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In the process of my numerical computations I have found a very special identity:

  • $\;\;1264483 + 1707789 \,\sqrt[3]{7} - 1238313\,\sqrt[3]{7^2} = 9.313225746154785 \times 10^{-10}$

  • $ -1500493 - 2026256\,\sqrt[3]{7} + 1469290\,\sqrt[3]{7^2} = 9.313225746154785 \times 10^{-10}$

Therefore if we subtract these two equations one should find the difference is zero. On my computer I found:

$$ 2764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} \stackrel{?}{=} \left\{ \begin{array}{cl} 0 & \text{by hand} \\ -1.862645149230957 \times 10^{-9} & \text{by computer} \end{array}\right. $$

Subtracting these two numbers - which might be the same - we have gotten twice the number!

Patrick Stevens
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cactus314
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  • We are showing $x = y \approx 10^{-9}$ and then showing $x - y \approx 2 \times 10^{-9}$ – cactus314 Aug 06 '16 at 13:07
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    The result of subtraction cannot be $0$ otherwise $\sqrt[3]{7}$ will be a quadratic surd, which is not the case. – Fabio Lucchini Aug 06 '16 at 13:08
  • @cactus314 How did you derive these numbers? What system did you use? – Patrick Stevens Aug 06 '16 at 13:18
  • @PatrickStevens I am using numpy. – cactus314 Aug 06 '16 at 13:22
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    If $u^3 = 7$ then $2764976 + 3734045u -2707603u^2$ $\approx 2.0876013027695663895820213961313224250067665378654174719507513689302335990341969812558072078446 \times 10^{-9}$. Calculated using mpmath with 400 decimal digits of precision. – PM 2Ring Aug 06 '16 at 13:26
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    To reiterate Fabio's point. The polynomial $m(x)=x^3-7$ is irreducible in $\Bbb{Q}[x]$ by Eisenstein's criterion (and Gauss' Lemma). Therefore $m(x)$ is the minimal polynomial of $\root3\of7$. Consequently $\root3\of7$ is not a zero of any quadratic polynomial with rational coefficients. – Jyrki Lahtonen Aug 06 '16 at 13:30
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    Note that your number $9.313225746154785 \times 10^{-10}$ is exactly equal to $2^{-30}$ which is a sure sign of a floating point catastrophe –  Aug 06 '16 at 13:47
  • On second look, no, it's not exact, but it's a truncation of the exact value, so I suspect you inadvertently calculated $2^{-30}$ and then printed it with less than full precision. –  Aug 06 '16 at 14:23

7 Answers7

7

To expand on the comments of Fabio and Jyrki:

You're claiming that $\sqrt[3]7$ is a root of the quadratic equation

$$2764976 + 3734045x -2707603x^2 = 0$$

but that leads to a contradiction.

The solutions of a quadratic with integer coefficients can be written as

$$x = \frac{p \pm \sqrt q}{r} $$

for some integers $p, q, r$.

If $x^3 = 7$ then

$$\begin{align} \left(\frac{p \pm \sqrt q}{r}\right)^3 = 7\\ \frac{p^3 \pm 3p^2\sqrt q + 3pq +q\sqrt q}{r^3} = 7\\ \frac{(p^3 + 3pq) \pm \sqrt q(3p^2 +q)}{r^3} = 7\\ \end{align}$$

Now the only way the expression on the LHS can be rational (let alone integer) is if $\sqrt q$ is rational. But $q$ is an integer, so its square root can only be rational if $q$ is a perfect square. But that would make $x = \frac{p \pm \sqrt q}{r}$ rational, and we know that $x = \sqrt[3]7$ is not rational.

PM 2Ring
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Assume $$ \tag 12764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} = 0$$ Multiply by $\sqrt[3]7$ to obtain $$ \tag 22764976\,\sqrt[3]7 + 3734045\,\sqrt[3]{7^2} -18953221 = 0.$$ The system of linear equations $$\begin{align} 3734045 x-2707603 y&=-12764976\\ 22764976 x+3734045 y&=18953221 \end{align} $$ has rational solutions, i.e., we are forced to conclude that $$\sqrt[3]7=\frac{-12764976\cdot 3734045+2707603\cdot 18953221}{3734045\cdot 3734045+2707603\cdot 22764976} =\frac{3652803231343}{75581609374553}$$ is rational, which is absurd! (Not to mention that this rational number is $\approx 0.05$).

Hagen von Eitzen
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Since you tagged it elementary number theory, let's give a proof at that level that no such equations can exist. If $\,w=\sqrt[3]7\,$ and $\,w^2 = a w + b\,$ and $\,w^2 = cw + d\,$ for rationals $\,a,b,c,d,\, a\ne c\,$ then subtracting we get $\, (a\!-\!c)w + b\!-\!d = 0\,$ so $\,w = (b\!-\!d)/(a\!-\!c)\,$ is rational, contra $\,w=\sqrt[2]7 $ is irrational, by $\,x^3-7\,$ has no rational roots via Rational Root Test.

