Spivak's Calculus: chapter 2, problem 18(c)

Question

In Spivak's calculus book, I cannot understand the solution proposed for question (c) of problem 18 in chapter 2:

Prove that $\sqrt{2}+\sqrt[3]{2}$ is irrational. Hint: Start by working out the first 6 powers of this number.

Working out the powers is quite easy:

• $(2^\frac{3}{6}+2^\frac{2}{6})^0 = 1$
• $(2^\frac{3}{6}+2^\frac{2}{6})^1 = 2^\frac{3}{6} + 2^\frac{2}{6}$
• $(2^\frac{3}{6}+2^\frac{2}{6})^2 = 2^\frac{6}{6} + 2 \cdot 2^\frac{3}{6} \cdot 2^\frac{2}{6} + 2^\frac{4}{6} = \\ 2 \cdot 2^\frac{0}{6} + 2^\frac{4}{6} + 2 \cdot 2^\frac{5}{6}$
• $(2^\frac{3}{6}+2^\frac{2}{6})^3 = 2^\frac{9}{6} + 3 \cdot 2^\frac{6}{6} \cdot 2^\frac{2}{6} + 3 \cdot 2^\frac{3}{6} \cdot 2^\frac{4}{6} + 2^\frac{6}{6} = \\ 2 \cdot 2^\frac{0}{6} + 6 \cdot 2^\frac{1}{6} + 6 \cdot 2^\frac{2}{6} + 2 \cdot 2^\frac{3}{6} $
• $(2^\frac{3}{6}+2^\frac{2}{6})^4 = 2^\frac{12}{6} + 4 \cdot 2^\frac{9}{6} \cdot 2^\frac{2}{6} + 6 \cdot 2^\frac{6}{6} \cdot 2^\frac{4}{6} + 4 \cdot 2^\frac{3}{6} \cdot 2^\frac{6}{6} + 2^\frac{8}{6} = \\ 4 \cdot 2^\frac{0}{6} + 2 \cdot 2^\frac{2}{6} + 8 \cdot 2^\frac{3}{6} + 12 \cdot 2^\frac{4}{6} + 8 \cdot 2^\frac{5}{6} $
• $(2^\frac{3}{6}+2^\frac{2}{6})^5 = 2^\frac{15}{6} + 5 \cdot 2^\frac{12}{6} \cdot 2^\frac{2}{6} + 10 \cdot 2^\frac{9}{6} \cdot 2^\frac{4}{6} + 10 \cdot 2^\frac{6}{6} \cdot 2^\frac{6}{6} + 5 \cdot 2^\frac{3}{6} \cdot 2^\frac{8}{6} + 2^\frac{10}{6} = \\ 40 \cdot 2^\frac{0}{6} + 40 \cdot 2^\frac{1}{6} + 20 \cdot 2^\frac{2}{6} + 4\cdot2^\frac{3}{6} + 2 \cdot 2^\frac{4}{6} + 10 \cdot 2^\frac{5}{6}$
• $(2^\frac{3}{6}+2^\frac{2}{6})^6 = 2^\frac{18}{6} + 6 \cdot 2^\frac{15}{6} \cdot 2^\frac{2}{6} + 15 \cdot 2^\frac{12}{6} \cdot 2^\frac{4}{6} + 20 \cdot 2^\frac{9}{6} \cdot 2^\frac{6}{6} + 15 \cdot 2^\frac{6}{6} \cdot 2^\frac{8}{6} + 6 \cdot 2^\frac{3}{6} \cdot 2^\frac{10}{6} + 2^\frac{12}{6} = \\ 12 \cdot 2^\frac{0}{6} + 24 \cdot 2^\frac{1}{6} + 60 \cdot 2^\frac{2}{6} + 80 \cdot 2^\frac{3}{6} + 60 \cdot 2^\frac{4}{6} + 24 \cdot 2^\frac{5}{6} $

I am at a loss as to how this can help towards the solution... The first question in problem 18 is asking to prove the "rational root theorem" but I don't see how I can combine that with this hint to solve the problem.

UPDATE: following Gerry Myerson's advice I create a polynomial $c_0 \cdot 2^\frac{0}{6} + c_1 \cdot 2^\frac{1}{6} + c_2 \cdot 2^\frac{2}{6} + c_3 \cdot 2^\frac{3}{6} + c_4 \cdot 2^\frac{4}{6} + c_5 \cdot 2^\frac{5}{6}$ where each $c_i$ multiplies the coeffficients of the respective powers I found above for $x^0, \dots, x^6$.

