Question

prove that root 6 is an irrational.

Answer 1

Let √6 be a rational number , then

√6 = p÷q , where p,q are integers , q not = 0 and p,q have no common factors ( except 1 )

=> 6 = p² ÷ q²

=> p² = 2q² ................(i)

As 2 divides 6q² , so 2 divides p² but 2 is a prime number

=> 2 divides p

Let p = 2m , where m is an integer .

Substituting this value of p in (i) , we get

(2m)² = 6q²

=> 2m² = 3q²

As 2 divides 2m² , 2 divides 3q²

=> 2 divides 3 or 2 divide q²

But 2 does not divide 3 , therefore , 2 divides q²

=> 2 divides q

Thus , p and q have a common factor 2 . This contradicts that p and q have no common factors ( except 1 ).

Hence , our supposition is wrong . Therefore , √6 is an irrational number.


Regards

-Steeven

Answer 2

√6 is an irrational number is proved

Given :

The number √6

To find :

To prove that √6 is an irrational number

Solution :

Solution :Step 1 of 2 :

Write down the given number

Here the given number is √6

Step 2 of 2 :

Prove that √6 is an irrational number

Let us assume that √6 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

⇒ √6 = p/q { where p and q are co- prime}

⇒ √6q = p

Now, by squaring both the side

we get,

(√6q)² = p²

⇒ 6q² = p² - - - - - - (1)

Now if 2 is the factor of p²

Then , 2 is also a factor of p - - - - - - (2)

⇒ Let p = 2m { where m is any integer }

squaring both sides

p² = (2m)²

⇒ p² = 4m²

Putting the value of p² in Equation 1 we get

6q² = p²

⇒ 6q² = 4m²

⇒ 3q² = 2m²

⇒ 3q² is even

⇒ q² is even

⇒q is even

So , 2 is also factor of q

Since

2 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

Hence √6 is an irrational number

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