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Prove that $6 + \sqrt 2 $ is irrational.

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Last updated date: 16th Jun 2024
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Answer
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Hint: Here if $6$ and $\sqrt 2 $ both are rational numbers then $6 + \sqrt 2 $ is rational number. If anyone is irrational then $6 + \sqrt 2 $ will be an irrational number. As $6$ is an integer so it is a rational number. So we have to check whether $\sqrt 2 $ is rational or irrational. For this we have to use the contrary method to prove. Rational numbers are represented in form of $\dfrac{a}{b}$ where a and b are co-primes means there will be no common factor between a and b. If we find any common factor of a and b then the number $\sqrt 2 $ will be an irrational number.

Complete step-by-step solution:
Here we have to prove $6 + \sqrt 2 $ is an irrational number.
We will use the contrary method to prove it. So let’s assume that $6 + \sqrt 2 $ is rational.
All rational numbers are represented in the form of $\dfrac{a}{b}$ where a and b are co-primes. There must be no common factor between a and b and also \[b \ne 0\].
So, $6 + \sqrt 2 = \dfrac{a}{b}$
Simplifying, $6 - \dfrac{a}{b} = \sqrt 2 $
So, $\dfrac{{6b - a}}{b} = \sqrt 2 $
Here a, b and -6 are integers. So the left side of the equation $\dfrac{{6b - a}}{b}$ is a rational number. We have to check whether the right side of the equation $\sqrt 2 $ is a rational or irrational number.
For this let’s assume that $\sqrt 2 $ is a rational number.
Similarly we can write $\sqrt 2 $ in the form of $\dfrac{p}{q}$ where p and q are co-primes and there must be no common factor between p and q and also \[q \ne 0\].
So, $\sqrt 2 = \dfrac{p}{q}$
So, $p = \sqrt 2 q$
Squaring on both side, ${(p)^2} = {(\sqrt 2 q)^2}$
So, ${p^2} = 2{q^2}$
We can say that ${p^2}$ is divisible by number 2.
So, p is also divisible by number 2 means 2 is one factor of number p.
As 2 is one factor of p we can write $p = 2r$
Putting this value in ${p^2} = 2{q^2}$ equation,
So, ${(2r)^2} = 2{q^2}$
So, $4{r^2} = 2{q^2}$
Simplifying, ${q^2} = 2{r^2}$.
So we can say that ${q^2}$ is divisible by number 2.
So, q is also divisible by number 2 means 2 is one factor of number q.
From above two we can say that 2 is factor of number p and q both. So, 2 is the common factor of number p and q, which means there is a common factor so the number $\sqrt 2 $ is an irrational number.
In the equation $\dfrac{{6b - a}}{b} = \sqrt 2 $, left side is rational number and right side is irrational number which is a contradiction. So our assumption that $6 + \sqrt 2 $ is rational is wrong.
So, the number $6 + \sqrt 2 $ is an irrational number.

Note: Using the same method we can prove that $\sqrt 3 $, $\sqrt 5 $, $\sqrt 7 $, $\sqrt[3]{2}$, $\sqrt[3]{3}$, $\sqrt[3]{4}$ are also irrational numbers. Similarly numbers like $\pi $, $e$, ${\pi ^2}$ are also irrational numbers. Square root of the number may be rational if the number is having the even power of its prime factor. Ex 4. As 4 is ${2^2}$ means it has even power 2 of prime factor 2 so $\sqrt 4 = 2$ is a rational number. Ex. 144 as 144 is \[{2^4} \times {3^2}\] so $\sqrt {144} = {2^2} \times 3$. So 144 is a rational number.