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How can I prove that the sum $\sqrt 2+\sqrt[3] 3$ is an irrational number ??

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4 Answers 4

up vote 21 down vote accepted

A variation on robjohn's proof, without the need of the notion of algebraic integer, and without computing the minimal polynomial of your algebraic number.

If $x=\sqrt{2}+\sqrt[3]3$, then

$$(x-\sqrt{2})^3=x^3-3\sqrt{2}x^2+6x-2\sqrt{2}=3$$

Thus

$$x^3+6x-3=\sqrt{2}(3x^2+2)$$

And

$$\frac{x^3+6x-3}{3x^2+2}=\sqrt{2}$$

But if $x$ is a rational, then so is the left hand side of the above equality. However we know $\sqrt{2}$ is not rational. Contradiction, so $x$ is irrational.

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The notion of an algebraic integer is not that complicated, and I think this implication is worth knowing. In this answer, I use only Bezout's Identity and some algebra to show that a rational algebraic integer is actually an integer. The proof is only a little more involved algebraically than the proof that $\sqrt2$ is irrational. –  robjohn Aug 31 at 2:18
    
Similarly, $\dfrac{2x^3-4x+3}{3x^2+2}=\sqrt[3]3$. –  columbus8myhw Aug 31 at 3:54

If $x=\sqrt2+\sqrt[3]3$, then $$ \begin{align} 3 &=(x-\sqrt2)^3\\ &=x^3-3\sqrt2x^2+6x-2\sqrt2\\ (x^3+6x-3)^2&=2(3x^2+2)^2\\ 0&=x^6-6x^4-6x^3+12x^2-36x+1 \end{align} $$ Thus, $x$ is an algebraic integer. Since $2\lt x\lt3$, $x\not\in\mathbb{Z}$, so $x\not\in\mathbb{Q}$. In this answer, it is shown that a rational algebraic integer is an integer.


Alternative approach

If $x=\sqrt2$ and $y=\sqrt[3]3$, then $x^2-2=0$ and $y^3-3=0$. Thus, both $x$ and $y$ are algebraic integers. Therefore, $x+y=\sqrt2+\sqrt[3]3$ is also an algebraic integer.

Note that this approach, while seeming simpler, requires that one knows that the sum of two algebraic integers is again an algebraic integer. This fact can be shown using some linear algebra.

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I am relatively new here, and you have much more reputation than I do, so I hesitate to ask--but I would really like to know, so this is a serious question. Is this answer giving the OP too much information? I tried in my answer to help a little but force the OP to do more work on his own. –  Rory Daulton Aug 30 at 14:26
    
@robjohn +1 For the last line, it's also possible to let $x=\frac pq$ and prove directly $p|1$ and $q|1$, thus a contradiction. –  Jean-Claude Arbaut Aug 30 at 14:32
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@RoryDaulton: Don't hesitate to ask because of anyone's reputation (see this meta thread). I thought about this a bit, and often if I feel a question is a homework problem, I leave a hint. However, sometimes I leave a more complete answer when either the hint I'd give would be too cryptic, or if a more complete answer does not give much more away. In this case, the hint would not give much more than cubing of squaring a polynomial. I may have misjudged. Thanks for your comment; it will affect my decisions in the future. –  robjohn Aug 30 at 14:34
    
Nice solution ! thank you very much.!!! –  Marianna Pap Aug 30 at 15:07
    
How does $x$ being algebraic and not an integer imply it's irrational? –  Darksonn Aug 30 at 16:03

The following is a solution using some (basic) notions from the theory of field extensions:

Assume $\sqrt{2}+\sqrt[3]{3}$ is rational. Then $\sqrt[3]{3}\in\mathbb{Q}(\sqrt{2})$, thus $\sqrt[3]{3}$ is a root of some quadratic polynomial over $\mathbb{Q}$ which leads to a contradiction, since the polynomial $x^3-3$ is irreducible over $\mathbb{Q}$.

The same idea can be expressed using only elementary terms. Let $a$ be a rational number, and note that $a-\sqrt{2}$ is a root of the rational polynomial $x^2-2ax+a^2-2$. Thus, if $\sqrt{2}+\sqrt[3]{3}$ is rational, it follows that $\sqrt[3]{3}$ is a root of a quadratic rational polynomial, which is a contradiction as $x^3-3$ is irreducible.

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Use these facts:

$$(\sqrt 2 + \sqrt[3]3) \cdot (-\sqrt 2 + \sqrt[3]3) = -2 + \sqrt[3]9$$

$$(-2 + \sqrt[3]9) \cdot [(-2)^2 - (-2)\sqrt[3]9 + (\sqrt[3]9)^2] = (-2)^3 + (\sqrt[3]9)^3 $$

That last number is rational.

Using those facts, find a polynomial of degree 6 with integral coefficients for which $\sqrt 2 + \sqrt[3]3$ is a root. Then, using the rational root theorem, show that any root of that polynomial is irrational.

Let us know if you need more help in finding that polynomial.


ADDED: Robjohn's answer gives a more straightforward way of finding an appropriate 6th-degree polynomial.

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