简化分数的平方根

Question

我发现此页面提供了用于简化整数平方根的宏：

简化平方根

例如，我们是否可以推广这种方法来简化分数的平方根\sqrt{4/3}=2/\sqrt{3}？

Answer 1

这是一个 LuaLaTeX 解决方案。有点过头了，但你可以将上述函数用于其他目的：

%!TEX program = lualatex
\documentclass{article}
\usepackage{luacode}
\begin{luacode*}
function IsPrime(d)
    -- Determines whether a number is prime or not
    local dummy = 0
    for i = 1, math.floor(math.sqrt(d)) do
        if d % i == 0 then 
            dummy = dummy + 1 
        end
    end
    if dummy == 1 then return true else return false end
end

--Prime factorization
--This function return two tables, the former with prime factors, the latter with their respective exponents
function PrimeFactors(d)
    local dummy1 = {}
    local dummy2 = {}
    --Search and collect prime factors
    for i= 2, math.ceil(math.sqrt(d)) do
        if d%i == 0 and IsPrime(i) == true then 
            table.insert(dummy1,i)
        end
    end
    --A void table means d is prime, so we add in that scenario.
    if #dummy1 == 0 then table.insert(dummy1,d) end
    --Now we search indices
    for k,v in ipairs(dummy1) do
        if (d%dummy1[k]==0) then 
            while (d%dummy1[k]==0) do
                if dummy2[k] == nil then dummy2[k] = 0 else
                    dummy2[k] = dummy2[k] + 1
                    d = d/dummy1[k]
                end
            end
        end
    end
    --We have both the factors and the indices
    local dummy3 = {dummy1, dummy2}
    return dummy3
end

--GCD
function GCD(a,b)
    if b ~= 0 then
        return GCD(b, a % b)
    else
        return math.abs(a)
    end
end

--This multiplies factors to d so it becomes a square
function CompleteSquare(d)
    local dummy = 1
    for i,v in ipairs(PrimeFactors(d)[1]) do
        dummy = dummy * PrimeFactors(d)[1][i]^(2*math.ceil(0.5*PrimeFactors(d)[2][i]))
    end
    return math.floor(dummy/d)
end

--Check
function IsSquare(d)
    if math.floor(d) - (math.floor(math.sqrt(d))^2) == 0 then return true else return false end
end

--Et voilà
function rsqrt(a,b)
    local A = (a/GCD(a,b))*CompleteSquare(b/GCD(a,b))
    local B = (b/GCD(a,b))*CompleteSquare(b/GCD(a,b))
    if IsSquare(A) == true then
        tex.print( [===[\frac{]===] .. math.floor(math.sqrt(A)) .. [===[}{]===] .. math.floor(math.sqrt(B)) .. [===[}]===])
    else
        tex.print( [===[\frac{\sqrt{]===] .. math.floor(A) .. [===[}}{]===] .. math.floor(math.sqrt(B)) .. [===[}]===])
    end
end
\end{luacode*}
\newcommand{\rsqrt}[2]{\directlua{rsqrt(#1,#2)}}
\begin{document}
%Some examples
$\rsqrt{1}{23}$, $\rsqrt{70}{45}$, $\rsqrt{50}{18}$, $\rsqrt{12}{16}$
\end{document}