Question

Prove that there is no rational number r such that r^3=2.

I know that I can solve this by contradiction which uses the same argument that proves that there is no rational r for r^2=2 but my professor wants this solved two different ways so what's another way I can solve this?

Answer 1

http://books.google.co.in/books?id=b4CKH4a7R6QC&pg=PA63&lpg=PA63&dq=Prove+that+there+is+no+rational+number+r+such+that+r%5E3%3D2.&source=bl&ots=kGho65jtte&sig=ZP_ZJlSyDIqbgAnaqnC03EHrPko&hl=en&sa=X&ei=e5wQVJiqBo7luQSvlYKYCg&ved=0CD0Q6AEwBA#v=onepage&q=Prove%20that%20there%20is%20no%20rational%20number%20r%20such%20that%20r%5E3%3D2.&f=false

Answer 2

hope this helps @kdekle

Answer 3

In all honesty it doesn't. I don't see how that helps me solve the problem.

Answer 4

i dont know if this makes sense but

1.25^3 = 1.953
1.26^3 = 2.000376

so the cube root of 2 lies between these 2 values
can they possibly be a decimal number which ends in 1 - 9 with a cube which is a whole number?

it seems that way - but i dont suppose this is a rigorous mathematical proof.

Answer 5

You could use the rational roots test. If r were a rational number such that $$r^{3}=2,$$then r would be a rational root of the polynomial $$x^3-2=0.$$However, by the rational roots test, the only possible rational roots of this polynomial are 2 and -2, both of which aren't roots. Hence cube root of 2 can't be rational.

Answer 6

Thank you that's awesome!