Prove that root 2 +root3 is irrational?

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Assume it's rational. Then there are integers m,n such that m/n = √2 + √3 Square both sides: m^2 / n^2 = 5 + 2√6 'Solve' for √6: √6 = (m^2 - 5n^2) / (2n^2) so if √2 + √3 is rational, then so is √6. then prove that √6 is irrational number thus if √6 is irrational number than . √2+√3 is also irrational hope u will like my technique

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Samjh nhi aa raha hai yah question

Answer

Proving that √2 * √3 is irrational:

To prove that √2 * √3 is irrational, we can use a proof by contradiction. We assume that √2 * √3 is rational and then show that it leads to a contradiction.

Assumption:

Let's assume that √2 * √3 is rational.

Expressing the assumption as an equation:

Since we assume that √2 * √3 is rational, we can express it as a fraction in the form p/q, where p and q are integers with no common factors and q is not equal to zero.

√2 * √3 = p/q

Manipulating the equation:

We can square both sides of the equation to eliminate the square roots:

(√2 * √3)^2 = (p/q)^2

2 * 3 = (p^2)/(q^2)

6 = (p^2)/(q^2)

Multiplying both sides by q^2, we get:

6q^2 = p^2

Conclusion:

From the equation 6q^2 = p^2, we can see that p^2 is divisible by 6, which implies that p is also divisible by 6.

Therefore, we can express p as p = 6k, where k is an integer.

Substituting p = 6k into the equation, we get:

6q^2 = (6k)^2

6q^2 = 36k^2

Dividing both sides of the equation by 6, we obtain:

q^2 = 6k^2

Contradiction:

From the equation q^2 = 6k^2, we can see that q^2 is also divisible by 6, which implies that q is also divisible by 6.

However, we initially assumed that p and q have no common factors, which contradicts the fact that both p and q are divisible by 6.

Final Conclusion:

Since assuming √2 * √3 to be rational leads to a contradiction, we can conclude that √2 * √3 is irrational.

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