Question

Prove that √6 is irrational

Answer 1

Hiii Abdul,


IF possible,let √6 be rational and let it's simplest form be a/b.
Then, a nd b are integers having no common factor other than 1, and b is not equal to 0.
Now,
✓6= a/b = 6= a²/b² [ on squaring both sides]
= 6b²= a².....................(1)
= 6 divides a² [ 6 divides 2b²]
= 6 divides a
Let,
a = 6c for some imteger c .
putting a = 6c in (1) , we get
6b²= 36c²= b²= 6c²
= 6 divides b². [ 6 divides 2c²]
= 6 divides b
Thus,
6 is common factor of a and b .
But,
this contradicts the fact that a and b have no common factor other than 1.
The contradicts arises by assuming that ✓6 is rational number.
Hence,
✓6 is irrational.


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Answer 2

Step-by-step explanation:

Method Used :-

Contradiction Method

Solution :-

Let's assume that the square root of 6 is rational.

That means there are two integers a and b where:

a / b = square root of 6.

Multiply both sides by themselves:

( a / b ) ( a / b ) = ( square root of 6 ) ( square root of 6 )

a²/b² = 6

a² = 6b²

This means the RHS is even, because it is a product of integers and one of those integers is even. So a² must be even but any odd number times itself is odd, so if a² is even, then a is even.

Since a is even, there is some integer c that is half of a, or in other words:

2c = a.

Now let's replace a with 2c:

a² = 6b²

(2c)² = (2)(3)b²

2c² = 3b²

But now we can argue the same thing for b, because the LHS is even, so the RHS must be even and that means b is even.

This is the contradiction that if a is even and b is even, then they have a common divisor (2). Then our initial assumption must be false, so the square root of 6 cannot be rational.