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I would like to know if there is any more elegant solution than the extraction of the root or the approximation.

I tried something like this:

$\sqrt{25}$ + $\sqrt{24}$ = $( \sqrt3 + \sqrt2 )^2$

But how can I get the fractional part from here?

Peter Phipps
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Akhtubir
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    Huh? Fractional part? Are you assuming that this is a rational number? Please explain. – David G. Stork Sep 28 '20 at 06:37
  • @DavidG.Stork The fractional part or decimal part[1] of a non‐negative real number x is the excess beyond that number's integer part. There you go. – player3236 Sep 28 '20 at 06:38
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    I assume OP means $x - \lfloor x \rfloor$ as the fractional part – Rob Sep 28 '20 at 06:39
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    What do you want to find out? The first few digits? A certain expression? – pancini Sep 28 '20 at 06:45
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    @DavidG.Stork Would you or the other 3 upvoters of the comment explain why you are badgering the OP? A look at any online source/ the tag fractional-part clearly shows the OP's wordings are valid. – player3236 Sep 28 '20 at 06:45
  • OK, then. $\sqrt{25} = 5$, and hence does not contribute to a "fractional part." $\sqrt{24} = 2 \sqrt{6}$, which has "fractional part" $0.8989794855663561963945681494117827839318949613133$. Is there any more to this question than that?? (To @player3236: So why did the OP include $\sqrt{25}$? And what does "root" have to do with anything?... or is asking that "badgering"?) – David G. Stork Sep 28 '20 at 06:45
  • @DavidG.Stork You should acknowledge that OP is asking a valid question, and textbook/teachers etc. like to add irrelevant information to confuse (and test) students. In this case: $\sqrt {25}$. – player3236 Sep 28 '20 at 06:50
  • And the OP didn't realize that?! And left it for solvers to wade through? And included $\sqrt{25}$ in his subsequent "derivation"? See https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question for that issue. – David G. Stork Sep 28 '20 at 06:52
  • @DavidG.Stork Sometimes students do mix up concepts. However, some accept their mistakes humbly, learn and move on. Some do not. – player3236 Sep 28 '20 at 06:54
  • Some of us with extensive textbook writing and teaching experience at world-class colleges and universities know that instilling in students the discipline and care of asking questions is essential for their intellectual development. Some do not. – David G. Stork Sep 28 '20 at 06:57
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    I upvoted because I suspect that the OP is asking a deep question, which I don't know the answer to.
    Let ${x}$ denote the fractional part of $x$. It seems that the OP is searching for an elegant way of computing (for example) $F = {\sqrt{25} + \sqrt{24}}.$ Searching, the OP discovered smaller positive integers $3$ and $2$ where $F$ happens to $= {~(\sqrt{3} + \sqrt{2})^2~}.$ What the OP seems to be asking is - has he accomplished anything in his attempt to compute $F$?
    – user2661923 Sep 28 '20 at 07:11
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    @user2661923 that's right. Sorry that I became a debate for such great teachers such as David G. Stork. – Akhtubir Sep 28 '20 at 07:50

4 Answers4

2

This response is a continuation of a comment that I made to the OP's query. Although this response can in no way be construed as an answer, I feel that the response is worthwhile. I also think that providing this response as a series of comments (instead) would have made this response much less legible.

To the OP:
If (after 24-48 hours) you find that no one has focused on the question that you are trying to ask then (regardless of whether the interpretation in my comment was accurate), I suggest that you do the following:

(1)
As another comment suggested, review How to ask a good question.

(2)
Create a 2nd mathSE query. In this 2nd query, provide a link to this query, and explain that you are re-posting your question as a new query, because the old query was unclear. Also, in this (original) query, add an Addendum that explains that you now realize that this query has caused confusion, and that you have consequently re-posted your question. In this Addendum, provide a link to the new mathSE query.


Let $\{x\}$ denote the fractional part of $x$.

(3)
Your new query should begin with a very precise goal.

One possible goal might be:
given any positive integers $n$ and $m$,
how does one simplify the computation of $\{\sqrt{n} + \sqrt{m}\}$?

If this is what you are attempting to accomplish, then you are going to have to make a serious effort to describe (perhaps) with examples exactly what you mean by the word simplify.

If this is not your actual goal, then (as I say) you must clearly indicate the mathematical problem (i.e. goal) that you are trying to solve.

Please try to make it impossible for anyone in your audience to be confused as to what you are asking.


The remainder of my response examines the confusion that I experienced when reading your query.

(A)
Where did the LHS of :
$[E_1]: ~\{\sqrt{25} + \sqrt{24}\} ~=~ \{~(\sqrt{3} + \sqrt{2})^2~\}$
originate?

Was the computation of $\{\sqrt{25} + \sqrt{24}\}$ an assigned problem from a book or a class, or a problem that you made up on your own?

(B)
How did you calculate the RHS of $[E_1].$
Although $[E_1]$ is accurate, that doesn't explain your thinking in associating the positive integers $3$ and $2$ with the positive integers $24$ and $25$.

Did you start with the positive integers $25$ and $24$, and employ some mathematical algorithm to compute the positive integers $3$ and $2$? If so, what algorithm?

Alternatively, did you simply presume that $\{\sqrt{25}\} = 0,$ notice that $\{\sqrt{24}\} = \{2\sqrt{6}\}$, and choose the positive integers $2$, and $3$ because you happened to notice that $2\sqrt{6} = 2\left(\sqrt{2}\sqrt{3}\right)$?

(C)
Your query finishes with the question:
"But how can I get the fractional part from here?"

$[Q_1]:~$ What type of expression are you ultimately looking for?

In your new query, this question, can be associated with your stating a clear mathematical goal. Further, if I were you, I would attack the confusion around $[Q_1]$ by providing two to three examples that illustrate exactly what you mean by the phrase:
get the fractional part.

Are you looking for a way of expressing tight lower and upper bounds on the fractional part?

Are you looking for a mathematical formula that expresses the fractional part exactly?
I am guessing that if this is your goal, then, given $x$,
you are looking for a mathematical expression for $y$,
where $0 \leq y < 1,$ and $y = \{x\}.$

If this is your goal, you should state this explicitly, and the examples that you provide should illustrate your (at least attempting to) compute $y$, given $x$.

I was very confused here.

user2661923
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0

$$\{5+\sqrt{24}\}=\sqrt{24}-4.$$ Is it the question?

0

$$x=\sqrt{25} + \sqrt{24},\\ [x]=[\sqrt{25} + \sqrt{24}]=5+4=9\\ \{x\}=x-[x]=5+\sqrt{24}-9$$

Khosrotash
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As described in other answers the number required is $\sqrt{24}-4$.

If we set $x+4 = \sqrt{24}$, squaring and rearranging gives $x^2+8x-8 = 0$.

This can be solved using numerical methods to your required level of accuracy.

The Newton-Raphson method gives the recursion $x_{n+1} = x_n - \dfrac{x_n^2-8x_n+8}{2x_n+8}$ or $x_{n+1} = \dfrac{x_n^2+8}{2x_n+8}$. Starting with $x_0=1$ we get $x_1 = \dfrac{9}{10}$, $x_2 = \dfrac{881}{980}$ and so on.

Note $\left (\dfrac{881}{980}+4 \right )^2 = 24.00000104 \dots$

Peter Phipps
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