Introduction:

In order to prove that 2/3√3 is an irrational number, we need to show that it cannot be expressed as the quotient of two integers. We will use a proof by contradiction to demonstrate this.

Proof by Contradiction:

Assume that 2/3√3 is a rational number. This means that it can be expressed as the quotient of two integers, let's say a and b, where b is not equal to zero and a and b have no common factors other than 1.

Assumption: 2/3√3 = a/b, where a and b are integers.

Squaring both sides: (2/3√3)^2 = (a/b)^2

Simplifying the equation, we get:

4/9 * 3 = a^2 / b^2
4/3 = a^2 / b^2

Observation: From the equation 4/3 = a^2 / b^2, we notice that a^2 must be divisible by 3, which means a must also be divisible by 3.

Assumption: Let a = 3c, where c is an integer.

Substituting this value of a into the equation, we get:

4/3 = (3c)^2 / b^2
4/3 = 9c^2 / b^2
4b^2 = 27c^2

Observation: From the equation 4b^2 = 27c^2, we notice that b^2 must be divisible by 3, which means b must also be divisible by 3.

Assumption: Let b = 3d, where d is an integer.

Substituting this value of b into the equation, we get:

4(3d)^2 = 27c^2
4(9d^2) = 27c^2
36d^2 = 27c^2

Observation: From the equation 36d^2 = 27c^2, we notice that c^2 must be divisible by 4, which means c must also be divisible by 2.

Assumption: Let c = 2e, where e is an integer.

Substituting this value of c into the equation, we get:

36d^2 = 27(2e)^2
36d^2 = 108e^2
d^2 = 3e^2

Observation: From the equation d^2 = 3e^2, we notice that d^2 must be divisible by 3, which means d must also be divisible by 3.

Contradiction: We have arrived at a contradiction. We initially assumed that a and b have no common factors other than 1, but we have shown that both a and b are divisible by 3. Therefore, our assumption that 2/3√3 is a rational number must be false.

Conclusion: Based on the proof by contradiction, we can conclude that 2/3√3 is an irrational number. It cannot be expressed as the quotient