## FANDOM

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• Proof that the √5 is irrational:

This will be a proof by contradiction.

1. Assume that the √5 is rational (can be expressed in the form $\frac{a}{b}$, where a and b are integers.
2. Then √5 can be written as an irreducible fraction (the fraction is reduced as much as possible) a / b such that a and b are coprime integers and (a / b)2 = 5.
3. It follows that a2 / b2 = 5 and a2 = 5 b2.
4. Therefore a2 is divisible by 5 because it is equal to 5 b2 which is obviously divisible by 5.
5. It follows that a must be divisible by 5.
6. Because a is divisible by 5, there exists an integer k that fulfills: a = 5k.
7. We insert the last equation of (3) in (6): 5b2 = (5k)2 is equivalent to 5b2 = 25k2 is equivalent to b2 = 5k2.
8. Because 5k2 is divisible by 5 it follows that b2 is also divisible by 5 which means that b is divisible by 5 because only numbers divisible by 5 have squares divisible by 5.
9. By (5) and (8) a and b are both divisible by 5, which contradicts that a / b is irreducible as stated in (2).

Therefore, √5 is irrational.