Proof that √5 is irrationalAssumption: √5 is a rational number.
Let √5 be expressed as a fraction in its simplest form, i.e.
√5 = a/b, where a and b are coprime integers (i.e. they have no common factors).
Step 1: Squaring both sides
Squaring both sides of the equation, we get:
5 = a^2/b^2
Step 2: Rearranging the equation
Multiplying both sides by b^2, we get:
5b^2 = a^2
Step 3: Analyzing the equation
Since a^2 is a multiple of 5, we can conclude that a must also be a multiple of 5.
Therefore, we can express a as 5k, where k is an integer.
Substituting this value of a in the equation, we get:
5b^2 = (5k)^2
5b^2 = 25k^2
b^2 = 5k^2
Step 4: Analyzing the equation
Now, we can see that b^2 is a multiple of 5, which means that b must also be a multiple of 5.
However, this contradicts our initial assumption that a and b are coprime.
Therefore, our assumption that √5 is a rational number must be false.
Conclusion:Thus, we have proved that √5 is an irrational number.