prove the difference between 2 rational numbers is a rational number.
Let’s assume x and y are 2 rational numbers.
$x=\dfrac{a}{b}$ and $y=\dfrac{p}{q}$ where b and q are not equal to 0.
$x-y=\dfrac{a}{b}-\dfrac{p}{q}$
$\Rightarrow $ $x-y=\dfrac{aq-pb}{pq}$ $=$ $\dfrac{e}{f}$
where $e=aq-bp$ and $f=pq$ where pq is not equal to 0.
$\left( p\ne 0;q\ne 0 \right)$
This proves that $x-y$ is a rational number.
In this case $3-\sqrt{5}$ , let’s assume $3-\sqrt{5}$ is a rational number
So, $3-\sqrt{5}$=$c$ where $c$ is a rational number.
$\Rightarrow $ $3-c=\sqrt{5}$
According to the above proof difference between 2 rational numbers is a rational number but $3-c=\sqrt{5}$ which is an irrational number.
So, $c$ is an irrational number .
That means $3-\sqrt{5}$ is an irrational number. (Proved)

Note: Always remember the theory that when a number ${{a}^{2}}$ is divisible by p then it implies that a will also divisible by p given that all the prime factors of p is different that means there is no repetitive prime factor. The sum, difference, product of 2 rational numbers is always a rational number. But in the case of irrational numbers, the sum, difference, product of 2 irrational numbers is not always irrational.