I
Icebreaker
How do I show that 2 has no rational roots?
Where p and q are coprime integers...Write square root of 2 = p / q, where p and q are integers.
Our assumption before was that 2^{1/n} where n was an integer greater than 1. However I don't think it is hard to extend if you let x = p/q.Icebreaker said:Yup, that's what I thought too. Thanks everyone.
On a side note, [tex]x^x = 2[/tex] implies that x cannot be rational because [tex]x = 2^{1/x}[/tex] and 2 has 2 rational roots, as we've shown. Is that correct?
If x is rational, then there are coprime p and q such that x = p/q. We get:Icebreaker said:Yup, that's what I thought too. Thanks everyone.
On a side note, [tex]x^x = 2[/tex] implies that x cannot be rational because [tex]x = 2^{1/x}[/tex] and 2 has 2 rational roots, as we've shown. Is that correct?