How to justify square root of 6 is irrational

sqrt(6) = sqrt(2*3) = sqrt(2) * sqrt(3)
2 and 3 are prime numbers. You've got two prime numbers without an even exponent. The square root of every prime number is an irrational number sqrt(6) is an irrational number.

Apfelkuchen:

sqrt(6) = sqrt(2*3) = sqrt(2) * sqrt(3)
2 and 3 are prime numbers. You've got two prime numbers without an even exponent. The square root of every prime number is an irrational number sqrt(6) is an irrational number.

actualy 6 is a whole number and whole numbers are always rational irrational numbers are not whole not a integer or natural
therefor it cant be irrational

So the square root of 2 would also be a rational number, because 2 is rational? Of course not.
6 is a rational number, thats correct.
But as we know, every square root of a number n is either an irrational number or an integer.

Claim: sqrt(6) is a rational number.
Therefore we could create an equation like sqrt(6) = n/k with n,k being integers > 1.
6 = (n/k)^2 = (n^2)/(k^2)
6*k^2 = n^2

We can see, that 6*k^2 is always even n^2 is also even n is even
So we rephrase the equation with 6*k^2 = n^2 = (2a)^2; n = 2a
(6*k^2)/2 = 2a^2 = 3*k^2

2a^2 is always even. Therefore 3*k^2 has to be even. This is only the case if k is even.
k = 2b
2a^2 = 3*k^2 = 3*(2b)^2 = 3*4b^2 = 12b^2
a^2 = 6b^2
etc.

sqrt(6) won't result in an integer. So there shouldn't be two even integers.