**Proof by Contradiction:**

To prove that √5 √7 is an irrational number, we will assume the opposite, that √5 √7 is a rational number.

**Assumption:**
Let's assume that √5 √7 is a rational number. By definition, a rational number can be expressed as a fraction p/q, where p and q are integers and q is not equal to zero.

**Simplifying the expression:**
√5 √7 = √(5 * 7) = √35

**Assumption in Fraction Form:**
Assuming √35 is rational, it can be expressed as p/q, where p and q have no common factors other than 1.

**Squaring Both Sides:**
(√35)^2 = (p/q)^2
35 = (p^2)/(q^2)

**Consequence and Proof:**
From the equation above, we can conclude that p^2 is divisible by 35, which means p must also be divisible by 5. Therefore, we can write p = 5k, where k is an integer.

Substituting p = 5k in the equation:
35 = (5k^2)/(q^2)

Simplifying the equation further:
7 = (k^2)/(q^2)

From this equation, we can deduce that q^2 is divisible by 7, which means q must also be divisible by 7. Therefore, we can write q = 7m, where m is an integer.

Substituting q = 7m in the equation:
7 = (k^2)/(7^2 * m^2)

Simplifying the equation further:
1 = (k^2)/(49m^2)

From the equation above, we can conclude that k^2 is divisible by 49, which means k must also be divisible by 7. However, this contradicts our assumption that p and q have no common factors other than 1.

**Contradiction:**
Since our assumption leads to a contradiction, we can conclude that √5 √7 is not a rational number. Hence, √5 √7 is an irrational number.

Therefore, we have proved that √5 √7 is an irrational number.

Let us assume to the contrary that 5+√7 is rationalSo,it can be written in the form a/b where a and b are co-primes,and b not equal to zero.5+√7=a/b√7=a-5b/bSince a,b and 5 are integers, therefore the are rationalBut this contradicts the fact that√7 is irrationalHence our assumption is incorrectTherefore 5+√7 is irrational... i hope u will like my answer