Worksheet07: Dickson 034
Current Work (editable)
is the left member of (7) with $${k} replaced by -$${k}, and hence the result is A - B $${k}. But A = B = 0. This shows that (8) x_{2} = e - f $${k} is a new root of our cubic equation. Since the sum of the three roots is equal to -\alpha by §20, the third root is (9) x_{3} = -\alpha - x_{1} - x_{2} = -\alpha - 2e. Now \alpha is rational. If also e is rational, x_{3} is a rational root and we have reached our goal. We next make the assumption that e is irrational and show that it leads to a contradiction. Since e is a component part of the constructible root (6), its only irrationalities are square roots. Let $${s} be one of the radicals of highest order in e. By the argument which led to (6), we may write e = e' + f' $${s}, whence, by (9), (9') x_{3} = g + h $${s}, (h $$ 0) where neither g nor h involves $${s}. Then by the argument which led to (8), g - h $${s} is a root, different from x_{3}, of our cubic equation, and hence is equal to x_{1} or x_{2} since there are only three roots (§16). Thus g - h $${s} = e ± f $${k}. By definition, $${s} is one of the radicals occurring in e. Also, by (9'), every radical occurring in g or h occurs in x_{3} and hence in e = 1/2(-\alpha - x_{3}), by (9), \alpha being rational. Hence $${k} is expressible rationally in terms of the remaining radicals occurring in e and f, and hence in x_{1}, whose value is given by (6). But this contradicts one of our agreements. 32. Trisection of an Angle. For a given angle A, we can construct with ruler and compasses a line of length cos A or -cos A, namely the adjacent leg of a right triangle, with hypotenuse unity, formed by dropping a perpendicular from a point in one side of A to the other, produced if necessary. If it were possible to trisect angle A, i.e., construct the angle A/3 with ruler and compasses, we could as before construct a line whose length is ±cos (A/3). Hence if we show that this last cannot be done when the only given geometric elements are the angle A and a line of unit length, we shall have proved that the angle A cannot be trisected. We shall give the proof for A = 120°. We employ the trigonometric identity cos A = 4 cos^{3} A/3 - 3 cos A/3.
Answer Key (non-editable)
is the left member of \Eq{(7)} with $\sqrt{k}$ replaced by $-\sqrt{k}$, and hence the result is $A - B\sqrt{k}$. But $A = B = 0$. This shows that \[ \Tag{(8)} x_{2} = e - f\sqrt{k} \] is a new root of our cubic equation. Since the sum of the three roots is equal to $-\alpha$ by §20, the third root is \[ \Tag{(9)} x_{3} = -\alpha - x_{1} - x_{2} = -\alpha - 2e. \] Now $\alpha$ is rational. If also $e$ is rational, $x_{3}$ is a rational root and we have reached our goal. We next make the assumption that $e$ is irrational and show that it leads to a contradiction. Since $e$ is a component part of the constructible root \Eq{(6)}, its only irrationalities are square roots. Let $\sqrt{s}$ be one of the radicals of highest order in $e$. By the argument which led to \Eq{(6)}, we may write $e = e' + f'\sqrt{s}$, whence, by \Eq{(9)}, \[ \Tag{(9')} %[** Attn alignment] x_{3} = g + h\sqrt{s}, \qquad (h \neq 0) \] where neither $g$ nor $h$ involves $\sqrt{s}$. Then by the argument which led to \Eq{(8)}, $g - h\sqrt{s}$ is a root, different from $x_{3}$, of our cubic equation, and hence is equal to $x_{1}$ or $x_{2}$ since there are only three roots (§16). Thus \[ g - h\sqrt{s} = e ± f\sqrt{k}. \] By definition, $\sqrt{s}$ is one of the radicals occurring in $e$. Also, by \Eq{(9')}, every radical occurring in $g$ or $h$ occurs in $x_{3}$ and hence in $e = \frac{1}{2}(-\alpha - x_{3})$, by (9), $\alpha$ being rational. Hence $\sqrt{k}$ is expressible rationally in terms of the remaining radicals occurring in $e$ and $f$, and hence in $x_{1}$, whose value is given by \Eq{(6)}. But this contradicts one of our agreements. \Par{32. Trisection of an Angle.} For a given angle $A$, we can construct with ruler and compasses a line of length $\cos A$ or $-\cos A$, namely the adjacent leg of a right triangle, with hypotenuse unity, formed by dropping a perpendicular from a point in one side of $A$ to the other, produced if necessary. If it were possible to trisect angle $A$, i.e., construct the angle $A/3$ with ruler and compasses, we could as before construct a line whose length is $±\cos(A/3)$. Hence if we show that this last cannot be done when the only given geometric elements are the angle $A$ and a line of unit length, we shall have proved that the angle $A$ cannot be trisected. We shall give the proof for $A = 120°$. We employ the trigonometric identity \[ \cos A = 4 \cos^{3} \frac{A}{3} - 3 \cos \frac{A}{3}. \]
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