Proving that √5 - √3 is irrational:

To prove that √5 - √3 is irrational, we will assume the opposite and assume that √5 - √3 is rational.

Assumption:
Let's assume that √5 - √3 is rational and can be expressed as a fraction p/q, where p and q are integers with no common factors other than 1.

Squaring both sides:
Now, let's square both sides of the equation (√5 - √3)^2 = (p/q)^2.

Expanding the equation, we get:
(√5 - √3)(√5 - √3) = (p/q)^2
5 - 2√15 + 3 = p^2/q^2

Simplifying further, we have:
8 - 2√15 = p^2/q^2

Isolating the irrational term:
Rearranging the equation, we get:
2√15 = 8 - p^2/q^2

Now, let's isolate the irrational term (√15) on one side:
2√15 = 8 - p^2/q^2

Adding p^2/q^2 to both sides, we get:
2√15 + p^2/q^2 = 8

Squaring both sides:
Now, let's square both sides of the equation (2√15 + p^2/q^2)^2 = 8^2.

Expanding the equation, we get:
(2√15)^2 + 2(2√15)(p^2/q^2) + (p^2/q^2)^2 = 64

Simplifying further, we have:
60 + 4(p^2/q^2) + (p^2/q^2)^2 = 64

Isolating the irrational term:
Rearranging the equation, we get:
(p^2/q^2)^2 + 4(p^2/q^2) + 60 = 64

Subtracting 64 from both sides, we get:
(p^2/q^2)^2 + 4(p^2/q^2) - 4 = 0

Applying the quadratic formula:
Now, let's solve the quadratic equation (p^2/q^2)^2 + 4(p^2/q^2) - 4 = 0 using the quadratic formula:

The quadratic formula states that for an equation ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 1, b = 4(p^2/q^2), and c = -4. Substituting these values into the quadratic formula, we get:
(p^2/q^2) = (-4(p^2/q^2) ± √((4(p^2/q^2))^2 - 4(1)(-4))) / (2(1))

Simplifying further, we have:
(p^2/q^2) = (-4(p^2/q^2

becoz we can't write them in p/q form that's why it is irrational numbers