The idea on the first floor is right.

(a 1-√3)*(a2-√3)

=(a 1-√3)*( 1+2/( 1+a 1)-√3)

=(a 1-√3)*(2+( 1-√3)*( 1+a 1))/( 1+a 1)

=(a 1-√3)*(3-√3+( 1-√3)* a 1)/( 1+a 1)

=(a 1-√3)*( 1-√3)*(a 1-√3)/( 1+a 1)

=-(√3- 1)*(a 1-√3)^2/( 1+a 1)<； 0

So the root number 3 is between a 1 and a2.

(2)

Separable a 1

Set a 1

Just prove A2-√ 3.

After substituting a2, the inequality can be sorted as follows

( 1-√3)*(a 1-√3)/( 1+a 1)+(a 1-√3)& lt； 0,

According to a 1-√ 3

Similarly, when a1>; √3,

It proves that √ 3-A2

After substituting a2, the inequality can be sorted as follows

( 1-√3)*(a 1-√3)/( 1+a 1)+(a 1-√3)>0,

According to a1-√ 3 >; 0，a 1 & gt； 0 inequality is clearly established.

(3)

Synthesis (1) and (2),

Let b(n) be a sequence of all even terms of a(n)

Let c(n) be the sequence of all odd terms of a(n)

It is easy to know that B (n) and C (n) approach √3 from the left and right respectively.

B (n) and c (n) are monotone bounded sequences,

Therefore, the limit of b(n) exists and is set to x; C(n) limit exists and is set to y.

Whether B (n) and C (n) approach √3 from the left or right depends on whether a 1 is on the left or right of √3.

It is known that A (n+2) =1+2/(1+A (n+1)) is used instead of A (n+1) =1+.

The recursive formula for finding a(n+2) from a (n) is as follows:

a(n+2)=(3+2a(n))/(2+a(n))

Therefore, the recurrence formula of the sequence b(n) is: b(n+ 1)=(3+2b(n))/(2+b(n)).

Because the limit of b(n) is x, you can get it by replacing both b(n+ 1) and b(n) in the above formula with X.

X=(3+2X)/(2+X)，

Where a 1 is the approximate value of √3 and a 1 >; 0, according to the conclusions in (1) and (2), it is not difficult to judge that a(n) is greater than 0.

So b(n) is greater than 0, and natural x cannot be equal to a negative number.

The above equation about x can be arranged as follows

X 2 = 3, so X=√3.

Similarly, you can get Y=√3.

Both the odd and even terms of the sequence a(n) are restricted by √3, so a(n) must also be restricted by √3.

The first term of a(n) is a rational number. By observing the recurrence formula of a(n), we can draw the conclusion that a(n) is rational.

Therefore, the sequence a(n) constitutes a method to find a rational near quasi-value with root number 3.