11 replies to this topic

[bvidinli]

As you know, 1/sqrt(6) = 0.4082482905

How can I convert 0.4082482905 back to 1/sqrt(6) ? (assuming I do not know 1/sqrt(6))
using casio calculators, fx 9850 or similar ?

Because, sometime, I find these numbers as result of calculations, however, these should be shortened...

or, similarly, 1/sqrt(3)

Edited by bvidinli, 03 March 2013 - 06:54 AM.

[Casimo]

Do you have the fx9860 Then you could try to pess the [F-D] button.

Edited by Casimo, 03 March 2013 - 07:06 AM.

[bvidinli]

F-D button works for value of 1/3, not for 1/sqrt(6) or 1/sqrt(3)

I actually remember that I have seen such a program or addin, however, I do not remember where.. I am not sure..

Edited by bvidinli, 03 March 2013 - 11:24 AM.

[Casimo]

Convert 0.4082482905 back to 1/sqrt(6) is impossible.

During calculations, you can work with the exact value using the key [F<->D].

I agree. Maybe this could be possible with a CAS but I don't know.


OFFTOPIC:

Idea how to link to Casiopeia: Hey flyingfisch, why don't link to Casiopeia with quotes like this?

Edited by Casimo, 03 March 2013 - 12:04 PM.

[helder7]

Casio Calculators give an approximate value:
1/sqrt(6)=0.4082482905

The (more) exact value (approximate) should be:
1/sqrt(6)≈0.40824829046386

So it is normal that the calculator does not recognize this value as a fraction. Maybe calculator only accept 10 decimal digits.

The site wolframalpha can discover the original value: http://www.wolframal...?i=0.4082482905

[bvidinli]

Thanks for responses. I need this in calculator, without internet connection, so, wolframalpha would not be useful.

[nsg]

Another thing you can do is to convert your mystery number into continuous fraction and look for repetitions. It is known that quadratic irrationalities have repeating continued fraction representation. For example, 1/sqrt(6)=[0;2,2,4,2,4,2,4...] (2,4 part repeats infinitely)
So, for example, if you start with approximation to 1/sqrt(6)=0.4082482905 (without knowing it is 1/sqrt(6)) and start obtaining its continued fraction elements you will get
[0; 2 2 4 2 4 2 4 2 4 2 3 19 1 16 4 2 1 2]
It is easy to suspect that [0; 2 (2 4)*] is a candidate and non-periodicity is due to inaccuracy of approximation.
Then using simple algebra you convert [0;2 (2 4)*] to a form (U+sqrt(D))/V, which is sqrt(6)/6, which is 1/sqrt(6)

Unfortunately this only works for mystery numbers in the form of (U+sqrt(D))/V. Say, sqrt(2)+sqrt(3) does not have repeating continued fraction and will need different method to guess tha form.

Edited by nsg, 03 March 2013 - 05:34 PM.

[Karan901]

There is a reason why these options are not available for most calculators. For example in examinations using the integral on the calculator to work out bounds will give an answer to 10dp. So the answer given may be 1.8137993642, but the markscheme may want an exact answer being (pi*(1/sqrt3)). So it is to stop people gaining answers the easy way. But unless the calculator can be modified to give more d.p i don't think it is possible.

[scientifix]

Hi everybody,
You can use this program :

?->N
Int N->I
Frac N->F
For 1->D To 1000
F×D->N
If Abs (Int (N+.5)-N)<1[size=1]E[/size]-10×D
Then Int (N+.5)->N
Break
IfEnd
Next
I[i][a b/c][/i]N[i][a b/c][/i]D[i][a b/c][/i]

[flyingfisch]

Looks good, why don't you upload as g1r?

[Karan901]

Would also appreciate a g1r upload.

[scientifix]

Can you explain me how ?