Suppose √3 is a rational number, then √ 3 = m/n, where m and n are coprime integers.

√ 3 = m/n, ∴√ 3n = m, ∴ 3N2 = m 2, ∴m is a multiple of 3. ①

Let m = 3t, where t is an integer, so:

3n 2 = (3t) 2 = 9t 2, ∴n 2 = 3t 2, ∴n is a multiple of 3. ②

From ① and ②, it is concluded that m and n have a common divisor of 3, which contradicts the assumed m and n coprime, and ∴√3 can only be an irrational number.

Question 2:

∫ 1/[(n- 1)n(n+ 1)]

=[ 1/(n- 1)- 1/n][ 1/(n+ 1)]

= 1/[(n- 1)(n+ 1)]- 1/[n(n+ 1)]

=( 1/2)[ 1/(n- 1)- 1/(n+ 1)]-[ 1/n- 1/(n+ 1)]

=( 1/2)[ 1/(n- 1)-2/n+ 1/(n+ 1)].

Let n = 10, 1 1, 12, ..., 99 be substituted into the above formula in turn, and we can get:

1/(9× 10× 1 1)=( 1/2)( 1/9-2/ 10+ 1/ 1 1),

1/( 10× 1 1× 12)=( 1/2)( 1/ 10-2/ 1 1+ 1/ 12),

1/( 1 1× 12× 13)=( 1/2)( 1/ 1 1-2/ 12+ 1/ 13),

,

1/(98×99× 100)=( 1/2)( 1/98-2/99+ 1/ 100),

∴ 1/ 1 1^3+ 1/ 12^3+ 1/ 13^3+ + 1/ 100^3

< 1/(9× 10× 1 1)+ 1/( 10× 1 1× 12)+ 1/( 1 1× 12× 13)+ + 1/(98×99× 100)

=( 1/2)[( 1/9-2/ 10+ 1/ 1 1)+( 1/ 10-2/ 1 1+ 1/ 12)+ ( 1/98-2/99+ 1/ 100)]

=( 1/2)[( 1/9- 1/ 10)-( 1/99- 1/ 100)]

=( 1/2)( 1/90- 1/9900)

=( 1/2)×[98 10/(90×9900)]

=( 1/2)×( 109/9900)

=( 1/2)×( 1/90)×( 109/ 1 10)

1/( 1 1× 12× 13)+ 1/( 12× 13× 14)+ 1/( 13× 14× 15)+ + 1/( 100× 10 1× 102)

=[( 1/ 1 1-2/ 12+ 1/ 13)+( 1/ 12-2/ 13+ 1/ 14)+ ( 1/ 100-2/ 10 1+ 1/ 102)]/2

=( 1/2)[( 1/ 1 1- 1/ 12)-( 1/ 10 1- 1/ 102)]

=( 1/2)( 1/ 132- 1/ 10302)

=( 1/2)×[ 10 170/( 132× 10302)]

=( 1/2)×( 1/ 134)×[( 134× 10 170)/( 132× 10302)]

=( 1/268)×[(67×5085)/(66×5 15 1)]

=( 1/268)×[(67×565)/(22× 17 17)]

=( 1/268)×(37855/37774)

> 1/268 ②

From ① and ②:1/268 <1113+123+1133+