This problem asks whether a square with the same area as a given circle can be constructed using a set of squares and compasses. It is one of the three great unconstructible problems of ancient Greece. Since the area of a circle with radius 1 is π, this problem is the same as the problem of whether a line segment of length 1 can be constructed using a set of squares and compasses. A complex number (including real numbers) that is the root of a polynomial with rational coefficients (except when all the coefficients are zero) is called an algebraic number, and a complex number (including real numbers) that is not an algebraic number is called a transcendental number. When a unit length is given, the length of a line segment that can be constructed from it using a set of rulers and a compass is an algebraic number. If can be constructed, then π can also be constructed, and therefore π must be an algebraic number. In 1882, CLF Lindemann (1852-1939) proved that π was a transcendental number, which showed that the problem of squaring a circle cannot be constructed. [Tsuneo Kanno] [Reference item] |Source: Shogakukan Encyclopedia Nipponica About Encyclopedia Nipponica Information | Legend |
与えられた円と同じ面積をもつ正方形を定木とコンパスで作図できるか、という問題をいう。古代ギリシアの三大作図不能問題のうちの一つである。半径1の円の面積はπであるから、この問題はの長さの線分を定木とコンパスで作図できるか、という問題と同じである。 有理数を係数とする多項式(係数が全部ゼロである場合を除く)の根(こん)になっている複素数(実数の場合も含む)を代数的数といい、代数的数でない複素数(実数の場合も含む)を超越数という。単位の長さが与えられたとき、これから定木とコンパスで作図できる線分の長さは代数的数である。もしが作図できれば、πも作図でき、したがってπは代数的数でなければならない。1882年リンデマンC. L. F. Lindemann(1852―1939)によってπは超越数であることが証明されたため、円積問題は作図不能であることが示された。 [菅野恒雄] [参照項目] |出典 小学館 日本大百科全書(ニッポニカ)日本大百科全書(ニッポニカ)について 情報 | 凡例 |
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