You'll find something interesting if you take the golden ratio, ϕ=1+52 \phi = \frac{1 + \sqrt{5}}{2} , to high powers:

ϕ6=17.944 \phi^6 = 17.944 ϕ10=122.992 \phi^{10} = 122.992 ϕ15=1364.0007 \phi^{15} = 1364.0007

It seems as if large exponents of ϕ\phi come arbitrarily close to integer values. In fact, ϕ\phi isn't the only number which does this. For example, (2+3)5=723.999 (2+ \sqrt{3})^{5} = 723.999 . This is no coincidence.

These numbers - known as Pisot-Vijayaraghavan numbers - produce "almost-integers" when raised to high powers. I'll give a quick sketch of why this is, as well as an easy way to find other numbers with this property:

We can take a number q=a+bq = a + \sqrt{b}, where aa and bb are integers, and consider how it behaves under an operation called Galois conjugation. Essentially, this is just a more general version of what you may already know as complex conjugation. To find the Galois conjugate of qq, simply change the sign of the square root term: q=abq* = a - \sqrt{b}. This is equivalent to setting qq to be the root of the polynomial x22ax+a2b=0x^2 - 2ax + a^2 - b = 0 and finding the second root, or qq*.

A few properties of Galois conjugation can be easily deduced, and are already implicit in the quadratic equation above:

q+q=2a q + q* = 2a qq=a2b qq* = a^2 - b q2+q2=2a2+2b q^2 + q*^2 = 2a^2 + 2b qn+qn=(nk)an2kb2k q^n + q*^n = \sum \binom{n}{k} a^{n-2k}b^{2k}

The last equality is important here. Notice especially that these four equations all produce integers results on the righthand side. Now, whenever q<1 |q*| < 1 , the second term of qn+qn q^n + q*^n becomes arbitrarily small for increasing nn. This immediately implies that

qn+qnqn(nk)an2kb2kn1 q^n + q*^n \approx q^n \approx \sum \binom{n}{k} a^{n-2k}b^{2k} \qquad n \gg 1

Thus, any number qq whose Galois conjugate satisfies q<1|q*| < 1 will produce near-integer values when raised to high powers. Here are some examples of such numbers:

(2+6)7=34528.0037 (2 + \sqrt{6})^7 = 34528.0037 (4+21)9=252672968.008 (4 + \sqrt{21})^9 = 252672968.008 (1+3)15=3526400.0093 (1 + \sqrt{3})^{15} = 3526400.0093

*It's worth noting that in the case that qq has a denominator of tt, as is the case with ϕ \phi where t=2t=2, then we also need the further restriction that t2qqt^2 | qq*, or that a2ba^2 -b is divisible by t2t^2 so that qqqq* remains an integer.

Note by Levi Walker
1 year, 3 months ago

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@Levi Adam Walker Yes, this is amazing right???!! I got to know about these numbers quite recently I must confess........But, check out "Heegner Number "...........Wikipedia has a page on it too...........when it comes to "Close" approximations, I prefer to quote them........:)

Aaghaz Mahajan - 1 year, 3 months ago

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Yeah, they're much cooler, but also much harder to explain :p

Levi Walker - 1 year, 3 months ago

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Does this also work with cuberoots? I suppose q q* has to be defined differently, right?

Henry U - 1 year, 3 months ago

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It won't generally work with this construction, since qn+qnq^n + q*^n must give an integer and cube roots tend to act erratically when raised to arbitrary powers.

Levi Walker - 1 year, 2 months ago

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Ah, right. Thanks!

Henry U - 1 year, 2 months ago

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chakravarthy b - 11 months, 3 weeks ago

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