So each upper-case $C_i$ is the sum of the coefficients from each of the powers $x_n$. For example $C_5 = 24 + 10c_5 + 8c_4 + 2c_2$. The expansion of this polynomial is:

$ \begin{aligned} C_0 \cdot 2^\frac{0}{6} + C_1 \cdot 2^\frac{1}{6} + C_2 \cdot 2^\frac{2}{6} + C_3 \cdot 2^\frac{3}{6} + C_4 \cdot 2^\frac{4}{6} + C_5 \cdot 2^\frac{5}{6} &&= \\ [ 12 + 40 \cdot c_5 + 4 \cdot c_4 + 2 \cdot c_3 + 2 \cdot c_2 + c_0] \cdot 2^\frac{0}{6} &+ &\\ [ 24 + 40 \cdot c_5 + 6 \cdot c_3] \cdot 2^\frac{1}{6} &+ & \\ [ 60 + 20 \cdot c_5 + 2 \cdot c_4 + 6 \cdot c_3 + 1 \cdot c_1] \cdot 2^\frac{2}{6} & + & \\ [ 80 + 4 \cdot c_5 + 8 \cdot c_4 + 2 \cdot c_3 + 1\cdot c_1 ] \cdot 2^\frac{3}{6}& + & \\ [ 60 + 2 \cdot c_5 + 12 \cdot c_4 + 1 \cdot c_2 ] \cdot 2^\frac{4}{6} & + & \\ [ 24 + 10 \cdot c_5 + 8 \cdot c_4 + 2 \cdot c_2 ] \cdot 2^\frac{5}{6} & & \\ \end{aligned} $

So as Danny Pak - Keung Chan explained, in order to use the rational root theorem I need to set each of these $C_i$ equal to zero and solve the system of equations:

$ \begin{aligned} 12 + 40 \cdot c_5 + 4 \cdot c_4 + 2 \cdot c_3 + 2 \cdot c_2 + c_0 = 0&\\ 24 + 40 \cdot c_5 + 6 \cdot c_3 = 0 \\ 60 + 20 \cdot c_5 + 2 \cdot c_4 + 6 \cdot c_3 + 1 \cdot c_1 = 0 \\ 80 + 4 \cdot c_5 + 8 \cdot c_4 + 2 \cdot c_3 + 1\cdot c_1 = 0 \\ 60 + 2 \cdot c_5 + 12 \cdot c_4 + 1 \cdot c_2 = 0\\ 24 + 10 \cdot c_5 + 8 \cdot c_4 + 2 \cdot c_2 = 0 \\ \end{aligned} $

Now this will hopefully yield integer solutions which I can replace into the polynomial and thus apply the rational root theorem (knowing that the value $\sqrt{2} + \sqrt[3]{2}$ is a root) which means it is either an integer or it is irrational.

At that point I guess I will just need to prove (easily verifying inequalities) that $\sqrt{2} + \sqrt[3]{2}$ is not an integer therefore it must be irrational.

This is a LONG way to solve this problem.

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Update 2

I had GNU Octave solve the equations and got:

$c_0=-4. c_1=-24, c_2=12, c_3=-4, c_4=-6, c_5=0$

So our polynomial is:

$x^6 + 0x^5 -6x^4 -4x^3 +12x^2 -24x^1 -4$

This really does match the accepted answer's "minimal polynomial"

So now as I mentioned one can use part (a) of the problem (integral root theorem) to claim that the roots are either integral or irrational.

At this point it is easy to check that:

$1.4 < \sqrt{2} < 1.5$ and $1.2 < \sqrt[3]{2} < 1.3$ so adding we have:

$2.6 < \sqrt{2} + \sqrt[3]{2} < 2.8$

Therefore $\sqrt{2} + \sqrt[3]{2}$ is not an integer so by (a) it must be irrational and the long proof has come to its end.

Answer 1

Let $a=\sqrt{2}+2^{\frac{1}{3}}.$ Prove by contradiction. Suppose the contrary that $a$ is rational. Observe that \begin{eqnarray*} 2 & = & (2^{\frac{1}{3}})^{3}\\ & = & (a-\sqrt{2})^{3}\\ & = & a^{3}-3a^{2}\sqrt{2}+3a\cdot2-2^{\frac{3}{2}}\\ & = & a^{3}+6a-\sqrt{2}(3a^{2}+2). \end{eqnarray*} Therefore, \begin{eqnarray*} \sqrt{2} & = & \frac{a^{3}+6a-2}{3a^{2}+2}\in\mathbb{Q}. \end{eqnarray*} It is well-known that $\sqrt{2}$ is irrational (Do I need to prove this fact?). Hence, we arrive a contradiction.

Answer 2

Hint: I guess the idea is to Try to write $(\sqrt{2}+\sqrt[3]{2})^6$ in terms of $\sqrt{2}+\sqrt[3]{2}$ then we get a something like $$(\sqrt{2}+\sqrt[3]{2})^6 = p(\sqrt{2}+\sqrt[3]{2})$$ where $p(x)$ is a polynomial then using the rational zeros theorem we can show that $p(x)-x^6$ has no rational roots but $p(\sqrt{2}+\sqrt[3]{2})-(\sqrt{2}+\sqrt[3]{2})^6=0$. Thus we get that $\sqrt{2}+\sqrt[3]{2}$ is irrational.

To check your work see the link: https://www.wolframalpha.com/input/?i=minimal+polynomial+2%5E%281%2F2%29+%2B+2%5E%281%2F3%29

Since you still looking for the answer and you want to use your work I will start to write what you did using $b=\sqrt[6]{2}$ and $a=\sqrt{2}+\sqrt[3]{2}$ the your work

$$a^2=2+2b^5+b^4$$ $$a^3=2+6b+6b^2+2b^3$$ And so on the eliminates the $b$ powers do you want me to continue?