Proof that the √5 is irrational

219,311pages on
this wiki
Add New Page
Discuss this page0 Share

  • 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.

See Also

Proof that √2 is irrational

Ad blocker interference detected!

Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.

Also on Fandom

Random wikia