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Prove: $\sqrt{2}+\sqrt[3]{5}$ is irrational

I have tried to look at $1+\sqrt{2}+\sqrt[3]{5}$ but I can see how to continue

gbox
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    Could you please add some context? In particular, it would help to know what is your background and where you found this exercise. – A.P. Oct 19 '15 at 20:44
  • first course in math for physicists – gbox Oct 19 '15 at 20:45
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    Do you know anything about algebraic numbers and field theory? – A.P. Oct 19 '15 at 20:48
  • One approach would be to compute a polynomial with integer coefficients such that $\sqrt{2} + \sqrt[3]{5}$ is a root. Then use the rational root theorem to show that either the polynomial you found has no rational roots, or the rational roots it has are not $\sqrt{2} + \sqrt[3]{5}$. – Alex Wertheim Oct 19 '15 at 20:52
  • @A.P. I know about algebraic numbers – gbox Oct 19 '15 at 20:53
  • See also http://math.stackexchange.com/questions/913934/process-to-show-that-sqrt-2-sqrt3-3-is-irrational?rq=1 – Nicky Hekster Oct 19 '15 at 20:57

3 Answers3

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Suppose that $\sqrt 2+\sqrt[3]{5}=r$ where $r$ is rational.

Then, we have $\sqrt[3]{5}=r-\sqrt 2\Rightarrow 5=(r-\sqrt 2)^3$.

Now, expand the RHS and you should be able to see how to get a contradiction.

$\begin{align}\sqrt[3]{5}=r-\sqrt 2&\Rightarrow 5=(r-\sqrt 2)^3\\&\Rightarrow 5=r^3-3\sqrt 2r^2+6r-2\sqrt 2\\&\Rightarrow (3r^2+2)\sqrt 2=r^3+6r-5\\&\Rightarrow \sqrt 2=\frac{r^3+6r-5}{3r^2+2}\end{align}$

The LHS is irrational and the RHS is rational. This is a contradiction.

mathlove
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One way to approach these kinds of problem in general is to constuct an integer polynomial that has your number as a root and then using the rational root theorem to make a list over all the possible rational roots. Then we can easily check that none of these alternatives are true roots which again proves that your number is irrational.

To construct the polynomial start with

$$x=\sqrt{2}+\sqrt[3]{5} \implies (x-\sqrt{2})^3=5$$

Now simplify and move the irrational factors to the right hand side $$x^3 -6x-5 = \sqrt{2}(3x^2+2)$$

A final squaring gives us that your number is a root of the integer polynomial

$$f(x) = x^6-6 x^4-10 x^3+12 x^2-60 x+17$$

for which you can apply the rational root theorem on to get two candiate roots $x=1$ and $x=17$. It's easy to check that none of these roots are real roots which implies that your number is irrational.

Winther
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S'pose not.

Then sqrt (2) + r = cube_root (5) for some rational r. Raise both to the 3rd power

You'll get that the square root of 2 is rattional. A contradiction.

fleablood
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