proof that is irrational
where and and are relatively prime. Then . Thus, . Therefore, . Since is prime, it must divide . Then for some . Thus, , yielding that . Therefore, . Since is prime, it must divide .
With a little bit of work, this argument can be generalized to any positive integer that is not a square. Let be such an integer. Then there must exist a prime and such that , where and is odd. Assume that , where and are relatively prime. Then . Thus, . From the fundamental theorem of arithmetic, it is clear that the maximum powers of that divides and are even. Since is odd and does not divide , the maximum power of that divides is also odd. Thus, the same should be true for . Hence, we have reached a contradiction and must be irrational.
The same argument can be generalized even more, for example to the case of nonsquare irreducible fractions and to higher order roots.
|Title||proof that is irrational|
|Date of creation||2013-03-22 12:39:13|
|Last modified on||2013-03-22 12:39:13|
|Last modified by||Wkbj79 (1863)|