■連続数のピタゴラス三角形(その27)
p/q→P/Q=(p+2q)/(p+q)
このとき,pをp+2q,qをp+qで置き換えています.
もう一度,pをp+2q,qをp+qで置き換えれば,
p/q→(p+2q)/(p+q)→(p+2q+2p+2q)/(p+2q+p+q)=(3p+4q)/(2p+3q)
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【1】√2の近似値とヘロン数列
p/q→(p+2q)/(p+q)
に次ぐ第3の近似分数は
p/q→(p+2q)/(p+q)→(3p+4q)/(2p+3q)
となります.
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[1]ペル数列
p^2−2q^2=±1が成り立つとき
p/q→P/Q=(p+2q)/(p+q)
Q=p+q,P=q+Q=p+2q
P^2−2Q^2=(p+2q)^2−2(p+q)^2
=−p^2+2q^2=±1
[2]ヘロン数列
p^2−2q^2=±1が成り立つとき
p/q→P/Q=(3p+4q)/(2p+3q)
Q=2p+3q,P=p+q+Q=3p+4q
P^2−2Q^2=(3p+4q)^2−2(2p+3q)^2
=p^2−2q^2=±1
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