秦九韶數書九章卷六環田三積題之一:環田外周三十步,虛徑八步,求積。這個問題的解法是很簡單的,因爲圓面積=π×徑2/4,徑=圓周/π,所以所求的環田積(設爲x)=1/4π[(30/π)~2-8~2]=225/π-16π………(1)但是此題按照其術與草來解,就要解一個特殊四次方程: -x~1+15265x~2=6262506.05………………(2)這是因爲秦氏在這個問題裏是用後漢張衡的圓周率10~(1/2),再將10~(1/2)有理化。由(1) x=225/10~(1/2)-16(10)~(1/2),按照我們現在解無理方程的方法(移項自乘)化之,即得(2)。在環田三積中,如求通徑內周等等,皆用10~(1/2)爲圓周率。我在此提出這一點,不在於解釋上述問題如何解決,而在於通過這一問題的解法,證明秦氏雖不是解無理方程,而已經得到了有理化的方法。因爲秦氏對於這樣簡單的 Qin Jiuxuan’s Nine Chapters, One of the Six-ring Fields, One of the three plot problems: Ring Tian’s thirty steps, the virtual path eight steps, and the quadrature. The solution to this problem is very simple, because the circle area = π × diameter 2/4, diameter = circumference / π, so the ring field product (set as x) = 1/4π [(30/π) ~ 2 -8~2] = 225/π-16π... (1) However, if this problem is solved according to its technique and grass, it is necessary to solve a special quartic equation: -x~1+15265x~2=6262506.05......... .........(2) This is because Qin Shi used the post-Han Zhang Heng’s pi of 10~(1/2), and then rationalized 10~(1/2). From (1) x = 225/10 ~ (1/2) -16 (10) ~ (1/2), according to our current solution of the irrational equation (shifting), we get (2). In the ring field three products, such as seeking the inner diameter of the path, etc., all use 10 ~ (1/2) for the pi. My point here is not to explain how the above problems are solved, but rather to prove that Qin’s solution to this problem has not been rationalized but has been rationalized. Because Qin’s is so simple