Community
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Bill Dubuque
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    Well, that's quite a bit shorter than my approach. :) – PM 2Ring Aug 06 '16 at 14:36
  • May be I am not the only person wondering where that second equation came from? Did you multiply the first by $w$, and then replace $w^3$ with $7$ and shuffle the terms or what? In other words, you took the first step in calculating the gcd of $x^2-ax-b$ and $x^3-7$? I realize that you may find it distracting for the reader to call such a straight forward step Euclidean algorithm even though it is built into the minimal polynomial concept. Anyway, a nice way of putting it. – Jyrki Lahtonen Aug 06 '16 at 15:42
  • @Jyrki I read the OP as having $2$ quadratics. But if instead the OP has only one (their difference) $,f(x),$ having root $,w=\sqrt[2]7,$ then the same idea works since dividing $,x^3!-7,$ by $,f(x),$ we still quickly reach the contradiction that $,x^3!-7,$ has a rational root (either a root of the linear remainder, if nonzero, else the linear quotient). Yes, of course it is a special case of the Euclidean algorithm, and that is the way to handle the general case. I avoided minimal polynomials etc in order to keep the argument as simple as possible. – Bill Dubuque Aug 06 '16 at 17:14
3

If $\sqrt[3]{7}$ and $\left(\sqrt[3]{7}\right)^2$ were linearly dependent over $\mathbb{Q}$, $\;\alpha=\sqrt[3]{7}$ would be an algebraic number over $\mathbb{Q}$ with degree $\leq 2$, hence $x^3-7$ would be a reducible polynomial over $\mathbb{Q}$. However, $x^3-7$ is an irreducible polynomial over $\mathbb{F}_{13}$, hence $x^3-7$ is an irreducible polynomial over $\mathbb{Q}$ and no linear combination of $1,\sqrt[3]{7},\sqrt[3]{49}$ with rational coefficients $\neq(0,0,0)$ can be zero.

However, $\sqrt[3]{7}$ is pretty close to $$ \frac{-25642+4 \sqrt{708733687}}{42263} $$ (I get that through the Mathematica command RootApproximant), hence, for instance, $$ 252756- 51284\sqrt[3]{7}+42263 \sqrt[3]{49}$$ is pretty close to zero, without actually being equal to zero (namely it is $\approx 2.91\cdot 10^{-11}$) and you may derive more accurate approximations by replacing the continued fraction of $\sqrt[3]{7}$ with a definitely periodic continued fraction, that is always associated to an algebraic number over $\mathbb{Q}$ with degree $2$. For instance, from: $$ \sqrt[3]{7}\approx [1; 1, 10, 2, 16, 2,1,4,2,1,21,1,3,5,\overline{1,2,1}] $$ we get:

$$ 646825196228418 - 676266038270144 \sqrt[3]{7} + 176761726809763 \sqrt[3]{49} \approx 3.9\cdot 10^{-13} $$

and

$$ \sqrt[3]{7}\approx \frac{324048498507980-\sqrt{10}}{169398931559230}-1.37\cdot 10^{-17}. $$

Jack D'Aurizio
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  • These floating point error are not unlike those discussed in http://math.stackexchange.com/q/986337/4997 where they discuss $[ a \sqrt{2}][b \sqrt{7} < [ab \sqrt{14}]$ for $a,b \in \mathbb{Z}$. – cactus314 Aug 06 '16 at 15:04
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    I like how you don't use Eisenstein to show irreducibility :) – Ege Erdil Aug 06 '16 at 15:17
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No. Mathematica tells us that its value is about $-2.0876013027695663896 \times 10^{-9}$.

$$1264483 + 1707789 \times 7^{1/3} - 1238313 \times 7^{2/3} = -3.1767789172657775703*10^{-10}$$ according to Mathematica.

$$-1500493 - 2026256 \times 7^{1/3} + 1469290 \times 7^{2/3} = 1.7699234110429886326 \times 10^{-9}$$ similarly.

These are calculated with Mathematica's arbitrary-precision arithmetic, so it is "guaranteed" not to have the rounding errors that floating-point arithmetic can often introduce.

Patrick Stevens
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The scale of integer factors in your expressions is $10^6$ while the roots themselves contain floating point errors in the region of $2^{-52}\approx 4·10^{-15}$ which gives an overall numerical uncertainty of a small multiple of $10^{-9}$.

The accumulated error of the sums is in the same scale as are the actual results reported in the answer of Patrick Stevens, so that your observations are a random result of error cancellations and the gaps between with 64bit floats representable numbers.

Lutz Lehmann
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$$ 276497+3734045x−2707603x^2 $$ is a quadratic with two roots: $$ x = \frac{-3734045\pm \sqrt{3734045^2 + 4 \cdot 2764976 \cdot 2707603}}{2 \cdot (−2707603)} $$ neither of which is a cube root of $7$, which can be verified by cubing them (or by knowing about the degrees of algebraic numbers, per @JackD'Aurizio).

John Hughes
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  • Have you verified this on a computer? And the equation should be: $2764976 + 3734045,x -2707603,x^2 = 0$ – cactus314 Aug 06 '16 at 13:11
  • Even then your equation is typed wrong... – Soham Aug 06 '16 at 13:22
  • How can that happen? $\sqrt[3]{7}$ is an algebraic number over $\mathbb{Q}$ with degree $3$, hence it cannot be a root of a non-zero polynomial $\in\mathbb{Q}[x]$ with degree $\leq 2$. – Jack D'Aurizio Aug 06 '16 at 14:29
  • Edits fixed. tatan's proposed edit replaces $-b$ in the numerator of the quadratic formula with $-a$, which is wrong. As for "did I verify this on a computer?" the answer is "No, I didn't have to...but you can verify this by hand if you really want to...it's just a bunch of algebra." But @JackD'Aurizio's remark was sufficient for me to make my claim with confidence. :) – John Hughes Aug 06 '16 at 15:46