If √6 is rational, then it can be expressed as a ratio of two integers a and b in the simplest form such that:

√6 = a/b

Where a and b have no common factors other than 1.


Squaring both sides of the equation, we get:

6 = a^2/b^2

Multiplying both sides by b^2, we get:

6b^2 = a^2

This means that a^2 is an even number because 6b^2 is even. Therefore, a must also be even.


We can express a as 2k, where k is an integer.

a = 2k

Substituting this in the equation above, we get:

6b^2 = (2k)^2

6b^2 = 4k^2

3b^2 = 2k^2

This means that 2k^2 is an even number because 3b^2 is odd. Therefore, k^2 is even and k must also be even.


We have now shown that both a and b are even, which means that they have a common factor of 2. This contradicts our assumption that a and b have no common factors other than 1. Therefore, our assumption that √6 is rational must be false.


We have shown that √6 is irrational because it cannot be expressed as a ratio of two integers in the simplest form.