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Given

t1 = {1/7 (4 + Sqrt[37]), 1/15 (3 + 2 Sqrt[7]), 1/2 (5 + 3 Sqrt[2])}

I do not want to simplify the square roots. I want to get

t2 = {1/7 (4 + Sqrt[37]), 1/15 (3 + Sqrt[28]), 1/2 (5 + Sqrt[18])}

I can cheat my way round by code:

t2 = t1[[All, 2, -1]]
t3 = t2^2
Defer[Sqrt[#]] & /@ t3

to get

{Sqrt[37], Sqrt[28], Sqrt[18]}

and fiddle it in again. What is an elegant way to obtain t2? How could I use Cases inserting two optional patterns? I tried with If inside Cases but that didn't do the job properly.

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    $\begingroup$ The problem is that Mathematica automatically likes to write Sqrt[28] as 2 Sqrt[7]. So you will fighting against the front end all the time trying to prevent it from doing this. But why exactly you want it displayed Sqrt[28] vs 2 Sqrt[7]? Is it just for display purposes? computationally there is no difference. $\endgroup$
    – Nasser
    Commented Oct 19, 2021 at 7:51
  • $\begingroup$ I kinda remember now this was asked before on this forum...may be if you search you might find it. $\endgroup$
    – Nasser
    Commented Oct 19, 2021 at 7:56
  • $\begingroup$ The roots show up in a sequence i am interested in. I found similar questions about this problem but structures where a bit easier. $\endgroup$
    – user57467
    Commented Oct 19, 2021 at 8:15
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    $\begingroup$ Related: mathematica.stackexchange.com/questions/101618/… $\endgroup$
    – Michael E2
    Commented Oct 20, 2021 at 4:12
  • $\begingroup$ I am reliably certain that it is not the Mathematica front end that converts Sqrt[28] to 2*Sqrt[7]. $\endgroup$
    – Daniel Lichtblau
    Commented Mar 20, 2022 at 14:56

3 Answers 3

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maybe replace Sqrt first?

{1/7 (4 + Sqrt[37]), 1/15 (3 + 2 Sqrt[7]), 1/2 (5 + 3 Sqrt[2])} //
#/. Sqrt[x_] :> f[x]& //
#/. a_ f[b_] :> f[a^2 b]& //
#/. f[x_] :> Defer[Sqrt[x]] &
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Is this helpful? It may be too fragile for your use-case but the desired output is obtained.

expr=Inactivate[{1/7 (4 + Sqrt[37]), 1/15 (3 + 2 Sqrt[7]), 1/2 (5 + 3 Sqrt[2])}, Sqrt] 

expr//.Times[a_, Inactive[Sqrt][b_]] -> Inactive[Sqrt][a^2*b]
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    $\begingroup$ Thank you, but the thing is, that I already get t1 in the simplified square root form with the square root sign (which I cannot copy here because it comes out as Sqrt). $\endgroup$
    – user57467
    Commented Oct 20, 2021 at 11:53
  • $\begingroup$ Oh, I think I understand you. When I define t1 and then try to inactivate stuff in it in later lines the method does not work. I did some looking in the documentation and it suggests that there is a variable $InactivateExclusions whose value determines that Inactivate should ignore certain heads. Probably those heads include stuff that is displayed in special ways by the front end like $$\sqrt{}$$ $\endgroup$
    – Diffycue
    Commented Oct 20, 2021 at 16:51
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I think I found a reasonable solution:

t1 = {1/7 (4 + Sqrt[37]), 1/15 (3 + 2 Sqrt[7]), 1/2 (5 + 3 Sqrt[2])};
Replace[#, a_*(b_ + c_*Sqrt[d_]) :> a*(b + Hold[Sqrt[c^2*d]])] & /@ t1;
% /. Hold[e_] :> Defer[e]

Which gives

{1/7 (4 + Sqrt[37]), 1/15 (3 + Sqrt[2^2 7]), 1/2 (5 + Sqrt[3^2 2])}

How would I evaluate just the number under the root and keep the rest?

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    $\begingroup$ replace Sqrt with f, eval the num, replace f[num] with defer[sqrt[num]] $\endgroup$
    – AsukaMinato
    Commented Mar 20, 2022 at 4:07
  • $\begingroup$ Thanks, I think that this is equivalent to my "cheating" method $\endgroup$
    – user57467
    Commented Mar 21, 2022 at 13:16

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