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小 （机器人：清理不当的来源、移除无用的模板参数） |
小 （机器人：清理不当的来源、移除无用的模板参数） |
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|publisher= 中国人民大学书报資料社 |
|publisher= 中国人民大学书报資料社 |
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|page = 49 |
|page = 49 |
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}}</ref>。勾股定理現約有400種[[数学证明|证明]]方法，是[[數學定理]]中證明方法最多的定理之一<ref>{{cite book |
}}</ref>。勾股定理現約有400種[[数学证明|证明]]方法，是[[數學定理]]中證明方法最多的定理之一<ref>{{cite book |
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|author= 李信明 |
|author= 李信明 |
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|ISBN = 9575671511 |
|ISBN = 9575671511 |
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|page = 106 |
|page = 106 |
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}}</ref>。 |
}}</ref>。 |
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{{Main|非欧几里得几何}} |
{{Main|非欧几里得几何}} |
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勾股定理是由[[欧几里得几何]]的公理推导出来的，其在非欧几里得几何中是不成立的<ref name=false>{{cite book |title=''cited work'' |author=Stephen W. Hawking |page=4 ||ISBN = 0-7624-1922-9 |year=2005}}</ref>。因为勾股定理的成立涉及到了[[平行公设]]。<ref name=Parallel>{{cite book |title=CRC concise encyclopedia of mathematics |author= Eric W. Weisstein ||page=2147 |quote=The parallel postulate is equivalent to the ''Equidistance postulate'', ''Playfair axiom'', ''Proclus axiom'', the ''Triangle postulate'' and the ''Pythagorean theorem''. |edition=2nd |isbn=1-58488-347-2 |year=2003}}</ref><ref name= Pruss>{{cite book |title=The principle of sufficient reason: a reassessment |author= Alexander R. Pruss |quote=We could include...the parallel postulate and derive the Pythagorean theorem. Or we could instead make the Pythagorean theorem among the other axioms and derive the parallel postulate. |ISBN = 0-521-85959-X |year=2006 |publisher=Cambridge University Press |page=11 |
勾股定理是由[[欧几里得几何]]的公理推导出来的，其在非欧几里得几何中是不成立的<ref name=false>{{cite book |title=''cited work'' |author=Stephen W. Hawking |page=4 ||ISBN = 0-7624-1922-9 |year=2005}}</ref>。因为勾股定理的成立涉及到了[[平行公设]]。<ref name=Parallel>{{cite book |title=CRC concise encyclopedia of mathematics |author= Eric W. Weisstein ||page=2147 |quote=The parallel postulate is equivalent to the ''Equidistance postulate'', ''Playfair axiom'', ''Proclus axiom'', the ''Triangle postulate'' and the ''Pythagorean theorem''. |edition=2nd |isbn=1-58488-347-2 |year=2003}}</ref><ref name= Pruss>{{cite book |title=The principle of sufficient reason: a reassessment |author= Alexander R. Pruss |quote=We could include...the parallel postulate and derive the Pythagorean theorem. Or we could instead make the Pythagorean theorem among the other axioms and derive the parallel postulate. |ISBN = 0-521-85959-X |year=2006 |publisher=Cambridge University Press |page=11 }}</ref> |
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== 参考文献 == |
== 参考文献 == |
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