154394

1 We want to find fractions in the form of p/q, such that
2 sqrt(65)<p/q<sqrt(67)
3 Squaring all parts of inequality and rearranging , we have
4 65(q^2)<p^2<67(q^2)
5 If this range lies between two consecutive square numbers, p cannot exist
6 (we are dealing with integers for p and q). For example, if 65(q^2)>16^2
7 but 67(q^2)<17^2 , there is no integer p which lies in between 16 and 17
8 Therefore, if p cannot exist, there has to be a number x such that
9 x^2<65(q^2)<p^2<67(q^2)<(x+1)^2
10 Breaking up the inequality, we have
11 x^2<65(q^2) and
12 (x+1)^2>67(q^2)
13 Square rooting both sides, of both inequalities and making the subject x,
14 we have
15 x<sqrt(65)*q
16 x>(sqrt(67)*q)-1
17 or (sqrt(67)*q)-1<x<sqrt(65)*q
18 So if there is an integer x that lies between these two bounds, then p cannot
19 exist
20 The inequality suggests that (sqrt(67)*q)-1<sqrt(65)*q but there is a q
21 such that (sqrt(67)*q)-1)>sqrt(65)*q
22 Subtracting sqrt(65)*q and adding 1 to both sides, we get
23 sqrt(67)q-sqrt(65)q>1
24 Grouping in terms of q, we get
25 (sqrt(67)-sqrt(65))q>1
26 Dividing both sides by sqrt(67)-sqrt(65), we get
27 q>1/((sqrt(67)-sqrt(65))
28 Multiplying numerator and denominator of fraction by sqrt(67)+sqrt(65),
29 q>(sqrt(67)+sqrt(65))/2 , which is approximately equal to 8.12
30 q>8.12 however this constraint applies for the first integer after 8.12, which 31is nine
32 Therefore, when q>=9, (sqrt(67)*q)-1)>sqrt(65)*q
33 When q>=9, it is impossible to find an x that is larger than sqrt(67)*q)-1
34 but smaller than sqrt(65)*q because (sqrt(67)*q)-1)>sqrt(65)*q
35 So fractions exist with a denominator of 9 or greater that lie between
36 sqrt(65) and sqrt(67)
37 For q's from 1 to 5, there exists an integer that lies between sqrt(67)*q)-1
38 and sqrt(65)*q, but for q's 6, 7 and 8, it is true that (sqrt(67)*q)-1<sqrt(65)*q, and there is an x between these bounds, but
39 sqrt(67)*q)-1 and sqrt(65)*q have the same digit in the units column, so
40 there is no integer x that lies between (sqrt(67)*q)-1 and sqrt(65)*q
41 so fractions exist with denominators 6, 7 and 8 between sqrt(65)and sqrt(67)
42 and there are no fractions that exist with denominators from 1 to 5
43 between sqrt(65) and sqrt(67)
44 For p/q to exist, the inequality
45 x^2<65(q^2)<p^2<67(q^2)<(x+a)^2, where a>=2, must exist
46 (sqrt(67)*q)-a<x<sqrt(65)*q
47 Also p=x+b where 0<b<a and b is an integer
48 Fraction with q=6 in the bounds and a=2
49 6*sqrt(65)=48.37
50 6sqrt(67)-2=47.11
51 Is there an integer x such that 47.11<x<48.37
52 x=48
53 because a=2 was picked p=x+b ,where 0<b<2 so b=1 (reference to inequality in line 47)
54 therefore p^2 =49^2 and p=49
55 49/6 is a fraction p/q that is within the bounds
56 Fraction with q=7 in the bounds and a=2
57 7*sqrt(65)=56.44
58 7sqrt(67)-2=55.29
59 Therefore x=56 and p=57
60 57/7 is a fraction lying within the bounds
61 (Note that when q is small, a can only be 2 or slightly greater than 2 but an
62 integer)
63 Let us pick a large q, say 3421 and let a=4
64 3421*sqrt(65)=27580.9
65 3421sqrt(67)-4=27998.1
66 We see that (sqrt(67)*q)-a>sqrt(65)*q and x cannot exist
67 Therefore we need to make constraints for what a can be when we pick an
68 arbitrary q. For x to exist,
69 (sqrt(67)*q)-a<sqrt(65)*q
70 Making a the subject through inverse operations,we get
71 a>(sqrt(67)-sqrt(65))q
72 So for q=3421, a>421 and p can be x+b ,where 0<b<421 and x=27580
73 An example of a p/q that exists within the bounds is 27582/3421
74 if q=555555, a>68386 and x can be 4479027 and p=x+b where 0<b<68386
75 An example of a p/q that exists within the bounds is 4479464/555555
76 if q=123456789, a>15196917 and x can be 995340453 and p=x+b where
77 0<b<15196917
78 995390453/123456789 lies within the bounds