$\sqrt{98} + 4^{\frac{1}{3}}$ is irrational

Question

Problem is to prove that $\sqrt{98} + 4^{\frac{1}{3}}$ is irrational.
I know that $\sqrt{98}$ and $4^{\frac{1}{3}}$ is irrational, but I don't know how to use it in this problem.

Do you know the rational root theorem? You should check out Mathologer's latest video. https://www.youtube.com/watch?v=D6AFxJdJYW4 — SmileyCraft, Jan 05 '19 at 18:12
That's an algebraic integer. An algebraic integer which is in $\Bbb Q$ must also be in $\Bbb Z$. Your calculator will tell you that's not in $\Bbb Z$. — Angina Seng, Jan 06 '19 at 04:36

score 7 · Answer 1 · answered Jan 05 '19 at 18:23

7

Let $$7\sqrt2+\sqrt[3]4=r\in\mathbb Q.$$ Thus, $$4=r^3-21\sqrt2r^2+294r-686\sqrt2$$ or $$\sqrt2=\frac{r^3+294r-4}{7(3r^2+98)},$$ which is a contradiction because $\sqrt2\notin\mathbb Q$..

answered Jan 05 '19 at 18:23

Michael Rozenberg

194,933

More generally it's true in any field where $2$ and $-6$ are not squares - see my answer. – Bill Dubuque Jan 05 '19 at 20:38
Yes, I saw. Nice! – Michael Rozenberg Jan 05 '19 at 20:50

score 1 · Answer 2 · answered Jan 05 '19 at 20:22

If it's rational $= a$ then cubing $\,4^{1/3} = a-7\sqrt 2\,$ yields the contradiction that $ \sqrt2$ or $\sqrt{-6}$ is rational, by applying the theorem below with $\,b,c = -7,2\ $ (where $\sqrt d\in F$ means $\ d = f^2$ for some $\,f\in F)$

Theorem $ $ If $\,a,b,c\in F$ a field, $b\neq 0,\,$ then $\,(a+b\sqrt{c})^3\in F\,\Rightarrow\, \sqrt c\in F\,$ or $\,\sqrt{-3c}\in F$

Proof $\ $ Cubing $\ (a+b\sqrt{c})^3 = b(3a^2+cb^2 ) \sqrt c + d = e\,$ for $\,d,e\in F$

When $\,3a^2+cb^2 \neq 0\,$ solving the above for $\,\sqrt c\,$ yields $\,\sqrt c \in F$.

Else $\,\ 9a^2 = -3cb^2\,\Rightarrow\, (3a/b)^2 = -3c\,\Rightarrow\,\sqrt{-3c}\in F$.

$\sqrt{98} + 4^{\frac{1}{3}}$ is irrational

2 Answers2