Final answer:
To ascertain √5's rationality, a proof by contradiction shows that assuming √5 is rational leads to 'a' and 'b' both being multiples of 5, despite being in lowest terms. This contradiction proves that √5 is irrational.
Step-by-step explanation:
Proving the Irrationality of √5
To prove that √5 is irrational, we can employ a proof by contradiction, which is a common method used to prove the irrationality of a number. Assuming √5 is rational, it can be expressed as a fraction of two integers (a/b) in its lowest terms, where 'a' and 'b' have no common factors besides 1, and 'b' is not zero.
According to our initial assumption: √5 = a/b. Squaring both sides gives us 5 = a²/b². Therefore, a² = 5b², implying that a² is a multiple of 5, so 'a' must also be a multiple of 5. This can be written as a = 5k, where 'k' is an integer. Substituting back, we get (5k)² = 5b², which simplifies to 25k² = 5b²; and further to b² = 5k². This conclusion indicates that 'b' would also have to be a multiple of 5.
This results in a contradiction because 'a' and 'b' are both multiples of 5, which contradicts our assumption that they have no common factors besides 1. Hence, √5 cannot be rational and must be irrational.