Let us assume that 15 +17√5 is a rational no. where 15 +17√5 =a/b,where a and v are Coprime, b is not equal to 0.



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> 15+17√5=a/b

>17√5=a/b -15

>√5=a-15b/17b


therefore a and b are integers.


so, a-15b/17b is a rational no. and so, √5 is rational



but this contradicts the fact that √5 is irrational. This contradiction has arisen because of our incorrect assumption that 15 +17√5 is rational.


hence, 15+17√5 is irrational.

Proof that 15 + 17√5 is irrational:

Assumption: Assume that 15 + 17√5 is rational.

Definition: A rational number can be expressed as a quotient of two integers, p and q, where q ≠ 0.

Equation: Let 15 + 17√5 = p/q, where p and q are integers and q ≠ 0.

Algebraic Manipulation: Multiply both sides by q to get 15q + 17√5q = p.

Squaring: Square both sides of the equation to get (15q + 17√5q)^2 = p^2.

Algebraic Manipulation: Expand the left-hand side of the equation to get 225q^2 + 510√5q^2 + 289q^2 * 5 = p^2.

Algebraic Simplification: Simplify the left-hand side of the equation to get 5(17^2 * q^2) + 15^2 * q^2 = p^2.

Conclusion: Since p^2 is divisible by 5, p must be divisible by 5. Therefore, we can write p as 5k, where k is an integer. Substituting p = 5k into the equation, we get 5(17^2 * q^2) + 15^2 * q^2 = (5k)^2. Simplifying the equation further, we get 17^2 * q^2 + 3^2 * q^2 = 5k^2. This implies that 17^2 * q^2 and 3^2 * q^2 must both be divisible by 5. However, this is impossible since 17^2 and 3^2 are not divisible by 5. Therefore, our assumption that 15 + 17√5 is rational is false.

Conclusion: Hence, 15 + 17√5 is irrational.