HP Prime & HP 49G Problem with square roots

[Gerald H]

Some (irrelevant to this thread) calculations produced these two expressions:

√1201+√(70+2*√1201)

&

√(35+2*√6)+√(1236-2*√6+2*√(42035-2402*√6))

which should be equal. I have checked with the Longfloat Lib on the HP 49G, & the two are equal to many decimal positions, but are they in fact exactly equal?

Assistance appreciated.

(Title edited to include HP Prime, 13:33, 26/6)

[Thomas Klemm]

(06-26-2014 10:25 AM)Gerald H Wrote:  are they in fact exactly equal?

True

[Gerald H]

(06-26-2014 10:45 AM)Thomas Klemm Wrote:  

(06-26-2014 10:25 AM)Gerald H Wrote:  are they in fact exactly equal?

True

Thank you.

I believe the result is correct & dislike relying on the authority of some cloud-computing; Why should I trust Wolframalpha if I have no means of checking the result?

My version of Maple is antiquated & can't deal with the question, nor can the HP 49G - & even if they did return an intelligible answer, I'd still want to know how.

[CosmicTruth]

(06-26-2014 10:25 AM)Gerald H Wrote:  <clipped>these two expressions:
√1201+√(70+2*√1201)
&
√(35+2*√6)+√(1236-2*√6+2*√(42035-2402*√6))
<clipped>but are they in fact exactly equal?
Assistance appreciated.

50G says no

[Gerald H]

(06-26-2014 11:01 AM)CosmicTruth Wrote:  

(06-26-2014 10:25 AM)Gerald H Wrote:  <clipped>these two expressions:
√1201+√(70+2*√1201)
&
√(35+2*√6)+√(1236-2*√6+2*√(42035-2402*√6))
<clipped>but are they in fact exactly equal?
Assistance appreciated.

50G says no

Thank you.

So now we have two authorities dsagreeing(see post #2)?

[Gerald H]

I have now tried the expressions on HP Prime CAS: for == a zero is returned & for - a value of -2.27373675443E-13.

The numerical value for - is certainly wrong.

[Claudio L.]

My 2 cents:

I ran it in the newRPL demo at 2007 digits precision, and the difference between both expressions came out 1e-2005, so I'd say they are equal at least up to the first 2000 digits.

Claudio

[Claudio L.]

To get an algebraic proof, I don't have the time but I think the key is:

Code:


1201 = 35^2 - 2^2*6

Claudio

[Gerald H]

(06-26-2014 04:08 PM)Claudio L. Wrote:  My 2 cents:

I ran it in the newRPL demo at 2007 digits precision, and the difference between both expressions came out 1e-2005, so I'd say they are equal at least up to the first 2000 digits.

Claudio

Thank you for the confirmation - I hadn't tested to such precision.

The means by which the two expressions arose implies, I believe, equality & I'm not bright enough to demonstrate this equality.

More precision will (hopefully) corroborate equality, but a convincing reasoning would settle the matter.

[Manolo Sobrino]

OK, first let's notice that 1201 is prime, 42035=35*1201 and 1236=35+1201. Now rewrite the longer expression:

\begin{equation} \sqrt{35+2\sqrt{6}}+\sqrt{1201+35-2\sqrt{6}+2\sqrt{\left(35-2\sqrt{6}\right)1201}}
\end{equation}
That is the square of a sum:\begin{equation} \sqrt{35+2\sqrt{6}}+\sqrt{\left(\sqrt{1201}+\sqrt{35-2\sqrt{6}}\right)^2}
\end{equation}
You don't need to worry about the absolute value, it's simply:\begin{equation}\sqrt{1201}+\sqrt{35+2\sqrt{6}}+\sqrt{35-2\sqrt{6}}
\end{equation}
If a>b it's trivial to prove that:\begin{equation}\sqrt{a+b}+\sqrt{a-b}=\sqrt{2a+2\sqrt{a^2-b^2}}\end{equation}
In this case: \begin{equation}\sqrt{70+2\sqrt{1225-4\cdot 6}}=\sqrt{70+2\sqrt{1201}}\end{equation}
There you go.

(You guys should use paper and pencil more often )

[Alberto Candel]

This reminds me of Dedekind's Theorem that \(\sqrt{2} \sqrt{3} = \sqrt{6}\). A very readable account is in the article "Dedekind's Theorem: ..." by Fowler in The American Mathematical Monthly Vol. 99, No. 8, Oct., 1992, p.725.

[Gerald H]

(06-26-2014 05:58 PM)Manolo Sobrino Wrote:  OK, first let's notice that 1201 is prime, 42035=35*1201 and 1236=35+1201. Now rewrite the longer expression:

\begin{equation} \sqrt{35+2\sqrt{6}}+\sqrt{1201+35-2\sqrt{6}+2\sqrt{\left(35-2\sqrt{6}\right)1201}}
\end{equation}
That is the square of a sum:\begin{equation} \sqrt{35+2\sqrt{6}}+\sqrt{\left(\sqrt{1201}+\sqrt{35-2\sqrt{6}}\right)^2}
\end{equation}
You don't need to worry about the absolute value, it's simply:\begin{equation}\sqrt{1201}+\sqrt{35+2\sqrt{6}}+\sqrt{35-2\sqrt{6}}
\end{equation}
If a>b it's trivial to prove that:\begin{equation}\sqrt{a+b}+\sqrt{a-b}=\sqrt{2a+2\sqrt{a^2-b^2}}\end{equation}
In this case: \begin{equation}\sqrt{70+2\sqrt{1225-4\cdot 6}}=\sqrt{70+\sqrt{1201}}\end{equation}
There you go.

(You guys should use paper and pencil more often )

The last calculation line is a typo?

[Manolo Sobrino]

(06-26-2014 06:51 PM)Gerald H Wrote:  The last calculation line is a typo?

Of course it was, fixed now. (LaTeX here seems to cause me dyslexia )

[Gerald H]

I take my hat off to you,Manolo Sobrino. Bravo!

[Manolo Sobrino]

Thank you Gerald!

[CosmicTruth]

HP50G calculator for sale or trade for good pencil and paper pad.


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