Question

prove that root 6 is irrational

Answer 1

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

By definition, that means there are two integers a and b with no common divisors where:

a/b = square root of 6.

So let's multiply both sides by themselves:

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

a2/b2 = 6

a2 = 6b2

But this last statement means the RHS (right hand side) is even, because it is a product of integers and one of those integers (at least) 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.

Now this is the contradiction: 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.

There you have it: a rational proof of irrationality.

....... hope it will help you ☺️☺️✌️✌️⭐⭐✅✅✅

Answer 2

Assume that sqrt(6) is rational.

Then sqrt(6) = p/q where p and q are coprime integers.

sqrt(6)^2 = 6 = p²/q²

p² = 6q²

therefore p² is an even number since an even number multiplied by any other integer is also an even number. If p² is even then p must also be even since if p were odd, an odd number multiplied by an odd number would also be odd.

So we can replace p with 2k where k is an integer.

(2k)² = 6q²

4k² = 6q²

2k² = 3q²

Now we see that 3q² is even. For 3q² to be even, q² must be even since 3 is odd and an odd times an even number is even. And by the same argument above, if q² is even then q is even.

So both p and q are even which means both are divisible by 2. But that means they are not coprime, contradicting our assumption so sqrt(6) is not rational.