Worksheet06: Dickson 033
Current Work (editable)
We agree to simplify x_{1} by making all possible replacements of certain types that are sufficiently illustrated by the following numerical examples. If x_{1} involves $${3}, $${5}, and $${15}, we agree to replace $${15} by $${3}·$${5}. If x_{1} = s - 7t, where s is given by (4) and t = 1/2 $${10 + 2 $${5}}, so that st = $${5}, we agree to write x_{1} in the form s - 7 $${5}/s, which involves a single radical of order 2 and no new radical of lower order. Finally, we agree to replace $${4 - 2 $${3}} by its simpler form $${3} - 1. After all possible simplifications of these types have been made, the resulting expressions have the following properties (to be cited as our agreements): no one of the radicals of highest order n in x_{1} is equal to a rational function, with rational coefficients, of the remaining radicals of order n and the radicals of lower orders, while no one of the radicals of order n - 1 is equal to a rational function of the remaining radicals of order n - 1 and the radicals of lower orders, etc. Let $${k} be a radical of highest order n in x_{1}. Then x_{1} = a + b $${k} / c + d $${k}, where a, b, c, d do not involve $${k}, but may involve other radicals. If d = 0, then c $$ 0 and we write e for a/c, f for b/c, and get (6) x_{1} = e + f $${k}, (f $$ 0) where neither e nor f involves $${k}. If d $$ 0, we derive (6) by multiplying the numerator and denominator of the fraction for x_{1} by c - d $${k}, which is not zero since $${k} = c/d would contradict our above agreements. By hypothesis, (6) is a root of equation (5). After expanding the powers and replacing the square of $${k} by k, we see that (7) (e + f $${k})^{3} + \alpha(e + f $${k})^{2} + \beta(e + f $${k}) + \gamma = A + B $${k}, where A and B are certain polynomials in e, f, k and the rational numbers \alpha, \beta, \gamma. Thus A + B $${k} = 0. If B $$ 0, $${k} = -A/B is a rational function, with rational coefficients, of the radicals, other than $${k}, in x_{1}, contrary to our agreements. Hence B = 0 and therefore A = 0. When e - f $${k} is substituted for x in the cubic function (5), the result
Answer Key (non-editable)
We agree to simplify $x_{1}$ by making all possible replacements of certain types that are sufficiently illustrated by the following numerical examples. If $x_{1}$ involves $\sqrt{3}$, $\sqrt{5}$, and $\sqrt{15}$, we agree to replace $\sqrt{15}$ by $\sqrt{3}·\sqrt{5}$. If $x_{1} = s - 7t$, where $s$ is given by \Eq{(4)} and \[ t = \tfrac{1}{2} \sqrt{10 + 2 \sqrt{5}}, \] so that $st = \sqrt{5}$, we agree to write $x_{1}$ in the form $s - 7 \sqrt{5}/s$, which involves a single radical of order $2$ and no new radical of lower order. Finally, we agree to replace $\sqrt{4 - 2 \sqrt{3}}$ by its simpler form $\sqrt{3} - 1$. After all possible simplifications of these types have been made, the resulting expressions have the following properties (to be cited as our agreements): no one of the radicals of highest order $n$ in $x_{1}$ is equal to a rational function, with rational coefficients, of the remaining radicals of order $n$ and the radicals of lower orders, while no one of the radicals of order $n - 1$ is equal to a rational function of the remaining radicals of order $n - 1$ and the radicals of lower orders, etc. Let $\sqrt{k}$ be a radical of highest order $n$ in $x_{1}$. Then \[ x_{1} = \frac{a + b \sqrt{k}}{c + d \sqrt{k}}, \] where $a$, $b$, $c$, $d$ do not involve $\sqrt{k}$, but may involve other radicals. If $d = 0$, then $c \neq 0$ and we write $e$ for $a/c$, $f$ for $b/c$, and get \[ \Tag{(6)} %[** Attn alignment] x_{1} = e + f\sqrt{k}, \qquad(f \neq 0) \] where neither $e$ nor $f$ involves $\sqrt{k}$. If $d \neq 0$, we derive \Eq{(6)} by multiplying the numerator and denominator of the fraction for $x_{1}$ by $c - d \sqrt{k}$, which is not zero since $\sqrt{k} = c/d$ would contradict our above agreements. By hypothesis, \Eq{(6)} is a root of equation \Eq{(5)}. After expanding the powers and replacing the square of $\sqrt{k}$ by $k$, we see that \[ \Tag{(7)} (e + f \sqrt{k})^{3} + \alpha(e + f \sqrt{k})^{2} + \beta(e + f \sqrt{k}) + \gamma = A + B\sqrt{k}, \] where $A$ and $B$ are certain polynomials in $e$, $f$, $k$ and the rational numbers $\alpha$, $\beta$, $\gamma$. Thus $A + B \sqrt{k} = 0$. If $B \neq 0$, $\sqrt{k} = -A/B$ is a rational function, with rational coefficients, of the radicals, other than $\sqrt{k}$, in $x_{1}$, contrary to our agreements. Hence $B = 0$ and therefore $A = 0$. When $e - f \sqrt{k}$ is substituted for $x$ in the cubic function \Eq{(5)}, the result
Comparison of Current Work and Answer Key
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