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RePoSS #11: The Mathematics of Niels Henrik Abel: Continuation ...

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8.4. Further ideas on the theory <strong>of</strong> equations 179<br />

radicals. In this way, the insolubility <strong>of</strong> fifth degree equations <strong>of</strong> the standard form 28<br />

x 5 + ax + b = 0 was demonstrated directly: If the equation had been solvable, ABEL<br />

possessed a solution formula, which he saw was not powerful enough to give the<br />

solution <strong>of</strong> arbitrary equations.<br />

In a letter to HOLMBOE from the same year, the result on the form <strong>of</strong> roots was<br />

given another twist.<br />

“Concerning equations <strong>of</strong> the 5 th degree I have found that whenever such an<br />

equation can be solved algebraically, the root must have the following form:<br />

x = A + 5√ R + 5√ R ′ + 5√ R ′′ + 5√ R ′′′<br />

where R, R ′ , R ′′ , R ′′′ are the 4 roots <strong>of</strong> an equation <strong>of</strong> the 4 th degree and have the<br />

property that they can be expressed with help <strong>of</strong> only square roots. — It has been<br />

a difficult task for me with respect to expressions and notation.” 29<br />

In this form, the statement is a refined version <strong>of</strong> EULER’S “conjecture” that the<br />

solution <strong>of</strong> the fifth degree equation should be <strong>of</strong> the form<br />

A + 5√ R + 5√ R ′ + 5√ R ′′ + 5√ R ′′′ (8.9)<br />

where R, R ′ , R ′′ , R ′′′ were solutions to an equation <strong>of</strong> the fourth degree (see section<br />

5.1).<br />

wo<br />

�<br />

a = m + n 1 + e2 +<br />

�<br />

a1 = m − n<br />

�<br />

a2 = m + n<br />

�<br />

a3 = m − n 1 + e2 +<br />

�<br />

�<br />

h 1 + e2 + �<br />

1 + e2 �<br />

,<br />

1 + e2 � �<br />

+ h 1 + e2 − �<br />

1 + e2 �<br />

,<br />

1 + e2 � �<br />

− h 1 + e2 + �<br />

1 + e2 �<br />

,<br />

� �<br />

h 1 + e2 − �<br />

1 + e2 �<br />

,<br />

A = K + K ′ a + K ′′ a2 + K ′′′ aa2,<br />

A1 = K + K ′ a1 + K ′′ a3 + K ′′′ a1a3,<br />

A2 = K + K ′ a2 + K ′′ a + K ′′′ aa2,<br />

A3 = K + K ′ a3 + K ′′ a1 + K ′′′ a1a3.<br />

Die Grössen c, b, e, m, n, K, K ′ , K ′′ , K ′′′ sind alle rationale Zahlen.<br />

Auf diese Weise lässt sich aber die Gleichung x 5 + ax + b = 0 nicht auflösen, so lange a und b<br />

beliebige Grössen sind.” (<strong>Abel</strong>→Crelle, Freyberg, 1826/03/14. N. H. <strong>Abel</strong>, 1902a, 21–22).<br />

28 If formulated in positive way, the researches <strong>of</strong> JERRARD (see section 6.9.1) demonstrated that every<br />

fifth degree equation could be transformed to this normal trinomial form. (W. R. Hamilton, 1839,<br />

251)<br />

29 “Med Hensyn til Ligninger af 5th Grad har jeg faaet at naar en saadan Ligning lader sig løse algebraisk<br />

maa Roden have følgende Form:<br />

x = A + 5√ R + 5√ R ′ + 5√ R ′′ + 5√ R ′′′<br />

hvor R, R ′ , R ′′ , R ′′′ ere de 4 Rødder af en Ligning af 4de Grad, og som lade sig udtrykke blot ved<br />

Hjelp af Qvadratrødder. — Det har været mig en vanskelig Opgave med Hensyn til Udtryk og<br />

Tegn.” (<strong>Abel</strong>→Holmboe, Paris, 1826/10/24. N. H. <strong>Abel</strong>, 1902a, 45).

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