Ask your own question, for FREE!
Mathematics 25 Online
OpenStudy (anonymous):

How can I use this theorem "Let \(\alpha\) be a root of the polynomial \(x^n+c_{n-1}x^{n-1}+...+c_1x+c_0\), where the coefficients \(c_0, c_1, ..., c_n\) are integers. Then \(\alpha\) is either an integer or an irrational number." to show that \(\sqrt[3]{5}\) is irrational?

OpenStudy (jamesj):

The solutions of \( x^3 -5 \) must therefore be integral or irrational, and it's not hard to show they are not integers.

OpenStudy (anonymous):

Ah! I see. I will try that now. Thank you.

OpenStudy (anonymous):

Proof: The solutions of \(x^3-5\) must be either integral or irrational. Thus, we are looking for an integer \(x\) such that \(x^3=5\). However, there is no such integer since \(1^3=1<5\) and \(2^3=8>5\). Therefore, \(x\) must be irrational, and \(\sqrt[3]{5}\) is irrational. \(\blacksquare\) Is this okay?

OpenStudy (jamesj):

Yes

OpenStudy (anonymous):

Thank you again.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
MakaylaChuck23: Help wats this
1 hour ago 2 Replies 0 Medals
MakaylaChuck23: uh oh
4 hours ago 1 Reply 0 Medals
Treeko: Pics I took w my friend !
6 hours ago 6 Replies 2 Medals
Aliciaa: Is anyone elses QC going wack? Mine keeps freezing and stuff
4 hours ago 26 Replies 1 Medal
autumnp: does anyone have bandlab?
6 hours ago 4 Replies 0 Medals
Jeromeccv: Art for @vanessad123
4 hours ago 49 Replies 4 Medals
ohhhhhhhhhhhhhh: how do i slim my waist fast
6 hours ago 4 Replies 1 Medal
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!