Proof that √5 is an irrational number:

Assumption: Let's assume that √5 is a rational number.

Definition of a rational number: A rational number can be expressed in the form of p/q, where p and q are integers and q is not equal to zero.

Representation of √5 as a rational number: Let's assume that √5 can be expressed as p/q, where p and q have no common factors other than 1.

Squaring both sides: If we square both sides of the equation, we get 5 = (p^2/q^2).

Manipulating the equation: Rearranging the equation, we get p^2 = 5q^2.

Observation: From the equation, we can see that p^2 is divisible by 5.

Consequence: If p^2 is divisible by 5, then p must also be divisible by 5.

Representation of p as a multiple of 5: We can write p as 5k, where k is an integer.

Substituting the value of p: Substituting the value of p in the equation p^2 = 5q^2, we get (5k)^2 = 5q^2, which simplifies to 25k^2 = 5q^2.

Simplifying the equation: Dividing both sides of the equation by 5, we get 5k^2 = q^2.

Observation: From the equation, we can see that q^2 is also divisible by 5.

Consequence: If q^2 is divisible by 5, then q must also be divisible by 5.

Contradiction: We have reached a contradiction because we initially assumed that p and q have no common factors other than 1, but we have now found that both p and q are divisible by 5.

Conclusion: Since assuming that √5 is a rational number leads to a contradiction, our assumption is incorrect. Therefore, √5 is an irrational number.

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