Introduction:
To prove that √5 - 3√2 is an irrational number, we need to show that it cannot be expressed as a ratio of two integers.

Assumption:
Assume that √5 - 3√2 is a rational number.

Proof by Contradiction:
Let √5 - 3√2 be expressed as a fraction p/q, where p and q are integers and q ≠ 0.

Step 1:
Now, we can write √5 = (3√2 + p)/q

Step 2:
Squaring both sides, we get
5 = (9*2 + p^2 + 2*3√2p)/q^2
p^2 + 18p + 36 = 5q^2

Step 3:
Since p^2 + 18p + 36 is even, 5q^2 must also be even. This implies that q^2 is even and hence q is even.

Step 4:
Let q = 2r, where r is an integer. Substituting this in the equation p^2 + 18p + 36 = 5q^2, we get
p^2 + 18p + 36 = 20r^2
p^2 + 18p + 16 = 20r^2 - 20
(p + 9)^2 = 5(2r)^2 - 1

Step 5:
Now, the left-hand side is a perfect square, but the right-hand side is 5 times a perfect square minus 1. This is a contradiction, since no perfect square can be 5 times another perfect square minus 1.

Conclusion:
Therefore, our assumption that √5 - 3√2 is a rational number is false. Hence, √5 - 3√2 is an irrational number.

Prove √2or√5 individually irrational and conclude that irrational +irrational=irrational