To prove: root6 is irrational Proof: Let us assume, to the contrary, that root6 is rational Then, root6=a/b ------eqn1 [a &b are coprimes , b is not equal to 0] squaring both sides of eqn1 6=a square /b square 6b square =a square or a square = 6b square -----eqn2i.e. 6 divides a square & also divides a (Theorem 1.3)----eqn3we may write a=6ma square = 36 m square putting the value of a square from eqn26b square = 36 m squareb square = 36 m square/6mb square = 6m square i.e. 6 divides b square & also divides b (Theorem 1.3)----eqn4therefore from eqn 3 & 4 6 divides both a & b Which implies that a & b are not coprimes Which contradicts our assumption that a & b are coprimes Therefore root6 is irrational.

Proof that √6 is an irrational number:

Assumption:
Let's assume that √6 is a rational number.

Definition of Rational Number:
A rational number can be expressed as a ratio of two integers, where the denominator is not zero.

Representation of √6 as a Rational Number:
If √6 is rational, then it can be written in the form of a/b, where a and b are integers with no common factors and b is not equal to zero.

Square of √6:
(√6)^2 = 6

Expressing √6 as a Ratio:
(√6)^2 = (a/b)^2
6 = a^2/b^2
6b^2 = a^2

Consequence:
This implies that a^2 is a multiple of 6, making a a multiple of 6.

Conclusion:
Since a is a multiple of 6, it can be expressed as 6c, where c is an integer.

Revisiting the Equation:
6b^2 = (6c)^2
6b^2 = 36c^2
b^2 = 6c^2

Implication:
This shows that b^2 is also a multiple of 6, making b a multiple of 6.

Contradiction:
Since both a and b are multiples of 6, they have a common factor, which contradicts our assumption that a and b have no common factors.

Final Verdict:
Therefore, our assumption that √6 is rational leads to a contradiction, proving that √6 is an irrational number.