I found that Q8 in the last passage was quite interesting so I'm going to discuss the question a bit.

1) Modolus approximation

Recall the four equation:

Find the smallest possible integer n>2 s.t.

i) {n^0.5-2^0.5}< 0.01

ii) {|n^0.5-2^0.5|}<0.01

iii) {n^0.5-2^0.5}<0.0001

iv) {|n^0.5-2^0.5|}<0.0001

Now let's modify the question to:

i) {n^0.5-2^0.5}<0.01

ii) {2^0.5-n^0.5}<0.01

iii) {n^0.5-2^0.5}<0.0001

iv) {2^0.5-n^0.5}<0.0001

Now consider the easier one {2^0.5-n^0.5}<0.01.

2^0.5\ < n^0.5 + 0.01 +a

1.4042 < n^0.5 + a

Square:

1.97 < (n+a^2) + 2an^0.5

This may help us to estimate the level of error (like the big O function)

Now the error level is 0.01 so after root squaring it becomes 0.1.

{n^0.5- 2^0.5} < 0.1 ~ 0

n = (a^2+2) + 2a2^0.5 where a is an integer

{0.8284a} < 0.1

Magifying the system, 8284a < 1000 mod 10000

Though they are all multiple of 4, it's still troublesome to get all solutions of a.

Here's the result with (=a mod 10000, a), readers can observe why only these numbers were selected.

(4,2331)

(8,2162)

(56,134)

(60,-35=2465)

(116,99)

(236,29)

(298,-6=2494)

(352,128)

(532,23)

(828,17)

The smallest a is 17, so we will have n ~ (17+2^0.5)^2 ~ 339

{2^0.5 - 339^0.5} = 0.00226... which is true.

Similarly, {2^0.5 - n^0.5} < 0.0001 is equivalent to 8284a < 100 mod 10000, which a = 134.

(134+2^0.5)^2 ~ 18337

{2^0.5-18337^0.5} = 0.00003 which is also true.

2) Differentiatial approximation

The formula is given by f(x+h) ~ f(x) + hf'(x)

Here's an simple example:

Estimate sin (pi/90).

we have sin (pi/90) ~ sin 0 + (pi/90)(sin 0)' = pi/90 ~ 0.3490

while sin (pi/90) ~ 0.03489, quite accurate!

Exercise

1) Complete {n^0.5-2^0.5}<0.01 and {n^0.5-2^0.5}<0.0001.

2) How many digits of 2^0.5 you should approximate to find n that {n^0.5-2^0.5}<10^-6?

3) Show that the answer obtained (339 and 18337) is really the smallest solution.

4) Show that there exist integer solution(s) in every interval [x^2,(x+1)^2] (x is real >2) which {n^0.5-2^0.5}<0.1.

5) Given tan (57pi/180) = 1.539864964..., find the smallest integer n such that {|tan (17n pi/180) - tan 57pi/180|} < 0.01 (four cases, same as {n^0.5-2^0.5})

6) Estimate 2^ 0.0101.

7) Estimate sin 79pi/180, suppose all value of sin x is NOT WELL KNOWN except sin 0 = 0 and sin pi/2 = 1. (Mean value theorem's approximation?)

8) Select integer p,q which is not bigger than 1000, in which (p/q)^2 is closet to 2. Now the n d.p. correction of (p/q) and 2^0.5 takes different value. Find the smallest n. (modified IMO prelim mock, 3M)

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