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Is this a new theorem

If $(n, m, k,)$ are three positive integers, where, ($k > 1$), $p$ is prime number satisfying the following inequality identity, $$m^k < p*(10)^{kn} < (m + 1)^k$$

Then, $m$ is an integer that represents the sequence digits of $k’th$ real arithmetical root of $p$ with $n$ number of accurate digits after the decimal notation

Example What is the cubic real arithmetical root of (7), $\sqrt[3]{7}$ to five digits of accuracy after the decimal notation?
Here, we have (k = 3), (p = 7), (n = 5), then (m) can be obtained from the following inequality identity:

$$m^3 < 7*(10)^{15} < (m+1)^3$$, where $ (m = 191293)$, just by too little inspection

And the real arithmetical cubic root of (7), $\sqrt[3]{7}$ is approximately (1.91293) to five digits of accuracy after the decimal notation as required

Any constructive comments are welcomed

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  • $\begingroup$ Is a 'prime' equation nessesary? $\endgroup$ – kotomord Dec 19 '16 at 14:35
  • $\begingroup$ Not really necessary, but I want to avoid the case when the roots are not irrational numbers now, there is of course more generalization to this, so let us see it here for $\sqrt[k]{p}$, first $\endgroup$ – bassam karzeddin Dec 19 '16 at 14:39
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It is not necesary that $p$ is prime here. I fact, for arbitrary positive $p$, $$m^n<p\cdot10^{kn}<(m+1)^n $$ is equivalent to (just take $n$th roots) $$ m<\sqrt[n]p\cdot 10^k<m+1$$ and after division by $10^k$ to $$\frac m{10^k}< \sqrt[n]p<\frac{m+1}{10^k}$$

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  • $\begingroup$ Unfortunately, I couldn't upvote your answer here due to my credit, but yes $p$ need not to be prime number, I actually wanted to avoid the case of real roots being for one or being rational numbers, but only to guarantee irrational roots as a result, thanks $\endgroup$ – bassam karzeddin Dec 19 '16 at 14:44
  • $\begingroup$ What actually you had proven here that any irrational number lies between two successive integers (right?), which is correct $\endgroup$ – bassam karzeddin Dec 19 '16 at 15:31
  • $\begingroup$ His original formulation has m^k not m^n, but thats not the error. Take p=1, and you will see that for example for k=2 and n=2, you wont find an m=100, unless you use =< sign for the first inequality. $\endgroup$ – Transfinite Numbers Dec 19 '16 at 16:11
  • $\begingroup$ @Hagen von Eitzen, so what exactly is your $\sqrt[q]{p}$ when you consider k as large as it tends to infinity?, where also (m) tends to integer with infinite sequence of digits, but this obviously not permissible in mathematics, thus our number does not exist except in mind only (right?) $\endgroup$ – bassam karzeddin Dec 26 '16 at 14:20
  • $\begingroup$ But may be a convenient rationale number would be sufficient as a magnitude for something assumed existing in mind only, which is exactly (unreal number) except the case when we make (q) as purely even positive integer of the form $2^n$, where then this is becomes as real existing number as constructible number, without its endless rational fake representation that tries hopelessly to replace the unique location of that constructible number, but always and forever unsuccessfully, however this had been proven elsewhere quite many times, that no number exists with nonzero endless terms. end $\endgroup$ – bassam karzeddin Dec 26 '16 at 14:34

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